Solution manual for introduction to electric circuits
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Solution manual for introduction to electric circuits

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Solucionario, dorf circuitos electricos 6 edicion

Solucionario, dorf circuitos electricos 6 edicion

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Solution manual for introduction to electric circuits Solution manual for introduction to electric circuits Document Transcript

  • Solution Manual to accompanyIntroduction to Electric Circuits, 6e By R. C. Dorf and J. A. Svoboda 1
  • Table of ContentsChapter 1 Electric Circuit VariablesChapter 2 Circuit ElementsChapter 3 Resistive CircuitsChapter 4 Methods of Analysis of Resistive CircuitsChapter 5 Circuit TheoremsChapter 6 The Operational AmplifierChapter 7 Energy Storage ElementsChapter 8 The Complete Response of RL and RC CircuitsChapter 9 The Complete Response of Circuits with Two Energy Storage ElementsChapter 10 Sinusoidal Steady-State AnalysisChapter 11 AC Steady-State PowerChapter 12 Three-Phase CircuitsChapter 13 Frequency ResponseChapter 14 The Laplace TransformChapter 15 Fourier Series and Fourier TransformChapter 16 Filter CircuitsChapter 17 Two-Port and Three-Port Networks 2
  • Errata for Introduction to Electric Circuits, 6th Edition Errata for Introduction to Electric Circuits, 6th EditionPage 18, voltage reference direction should be + on the right in part B:Page 28, caption for Figure 2.3-1: "current" instead of "cuurent"Page 41, line 2: "voltage or current" instead of "voltage or circuit"Page 41, Figure 2.8-1 b: the short circuit is drawn as an open circuit.Page 42, line 11: "Each dependent source ..." instead of "Each dependent sources..."Page 164, Table 5.5-1: method 2, part c, one should insert the phrase "Zero all independent sources,then" between the "(c)" and "Connect a 1-A source. . ." The edited phrase will read:"Zero all independent sources, then connect a 1-A source from terminal b to terminal a. Determine Vab. Then Rt = Vab/1."Page 340, Problem P8.3-5: The answer should be .Page 340, Problem P8.3-6: The answer should be .Page 341, Problem P.8.4-1: The answer should bePage 546, line 4: The angle is instead of .Page 554, Problem 12.4.1 Missing parenthesis:Page 687, Equation 15.5-2: Partial t in exponent: http://www.clarkson.edu/~svoboda/errata/6th.html (1 of 2)5/10/2004 7:41:43 PM
  • Errata for Introduction to Electric Circuits, 6th EditionPage 757, Problem 16.5-7: Hb(s) = V2(s) / V1(s) and Hc(s) = V2(s) / Vs(s) instead of Hb(s) = V1(s) / V2(s) and Hc(s) = V1(s) / Vs(s). http://www.clarkson.edu/~svoboda/errata/6th.html (2 of 2)5/10/2004 7:41:43 PM
  • Chapter 1 – Electric Circuit VariablesExercisesEx. 1.3-1 i (t ) = 8 t 2 − 4 t A t t 8 t 8 q(t ) = ∫ 0 i dτ + q(0) = ∫ 0 (8τ 2 − 4τ ) dτ + 0 = τ 3 −2τ 2 = t 3 − 2 t 2 C 3 0 3Ex. 1.3-3 t t 4 4 4 q ( t ) = ∫ i (τ ) dτ + q ( 0 ) = ∫ 4sin 3τ dτ + 0 = − t cos 3τ 0 = − cos 3 t + C 0 0 3 3 3Ex. 1.3-4 0 t <0 dq ( t ) i (t ) = i (t ) = 2 0< t < 2 dt  −2( t − 2 ) −2e t >2Ex. 1.4-1 i1 = 45 µA = 45 × 10-6 A < i2 = 0.03 mA = .03 × 10-3 A = 3 × 10-5 A < i3 = 25 × 10-4 AEx. 1.4-2 ∆ q = i∆ t = ( 4000 A )( 0.001 s ) = 4 CEx. 1.4-3 ∆ q 45 × 10−9 i= = −3 = 9 × 10−6 = 9 µA ∆t 5 × 10Ex. 1.4-4  electron   −19 C   9 electron   −19 C  i = 10 billion  s  1.602 ×10 electron  =   10×10  s  1.602 × 10 electron    electron C = 1010 × 1.602 ×10−19 s electron C = 1.602 × 10−9 = 1.602 nA s 1-1
  • Ex. 1.6-1 (a) The element voltage and current do not adhere to the passive convention in Figures 1.6-1B and 1.6-1C so the product of the element voltage and current is the power supplied by these elements. (b) The element voltage and current adhere to the passive convention in Figures 1.6-1A and 1.6-1D so the product of the element voltage and current is the power delivered to, or absorbed by these elements. (c) The element voltage and current do not adhere to the passive convention in Figure 1.6-1B, so the product of the element voltage and current is the power delivered by this element: (2 V)(6 A) = 12 W. The power received by the element is the negative of the power delivered by the element, -12 W. (d) The element voltage and current do not adhere to the passive convention in Figure 1.6-1B, so the product of the element voltage and current is the power supplied by this element: (2 V)(6 A) = 12 W. (e) The element voltage and current adhere to the passive convention in Figure 1.6-1D, so the product of the element voltage and current is the power delivered to this element: (2 V)(6 A) = 12 W. The power supplied by the element is the negative of the power delivered to the element, -12 W.ProblemsSection 1-3 Electric Circuits and Current FlowP1.3-1 d i (t ) = dt ( ) 4 1 − e −5t = 20 e −5t AP1.3-2 4 4 ( ) t t t t q ( t ) = ∫ i (τ ) dτ + q ( 0 ) = ∫ 4 1 − e −5τ dτ + 0 = ∫ 4 dτ − ∫ 4 e−5τ dτ = 4 t + e−5t − C 0 0 0 0 5 5P1.3-3 t tq ( t ) = ∫ i (τ ) dτ = ∫ 0 dτ = 0 C for t ≤ 2 so q(2) = 0. −∞ −∞ t tq ( t ) = ∫ i (τ ) dτ + q ( 2 ) = ∫ 2 dτ = 2 τ 2 = 2 t − 4 C for 2 ≤ t ≤ 4. In particular, q(4) = 4 C. t 2 2 t tq ( t ) = ∫ i (τ ) dτ + q ( 4 ) = ∫ −1 dτ + 4 = − τ 4 + 4 = 8 − t C for 4 ≤ t ≤ 8. In particular, q(8) = 0 C. t 4 4 t tq ( t ) = ∫ i (τ ) dτ + q ( 8 ) = ∫ 0 dτ + 0 = 0 C for 8 ≤ t . 8 8 1-2
  • P1.3-4 C i = 600 A = 600 s C s mg Silver deposited = 600 ×20 min×60 ×1.118 = 8.05×105 mg=805 g s min CSection 1-6 Power and EnergyP1.6-1 a.) q = ∫ i dt = i∆t = (10 A )( 2 hrs )( 3600s/hr ) = 7.2×10 4 C b.) P = v i = (110 V )(10 A ) = 1100 W 0.06$ c.) Cost = × 1.1kW × 2 hrs = 0.132 $ kWhrP1.6-2 P = ( 6 V )(10 mA ) = 0.06 W ∆w 200 W⋅s ∆t = = = 3.33×103 s P 0.06 WP1.6-3 30 for 0 ≤ t ≤ 10 s: v = 30 V and i = t = 2t A ∴ P = 30(2t ) = 60t W 15 25 for 10 ≤ t ≤ 15 s: v ( t ) = − t + b ⇒ v (10 ) = 30 V ⇒ b = 80 V 5 v(t ) = −5t + 80 and i (t ) = 2t A ⇒ P = ( 2t )( −5t +80 ) = −10t 2 +160t W 30 for 15 ≤ t ≤ 25 s: v = 5 V and i (t ) = − t +b A 10 i (25) = 0 ⇒ b = 75 ⇒ i (t ) = −3t + 75 A ∴ P = ( 5 )( −3t + 75 ) = −15t + 375 W 1-3
  • 60t dt + ∫10 (160t −10t 2 ) dt + ∫15 ( 375−15t ) dt 10 15 25 Energy = ∫ P dt = ∫0 15 25 + 80t 2 − 10 t 3 + 375t − 15 t 2 = 5833.3 J 10 = 30t 2 0 3 10 2 15P1.6-4 a.) Assuming no more energy is delivered to the battery after 5 hours (battery is fully charged). 5( 3600 ) t 5 ( 3600 )  0.5 τ  0.5 2 w = ∫ Pdt = ∫0 vi dτ = ∫0 2 11 +  dτ = 22 t + 3600 τ  3600  0 = 441× 103 J = 441 kJ 1 hr 10¢ b.) Cost = 441kJ × × = 1.23¢ 3600s kWhrP1.6-5 1 1 p (t ) = ( cos 3 t )( sin 3 t ) = sin 6 t 3 6 1 p ( 0.5 ) = sin 3 = 0.0235 W 6 1 p (1) = sin 6 = −0.0466 W 6 1-4
  • Here is a MATLAB program to plot p(t):cleart0=0; % initial timetf=2; % final timedt=0.02; % time incrementt=t0:dt:tf; % timev=4*cos(3*t); % device voltagei=(1/12)*sin(3*t); % device currentfor k=1:length(t) p(k)=v(k)*i(k); % powerendplot(t,p)xlabel(time, s);ylabel(power, W)P1.6-6 p ( t ) = 16 ( sin 3 t )( sin 3 t ) = 8 ( cos 0 − cos 6 t ) = 8 − 8cos 6 t WHere is a MATLAB program to plot p(t):cleart0=0; % initial timetf=2; % final timedt=0.02; % time incrementt=t0:dt:tf; % timev=8*sin(3*t); % device voltagei=2*sin(3*t); % device currentfor k=1:length(t) p(k)=v(k)*i(k); % powerendplot(t,p)xlabel(time, s);ylabel(power, W) 1-5
  • P1.6-7 ( ) ( ) p ( t ) = 4 1 − e −2 t × 2 e −2 t = 8 1 − e−2 t e −2tHere is a MATLAB program to plot p(t):cleart0=0; % initial timetf=2; % final timedt=0.02; % time incrementt=t0:dt:tf; % timev=4*(1-exp(-2*t)); % device voltagei=2*exp(-2*t); % device currentfor k=1:length(t) p(k)=v(k)*i(k); % powerendplot(t,p)xlabel(time, s);ylabel(power, W)P1.6-8 P = V I =3 × 0.2=0.6 W w = P ⋅ t = 0.6 × 5 × 60=180 J 1-6
  • Verification ProblemsVP 1-1Notice that the element voltage and current of each branch adhere to the passive convention. Thesum of the powers absorbed by each branch are:(-2 V)(2 A)+(5 V)(2 A)+(3 V)(3 A)+(4 V)(-5 A)+(1 V)(5 A) = -4 W + 10 W + 9 W -20 W + 5 W =0WThe element voltages and currents satisfy conservation of energy and may be correct.VP 1-2Notice that the element voltage and current of some branches do not adhere to the passiveconvention. The sum of the powers absorbed by each branch are:-(3 V)(3 A)+(3 V)(2 A)+ (3 V)(2 A)+(4 V)(3 A)+(-3 V)(-3 A)+(4 V)(-3 A) = -9 W + 6 W + 6 W + 12 W + 9 W -12 W ≠0WThe element voltages and currents do not satisfy conservation of energy and cannot be correct.Design ProblemsDP 1-1The voltage may be as large as 20(1.25) = 25 V and the current may be as large as (0.008)(1.25)= 0.01 A. The element needs to be able to absorb (25 V)(0.01 A) = 0.25 W continuously. AGrade B element is adequate, but without margin for error. Specify a Grade B device if you trustthe estimates of the maximum voltage and current and a Grade A device otherwise. 1-7
  • DP1-2 ( ) ( ) p ( t ) = 20 1 − e −8 t × 0.03 e −8 t = 0.6 1 − e−8t e−8tHere is a MATLAB program to plot p(t):cleart0=0; % initial timetf=1; % final timedt=0.02; % time incrementt=t0:dt:tf; % timev=20*(1-exp(-8*t)); % device voltagei=.030*exp(-8*t); % device currentfor k=1:length(t) p(k)=v(k)*i(k); % powerendplot(t,p)xlabel(time, s);ylabel(power, W)Here is the plot:The circuit element must be able to absorb 0.15 W. 1-8
  • Chapter 2 - Circuit ElementsExercisesEx. 2.3-1 m ( i1 + i 2 ) = mi1 + mi 2 ⇒ superposition is satisfied m ( a i1 ) = a ( mi1 ) ⇒ homogeneity is satisfied Therefore the element is linear.Ex. 2.3-2 m ( i1 + i 2 ) + b = mi1 + mi 2 + b ≠ ( mi1 + b ) + ( mi 2 + b ) ⇒ superposition is not satisfied Therefore the element is not linear.Ex. 2.5-1 v 2 (10 ) 2 P= = =1 W R 100Ex. 2.5-2 v 2 (10 cos t ) 2 P= = = 10 cos 2 t W R 10Ex. 2.8-1 ic = − 1.2 A, v d = 24 V id = 4 ( − 1.2) = − 4.8 A id and vd adhere to the passive convention so P = vd id = (24) (−4.8) = −115.2 W is the power received by the dependent source 2-1
  • Ex. 2.8-2 vc = −2 V, id = 4 vc = −8 A and vd = 2.2 V id and vd adhere to the passive convention so P = vd id = (2.2) (−8) = −17.6 W is the power received by the dependent source. The power supplied by the dependent source is 17.6 W.Ex. 2.8-3 ic = 1.25 A, vd = 2 ic = 2.5 V and id = 1.75 A id and vd adhere to the passive convention so P = vd id = (2.5) (1.75) = 4.375 W is the power received by the dependent source. 2-2
  • Ex. 2.9-1 θ = 45° , I = 2 mA, R p = 20 kΩ θ 45 a= ⇒ aR = (20 kΩ) = 2.5 kΩ 360 p 360 vm = (2 ×10−3 )(2.5 ×103 ) = 5 VEx. 2.9-2 µA v = 10 V, i = 280 µA, k = 1 for AD590 °K i  °K  i = kT ⇒ T = = (280µA)1  = 280° K k  µA   Ex. 2.10-1 At t = 4 s both switches are open, so i = 0 A.Ex. 2.10.2 At t = 4 s the switch is in the up position, so v = i R = (2 mA)(3 kΩ) = 6V . At t = 6 s the switch is in the down position, so v = 0 V.ProblemsSection 2-3 Engineering and Linear ModelsP2.3-1The element is not linear. For example, doubling the current from 2 A to 4 A does not double thevoltage. Hence, the property of homogeneity is not satisfied.P2.3-2 (a) The data points do indeed lie on a straight line. The slope of the line is 0.12 V/A andthe line passes through the origin so the equation of the line is v = 0.12 i . The element is indeedlinear. (b) When i = 40 mA, v = (0.12 V/A)×(40 mA) = (0.12 V/A)×(0.04 A) = 4.8 mV 4 (c) When v = 4 V, i = = 33 A = 33 A. 0.12 2-3
  • P2.3-3 (a) The data points do indeed lie on a straight line. The slope of the line is 256.5 V/A andthe line passes through the origin so the equation of the line is v = 256.5i . The element is indeedlinear. (b) When i = 4 mA, v = (256.5 V/A)×(4 mA) = (256.5 V/A)×(0.004 A) = 1.026 V 12 (c) When v = 12 V, i = = 0.04678 A = 46.78 mA. 256.5P2.3-4 Let i = 1 A , then v = 3i + 5 = 8 V. Next 2i = 2A but 16 = 2v ≠ 3(2i) + 5 = 11.. Hence,the property of homogeneity is not satisfied. The element is not linear.Section 2-5 ResistorsP2.5-1 i = is = 3 A and v = Ri = 7 × 3 = 21 V v and i adhere to the passive convention ∴ P = v i = 21 × 3 = 63 W is the power absorbed by the resistor.P2.5-2 i = is = 3 mA and v = 24 V v 24 R = = = 8000 = 8 k Ω i .003 P = (3×10 −3 )× 24 = 72×10 −3 = 72 mWP2.5-3 v = vs =10 V and R = 5 Ω v 10 i = = =2 A R 5 v and i adhere to the passive convention ∴ p = v i = 2⋅10 = 20 W is the power absorbed by the resistor 2-4
  • P2.5-4 v = vs = 24 V and i = 2 A v 24 R= = = 12 Ω i 2 p = vi = 24⋅2 = 48 WP2.5-5 v1 = v 2 = vs = 150 V; R1 = 50 Ω; R2 = 25 Ω v 1 and i1 adhere to the passive convention so v 1 150 i1 = = =3 A R 1 50 v 150v2 and i 2 do not adhere to the passive convention so i 2 = − 2 = − = −6 A R2 25The power absorbed by R1 is P = v1 i1 = 150 ⋅ 3 = 450 W 1The power absorbed by R 2 is P 2 = − v 2i 2 = −150(−6) = 900 WP2.5-6 i1 = i 2 = is = 2 A ; R1 =4 Ω and R2 = 8 Ω v 1 and i 1 do not adhere to the passive convention so v 1 =− R 1 i 1 =−4⋅2=−8 V. The power absorbed by R 1 is P1 =−v 1i 1 =−(−8)(2) = 16 W.v2 and i 2 do adhere to the passive convention so v2 = R 2 i 2 = 8 ⋅ 2 = 16 V .The power absorbed by R 2 is P 2 = v 2i 2 = 16 ⋅ 2 = 32 W.P2.5-7 Model the heater as a resistor, then v2 v2 (250) 2 with a 250 V source: P = ⇒ R = = = 62.5 Ω R P 1000 v 2 (210) 2 with a 210 V source: P = = = 705.6 W R 62.5 2-5
  • P2.5-8 P 5000 125 The current required by the mine lights is: i = = = A v 120 3 Power loss in the wire is : i 2 R Thus the maximum resistance of the copper wire allowed is 0.05P 0.05×5000 R= = = 0.144 Ω i2 (125/3) 2 now since the length of the wire is L = 2×100 = 200 m = 20,000 cm thus R = ρ L / A with ρ = 1.7×10−6 Ω⋅ cm from Table 2.5−1 ρL 1.7×10−6 ×20,000 A= = = 0.236 cm 2 R 0.144Section 2-6 Independent SourcesP2.6-1 v s 15 = 3 A and P = R i 2 = 5 ( 3 ) = 45 W 2 (a) i = = R 5 (b) i and P do not depend on is . The values of i and P are 3 A and 45 W, both when i s = 3 A and when i s = 5 A.P2.6-2 v 2 102 (a) v = R i s = 5 ⋅ 2 = 10 V and P = = = 20 W R 5 (b) v and P do not depend on v s . The values of v and P are 10V and 20 W both when v s = 10 V and when v s = 5 V 2-6
  • P2.6-3 Consider the current source: i s and v s do not adhere to the passive convention, so Pcs =i s v s =3⋅12 = 36 W is the power supplied by the current source. Consider the voltage source: i s and v s do adhere to the passive convention, so Pvs = i s vs =3 ⋅12 = 36 W is the power absorbed by the voltage source. ∴ The voltage source supplies −36 W.P2.6-4 Consider the current source: i s and vs adhere to the passive convention so Pcs = i s vs =3 ⋅12 = 36 W is the power absorbed by the current source. Current source supplies − 36 W. Consider the voltage source: i s and vs do not adhere to the passive convention so Pvs = i s vs = 3 ⋅12 =36 W is the power supplied by the voltage source.P2.6-5 (a) P = v i = (2 cos t ) (10 cos t ) = 20 cos 2 t mW 1 1 1 1 1  (b) w = ∫0 P dt = ∫0 20 cos t dt = 20 t + sin 2t  = 10 + 5 sin 2 mJ 2 2 4 0 2-7
  • Section 2-7 Voltmeters and AmmetersP2.7-1 v 5 (a) R = = = 10 Ω i 0.5 (b) The voltage, 12 V, and the current, 0.5 A, of the voltage source adhere to the passive convention so the power P = 12 (0.5) = 6 W is the power received by the source. The voltage source delivers -6 W.P2.7-2 The voltmeter current is zero so the ammeter current is equal to the current source current except for the reference direction: i = -2 A The voltage v is the voltage of the current source. The power supplied by the current source is 40 W so 40 = 2 v ⇒ v = 20 V 2-8
  • Section 2-8 Dependent SourcesP2.8-1 vb 8 r = = =4 Ω ia 2P2.8-2 ia 2 A vb = 8 V ; g v b = i a = 2 A ; g = = = 0.25 vb 8 VP2.8-3 i a 32 A i b = 8 A ; d i b = i a = 32A ; d = = =4 ib 8 AP2.8-4 vb 8 V va = 2 V ; b va = vb = 8 V ; b = = =4 va 2 VSection 2-9 TransducersP2.9-1 θ 360 vm a= , θ = 360 Rp I (360)(23V) θ = = 75.27° (100 kΩ)(1.1 mA)P2.9-2 µA AD590 : k =1 ° , K v =20 V (voltage condition satisfied) 4 µ A < i < 13 µ A   i  ⇒ 4 ° K< T <13° K T =  k  2-9
  • Section 2-10 SwitchesP2.10-1 At t = 1 s the left switch is open and the right switch is closed so the voltage across the resistor is 10 V. v 10 i= = = 2 mA R 5×103At t = 4 s the left switch is closed and the right switch is open so the voltage across the resistor is15 V. v 15 i= = = 3 mA R 5×103P2.10-2 At t = 1 s the current in the resistor is 3 mA so v = 15 V. At t = 4 s the current in the resistor is 0 A so v = 0 V.Verification ProblemsVP2-1 vo =40 V and i s = − (−2) = 2 A. (Notice that the ammeter measures − i s rather than i s .) vo 40 V So = = 20 is 2 A Your lab partner is wrong.VP2-2 vs 12 We expect the resistor current to be i = = = 0.48 A. The power absorbed by R 25 this resistor will be P = i vs = (0.48) (12) = 5.76 W. A half watt resistor cant absorb this much power. You should not try another resistor. 2-10
  • Design ProblemsDP2-1 10 10 1.) > 0.04 ⇒ R < = 250 Ω R 0.04 102 1 2.) < ⇒ R > 200 Ω R 2 Therefore 200 < R < 250 Ω. For example, R = 225 Ω.DP2-2 1.) 2 R > 40 ⇒ R > 20 Ω 15 2.) 2 2 R < 15 ⇒ R < = 3.75 Ω 4 Therefore 20 < R < 3.75 Ω. These conditions cannot satisfied simultaneously.DP2-3 P = ( 30 mA ) ⋅ (1000 Ω ) = (.03) (1000 ) = 0.9 W < 1 W 2 2 1 P2 = ( 30 mA ) ⋅ ( 2000 Ω ) = (.03) ( 2000 ) = 1.8 W < 2 W 2 2 P3 = ( 30 mA ) ⋅ ( 4000 Ω ) = (.03) ( 4000 ) = 3.6 W < 4 W 2 2 2-11
  • Chapter 3 – Resistive CircuitsExercisesEx 3.3-1Apply KCL at node a to get 2 + 1 + i3 = 0 ⇒ i3 = -3 AApply KCL at node c to get 2 + 1 = i4 ⇒ i4 = 3 AApply KCL at node b to get i3 + i6 = 1 ⇒ -3 + i6 = 1 ⇒ i6 = 4 AApply KVL to the loop consisting of elements A and B to get -v2 – 3 = 0 ⇒ v2 = -3 VApply KVL to the loop consisting of elements C, E, D, and A to get 3 + 6 + v4 – 3 = 0 ⇒ v4 = -6 VApply KVL to the loop consisting of elements E and F to get v6 – 6 = 0 ⇒ v6 = 6 VCheck: The sum of the power supplied by all branches is-(3)(2) + (-3)(1) – (3)(-3) + (-6)(3) – (6)(1) + (6)(4) = -6 - 3 + 9 - 18 - 6 + 24 = 0 3-1
  • Ex 3.3-2 Apply KCL at node a to determine the current in the horizontal resistor as shown. Apply KVL to the loop consisting of the voltages source and the two resistors to get -4(2-i) + 4(i) - 24 = 0 ⇒ i = 4 A 2Ex 3.3-3 −18 + 0 − 12 − va = 0 ⇒ va = −30 V and im = va + 3 ⇒ im = 9 A 5 18Ex 3.3-4 −va − 10 + 4va − 8 = 0 ⇒ va = = 6 V and vm = 4 va = 24 V 3Ex 3.4-1 From voltage division  3  v3 = 12   = 3V  3+9  then v i = 3 = 1A 3The power absorbed by the resistors is: (12 ) ( 6 ) + (12 ) ( 3) + (12 ) ( 3) = 12 WThe power supplied by the source is (12)(1) = 12 W. 3-2
  • Ex 3.4-2 P = 6 W and R1 = 6 Ω P 6 i2 = = = 1 or i =1 A R1 6 v0 = i R1 =(1) (6)=6V from KVL: − v+ i (2 + 4 + 6 + 2) = 0 s ⇒ v = 14 i = 14 V s 25Ex 3.4-3 From voltage division ⇒ v = m 25+75 (8) = 2 V 25Ex 3.4-4 From voltage division ⇒ v = m 25+75 ( −8 ) = −2 VEx. 3.5-1 1 1 1 1 1 4 103 1 = + 3+ 3+ 3= 3 ⇒ R = = kΩ R 3 10 10 10 10 10 eq 4 4 eq 1 -3 1 By current division, the current in each resistor = (10 ) = mA 4 4Ex 3.5-2 10 From current division ⇒ i = m 10+40 ( −5 ) = − 1 A 3-3
  • ProblemsSection 3-3 Kirchoff’s LawsP3.3-1Apply KCL at node a to get 2 + 1 = i + 4 ⇒ i = -1 AThe current and voltage of element B adhere to the passive convention so (12)(-1) = -12 W ispower received by element B. The power supplied by element B is 12 W.Apply KVL to the loop consisting of elements D, F, E, and C to get 4 + v + (-5) – 12 = 0 ⇒ v = 13 VThe current and voltage of element F do not adhere to the passive convention so (13)(1) = 13 Wis the power supplied by element F.Check: The sum of the power supplied by all branches is -(2)(-12) + 12 – (4)(12) + (1)(4) + 13 – (-1)(-5) = 24 +12 – 48 + 4 +13 –5 = 0 3-4
  • P3.3-2Apply KCL at node a to get 2 = i2 + 6 = 0 ⇒ i2 = -4 AApply KCL at node b to get 3 = i4 + 6 ⇒ i4 = -3 AApply KVL to the loop consisting of elements A and B to get -v2 – 6 = 0 ⇒ v2 = -6 VApply KVL to the loop consisting of elements C, D, and A to get -v3 – (-2) – 6 = 0 ⇒ v4 = -4 VApply KVL to the loop consisting of elements E, F and D to get 4 – v6 + (-2) = 0 ⇒ v6 = 2 VCheck: The sum of the power supplied by all branches is-(6)(2) – (-6)(-4) – (-4)(6) + (-2)(-3) + (4)(3) + (2)(-3) = -12 - 24 + 24 + 6 + 12 – 6 = 0 3-5
  • P3.3-3 KVL : −12 − R 2 (3) + v = 0 (outside loop) v − 12 v = 12 + 3R 2 or R 2 = 3 12 KCL i+ − 3 = 0 (top node) R1 12 12 i = 3− or R1 = R1 3−i(a) v = 12 + 3 ( 3) = 21 V 12 i = 3− =1 A 6(b) 2 − 12 10 12 R2 = = − Ω ; R1 = =8Ω 3 3 3 − 1.5 (checked using LNAP 8/16/02)(c) 24 = − 12 i, because 12 and i adhere to the passive convention. 12 ∴ i = − 2 A and R1 = = 2.4 Ω 3+ 2 9 = 3v, because 3 and v do not adhere to the passive convention 3 − 12 ∴ v = 3V and R 2 = = −3 Ω 3The situations described in (b) and (c) cannot occur if R1 and R2 are required to be nonnegative. 3-6
  • P3.3-4 12 i = =2A 1 6 20 i = = 5A 2 4 i = 3−i = − 2 A 3 2 i = i +i = 3A 4 2 3Power absorbed by the 4 Ω resistor = 4 ⋅ i 2 = 100 W 2Power absorbed by the 6 Ω resistor = 6 ⋅ i 2 = 24 W 1Power absorbed by the 8 Ω resistor = 8 ⋅ i 2 = 72 W 4 (checked using LNAP 8/16/02)P3.3-5 v1 = 8 V v2 = −8 + 8 + 12 = 12 V v3 = 2⋅ 4 = 8 V v2 4Ω : P = 3 = 16 W 4 2 v2 6Ω : P = = 24 W 6 v2 8Ω : P = 1 = 8 W (checked using LNAP 8/16/02) 8P3.3-6 P2 mA = − 3 × ( 2 ×10−3 )  = −6 × 10−3 = −6 mW   P1 mA = −  −7 × (1× 10−3 )  = 7 × 10−3 = 7 mW   (checked using LNAP 8/16/02) 3-7
  • P3.3-7 P2 V = +  2 × (1× 10−3 )  = 2 × 10−3 = 2 mW   P3 V = + 3 × ( −2 × 10 )  = −6 × 10−3 = −6 mW  −3  (checked using LNAP 8/16/02)P3.3-8 KCL: iR = 2 + 1 ⇒ iR = 3 A KVL: vR + 0 − 12 = 0 ⇒ vR = 12 V vR 12 ∴ R= = =4Ω iR 3 (checked using LNAP 8/16/02)P3.3-9 KVL: vR + 56 + 24 = 0 ⇒ vR = −80 V KCL: iR + 8 = 0 ⇒ iR = −8 A vR −80 ∴ R= = = 10 Ω iR −8 (checked using LNAP 8/16/02) 3-8
  • P3.3-10 5.61 3.71 − 5.61 12 − 5.61 −1.9KCL at node b: = + ⇒ 0.801 = + 1.278 7 R1 5 R1 1.9 ⇒ R1 = = 3.983 ≈ 4 Ω 1.278 − 0.801 3.71 3.71 − 5.61 3.71 − 12 −8.29KCL at node a: + + = 0 ⇒ 1.855 + ( −0.475 ) + =0 2 4 R2 R2 8.29 ⇒ R2 = = 6.007 ≈ 6 Ω 1.855 − 0.475 (checked using LNAP 8/16/02) 3-9
  • Section 3-4 A Single-Loop Circuit – The Voltage DividerP3.4-1 6 6 v = 12 = 12 = 4 V 1 6+3+5+ 4 18 3 5 10 v = 12 = 2 V ; v = 12 = V 2 18 3 18 3 4 8 v = 12 = V 4 18 3 (checked using LNAP 8/16/02)P3.4-2 (a) R = 6 + 3 + 2 + 4 = 15 Ω 28 28 (b) i = = = 1.867 A R 15 ( c ) p = 28 ⋅ i =28(1.867)=52.27 W (28 V and i do not adhere to the passive convention.) (checked using LNAP 8/16/02) 3-10
  • P3.4-3 i R2 = v = 8 V 12 = i R1 + v = i R1 + 8 ⇒ 4 = i R1 8 8 4 4 ⋅ 100(a) i= = ; R1 = = = 50 Ω R 2 100 i 8 4 4 8 8 ⋅ 100(b) i = = ; R2 = = = 200 Ω R1 100 i 4 4 8( c ) 1.2 = 12 i ⇒ i = 0.1 A ; R1 = = 40 Ω; R2 = = 80 Ω i i (checked using LNAP 8/16/02)P3.4-4 Voltage division 16 v1 = 12 = 8 V 16 + 8 4 v3 = 12 = 4 V 4+8 KVL: v3 − v − v1 = 0 v = −4 V (checked using LNAP 8/16/02)P3.4-5  100  v  using voltage divider: v =  ⇒ R = 50  s − 1 v 0  100 + 2 R  s  v   o  with v = 20 V and v > 9 V, R < 61.1 Ω  s 0   R = 60 Ω with v = 28 V and v < 13 V, R > 57.7 Ω  s 0  3-11
  • P3.4-6  240 a.)   18 = 12 V  120 + 240   18 b.) 18   = 0.9 W  120 + 240   R c.)   18 = 2 ⇒ 18 R = 2 R + 2 (120 ) ⇒ R = 15 Ω  R + 120  Rd.) 0.2 = ⇒ ( 0.2 )(120 ) = 0.8 R ⇒ R = 30 Ω R + 120 (checked using LNAP 8/16/02) 3-12
  • Section 3-5 Parallel Resistors and Current DivisionP3.5-1 1 6 1 1 i = 4= 4= A 1 1 + 1 + 1 +1 1+ 2 + 3 + 6 3 6 3 2 1 1 3 2 i = 4 = A; 2 1 + 1 + 1 +1 3 6 3 2 1 1 i = 2 4 =1 A 3 1 + 1 + 1 +1 6 3 2 1 1 i = 4=2 A 4 1 + 1 + 1 +1 6 3 2P3.5-2 1 1 1 1 1 (a) = + + = ⇒ R = 2Ω R 6 12 4 2 (b) v = 6 ⋅ 2 = 12 V (c) p = 6 ⋅12 = 72 WP3.5-3 8 8 i= or R1 = R1 i 8 8 8 = R 2 (2 − i ) ⇒ i = 2 − or R 2 = R2 2−i 8 4 8 (a) i = 2− = A ; R1 = 4 =6Ω 12 3 3 8 2 8 (b) i = = A ; R2 = 2 =6Ω 12 3 2− 3 3-13
  • 1 ( c ) R1 = R 2 will cause i= 2 = 1 A. The current in both R1 and R 2 will be 1 A. 2 R1 R 2 1 2 ⋅ = 8 ; R1 = R 2 ⇒ 2 ⋅ R1 = 8 ⇒ R1 = 8 ∴ R1 = R 2 = 8 Ω R1 + R 2 2P3.5-4 Current division: 8 1 16 + 8 ( ) i = −6 = −2 A 8 2 8+8( ) i = −6 = −3 A i = i −i = +1 A 1 2P3.5-5  R  current division: i =  1  i and 2 R + R  s  1 2 Ohms Law: v = i R yields o 2 2  v  R + R  i =  o  1 2 s  R  R   2  1  plugging in R = 4Ω, v > 9 V gives i > 3.15 A 1 o s and R = 6Ω, v < 13 V gives i < 3.47 A 1 o s So any 3.15 A < i < 3.47 A keeps 9 V < v < 13 V. s o 3-14
  • P3.5-6  24  a)   1.8 = 1.2 A  12 + 24   R  b)   2 = 1.6 ⇒ 2 R = 1.6 R + 1.6 (12 ) ⇒ R = 48 Ω  R + 12  R c) 0.4 = ⇒ ( 0.4 )(12 ) = 0.6 R ⇒ R = 8 Ω R + 12Section 3-7 Circuit AnalysisP3.7-1 48 ⋅ 24 (a) R = 16 + = 32 Ω 48 + 24 32 ⋅ 32 (b) v = 32 + 32 24 = 16 V ; 32 ⋅ 32 8+ 32 + 32 16 1 i= = A 32 2 48 1 1 (c) i2 = ⋅ = A 48 + 24 2 3 3-15
  • P3.7-2 3⋅ 6(a) R1 = 4 + =6Ω 3+ 6 1 1 1 1(b) = + + ⇒ R p = 2.4 Ω then R 2 = 8 + R p = 10.4 Ω Rp 12 6 6(c) KCL: i2 + 2 = i1 and − 24 + 6 i2 + R 2i1 = 0 ⇒ −24+6 (i1 −2)+10.4i1 = 0 36 ⇒ i1 = =2.2 A ⇒ v1 =i1 R 2 =2.2 (10.4)=22.88 V 16.4 1(d ) i2 = 6 ( 2.2 ) = 0.878 A, 1 1 1 + + 6 6 12 v2 = ( 0.878 ) (6) = 5.3 V 6 2(e) i3 = i2 = 0.585 A ⇒ P = 3 i3 = 1.03 W 3+ 6 3-16
  • P3.7-3Reduce the circuit from the right side by repeatedly replacing series 1 Ω resistors in parallel witha 2 Ω resistor by the equivalent 1 Ω resistorThis circuit has become small enough to be easily analyzed. The vertical 1 Ω resistor isequivalent to a 2 Ω resistor connected in parallel with series 1 Ω resistors: 1+1 i1 = (1.5 ) = 0.75 A 2 + (1 + 1) 3-17
  • P3.7-4(a) 1 1 1 1 (10 + 8) ⋅ 9 = + + ⇒ R2 = 4 Ω and R1 = = 6Ω R2 24 12 8 b g 10 + 8 + 9(b) First, apply KVL to the left mesh to get −27 + 6 ia + 3 ia = 0 ⇒ ia = 3 A . Next, apply KVL to the left mesh to get 4 ib − 3 ia = 0 ⇒ ib = 2.25 A .(c) 1 i2 = 8 2.25 = 1125 A . b gLM b10 +98g + 9 3OP = −10 V and v1 = − 10 1 1 1 + + 24 8 12 N Q 3-18
  • P3.7-5 30 v1 = 6 ⇒ v1 = 8 V 10 + 30 R2 12 = 8 ⇒ R2 = 20 Ω R2 + 10 20 = b R1 10 + 30 g ⇒ R1 = 40 Ω b R1 + 10 + 30 gAlternate values that can be used to change the numbers in this problem:meter reading, V Right-most resistor, Ω R1, Ω 6 30 40 4 30 10 4 20 15 4.8 20 30 3-19
  • P3.7-6P3.7-7 24 1× 10−3 = ⇒ R p = 12 ×103 = 12 kΩ 12 ×103 + R p 12 × 10 = R p 3 = ( 21×10 ) R3 ⇒ R = 28 kΩ ( 21×10 ) + R 3P3.7-8  130 500  Voltage division ⇒ v = 50  = 15.963 V  130 500 + 200 + 20     100   10  ∴v = v  h  = (15.963)   = 12.279 V  100 + 30   13  v ∴ i = h = .12279 A h 100 3-20
  • P3.7-9 3-21
  • P3.7-10 15 ( 20 + 10 )Req = = 10 Ω 15 + ( 20 + 10 ) 60  30   60   20 ia = − = −6 A, ib =  R   = 4 A, vc =   ( −60 ) = −40 V Req  30 + 15   eq   20 + 10 P3.7-11 a) (24)(12) Req = 24 12 = =8Ω 24 + 12 b) from voltage division: 100  20  100 5 v = 40  = V∴ i = 3 = A x  20 + 4  3 x 20 3  8  5 from current division: i = i = A x 8+8   6 3-22
  • P3.7-12 9 + 10 + 17 = 36 Ω 36 (18 ) a.) = 12 Ω 36+18 36 R b.) = 18 ⇒ 18 R = (18 )( 36 ) ⇒ R = 36 Ω 36+RP3.7-13 2 R( R ) 2 Req = = R 2R + R 3 v 2 240 Pdeliv. = = =1920 W to ckt Req 2 R 3 Thus R =45 ΩP3.7-14 R = 2 + 1 + ( 6 12 ) + ( 2 2 ) = 3 + 4 + 1 = 8 Ω eq 40 40 ∴i = = =5 A Req 8  6  i1 = i   6 + 12  ( )  = ( 5) 3 = 3 A 1 5 from current division i2 = i   2   2+2 ( )  = ( 5) 2 = 2 A 1 5 3-23
  • Verification ProblemsVP3-1 KCL at node a: i = i + i 3 1 2 − 1.167 = − 0.833 + ( −0.333) − 1.167= − 1.166 OK KVL loop consisting of the vertical 6 Ω resistor, the 3 Ω and4Ω resistors, and the voltage source: 6i + 3i + v + 12 = 0 3 2 yields v = −4.0 V not v = −2.0 VVP3-2 reduce circuit: 5+5=10 in parallel with 20 Ω gives 6.67Ω  6.67  by current division: i =   5 = 1.25 A  20 + 6.67  ∴Reported value was correct.VP3-3  320  v = o  320 + 650 + 230  ( 24 ) = 6.4 V ∴Reported value was incorrect.  3-24
  • VP3-4 KVL bottom loop: − 14 + 0.1iA + 1.2iH = 0 KVL right loop: − 12 + 0.05iB + 1.2iH = 0 KCL at left node: iA + iB = iH This alone shows the reported results were incorrect. Solving the three above equations yields: iA = 16.8 A iH = 10.3 A iB = −6.49 A ∴ Reported values were incorrect.VP3-5  1 Top mesh: 0 = 4 i a + 4 i a + 2  i a + − i b  = 10 ( −0.5 ) + 1 − 2 ( −2 )  2 Lower left mesh: vs = 10 + 2 ( i a + 0.5 − i b ) = 10 + 2 ( 2 ) = 14 VLower right mesh: vs + 4 i a = 12 ⇒ vs = 12 − 4 (−0.5) = 14 VThe KVL equations are satisfied so the analysis is correct. 3-25
  • VP3-6Apply KCL at nodes b and c to get: KCL equations: Node e: −1 + 6 = 0.5 + 4.5 Node a: 0.5 + i c = −1 ⇒ i c = −1.5 mA Node d: i c + 4 = 4.5 ⇒ i c = 0.5 mA Thats a contradiction. The given values of ia and ib are not correct.Design ProblemsDP3-1 Using voltage division: R 2 + aR p R 2 + aR p vm = 24 = 24 R1 + (1 − a ) R p + R 2 + aR p R1 + R 2 + R p vm = 8 V when a = 0 ⇒ R2 1 = R1 + R 2 + R p 3 vm = 12 V when a = 1 ⇒ R2 + R p 1 = R1 + R 2 + R p 2The specification on the power of the voltage source indicates 242 1 ≤ ⇒ R1 + R 2 + R p ≥ 1152 Ω R1 + R 2 + R p 2Try Rp = 2000 Ω. Substituting into the equations obtained above using voltage division gives3R 2 = R1 + R 2 + 2000 and 2 ( R 2 + 2000 ) = R1 + R 2 + 2000 . Solving these equations givesR1 = 6000 Ω and R 2 = 4000 Ω .With these resistance values, the voltage source supplies 48 mW while R1, R2 and Rp dissipate24 mW, 16 mW and 8 mW respectively. Therefore the design is complete. 3-26
  • DP3-2Try R1 = ∞. That is, R1 is an open circuit. From KVL, 8 V will appear across R2. Using voltage 200division, 12 = 4 ⇒ R 2 = 400 Ω . The power required to be dissipated by R2 R 2 + 200 82 1is = 0.16 W < W . To reduce the voltage across any one resistor, let’s implement R2 as the 400 8series combination of two 200 Ω resistors. The power required to be dissipated by each of these 42 1resistors is = 0.08 W < W . 200 8Now let’s check the voltage: 190 210 11.88 < v < 12.12 190 + 420 0 210 + 380 3.700 < v0 < 4.314 4 − 7.5% < v0 < 4 + 7.85%Hence, vo = 4 V ± 8% and the design is complete.DP3-3 Vab ≅ 200 mV 10 10 v= 120 Vab = (120) (0.2) 10 + R 10 + R 240 let v = 16 = ⇒ R=5Ω 10 + R 162 ∴P= = 25.6W 10DP3-4 N N 1 1 i = G v = v where G = ∑ = N  T R T n = 1 Rn  R iR ( 9 )(12 ) ∴N= = = 18 bulbs v 6 3-27
  • 28
  • Chapter 4 – Methods of Analysis of Resistive CircuitsExercisesEx. 4.3-1 v v −v a a b KCL at a: + + 3 = 0 ⇒ 5 v − 3 v = −18 3 2 a b v −v b a KCL at b: − 3 −1 = 0 ⇒ v − v = 8 2 b a Solving these equations gives: va = 3 V and vb = 11 VEx. 4.3-2 KCL at a: v v −v a a b + + 3 = 0 ⇒ 3 v − 2 v = −12 4 2 a b v v −v b a b − −4=0 KCL at a: 3 2 ⇒ − 3 v + 5 v = 24 a b Solving: va = −4/3 V and vb = 4 VEx. 4.4-1Apply KCL to the supernode to get v + 10 v 2+ b + b =5 20 30Solving: v = 30 V and v = v + 10 = 40 V b a b 4-1
  • Ex. 4.4-2 ( vb + 8) − ( −12) + vb = 3 ⇒ v = 8 V and v = 16 V 10 40 b aEx. 4.5-1Apply KCL at node a to express ia as a function of the node voltages. Substitute the result intovb = 4 ia and solve for vb . 6 vb  9 + vb  + =i ⇒ v = 4i = 4   ⇒ v = 4.5 V 8 12 a b a  12  b  Ex. 4.5-2The controlling voltage of the dependent source is a node voltage so it is already expressed as afunction of the node voltages. Apply KCL at node a. v −6 v −4v a + a a = 0 ⇒ v = −2 V 20 15 aEx. 4.6-1Mesh equations: −12 + 6 i + 3  i − i  − 8 = 0 ⇒ 9 i − 3 i = 20  1 2 1   1 2 8 − 3  i − i  + 6 i = 0 ⇒ − 3 i + 9 i = −8  1 2   2 1 2Solving these equations gives: 13 1 i = A and i = − A 1 6 2 6The voltage measured by the meter is 6 i2 = −1 V. 4-2
  • Ex. 4.7-1  3 −12Mesh equation: 9 + 3 i + 2 i + 4  i +  = 0 ⇒ ( 3 + 2 + 4 ) i = −9 − 3 ⇒ i= A  4 9The voltmeter measures 3 i = −4 VEx. 4.7-2 −33 2Mesh equation: 15 + 3 i + 6 ( i + 3) = 0 ⇒ ( 3 + 6 ) i = −15 − 6 ( 3) ⇒ i= = −3 A 9 3Ex. 4.7-3 3 3Express the current source current in terms of the mesh currents: = i1 − i 2 ⇒ i1 = + i 2 . 4 4 3 Apply KVL to the supermesh: −9 + 4i1 + 3 i 2 + 2 i 2 = 0 ⇒ 4  + i 2  + 5 i 2 = 9 ⇒ 9 i 2 = 6 4  2 4so i 2 = A and the voltmeter reading is 2 i 2 = V 3 3 4-3
  • Ex. 4.7-4Express the current source current in terms of the mesh currents: 3 = i1 − i 2 ⇒ i1 = 3 + i 2 .Apply KVL to the supermesh: −15 + 6 i1 + 3 i 2 = 0 ⇒ 6 ( 3 + i 2 ) + 3 i 2 = 15 ⇒ 9 i 2 = −3 1Finally, i 2 = − A is the current measured by the ammeter. 3ProblemsSection 4-3 Node Voltage Analysis of Circuits with Current SourcesP4.3-1 KCL at node 1: v v −v 1 1 2 −4 − 4 − 2 0= + +i = + + i = −1.5 + i ⇒ i = 1.5 A 8 6 8 6 (checked using LNAP 8/13/02) 4-4
  • P4.3-2 KCL at node 1: v −v v 1 2 1 + + 1 = 0 ⇒ 5 v − v = −20 20 5 1 2 KCL at node 2: v −v v −v 1 2 2 3 +2= ⇒ − v + 3 v − 2 v = 40 20 10 1 2 3 KCL at node 3: v −v v 2 3 3 +1 = ⇒ − 3 v + 5 v = 30 10 15 2 3 Solving gives v1 = 2 V, v2 = 30 V and v3 = 24 V. (checked using LNAP 8/13/02)P4.3-3 KCL at node 1: v −v v 1 2 1 4 − 15 4 + =i ⇒ i = + = −2 A 5 20 1 1 5 20 KCL at node 2: v −v v −v 1 2 2 3 +i = 5 2 15  4 − 15  15 − 18 ⇒ i = − + =2A 2  5  15 (checked using LNAP 8/13/02) 4-5
  • P4.3-4Node equations: v1 v1 − v2 −.003 + + =0 R1 500 v1 − v2 v2 − + − .005 = 0 500 R2When v1 = 1 V, v2 = 2 V 1 −1 1 −.003 + + = 0 ⇒ R1 = = 200 Ω R1 500 1 .003 + 500 −1 2 2 − + − .005 = 0 ⇒ R2 = = 667 Ω 500 R2 1 .005 − 500 (checked using LNAP 8/13/02)P4.3-5 Node equations: v1 v − v 2 v1 − v3 + 1 + =0 500 125 250 v − v2 v − v3 − 1 − .001 + 2 =0 125 250 v − v3 v1 − v3 v3 − 2 − + =0 250 250 500 Solving gives: v1 = 0.261 V, v2 = 0.337 V, v3 = 0.239 V Finally, v = v1 − v3 = 0.022 V (checked using LNAP 8/13/02) 4-6
  • Section 4-4 Node Voltage Analysis of Circuits with Current and Voltage SourcesP4.4-1Express the branch voltage of the voltage source in terms of its node voltages: 0 − va = 6 ⇒ va = −6 VKCL at node b: va − vb v −v −6 − vb v −v vb v −v +2= b c ⇒ +2= b c ⇒ −1− +2= b c ⇒ 30 = 8 vb − 3 vc 6 10 6 10 6 10 vb − vc vc 9KCL at node c: = ⇒ 4 vb − 4 vc = 5 vc ⇒ vb = vc 10 8 4 9 Finally: 30 = 8  vc  − 3 vc ⇒ vc = 2 V 4  (checked using LNAP 8/13/02)P4.4-2Express the branch voltage of each voltage source in terms of its node voltages to get: va = −12 V, vb = vc = vd + 8 4-7
  • KCL at node b: vb − va vb − ( −12 ) = 0.002 + i ⇒ = 0.002 + i ⇒ vb + 12 = 8 + 4000 i 4000 4000KCL at the supernode corresponding to the 8 V source: v 0.001 = d + i ⇒ 4 = vd + 4000 i 4000so vb + 4 = 4 − vd ⇒ ( vd + 8 ) + 4 = 4 − vd ⇒ vd = −4 V 4 − vdConsequently vb = vc = vd + 8 = 4 V and i = = 2 mA 4000 (checked using LNAP 8/13/02)P4.4-3Apply KCL to the supernode: va − 10 va va − 8 + + − .03 = 0 ⇒ va = 7 V 100 100 100 (checked using LNAP 8/13/02)P4.4-4 Apply KCL to the supernode: va + 8 ( va + 8 ) − 12 va − 12 va + + + =0 500 125 250 500 Solving yields va = 4 V (checked using LNAP 8/13/02) 4-8
  • P4.4-5The power supplied by the voltage source is v −v v −v   12 − 9.882 12 − 5.294  va ( i1 + i 2 ) = va  a b + a c  = 12  +   4 6   4 6  = 12(0.5295 + 1.118) = 12(1.648) = 19.76 W (checked using LNAP 8/13/02)P4.4-6Label the voltage measured by the meter. Notice that this is a node voltage. Write a node equation at the node at which the node voltage is measured.  12 − v m  v m v −8 − + + 0.002 + m =0  6000  R 3000 That is  6000  6000 3 +  v m = 16 ⇒ R = 16  R  −3 vm(a) The voltage measured by the meter will be 4 volts when R = 6 kΩ.(b) The voltage measured by the meter will be 2 volts when R = 1.2 kΩ. 4-9
  • Section 4-5 Node Voltage Analysis with Dependent SourcesP4.5-1 Express the resistor currents in terms of the node voltages: va − vc i 1= = 8.667 − 10 = −1.333 A and 1 v −v 2 − 10 i 2= b c = = −4 A 2 2 Apply KCL at node c: i1 + i 2 = A i1 ⇒ − 1.333 + ( −4 ) = A (−1.333) −5.333 ⇒ A= =4 −1.333 (checked using LNAP 8/13/02)P4.5-2 Write and solve a node equation: va − 6 v v − 4va + a + a = 0 ⇒ va = 12 V 1000 2000 3000 va − 4va ib = = −12 mA 3000 (checked using LNAP 8/13/02)P4.5-3 First express the controlling current in terms of the node voltages: 2 − vb i = a 4000 Write and solve a node equation: 2 − vb v  2 − vb  − + b − 5  = 0 ⇒ vb = 1.5 V 4000 2000  4000  (checked using LNAP 8/14/02) 4-10
  • P4.5-4 Apply KCL to the supernode of the CCVS to get 12 − 10 14 − 10 1 + − + i b = 0 ⇒ i b = −2 A 4 2 2 Next 10 − 12 1 ia = =−  −2 V 4 2 ⇒ r = =4 1 A r i a = 12 − 14   − 2 (checked using LNAP 8/14/02)P4.5-5 First, express the controlling current of the CCVS in v2 terms of the node voltages: i x = 2 Next, express the controlled voltage in terms of the node voltages: v2 24 12 − v 2 = 3 i x = 3 ⇒ v2 = V 2 5 so ix = 12/5 A = 2.4 A. (checked using ELab 9/5/02) 4-11
  • Section 4-6 Mesh Current Analysis with Independent Voltage SourcesP 4.6-1 2 i1 + 9 (i1 − i 3 ) + 3(i1 − i 2 ) = 0 15 − 3 (i1 − i 2 ) + 6 (i 2 − i 3 ) = 0 −6 (i 2 − i 3 ) − 9 (i1 − i 3 ) − 21 = 0 or 14 i1 − 3 i 2 − 9 i 3 = 0 −3 i1 + 9 i 2 − 6 i 3 = −15 −9 i1 − 6 i 2 + 15 i 3 = 21 so i1 = 3 A, i2 = 2 A and i3 = 4 A. (checked using LNAP 8/14/02)P 4.6-2 Top mesh: 4 (2 − 3) + R(2) + 10 (2 − 4) = 0 so R = 12 Ω. Bottom, right mesh: 8 (4 − 3) + 10 (4 − 2) + v 2 = 0 so v2 = −28 V. Bottom left mesh −v1 + 4 (3 − 2) + 8 (3 − 4) = 0 so v1 = −4 V. (checked using LNAP 8/14/02) 4-12
  • P 4.6-3 −6 Ohm’s Law: i 2 = = −0.75 A 8 KVL for loop 1: R i1 + 4 ( i1 − i 2 ) + 3 + 18 = 0 KVL for loop 2 + (−6) − 3 − 4 ( i1 − i 2 ) = 0 ⇒ − 9 − 4 ( i1 − ( −0.75 ) ) = 0 ⇒ i 1 = −3 A R ( −3) + 4 ( −3 − ( −0.75 ) ) + 21 = 0 ⇒ R = 4 Ω (checked using LNAP 8/14/02)P4.6-4 KVL loop 1: 25 ia − 2 + 250 ia + 75 ia + 4 + 100 (ia − ib ) = 0 450 ia −100 ib = −2 KVL loop 2: −100(ia − ib ) − 4 + 100 ib + 100 ib + 8 + 200 ib = 0 −100 ia + 500 ib = − 4 ⇒ ia = − 6.5 mA , ib = − 9.3 mA (checked using LNAP 8/14/02)P4.6-5 Mesh Equations: mesh 1 : 2i1 + 2 (i1 − i2 ) + 10 = 0 mesh 2 : 2(i2 − i1 ) + 4 (i2 − i3 ) = 0 mesh 3 : − 10 + 4 (i3 − i2 ) + 6 i3 = 0 Solving: 5 i = i2 ⇒ i = − = −0.294 A 17 (checked using LNAP 8/14/02) 4-13
  • Section 4-7 Mesh Current Analysis with Voltage and Current SourcesP4.7-1 1 mesh 1: i1 = A 2 mesh 2: 75 i2 + 10 + 25 i2 = 0 ⇒ i2 = − 0.1 A ib = i1 − i2 = 0.6 A (checked using LNAP 8/14/02)P4.7-2 mesh a: ia = − 0.25 A mesh b: ib = − 0.4 A vc = 100(ia − ib ) = 100(0.15) =15 V (checked using LNAP 8/14/02)P4.7-3Express the current source current as a function of the mesh currents: i1 − i2 = − 0.5 ⇒ i1 = i2 − 0.5Apply KVL to the supermesh: 30 i1 + 20 i2 + 10 = 0 ⇒ 30 (i2 − 0.5) + 20i2 = − 10 5 50 i2 − 15 = − 10 ⇒ i2 = = .1 A 50 i1 =−.4 A and v2 = 20 i2 = 2 V (checked using LNAP 8/14/02) 4-14
  • P4.7-4Express the current source current in termsof the mesh currents: ib = ia − 0.02Apply KVL to the supermesh: 250 ia + 100 (ia − 0.02) + 9 = 0 ∴ ia = − .02 A = − 20 mA vc = 100(ia − 0.02) = −4 V (checked using LNAP 8/14/02)P4.7-5Express the current source current in terms of the mesh currents: i 3 − i 1 = 2 ⇒ i1 = i 3 − 2Supermesh: 6 i1 + 3 i 3 − 5 ( i 2 − i 3 ) − 8 = 0 ⇒ 6 i1 − 5 i 2 + 8 i 3 = 8Lower, left mesh: −12 + 8 + 5 ( i 2 − i 3 ) = 0 ⇒ 5 i 2 = 4 + 5 i 3Eliminating i1 and i2 from the supermesh equation: 6 ( i 3 − 2 ) − ( 4 + 5 i 3 ) + 8 i 3 = 8 ⇒ 9 i 3 = 24  24 The voltage measured by the meter is: 3 i 3 = 3   = 8 V  9  (checked using LNAP 8/14/02) 4-15
  • P4.7-6Mesh equation for right mesh: 10 5 4 ( i − 2 ) + 2 i + 6 ( i + 3) = 0 ⇒ 12 i − 8 + 18 = 0 ⇒ i = − A=− A 12 6 (checked using LNAP 8/14/02)P 4.7-7 i2 = −3 A i1 − i2 = 5 ⇒ i1 − ( −3) = 5 ⇒ i1 = 2 A 2 ( i3 − i1 ) + 4 i3 + R ( i3 − i2 ) = 0 ⇒ 2 ( −1 − 2 ) + 4 ( −1) + R ( −1 − ( −3) ) = 0 ⇒ R=5 Ω (checked using LNAP 8/14/02) 4-16
  • P 4.7-8Express the controlling voltage of thedependent source as a function of themesh current v2 = 50 i1Apply KVL to the right mesh: −100 (0.04(50i1 ) − i1 ) + 50i1 + 10 = 0 ⇒ i1 = 0.2 A v2 = 50 i1 = 10 V (checked using LNAP 8/14/02)P 4.7-9 1 ib = 4ib − ia ⇒ ib = ia 3 1  −100  ia  + 200ia + 8 = 0 3  ⇒ ia = − 0.048 A (checked using LNAP 8/14/02)P4.7-10 Express the controlling current of the dependent source as a function of the mesh current: ib = .06 − iaApply KVL to the right mesh: −100 (0.06 − i a ) + 50 (0.06 − i a ) + 250 i a = 0 ⇒ ia = 10 mAFinally: vo = 50 i b = 50 (0.06 − 0.01) = 2.5 V (checked using LNAP 8/14/02) 4-17
  • P4.7-11 Express the controlling voltage of the dependent source as a function of the mesh current: vb = 100 (.006 − ia )Apply KVL to the right mesh: −100 (.006 − ia ) + 3[100(.006 − ia )] + 250 ia = 0 ⇒ ia = −24 mA (checked using LNAP 8/14/02)P4.7-12apply KVL to left mesh : − 3 + 10 × 103 i1 + 20 × 103 ( i1 − i2 ) = 0 ⇒ 30 × 103 i1 − 20 × 103 i2 = 3 (1)apply KVL to right mesh : 5 ×103 i1 + 100 × 103 i2 + 20 × 103 ( i2 − i1 ) = 0 ⇒ i1 = 8i2 ( 2) 6 3Solving (1) & ( 2 ) simultaneously ⇒ i1 = mA, i2 = mA 55 220Power delevered to cathode = ( 5 i1 ) ( i2 ) + 100 ( i2 )2 ( 55)( 3 220) + 100 ( 3 220) 2 = 5 6 = 0.026 mW∴ Energy in 24 hr. = Pt = ( 2.6 ×10−5 W ) ( 24 hr ) (3600 s hr ) = 2.25 J 4-18
  • P4.7-13 R2 vo RL R2(a) vo = − g R L v and v = vi ⇒ = −g R1 + R 2 vi R1 + R 2(b) ∴ vo = −g (5 ×103 )(103 ) = −170 ⇒ g = 0.0374 S vi 1.1×103PSpice ProblemsSP 4-1 4-19
  • SP 4-2From the PSpice output file: VOLTAGE SOURCE CURRENTS NAME CURRENT V_V1 -3.000E+00 V_V2 -2.250E+00 V_V3 -7.500E-01The voltage source labeled V3 is a short circuit used to measure the mesh current. The meshcurrents are i1 = −3 A (the current in the voltage source labeled V1) and i2 = −0.75 A (the currentin the voltage source labeled V3).SP 4-3The PSpice schematic after running the simulation:The PSpice output file: **** INCLUDING sp4_2-SCHEMATIC1.net **** * source SP4_2 V_V4 0 N01588 12Vdc 4-20
  • R_R4 N01588 N01565 4k V_V5 N01542 N01565 0Vdc R_R5 0 N01516 4k V_V6 N01542 N01516 8Vdc I_I1 0 N01565 DC 2mAdc I_I2 0 N01542 DC 1mAdc VOLTAGE SOURCE CURRENTS NAME CURRENT V_V4 -4.000E-03 V_V5 2.000E-03 V_V6 -1.000E-03From the PSpice schematic: va = −12 V, vb = vc = 4 V, vd = −4 V. From the output file: i = 2 mA.SP 4-4The PSpice schematic after running the simulation:The PSpice output file: VOLTAGE SOURCE CURRENTS NAME CURRENT V_V7 -5.613E-01 V_V8 -6.008E-01The current of the voltage source labeled V7 is also the current of the 2 Ω resistor at the top ofthe circuit. However this current is directed from right to left in the 2 Ω resistor while the currenti is directed from left to right. Consequently, i = +5.613 A. 4-21
  • Verification ProblemsVP 4-1 Apply KCL at node b: vb − va 1 v −v − + b c = 0 4 2 5 −4.8 − 5.2 1 − 4.8 − 3.0 − + ≠0 4 2 5 The given voltages do not satisfy the KCL equation at node b. They are not correct.VP 4-2 Apply KCL at node a: v −v  v − b a  − 2 + a = 0  4  2  20 − 4  4 − −2+ = −4≠ 0  4  2 The given voltages do not satisfy the KCL equation at node a. They are not correct. 4-22
  • VP 4-3 Writing a node equation:  12 − 7.5  7.5 7.5 − 6 − + + =0  R1  R3 R2 so 4.5 7.5 1.5 − + + =0 R1 R3 R2 There are only three cases to consider. Suppose R1 = 5 kΩ and R 2 = R 3 = 10 kΩ. Then 4.5 7.5 1.5 −0.9 + 0.75 + 0.15 − + + = = 0 R1 R3 R2 1000 This choice of resistance values corresponds to branch currents that satisfy KCL. Therefore, it is indeed possible that two of the resistances are 10 kW and the other resistance is 5 kW. The 5 kW is R1.VP 4-4KCL at node 1: v1 − v 2 v1 −8 − ( −20 ) −8 0= + +1 ⇒ + +1 = 0 20 5 20 5KCL at node 2:v1 − v 2 v 2 − v3 −8 − ( −20 ) −20 − ( −6 ) = 2+ ⇒ = 2+ 20 10 20 10 12 6 ⇒ = 20 10 v2 − v3 v3 −20 − ( −6 ) −6 −4 −6KCL at node 3: +1 = ⇒ +1 = ⇒ = 10 15 10 15 10 15KCL is satisfied at all of the nodes so the computer analysis is correct. 4-23
  • VP 4-5Top mesh: 10 (2 − 4) + 12(2) + 4 (2 − 3) = 0Bottom right mesh 8 (3 − 4) + 4 (3 − 2) + 4 = 0Bottom, left mesh: 28 + 10 (4 − 2) + 8 (4 − 3) ≠ 0 (Perhaps the polarity of the 28 V source wasentered incorrectly.)KVL is not satified for the bottom, left mesh so the computer analysis is not correct. 4-24
  • Design ProblemsDP 4-1Model the circuit as:a)We need to keep v2 across R2 as 4.8 ≤ v2 ≤ 5.4 0.3 A display is activeFor I =   0.1 A display is not active v2 − 15 v2 KCL at a: + +I =0 R1 R2 Assumed that maximum I results in minimum v2 and visa-versa. Then 4.8 V when I = 0.3 A v2 =  5.4 V when I = 0.1 ASubstitute these corresponding values of v2 and I into the KCL equation and solve for the resistances 4.8 − 15 4.8 + + 0.3 = 0 R1 R2 5.4 − 15 5.4 + + 0.1 = 0 R1 R2 ⇒ R1 = 7.89 Ω, R2 = 4.83 Ωb) 15 − 4.8 IR = = 1.292 A ⇒ PR = (1.292)2 (7.89) = 13.17 W 1max 7.89 1max IR = 5.4 = 1.118 A ⇒ PR = ( 5.4 )2 = 6.03 W 2max 4.83 2 max 4.83 maximum supply current = I R = 1.292 A 1maxc) No; if the supply voltage (15V) were to rise or drop, the voltage at the display would dropbelow 4.8V or rise above 5.4V.The power dissipated in the resistors is excessive. Most of the power from the supply isdissipated in the resistors, not the display. 4-25
  • DP 4-2Express the voltage of the 8 V source in terms of its node voltages to get vb − va = 8 . Apply KCLto the supernode corresponding to the 8 V source: va − v1 va vb vb − ( −v 2 ) + + + = 0 ⇒ 2 va − v1 + 2 vb + v 2 = 0 R R R R ⇒ 2 va − v1 + 2 ( va + 8 ) + v 2 = 0 ⇒ 4 va − v1 + v 2 + 16 = 0 v1 − v 2 ⇒ va = −4 4Next set va = 0 to get v1 − v 2 0= − 4 ⇒ v1 − v 2 = 16 V 4For example, v1 = 18 V and v2 = 2 V. 4-26
  • DP 4-3a)pply KCL to left mesh: −5 + 50 i1 + 300 (i1 − I ) = 0Apply KCL to right mesh: ( R + 2) I + 300 ( I − i1 ) = 0 150Solving for I: I= 1570 + 35 RWe desire 50 mA ≤ I ≤ 75 mA so if R = 100 Ω, then I = 29.59 mA fi l amp so the lamp will notlight.b) From the equation for I, we see that decreasing R increases I: try R = 50 Ω ⇒ I = 45 mA (wont light) try R = 25Ω ⇒ I = 61 mA ⇒ will lightNow check R±10% to see if the lamp will light and not burn out: −10% → 22.5Ω → I = 63.63 mA  lamp will  +10% → 27.5Ω → I = 59.23 mA  stay onDP 4-4Equivalent resistance: R = R1 || R 2 || ( R 3 + R 4 ) RVoltage division in the equivalent circuit: v1 = ( 25 ) 10 + RWe require vab = 10 V. Apply the voltage division principle in the left circuit to get: 4-27
  • 10 = R4 v1 = R4 × ( R1 R 2 ( R3 + R4 ) × 25 ) R3 + R4 ( R3 + R4 10 + R1 R 2 ( R3 + R4 ) )This equation does not have a unique solution. Here’s one solution: choose R1 = R2 = 25 Ω and R3 + R4 = 20 Ω then 10 = R4 × (12.5 20 ) × 25 ⇒ R = 18.4Ω 20 10 + (12.5 20 ) 4 and R3 + R4 = 20 ⇒ R3 = 1.6 ΩDP 4-5 Apply KCL to the left mesh: (R 1 + R 3 ) i1 − R 3 i2 − v1 = 0 Apply KCL to the left mesh: − R 3 i1 + (R 2 + R 3 ) i2 + v2 = 0Solving for the mesh currents using Cramer’s rule:  v1 − R3  ( R1 + R 3 ) v1     −v2 ( R 2 + R 3 )    i1 =   and i =  − R 3 − v2  2 ∆ ∆ where ∆ = ( R1 + R 3 ) ( R 2 + R 3 ) − R 3 2Try R1 = R2 = R3 = 1 kΩ = 1000 Ω. Then ∆ = 3 MΩ. The mesh currents will be given by i = [ 2v1 − v2 ] 1000 and i 2 = [ −2v2 + v1 ] 1000 ⇒ i = i1 − i2 = v1 + v2 1 3 × 10 6 3 × 10 6 3000Now check the extreme values of the source voltages: if v1 = v2 = 1 V ⇒ i = 2 mA okay 3 if v1 = v2 = 2 V ⇒ i = 4 mA okay 3 4-28
  • Chapter 5 Circuit TheoremsExercisesEx 5.3-1 R = 10 Ω and is = 1.2 A.Ex 5.3-2 R = 10 Ω and is = −1.2 A.Ex 5.3-3 R = 8 Ω and vs = 24 V.Ex 5.3-4 R = 8 Ω and vs = −24 V.Ex 5.4-1 20  10  2 vm = (15) + 20  − ( 2 ) = 6 + 20(− ) = −2 V 10 + 20 + 20  10 + (20 + 20)  5Ex 5.4-2 25 3 im = − ( 5) = 5 − 3 = 2 A 3+ 2 2+3Ex 5.4-3  3  3 vm = 3  ( 5) − (18 ) = 5 − 6 = −1 A  3 + (3 + 3)  3 + (3 + 3) 5-1
  • Ex 5.5-1 5-2
  • Ex 5.5-2 2 i a − 12 ia = ⇒ i a = −3 A 6 voc = 2 i a = −6 V 12 + 6 i a = 2 i a ⇒ i a = −3 A 2 3 i sc = 2 i a ⇒ i sc = ( −3 ) = − 2 A 3 −6 Rt = =3Ω −2Ex 5.6-1 5-3
  • Ex 5.6-2 2 i a − 12 ia = ⇒ i a = −3 A 6 voc = 2 i a = −6 V 12 + 6 i a = 2 i a ⇒ i a = −3 A 2 3 i sc = 2 i a ⇒ i sc = ( −3 ) = − 2 A 3 −6 Rt = =3Ω −2 5-4
  • Ex 5.6-3 12 × 24 12 × 24 Rt = = = 8Ω 12 + 24 36 24 voc = ( 30 ) = 20 V 12 + 24 20 i= 8+ REx 5.7-1 6 voc = (18) = 12 V 6+3 Rt = 2 + ( 3)( 6 ) = 4 Ω 3+ 6 For maximum power, we require R L = Rt = 4 Ω Then 2 voc 122 pmax = = =9 W 4 Rt 4 ( 4 ) 5-5
  • Ex 5.7-2 1 3 50 i sc = 1 1 1 ( 5.6 ) = ( 5.6 ) = 5 A + + 50 + 1 + 5 3 150 30 150 ( 30 ) Rt = 3 + = 3 + 25 = 28 Ω 150 + 30 pmax = R t i sc 2 = ( 28 ) 52 = 175 W 4 4Ex 5.7-3  10   R L  100 R L p=iv=    (10 ) =  Rt + R L   Rt + R L  ( Rt + R L ) 2 The power increases as Rt decreases so choose Rt = 1 Ω. Then 100 ( 5 ) pmax = i v = = 13.9 W (1 + 5) 2Ex 5.7-4From the plot, the maximum power is 5 W when R = 20 Ω. Therefore: Rt = 20 Ωand 2 v pmax = oc ⇒ voc = pmax 4 Rt = 5 ( 4 ) 20 = 20 V 4 Rt 5-6
  • ProblemsSection 5-3: Source TransformationsP5.3-1(a) 5-7
  • ∴ Rt = 2 Ω vt = − 0.5 V(b) −9 − 4i − 2i + (−0.5) = 0 −9 + (−0.5) i = = −1.58 A 4+2 v = 9 + 4 i = 9 + 4(−1.58) = 2.67 V(c) ia = i = − 1.58 A (checked using LNAP 8/15/02)P5.3-2 16Finally, apply KVL: −10 + 3 ia + 4 ia − =0 ∴ ia = 2.19 A 3 (checked using LNAP 8/15/02) 5-8
  • P5.3-3Source transformation at left; equivalent resistor for parallel 6 and 3 Ω resistors:Equivalents for series resistors, series voltage source at left; series resistors, then sourcetransformation at top:Source transformation at left; series resistors at right:Parallel resistors, then source transformation at left: 5-9
  • Finally, apply KVL to loop − 6 + i (9 + 19) − 36 − vo = 0 i = 5 / 2 ⇒ vo = −42 + 28 (5 / 2) = 28 V (checked using LNAP 8/15/02)P5.3-4 − 4 − 2000 ia − 4000 ia + 10 − 2000 ia − 3 = 0 ∴ ia = 375 µ A (checked using LNAP 8/15/02) 5-10
  • P5.3-5 −12 − 6 ia + 24 − 3 ia − 3 = 0 ⇒ ia = 1 A (checked using LNAP 8/15/02)P5.3-6A source transformation on the right side of the circuit, followed by replacing series resistorswith an equivalent resistor:Source transformations on both the right side and the left side of the circuit: 5-11
  • Replacing parallel resistors with an equivalent resistor and also replacing parallel current sourceswith an equivalent current source: 50 (100 ) 100Finally, va = ( 0.21) = ( 0.21) = 7 V 50 + 100 3 (checked using LNAP 8/15/02) 5-12
  • Section 5-4 SuperpositionP5.4–1Consider 6 A source only (open 9 A source) Use current division: v1  15  = 6  ⇒ v1 = 40 V 20 15 + 30  Consider 9 A source only (open 6 A source) Use current division: v2  10  = 9  ⇒ v2 = 40 V 20 10 + 35   ∴ v = v1 + v2 = 40 + 40 = 80 V (checked using LNAP 8/15/02)P5.4-2Consider 12 V source only (open both current sources) KVL: 20 i1 + 12 + 4 i1 + 12 i1 = 0 ⇒ i1 = −1/ 3 mAConsider 3 mA source only (short 12 V and open 9mA sources) Current Division:  16  4 i2 = 3   = 3 mA 16 + 20  5-13
  • Consider 9 mA source only (short 12 V and open 3mA sources) Current Division:  12  i3 = −9  = −3 mA  24 + 12   ∴ i = i1 + i2 + i3 = − 1/ 3 + 4 / 3 − 3 = − 2 mA (checked using LNAP 8/15/02)P5.4–3Consider 30 mA source only (open 15 mA and short 15 V sources). Let i1 be the part of i due tothe 30 mA current source.  2   6  ia = 30   = 6 mA ⇒ i1 = ia   = 2 mA  2+8  6 + 12 Consider 15 mA source only (open 30 mA source and short 15 V source) Let i2 be the part of idue to the 15 mA current source.  4   6  ib = 15   = 6 mA ⇒ i2 = ib   = 2 mA  4+6  6 + 12  5-14
  • Consider 15 V source only (open both current sources). Let i3 be the part of i due to the 15 Vvoltage source.  6 || 6   3  i3 = − 2.5   ( 6 || 6 ) + 12  = − 10  3 + 12  = −0.5 mA     Finally, i = i1 + i2 + i3 = 2 + 2 − 0.5 = 3.5 mA (checked using LNAP 8/15/02)P5.4–4Consider 10 V source only (open 30 mA source and Let v1 be the part of va due to theshort the 8 V source) 10 V voltage source. 100 ||100 v1 = (10 ) (100 ||100 ) + 100 50 10 = (10 ) = V 150 3Consider 8 V source only (open 30 mA source and Let v2 be the part of va due to theshort the 10 V source) 8 V voltage source. 100 ||100 v1 = (8) (100 ||100 ) + 100 50 8 = (8) = V 150 3 5-15
  • Consider 30 mA source only (short both the 10 Vsource and the 8 V source) Let v2 be the part of va due to the 30 mA current source. v3 = (100 ||100 ||100)(0.03) 100 = (0.03) = 1 V 3 10 8Finally, va = v1 + v2 + v3 = + +1 = 7 V 3 3 (checked using LNAP 8/15/02)P5.4-5Consider 8 V source only (open the 2 A source) Let i1 be the part of ix due to the 8 V voltage source. Apply KVL to the supermesh: 6 ( i1 ) + 3 ( i 1 ) + 3 ( i 1 ) − 8 = 0 8 2 i1 = = A 12 3Consider 2 A source only (short the 8 V source) Let i2 be the part of ix due to the 2 A current source. Apply KVL to the supermesh: 6 ( i 2 ) + 3 ( i 2 + 2 ) + 3 i2 = 0 −6 1 i2 = =− A 12 2 2 1 1Finally, i x = i1 + i 2 = − = A 3 2 6 5-16
  • Section 5-5: Thèvenin’s TheoremP5.5-1 (checked using LNAP 8/15/02) 5-17
  • P5.5-2The circuit from Figure P5.5-2a can be reduced to its Thevenin equivalent circuit in four steps: (a) (b) (c) (d)Comparing (d) to Figure P5.5-2b shows that the Thevenin resistance is Rt = 16 Ω and the opencircuit voltage, voc = −12 V. 5-18
  • P5.5-3The circuit from Figure P5.5-3a can be reduced to its Thevenin equivalent circuit in five steps: (a) (b) (c) (d) (e)Comparing (e) to Figure P5.5-3b shows that the Thevenin resistance is Rt = 4 Ω and the opencircuit voltage, voc = 2 V. (checked using LNAP 8/15/02) 5-19
  • P5.5-4Find Rt: 12 (10 + 2 ) Rt = =6Ω 12 + (10 + 2 )Write mesh equations to find voc: Mesh equations: 12 i1 + 10 i1 − 6 ( i2 − i1 ) = 0 6 ( i2 − i1 ) + 3 i 2 − 18 = 0 28 i1 = 6 i 2 9 i 2 − 6 i 1 = 18 1 36 i1 = 18 ⇒ i1 = A 2 14  1  7 i2 =  = A 3 2 3 7 1Finally, voc = 3 i 2 + 10 i1 = 3   + 10   = 12 V  3  2 (checked using LNAP 8/15/02) 5-20
  • P5.5-5Find voc:Notice that voc is the node voltage at node a. Expressthe controlling voltage of the dependent source as afunction of the node voltage: va = −vocApply KCL at node a:  6 − voc  voc  3  − + +  − voc  = 0  8  4  4  −6 + voc + 2 voc − 6 voc = 0 ⇒ voc = −2 VFind Rt:We’ll find isc and use it to calculate Rt. Notice thatthe short circuit forces va = 0Apply KCL at node a:  6−0 0  3  −  + +  − 0  + i sc = 0  8  4  4  6 3 i sc = = A 8 4 voc −2 8 Rt = = =− Ω i sc 3 4 3 (checked using LNAP 8/15/02) 5-21
  • P5.5-6Find voc: 2 va − va vaApply KCL at the top, middle node: = + 3 + 0 ⇒ va = 18 V 3 6The voltage across the right-hand 3 Ω resistor is zero so: va = voc = 18 VFind isc: 2 va − va va vApply KCL at the top, middle node: = + 3 + a ⇒ va = −18 V 3 6 3 va −18Apply Ohm’s law to the right-hand 3 Ω resistor : i sc = = = −6 V 3 3 v 18Finally: R t = oc = = −3 Ω i sc −6 (checked using LNAP 8/15/02) 5-22
  • P5.5-7 (a) −vs + R1 ia + ( d + 1) R 2 ia = 0 vs ia = R1 + ( d + 1) R 2 ( d + 1) R 2vs v oc = R1 + ( d + 1) R 2 vs ia = R1 i sc = ( d + 1) ia = ( d + 1) vs R1 vT −ia − d ia + − iT = 0 R2 R1 ia = −vT vT vT R 2 ( d + 1) + R1 iT = ( d + 1) + = × vT R1 R 2 R1 R 2 vT R1 R 2 Rt = = iT R1 + ( d + 1) R 2(b) Let R1 = R2 = 1 kΩ. Then 1000 1000 625 Ω = R t = ⇒ d= − 2 = −0.4 A/A d +2 625and 5 = voc = ( d + 1) vs ⇒ vs = −0.4 + 2 5 = 13.33 V d +2 −0.4 + 1 (checked using LNAP 8/15/02) 5-23
  • P5.5-8 From the given data: 2000  6= voc  R t + 2000   voc = 1.2 V  ⇒  2= 4000 voc   R t = −1600 Ω R t + 4000   When R = 8000 Ω, R v= voc 8000 Rt + R v= (1.2 ) = 1.5 V −1600 + 8000P5.5-9 From the given data: voc  0.004 = R t + 2000    voc = 24 V  ⇒  0.003 = voc   R t = 4000 Ω R t + 4000   (a) When i = 0.002 A: 24 voc 0.002 = ⇒ R = 8000 Ω i= 4000 + R Rt + R (b) Maximum i occurs when R = 0: 24 = 0.006 = 6 mA ⇒ i ≤ 6 mA 4000P5.5-10The current at the point on the plot where v = 0 is the short circuit current, so isc = 20 mA.The voltage at the point on the plot where i = 0 is the open circuit voltage, so voc = −3 V.The slope of the plot is equal to the negative reciprocal of the Thevenin resistance, so 1 0 − 0.002− = ⇒ R t = −150 Ω Rt −3 − 0 5-24
  • P5.5-11 −12 + 6000 ia + 2000 ia + 1000 ia = 0 ia = 4 3000 A 4 voc = 1000 ia = V 3 ia = 0 due to the short circuit −12 + 6000 isc = 0 ⇒ isc = 2 mA 4 voc Rt = = 3 = 667 Ω isc .002 4 ib = 3 667 + R ib = 0.002 A requires 4 R = 3 − 667 = 0 0.002 (checked using LNAP 8/15/02) 5-25
  • P5.5-12 10 = i + 0 ⇒ i = 10 A voc + 4 i − 2 i = 0 ⇒ voc = −2 i = −20 V i + i sc = 10 ⇒ i = 10 − i sc 4 i + 0 − 2 i = 0 ⇒ i = 0 ⇒ i sc = 10 A voc −20 Rt = = = −2 Ω isc 10 −20 −2 = iL = ⇒ RL = 12 Ω RL − 2 (checked using LNAP 8/15/02) 5-26
  • Section 5-6: Norton’s TheoremP5.6-1When the terminals of the boxes are open-circuited, no current flows in Box A, but the resistor inBox B dissipates 1 watt. Box B is therefore warmer than Box A. If you short the terminals ofeach box, the resistor in Box A will draw 1 amp and dissipate 1 watt. The resistor in Box B willbe shorted, draw no current, and dissipate no power. Then Box A will warm up and Box B willcool off.P5.6-2 (checked using LNAP 8/16/02) 5-27
  • P5.6-3P5.6-4To determine the value of the short circuit current, isc, we connect a short circuit across theterminals of the circuit and then calculate the value of the current in that short circuit. Figure (a)shows the circuit from Figure 5.6-4a after adding the short circuit and labeling the short circuitcurrent. Also, the meshes have been identified and labeled in anticipation of writing meshequations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively. In Figure (a), mesh current i2 is equal to the current in the short circuit. Consequently,i2 = isc . The controlling current of the CCVS is expressed in terms of the mesh currents as i a = i1 − i 2 = i1 − iscApply KVL to mesh 1 to get 3 i1 − 2 ( i1 − i 2 ) + 6 ( i1 − i 2 ) − 10 = 0 ⇒ 7 i1 − 4 i 2 = 10 (1)Apply KVL to mesh 2 to get 11 5 i 2 − 6 ( i1 − i 2 ) = 0 ⇒ − 6 i1 + 11 i 2 = 0 ⇒ i1 = i2 6Substituting into equation 1 gives  11  7  i 2  − 4 i 2 = 10 ⇒ i 2 = 1.13 A ⇒ i sc = 1.13 A 6  5-28
  • Figure (a) Calculating the short circuit current, isc, using mesh equations. To determine the value of the Thevenin resistance, Rt, first replace the 10 V voltagesource by a 0 V voltage source, i.e. a short circuit. Next, connect a current source across theterminals of the circuit and then label the voltage across that current source as shown in Figure(b). The Thevenin resistance will be calculated from the current and voltage of the current sourceas v Rt = T iTIn Figure (b), the meshes have been identified and labeled in anticipation of writing meshequations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively. In Figure (b), mesh current i2 is equal to the negative of the current source current.Consequently, i2 = i T . The controlling current of the CCVS is expressed in terms of the meshcurrents as i a = i1 − i 2 = i1 + i TApply KVL to mesh 1 to get 4 3 i1 − 2 ( i1 − i 2 ) + 6 ( i1 − i 2 ) = 0 ⇒ 7 i1 − 4 i 2 = 0 ⇒ i1 = i2 (2) 7Apply KVL to mesh 2 to get 5 i 2 + vT − 6 ( i1 − i 2 ) = 0 ⇒ − 6 i1 + 11 i 2 = −vTSubstituting for i1 using equation 2 gives 4  −6  i 2  + 11 i 2 = −vT ⇒ 7.57 i 2 = −vT 7 Finally, vT −vT −vT Rt = = = = 7.57 Ω iT −iT i2 5-29
  • vT Figure (b) Calculating the Thevenin resistance, R t = , using mesh equations. iT To determine the value of the open circuit voltage, voc, we connect an open circuit acrossthe terminals of the circuit and then calculate the value of the voltage across that open circuit.Figure (c) shows the circuit from Figure 4.6-4a after adding the open circuit and labeling theopen circuit voltage. Also, the meshes have been identified and labeled in anticipation of writingmesh equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively. In Figure (c), mesh current i2 is equal to the current in the open circuit. Consequently,i2 = 0 A . The controlling current of the CCVS is expressed in terms of the mesh currents as i a = i1 − i 2 = i1 − 0 = i1Apply KVL to mesh 1 to get 3 i1 − 2 ( i1 − i 2 ) + 6 ( i1 − i 2 ) − 10 = 0 ⇒ 3 i1 − 2 ( i1 − 0 ) + 6 ( i1 − 0 ) − 10 = 0 10 ⇒ i1 = = 1.43 A 7Apply KVL to mesh 2 to get 5 i 2 + voc − 6 ( i1 − i 2 ) = 0 ⇒ voc = 6 ( i1 ) = 6 (1.43) = 8.58 V Figure (c) Calculating the open circuit voltage, voc, using mesh equations.As a check, notice that R t isc = ( 7.57 )(1.13) = 8.55 ≈ voc (checked using LNAP 8/16/02) 5-30
  • P5.6-5 To determine the value of the short circuit current, Isc, we connect a short circuit across theterminals of the circuit and then calculate the value of the current in that short circuit. Figure (a)shows the circuit from Figure 4.6-5a after adding the short circuit and labeling the short circuitcurrent. Also, the nodes have been identified and labeled in anticipation of writing nodeequations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively. In Figure (a), node voltage v1 is equal to the negative of the voltage source voltage.Consequently, v1 = −24 V . The voltage at node 3 is equal to the voltage across a short, v3 = 0 .The controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. va = v2 . Thevoltage at node 3 is equal to the voltage across a short, i.e. v3 = 0 . Apply KCL at node 2 to get v1 − v 2 v 2 − v3 = ⇒ 2 v1 + v 3 = 3 v 2 ⇒ − 48 = 3 v a ⇒ v a = −16 V 3 6Apply KCL at node 3 to get v2 − v3 4 9 9 + v 2 = isc ⇒ v a = isc ⇒ isc = ( −16 ) = −24 A 6 3 6 6 Figure (a) Calculating the short circuit current, Isc, using mesh equations. To determine the value of the Thevenin resistance, Rth, first replace the 24 V voltagesource by a 0 V voltage source, i.e. a short circuit. Next, connect a current source circuit acrossthe terminals of the circuit and then label the voltage across that current source as shown inFigure (b). The Thevenin resistance will be calculated from the current and voltage of the currentsource as v R th = T iTAlso, the nodes have been identified and labeled in anticipation of writing node equations. Letv1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively. 5-31
  • In Figure (b), node voltage v1 is equal to the across a short circuit, i.e. v1 = 0 . Thecontrolling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. va = v2 . Thevoltage at node 3 is equal to the voltage across the current source, i.e. v3 = vT . Apply KCL at node 2 to get v1 − v 2 v 2 − v3 = ⇒ 2 v1 + v 3 = 3 v 2 ⇒ vT = 3 v a 3 6Apply KCL at node 3 to get v2 − v3 4 + v 2 + iT = 0 ⇒ 9 v 2 − v3 + 6 iT = 0 6 3 ⇒ 9 v a − vT + 6 iT = 0 ⇒ 3 v T − vT + 6 iT = 0 ⇒ 2 vT = −6 iTFinally, vT Rt = = −3 Ω iT vT Figure (b) Calculating the Thevenin resistance, R th = , using mesh equations. iT To determine the value of the open circuit voltage, voc, we connect an open circuit acrossthe terminals of the circuit and then calculate the value of the voltage across that open circuit.Figure (c) shows the circuit from Figure P 4.6-5a after adding the open circuit and labeling theopen circuit voltage. Also, the nodes have been identified and labeled in anticipation of writingnode equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively. In Figure (c), node voltage v1 is equal to the negative of the voltage source voltage.Consequently, v1 = −24 V . The controlling voltage of the VCCS, va, is equal to the node voltageat node 2, i.e. va = v2 . The voltage at node 3 is equal to the open circuit voltage, i.e. v3 = voc . Apply KCL at node 2 to get 5-32
  • v1 − v 2 v 2 − v3 = ⇒ 2 v1 + v 3 = 3 v 2 ⇒ − 48 + v oc = 3 v a 3 6Apply KCL at node 3 to get v2 − v3 4 + v 2 = 0 ⇒ 9 v 2 − v 3 = 0 ⇒ 9 v a = v oc 6 3Combining these equations gives 3 ( −48 + voc ) = 9 v a = voc ⇒ voc = 72 VFigure (c) Calculating the open circuit voltage, voc, using node equations.As a check, notice that R th I sc = ( −3)( −24 ) = 72 = Voc (checked using LNAP 8/16/02)Section 5-7: Maximum Power TransferP5.7-1a) For maximum power transfer, set RL equalto the Thevenin resistance: R L = R t = 100 + 1 = 101 Ωb) To calculate the maximum power, first replace the circuit connected to RL be its Theveninequivalent circuit: 5-33
  • 101The voltage across RL is vL = (100 ) = 50 V 101 + 101 2 v 502Then pmax = L = = 24.75 W R L 101P5.7-2Reduce the circuit using source transformations:Then (a) maximum power will be dissipated in resistor R when: R = Rt = 60 Ω and (b) the valueof that maximum power is P = i 2 ( R) = (0.03)2 (60) = 54 mW max R 5-34
  • P5.7-3  RL  v L = vS    RS + R L    2 2 vL vS RL ∴ pL = = R L ( RS + R L )2 By inspection, pL is max when you reduce RS to get the smallest denominator. ∴ set RS = 0P5.7-4Find Rt by finding isc and voc:The current in the 3 Ω resistor is zero because of the short circuit. Consequently, isc = 10 ix.Apply KCL at the top-left node to get 0.9 ix + 0.9 = 10 ix ⇒ ix = = 0.1 A 9so isc = 10 ix = 1ANextApply KCL at the top-left node to get 5-35
  • 0.9 ix + 0.9 = 10 ix ⇒ ix = = 0.1 A 9Apply Ohm’s law to the 3 Ω resistor to get voc = 3 (10 ix ) = 30 ( 0.1) = 3 VFor maximum power transfer to RL: voc 3 R L = Rt = = =3Ω isc 1The maximum power delivered to RL is given by 2 voc 32 3 pmax = = = W 4 R t 4 ( 3) 4P5.7-5  170 ( 20 )  10 −   30The required value of R is voc =  170 + 30   170 + 30 ( 20 )  50   ( 20 + 120 ) (10 + 50 ) = 50 Ω 170(20)(10) − 30(20)(50) 4000R = Rt = 8 + = = = 20 V 200 200 ( 20 + 120 ) + (10 + 50 ) The maximum power is given by 2 v 202 pmax = oc = =2W 4 R t 4 ( 50 ) 5-36
  • PSpice ProblemsSP5-1 a = 0.3333 b = 0.3333 c =33.33 V/A(a) vo = 0.3333 v1 + 0.3333 v2 + 33.33 i 3 18 7−(b) 7 = 0.3333 (10 ) + 0.3333 ( 8 ) + 33.33 i 3 ⇒ i3 = 3 = 3 = 30 mA 100 100 3 5-37
  • SP5-2Before the source transformation: VOLTAGE SOURCE CURRENTS NAME CURRENT V_V1 -3.000E-02 V_V2 -4.000E-02After the source transformation: VOLTAGE SOURCE CURRENTS NAME CURRENT V_V2 -4.000E-02 5-38
  • SP5-3 voc = −2 V VOLTAGE SOURCE CURRENTS NAME CURRENT V_V3 -7.500E-01 V_V4 7.500E-01 isc = 0.75 A Rt = −2.66 Ω 5-39
  • SP5-4 voc = 8.571 V VOLTAGE SOURCE CURRENTS NAME CURRENT V_V5 -2.075E+00 V_V6 1.132E+00 X_H1.VH_H1 9.434E-01 isc = 1.132 A Rt = 7.571 Ω 5-40
  • Verification ProblemsVP5-1 Use the data in the first two lines of the table to determine voc and Rt: voc  0.0972 =  Rt + 0  voc = 39.9 V  ⇒  0.0438 = voc   R t = 410 Ω R t + 500   Now check the third line of the table. When R= 5000 Ω: v 39.9 i = oc = = 7.37 mA R t + R 410 + 5000 which disagree with the data in the table. The data is not consistent.VP5-2 Use the data in the table to determine voc and isc: voc = 12 V (line 1 of the table) isc = 3 mA (line 3 of the table) voc so Rt = = 4 kΩ isc Next, check line 2 of the table. When R = 10 kΩ: v 12 i = oc = = 0.857 mA R t + R 10 (10 ) + 5 (103 ) 3 which agrees with the data in the table. v 12To cause i = 1 mA requires 0.001 = i = oc = ⇒ R = 8000 Ω R t + R 10 (103 ) + RI agree with my lab partner’s claim that R = 8000 causes i = 1 mA. 5-41
  • VP5-3 60 60 voc 11 i= = = 11 = 54.55 mA R t + R 6 110 + 40 60 + 40 ( ) 11 The measurement supports the prelab calculation.Design ProblemsDP5-1The equation of representing the straight line in Figure DP 5-1b is v = − R t i + voc . That is, theslope of the line is equal to -1 times the Thevenin resistance and the "v - intercept" is equal to the 0−5open circuit voltage. Therefore: R t = − = 625 Ω and voc = 5 V. 0.008 − 0 Try R1 = R 2 = 1 kΩ . (R1 || R2 must be smaller than Rt = 625 Ω.) Then R2 1 5= vs = vs ⇒ vs = 10 V R1 + R 2 2and R1 R2 625 = R 3 + = R3 + 500 ⇒ R3 = 125 Ω R1 + R2Now vs, R1, R2 and R3 have all been specified so the design is complete.DP5-2The equation of representing the straight line in Figure DP 5-2b is v = − R t i + voc . That is, theslope of the line is equal to -1 times the Thevenin resistance and the "v - intercept" is equal to the 0 − ( −3 )open circuit voltage. Therefore: R t = − = 500 Ω and voc = −3 V. −0.006 − 0 From the circuit we calculate R 3 ( R1 + R 2 ) R1 R 3 Rt = and voc = − is R1 + R 2 + R 3 R1 + R 2 + R 3so R 3 ( R1 + R 2 ) R1 R 3 500 Ω = and −3 V = − is R1 + R 2 + R 3 R1 + R 2 + R 3 5-42
  • Try R 3 = 1kΩ and R1 + R 2 = 1kΩ . Then R t = 500 Ω and 1000 R1 R1 −3 = − is = − i s ⇒ 6 = R1 i s 2000 2This equation can be satisfied by taking R1 = 600 Ω and is = 10 mA. Finally, R2 = 1 kΩ - 400 Ω =400 Ω. Now is, R1, R2 and R3 have all been specified so the design is complete.DP5-3The slope of the graph is positive so the Thevenin resistance is negative. This would require R1 R 2R3 + < 0 , which is not possible since R1, R2 and R3 will all be non-negative. R1 + R 2 Is it not possible to specify values of vs, R1, R2 and R3 that cause the current i and thevoltage v in Figure DP 5-3a to satisfy the relationship described by the graph in Figure DP 5-3b.DP5-4The equation of representing the straight line in Figure DP 5-4b is v = − R t i + voc . That is, theslope of the line is equal to the Thevenin impedance and the "v - intercept" is equal to the open −5 − 0circuit voltage. Therefore: R t = − = −625 Ω and voc = −5 V. 0 − 0.008 The open circuit voltage, voc, the short circuit current, isc, and the Thevenin resistance, Rt,of this circuit are given by R 2 ( d + 1) voc = vs R1 + ( d + 1) R 2 , isc = ( d + 1) v s R1and R1 R 2 Rt = R1 + ( d + 1) R 2Let R1 = R2 = 1 kΩ. Then 1000 1000 −625 Ω = R t = ⇒ d= − 2 = −3.6 A/A d +2 −625and ( d + 1) vs −3.6 + 2 −5 = ⇒ vs = ( − 5 ) = −3.077 V d +2 −3.6 + 1Now vs, R1, R2 and d have all been specified so the design is complete. 5-43
  • Chapter 6: The Operational AmplifierExercisesEx. 6.4-1 vs vs − vo + +0 = 0 R1 R2 vo R = 1+ 2 vs R1Ex. 6.4-2 a) R2 va = vs R1 + R2 va va − v0 + +0 =0 R3 R4 vo R vo  R2   R4  = 1+ 4 ⇒ =  1 +  va R3 vs  R1 + R2  R3  R2 R vo Rb) When R2 >> R1 then − 2 = 1 and − 1+ 4 R1 + R2 R2 vs R3 6-1
  • Ex. 6.5-1 vo vo − vs + +0= 0 R2 R1 vo R2 = vs R1 + R2Ex. 6.6–1vin − vout vin  R  + + 0 = 0 ⇒ vout = 1 + f  vin Rf R1  R1   when R f = 100 kΩ and R1 = 25 kΩ then vout  100 ⋅103  = 1 +  = 5 vin  25 ⋅103  6-2
  • Ex. 6.7-1  10 × 103   R2   10 ×103   R1  v3 =  −   v2 +  1 +  1 + v 3  1  10 × 10   R 2 + 10 × 10   10 × 10   10 × 10  3 3 3    R2   R1  = − v + 2 1 + 3  2 v 3  1    R 2 + 10 × 10   10 × 10  1We require v3 = ( 4 ) v1 −   v2 , so 5  R1  4 = 2 1 +  ⇒ R1 = 10 × 10 = 10 kΩ 3  10 ×103 and 1 R2 = ⇒ R 2 + 10 × 103 = 5 R 2 ⇒ R 2 = 2.5 kΩ 5 R 2 + 10 × 103 6-3
  • Ex. 6.7-2As in Ex 6.7-1  R2   R1  v3 = −  v + 2 1 + 3  2 v 3  1  R 2 + 10 × 10   10 × 10     4We require v3 = ( 6 ) v1 −   v2 , so 5  R1  6 = 2 1 +  ⇒ R1 = 20 ×10 = 20 kΩ 3  10 ×103 and 4 R2 = ⇒ 4 R 2 + 40 × 103 = 5 R 2 ⇒ R 2 = 40 kΩ 5 R 2 + 10 × 103Ex. 6.8-1Analysis of the circuit in Section 6.7 showed that output offset voltage = 6 vos + (50 × 103 ) ib1For a µ A741 op amp, vos ≤ 1 mV and ib1 ≤ 80 nA sooutput offset voltage = 6 vos + (50 × 103 )ib1 ≤ 6 (10−3 )+(50.103 )(80×10−9 ) = 10 mVEx. 6.8-2 R2  R  vo = − vin + 1 + 2  vos + R2ib1 R1  R1 When R2 = 10 kΩ, R1 = 2 kΩ, vos ≤ 5 mV and ib1 ≤ 500 nA then ( ) (output offset voltage ≤ 6 5 × 10−3 + 10 × 103 ) ( 500.10 ) ≤ 35×10 −9 −3 = 35 mV 6-4
  • Ex. 6.8-3Analysis of this circuit in Section 6.7 showed that output offset voltage = 6 vos + ( 50 ×103 ) ib1For a typical OPA1O1AM, vos = 0.1 mV and ib = 0.012 nA so output offset voltage ≤ 6 0.1×10−3  + ( 50 ×103 ) 0.012 ×10−9      −3 −6 −3 ≤ 0.6 ×10 + 0.6 ×10 − 0.6 ×10 = 0.6 mVEx. 6.8-4Writing node equations v− − vs v− − vo v− + + =0 Ra Rb Ri + Rs  R  vo −  − A i v  −  R +R   i s  + vo − v− = 0 R0 RbAfter some algebra vo R0 ( Ri + Rs ) + ARi R f Av = = vs ( R f + R0 ) ( Ri + Rs ) + Ra ( R f + R0 + Ri + Rs ) − ARi RaFor the given values, Av = −2.00006 V/V. 6-5
  • ProblemsSection 6-4: The Ideal Operational AmplifierP6.4-1 (checked using LNAP 8/16/02)P6.4-2 Apply KVL to loop 1: − 12 + 3000 i1 + 0 + 2000 i1 = 0 12 ⇒ i1 = = 2.4 mA 5000 The currents into the inputs of an ideal op amp are zero so io = i1 = 2.4 mA i2 = − i1 = − 2.4 mA va = i2 (1000 ) + 0 = −2.4 V Apply Ohm’s law to the 4 kΩ resistor vo = va − io ( 4000 ) = −2.4 − ( 2.4 ×10−3 ) ( 4000 ) = −12 V (checked using LNAP 8/16/02) 6-6
  • P6.4-3 The voltages at the input nodes of an ideal op amp are equal so va = −2 V . Apply KCL at node a: vo − ( −2 ) 12 − ( −2 ) + = 0 ⇒ vo = −30 V 8000 4000 Apply Ohm’s law to the 8 kΩ resistor −2 − vo io = = 3.5 mA 8000 (checked using LNAP 8/16/02)P6.4-4 The voltages at the input nodes of an ideal op amp are equal so v = 5 V . Apply KCL at the inverting input node of the op amp:  v −5  −3 − a  − 0.1× 10 − 0 = 0 ⇒ va = 4 V  10000  Apply Ohm’s law to the 20 kΩ resistor va 1 i = = mA 20000 5 (checked using LNAP 8/16/02)P6.4-5The voltages at the input nodes ofan ideal op amp are equal sova = 0 V . Apply KCL at node a:  v − 0   12 − 0  −3 − o −  − 2 ⋅10 = 0  3000   4000  ⇒ vo = − 15 VApply KCL at the output node ofthe op amp: v vio + o + o = 0 ⇒ io = 7.5 mA 6000 3000 (checked using LNAP 8/16/02) 6-7
  • P6.4-6The currents into the inputs of an ideal op ampare zero and the voltages at the input nodes ofan ideal op amp are equal so va = 2.5 V .Apply Ohm’s law to the 4 kΩ resistor: v 2.5 ia = a = = 0.625 mA 4000 4000Apply KCL at node a: ib = ia = 0.625 mAApply KVL: vo = 8000 ib + 4000 ia = (12 × 103 )( 0.625 × 10−3 ) = 7.5 V (checked using LNAP 8/16/02)P6.4-7  v − 0   va − 0  R2 − s −  + 0 = 0 ⇒ va = − vs  R1   R 2    R1 0 − va 0 − va R 2 + R3  R 2 + R3  io = + =− va =   R1 R 3  s v R2 R3 R 2 R3     v − 0   va − 0  R4 R2 R4 − o −  + 0 = 0 ⇒ vo = − va = vs  R 4   R3  R3 R1 R 3     6-8
  • P6.4-8The node voltages have beenlabeled using:1. The currents into theinputs of an ideal op amp arezero and the voltages at theinput nodes of an ideal opamp are equal.2. KCL3. Ohm’s lawThen v0 = 11.8 − 1.8 = 10 Vand 10 io = = 2.5 mA 4000 (checked using LNAP 8/16/02)P6.4-9KCL at node a:va − ( −18 ) v + a + 0 = 0 ⇒ va = −12 V 4000 8000The node voltages at the input nodes ofideal op amps are equal so vb = va .Voltage division: 8000vo = vb = −8 V 4000 + 8000 (check using LNAP 8/16/02) 6-9
  • Section 6-5: Nodal Analysis of Circuits Containing Ideal Operational AmplifiersP6.5-1 vb − 2 vb v +5 1 KCL at node b: + + b = 0 ⇒ vb = − V 20000 40000 40000 4 1 The node voltages at the input nodes of an ideal op amp are equal so ve = vb = − V. 4 ve v −v 10 KCL at node e: + e d = 0 ⇒ vd = 10 ve = − V 1000 9000 4 (checked using LNAP 8/16/02)P6.5-2Apply KCL at node a: va − 12 v v −00= + a + a ⇒ va = 4 V 6000 6000 6000Apply KCL at the inverting input of theop amp: v −0  0 − vo  − a +0+  = 0  6000   6000  ⇒ vo = −va = −4 VApply KCL at the output of the op amp:  0 − vo  vo io −  + = 0  6000  6000 (checked using LNAP 8/16/02) v ⇒ io = − o = 1.33 mA 3000 6-10
  • P6.5-3 Apply KCL at the inverting input of the op amp:  v − 0   vs − 0  − a −  = 0  R2   R1  R ⇒ va = − 2 vs R1Apply KCL at node a: va −v0 va va − 0  1 1 1  R R +R R +R R + + = 0 ⇒ v0 = R4  + +  va = 2 3 2 4 3 4 va R4 R3 R2  R4 R3 R2  R2 R3 R2 R3 + R2 R4 + R3 R4 =− vs R1 R3 vo 30 + 900 + 30Plug in values ⇒ yields =− = −200 V/V vs 4.8P6.5-4Ohm’s law: v1 − v2 i= R2KVL: R1 + R2 + R3 v0 = ( R1 + R2 + R3 ) i = ( v1 − v2 ) R2 6-11
  • P6.5-5 v1 −va v1 −v2  R  R + + 0 = 0 ⇒ va = 1+ 1  v1 − 1 v2 R1 R7  R7  R7 v2 − vb v −v  R  R − 1 2 + 0 = 0 ⇒ vb =  1+ 2  v2 − 2 v1 R2 R7  R7  R7  v −v  v −0 R6 −  b c  + c + 0 = 0 ⇒ vc = vb  R4  R6 R4 + R6  v −v   v −v  R R − a c  + c 0  + 0 = 0 ⇒ v0 = − 5 va + (1+ 5 )vc  R3   R5  R3 R3  R R R (R +R ) R  R R R (R +R ) R  v0 =  5 1 + 6 3 5 (1+ 2 )  v2 −  5 (1+ 1 ) + 6 3 5 2  v1  R3 R7 R3 ( R4 + R6 ) R7   R3 R7 R3 ( R4 + R6 ) R7  v −v i0 = c 0 = R5 6-12
  • P6.5-6 va vc 5KCL at node b: + = 0 ⇒ vc = − va 20 × 10 25 × 10 3 3 4  5  va −  − va  va − ( −12 ) va va + 0  4  = 0 ⇒ v = − 12 VKCL at node a: + + + a 40 × 10 3 10 × 10 20 ×10 3 3 10 ×103 13 5 15So vc = − va = − . 4 13 6-13
  • P6.5-7Apply KCL at the inverting inputnode of the op amp  v −0   ( va + 6 ) − 0  − a +0−  = 0  10000   30000  ⇒va = −1.5 VApply KCL to the super nodecorresponding the voltage source: va −0 va + 6− 0 + 10000 30000 v −v + a b + a ( v + 6 )−vb = 0 30000 10000 ⇒ 3va + va + 6 + va − vb + 3 ( va + 6 )− vb  = 0   ⇒ vb = 2va + 6 = 3 VApply KCL at node b: vb v − v  v − v   ( v + 6 ) − vb  + b 0 − a b − a  = 0 10000 30000  30000   10000  ⇒ 3vb +( vb − v0 )−( va − vb )−3( va + 6 ) − vb  = 0   ⇒ v0 = 8vb − 4va −18 = 12 VApply KCL at the output node of the op amp: v0 v −v i0 + + 0 b = 0 ⇒ i0 = − 0.7 mA 30000 30000 6-14
  • P6.5-8Apply KVL to the bottom mesh: −i0 (10000) − i0 (20000) + 5 = 0 1 ⇒ i0 = mA 6The node voltages at the input nodes of anideal op amp are equal. Consequently 10 va = 10000 i0 = V 6Apply KCL at node a: va v −v + a 0 = 0 ⇒ v0 = 3va = 5 V 10000 20000P6.5-9 vb + 12 vbKCL at node b: + = 0 ⇒ vb = −4 V 40000 20000The node voltages at the input nodes of an ideal op amp are equal, so vc = vb = −4 V .The node voltages at the input nodes of an ideal op amp are equal, so vd = vc + 0 × 10 4 = −4 V .  v f − vg  vg 2KCL at node g: − 3  + = 0 ⇒ vg = v f  20 × 10  40 ×10 3 3 6-15
  • 2The node voltages at the input nodes of an ideal op amp are equal, so ve = vg = vf . 3 2 vd − v f vd − ve vd − v f vd − 3 v f 6 24KCL at node d: 0 = + = + ⇒ v f = vd = − V 20 ×103 20 ×103 20 ×103 20 ×103 5 5 2 16Finally, ve = vg = vf = − V. 3 5P6.5-10By voltage division (or by applying KCL atnode a) R0 va = vs R1 + R0Applying KCL at node b: vb − vs vb −v0 + = 0 R1 R0 +∆R R0 +∆R ⇒ ( vb −vs )+ vb = v0 R1The node voltages at the input nodes of anideal op amp are equal so vb = va .  R +∆R  R0 R +∆R  ∆R  R0  ∆R v0 =  0 +1 − 0  vs = − vs =  −vs   R1  R1 + R0 R1  R1 + R0  R1 + R0  R0 6-16
  • Section 6-6: Design Using Operational AmplifierP6.6-1Use the current-to-voltage converter, entry (g) in Figure 6.6-1.P6.6-2Use the voltage –controlled current source, entry (i) in Figure 6.6-1.P6.6-3Use the noninverting summing amplifier, entry (e) in Figure 6.6-1. 6-17
  • P6.6-4Use the difference amplifier, entry (f) in Figure 6.6-1.P6.6-5Use the inverting amplifier and the summing amplifier, entries (a) and (d) in Figure 6.6-1.P6.6-6Use the negative resistance converter, entry (h) in Figure 6.6-1. 6-18
  • P6.6-7Use the noninverting amplifier, entry (b) in Figure 6.6-1. Notice that the ideal op amp forces thecurrent iin to be zero.P6.6-8 Summing Amplifier: va = − ( 6 v1 + 2 v2 )   ⇒ vo = 6 v1 + 2 v2 Inverting Amplifier: vo = −va P6.6-9 6-19
  • Using superposition, vo = v1 + v2 + v3 = −9 − 16 + 32 = 7 VP6.6-10 R1 6 12 24 6||12 6||24 R2 12||12||24 6||12||24 6||12||12 12||24 12||12 -vo/vs 0.8 0.286 0.125 2 1.25 R1 12||12 12||24 6||12||12 6||12||24 12||12||24 R2 6||24 6||12 24 12 6 -vo/vs 0.8 0.5 8 3.5 1.25 6-20
  • Section 6-7: Operational Amplifier Circuits and Linear Algebraic EquationsP6.7-1 6-21
  • P6.7-2 6-22
  • Section 6-8: Characteristics of the Practical Operational AmplifierP6.8-1 vout − vos vosThe node equation at node a is: = + ib1 100×10 3 10×103Solving for vout:  100×103  vout = 1+  vos + (100 × 10 ) ib1 = 11vos + (100 × 10 ) ib1 3 3  10×103  ( ) = 11 ( 0.03×10−3 )+(100×103 ) 1.2×10−9 = 0.45 mVP6.8-2 vos v −vThe node equation at node a is: + ib1 = 0 os 10000 90000Solving for vo:  90×103 vo = 1+ v + ( 90 × 103 ) ib1 = 10 vos + ( 90 × 103 ) i b1 3  os  10×10  = 10(5 ×10−3 )+ ( 90 ×103 ) (.05 × 10−9 ) = 50.0045 × 10−3 − 50 mV 6-23
  • P6.8-3 v1 −vin v1 v1 −v0  + + = 0 R1 Rin R2  v0 Rin ( R0 − AR2 )  ⇒ = v0 + Av1 v0 − v1 vin ( R1 + Rin )( R0 + R2 ) + R1 Rin (1+ A) + =0  R0 R2  P6.8-4a) v0 R 49×103 = − 2 = − = −9.6078 vin R1 5.1×103b) v0 = ( ( 2×106 ) 75−( 200,000)( 50×103 ) ) = −9.9957 vin (5×103 + 2×106 )(75+50×103 ) + (5×103 )(2×106 )(1+ 200,000)c) v0 2×106 (75− (200,000)(49×103 )) = = −9.6037 vin (5.1×103 +2×106 )(75+49×103 )+(5.1×103 )(2×106 )(1+200,000) 6-24
  • P6.8-5 Apply KCL at node b: R3 vb = (vcm − v p ) R2 + R3 Apply KCL at node a: va −v0 va −(vcm + vn ) + = 0 R4 R1 The voltages at the input nodes of an ideal op amp are equal so va = vb . R R +R v0 = − 4 (vcm + vn ) + 4 1 va R1 R1 R v0 = − 4 (vcm + vn ) + R1 ( R4 + R1 ) R3 (vcm − v p ) R1 ( R2 + R3 ) R4 +1 R4 R3 ( R4 + R1 ) R3 R1 R Rwhen = then = × 3 = 4 R1 R2 R1 ( R2 + R3 ) R3 +1 R2 R1 R2so R4 R R v0 = − (vcm + vn ) + 4 (vcm − v p ) = − 4 (vn + v p ) R1 R1 R1 6-25
  • PSpice ProblemsSP6-1:(a) v z = a vw + b v x + c v yThe following three PSpice simulations show 1 V = vz = a when vw= 1 V, vx = 0 V and vy = 0 V 4 V = vz = b when vw= 0 V, vx = 1 V and vy = 0 V -5 V = vz = c when vw= 0 V, vx = 0 V and vy = 1 VTherefore v z = vw + 4 v x − 5 v y(b) When vw= 2 V: vz = 4 vx − 5 v y + 2 1 V = vz = a when vw= 1 V, vx = 0 V and vy = 0 V: 6-26
  • 4 V = vz = b when vw= 0 V, vx = 1 V and vy = 0 V:-5 V = vz = c when vw= 0 V, vx = 0 V and vy = 1 V: 6-27
  • SP6-2 a) Using superposition and recognizing the inverting and noninverting amplifiers: 80  80  vo = − vs +  1 +  ( −2 ) = −3.2 vs − 8.4 25  25  b) Using the DC Sweep feature of PSpice produces the plot show below. Two points have been labeled in anticipation of c). c) Notice that the equation predicts ( −3.2 ) ( −5 ) − 8.4 = 7.6 and ( −3.2 ) ( 0 ) − 8.4 = −8.4 Both agree with labeled points on the plot. 6-28
  • SP6-3 VOLTAGE SOURCE CURRENTS NAME CURRENT V_V1 -3.000E-04 V_V2 -7.000E-04 v34 = −1.5 − −12 × 10−6 ≅ −1.5 V v23 = 4.5 − ( −1.5 ) = 6 V v50 = 12 − 0 = 12 V io = −7 ×10−4 = −0.7 mA 6-29
  • SP6-4V4 is a short circuit used to measure io.The input of the VCCS is the voltage of theleft-hand voltage source. (The nominal valueof the input is 20 mV.) The output of theVCCS is io.A plot of the output of the VCCS versus theinput is shown below.The gain of the VCVS is 50 × 10−6 − ( −50 × 10−6 ) 1 A gain = = ×10−3 100 × 10 − ( −100 × 10 ) 2 −3 −3 V(The op amp saturates for the inputs larger than0.1 V, limiting the operating range of thisVCCS.) 6-30
  • Verification ProblemsVP6-1 Apply KCL at the output node of the op amp vo v − ( −5 ) io = + o =0 10000 4000 Try the given values: io =−1 mA and vo = 7 V 7 7 − ( −5 ) −1×10−3 ≠ 3.7 × 10−3 = + 10000 4000 KCL is not satisfied. These cannot be the correct values of io and vo.VP6-2 va = ( 4 ×103 )( 2 ×10−3 ) = 8 V 12 ×103 vo = − va = −1.2 ( 8 ) = −9.6 V 10 ×103 So vo = −9.6 V instead of 9.6 V. 6-31
  • VP6-3First, redraw the circuit as:Then using superposition, and recognizing of the inverting and noninverting amplifiers:  6  4   4 vo =  −  −  ( −3) + 1 +  ( 2 ) = −18 + 6 = −12 V  2  2   2The given answer is correct.VP6-4First notice that ve = v f = vc is required by the ideal op amp. (There is zero current into the inputlead of an ideal op amp so there is zero current in the 10 kΩ connected between nodes e and f,hence zero volts across this resistor. Also, the node voltages at the input nodes of an ideal opamp are equal.)The given voltages satisfy all the node equations at nodes b, c and d: 0− (−5) 0 0− 2node b: + + =0 10000 40000 4000 0− 2 2 −5node c: = +0 4000 6000 2 −5 5 5−11node d: = + 6000 5000 4000Therefore, the analysis is correct. 6-32
  • VP6-5The given voltages satisfy the node equations at nodes b and e: −.25− 2 −.25 −.25−( −5 )node b: + + =0 20000 40000 40000 −2.5−( −0.25 ) −0.25node e: ≠ +0 9000 1000Therefore, the analysis is not correct. −2.5−( +0.25 ) +0.25Notice that = +0 9000 1000So it appears that ve = +0.25 V instead of ve = −0.25 V. Also, the circuit is an noninverting summer with Ra = 10 kΩ and Rb = 1 kΩ, K1 =1/ 2, K2= 1/ 4 and K4 = 9. The given node voltages satisfy the equation −2.5 = vd = K 4 ( K v + K v ) = 10  1 ( 2 )+ 1 ( −5)  1 a 2 c  2 4  None-the-less, the analysis is not correct. 6-33
  • Design ProblemsDP6-1From Figure 6.6-1g, this circuit vo 1 i out Rfis described by vo = R f i in . Since i out = , we require = = , or Rf = 1250 Ω 5000 4 i in 5000 (1250 ) i in 5 5Notice that i oa = i in + = i in . To avoid current saturation requires i in < i sat or 5000 4 4 4 i in < i sat . For example, if isat = 2 mA, then iin < 1.6 mA is required to avoid current saturation. 5DP6-2 3 3  3   12  3  3   12  vo = − vi + 3 = − vi + 1 +    5 − vi +  1 +   5 4 4  4   35  4  4   12 + 23  6-34
  • DP6-3 1 1  1 1(a) 12 v i + 6 = 24  v i + ( 5 )  ⇒ K 4 = 24, K 1 = , and K 2 = . Take Ra = 18 kΩ and 2 20  2 20Rb = 1 kΩ to get(b)(c)(d) 6-35
  • DP6-4 6-36
  • DP6-5 4We require a gain of = 200 . Using an inverting amplifier: 20 ×10−3 R2Here we have 200 = − . For example, let R1 = 0 and R2 = 1 MΩ. Next, using the 10 ×103 + R1noninverting amplifier: R2Here we have 200 = 1 + . For example, let R1 = 1 kΩ and R2 = 199 kΩ. R1The gain of the inverting amplifier circuit does not depend on the resistance of the microphone.Consequently, the gain does not change when the microphone resistance changes. 6-37
  • Chapter 7 Energy Storage ElementsExercisesEx. 7.3-1 2 2<t <4 2 t − 4 2 < t < 4 d  i C ( t ) = 1 v s ( t ) = −1 4 < t < 8 and i R (t ) = 1 v s (t ) =  8 − t 4<t <8 dt  0 otherwise  0   otherwise 2 t − 2 2 < t < 4 so i (t ) = i C ( t ) + i R ( t ) =  7 − t 4<t <8  0 otherwise Ex. 7.3-2 1 t 1 tv(t ) = ∫ i s (τ ) dτ + v(t 0 ) = ∫ i s (τ ) dτ − 12 C t0 1 0 3 tv(t ) = 3∫ 4 dτ − 12 = 12 t − 12 for 0 < t < 4 In particular, v(4) = 36 V. 0 tv(t ) = 3∫ ( −2 ) dτ + 36 = 60 − 6 t for 4 < t < 10 In particular, v(10) = 0 V. 4 tv(t ) = 3∫ 0 dτ + 0 = 0 for 10 < t 10Ex. 7.4-1 Cv 2 1 = ( 2×10 −4 ) (100 ) = 1 J 2 W = 2 2 vc ( 0 ) = vc ( 0 ) = 100 V + − 7-1
  • Ex. 7.4-2 a) W ( t ) = W ( 0 ) + ∫ 0 vi dt t First, W ( 0 ) = 0 since v ( 0 ) = 0 1 t Next, v( t ) = v( 0 ) + 4 t ∫ 0 i dt = 10 ∫ 0 2 dt = 2×10 t 4 C ∴ W (t ) = ∫ ( 2×10 ) t ( 2 )dt t 0 4 = 2 × 104 t 2 W (1s ) = 2 ×104 J = 20 kJ b) W (100s ) = 2 × 104 (100 ) 2 = 2 × 108 J = 200 MJEx. 7.4-3 We have v (0+ ) = v (0− ) = 3 V 1 t vc ( t ) = ∫ 0 i(t ) dt + vc (0) = 5 ∫ 0 3 e dt + 3 = 3 ( e −1)+3 = 3 e V, 0<t <1 t 5t 5t 5t C a) v(t ) = vR ( t ) + vc ( t ) = 5 i ( t ) + vc ( t ) = 15 e5t + 3 e5t = 18 e5t V, 0 < t < 1  W (t ) t =0.2 s = 6.65 J  W ( t ) = 1 Cvc2 ( t ) = 1 × 0.2 ( 3e5t ) = 0.9e10t J ⇒  2 b) W ( t ) t =0.8 s = 2.68 kJ 2 2 Ex. 7.5-1 7-2
  • Ex. 7.5-2 dv1 dv2 i i C v1 = v2 ⇒ = ⇒ 1 = 2 ⇒ i1 = 1 i2 dt dt C1 C2 C2 C  C2 KCL: i = i1 + i2 =  1 + 1 i2 ⇒ i2 = i  C2  C1 +C2Ex. 7.5-3 1 1 1 10(a) to (b) : = mF , (b) to (c) : 1+ = mF , 1 1 1 9 9 9 + + 1 1 1 3 3 3 1 1 1 1 10(c) to (d) : = + + ⇒ Ceq = mF Ceq 2 2 10 19 9 7-3
  • Ex. 7.6-1  2 2<t <4 2 t − 4 2 < t < 4 d  v L ( t ) = 1 i s ( t ) = −1 4 < t < 8 and v R (t ) = 1 is (t ) =  8 − t 4<t <8 dt  0 otherwise  0   otherwise 2 t − 2 2 < t < 4 so v(t ) = v L ( t ) + v R ( t ) =  7 − t 4<t <8  0 otherwise Ex. 7.6-2 1 t 1 t i (t ) = L ∫t 0 v s (τ ) dτ + i(t 0 ) = 1 ∫0 v s (τ ) dτ − 12 3 t i (t ) = 3∫ 4 dτ − 12 = 12 t − 12 for 0 < t < 4 In particular, i(4) = 36 A. 0 t i (t ) = 3∫ ( −2 ) dτ + 36 = 60 − 6 t for 4 < t < 10 In particular, i(10) = 0 A. 4 t i (t ) = 3∫ 0 dτ + 0 = 0 for 10 < t 10Ex. 7.7-1 di 1 d v = L dt =    4  dt ( 4 t e−t ) = (1−t ) e−t V P = vi =  (1−t ) e− t  ( 4 t e− t ) = 4 t (1−t ) e −2t W   1 11 W = Li 2 =   ( 4 t e− t ) = 2 t 2 e − 2t J 2 2 24 7-4
  • Ex. 7.7-2  0 t <0  0 t <0  2t 0<t <1  di 1 di   1 0<t <1v (t ) = L = and i ( t ) =  ⇒ v( t ) =  dt 2 dt −2( t − 2 ) 1<t < 2  −1 1<t < 2  0 t >2  0  t >2   0 t <0 2t 0<t <1 p (t ) = v (t ) i (t ) =  2( t − 2 ) 1<t < 2  0 t >2 W ( t ) = W ( t0 ) + ∫ tt p( t ) dt 0i (t ) = 0 for t < 0 ⇒ p ( t ) = 0 for t < 0 ⇒ W ( t0 ) = 00 < t < 1: W ( t ) = t ∫ 2 t dt = t 2 01<t < 2 : W ( t ) = W (1)+ ∫ 2 ( t − 2 ) dt = t t 1 2 − 4t + 4 t >2 : W (t ) = W ( 2) = 0Ex. 7.8-1 7-5
  • Ex. 7.8-2Ex. 7.8-3 1 t 1 t i1 = ∫ v dt + i1 ( t0 ) , i2 = ∫ v dt + i 2 ( t0 ) but i1 ( t0 ) = 0 and i 2 ( t0 ) = 0 L1 t 0 L 2 t0 1 t t 1 1  t 1 t i = i1 + i2 = ∫ t v dt + ∫ t v dt =  +  ∫ t v dt = ∫ v dt L1 0 0  L1 L2  0 LP t 0 1 t 1 ∫ t v dt i L 0 L1 L2 ∴1 = 1 = = i 1 t 1 1 L1 + L2 ∫ t v dt + LP 0 L1 L2 7-6
  • ProblemsSection 7-3: CapacitorsP7.3-1 1 t v (t ) = v (0) + i (τ ) dτ C ∫0 and q = Cv t In our case, the current is constant so ∫ i (τ ) dτ . 0 ∴ Cv ( t ) = Cv ( 0 ) + i t q −Cv( 0 ) 150×10 −(15×10 )( 5 ) −6 −6 ∴ t= = = 3 ms i 25×10−3P7.3-2 d 1d 1 i (t ) = C v (t ) = 12 cos ( 2t + 30° ) = (12 )( −2 ) sin ( 2t + 30° ) = 3cos ( 2t + 120° ) A dt 8 dt 8P7.3-3 d ( 3×10 ) cos ( 500t + 45 ) = C dt −3 ° 12 cos ( 500t − 45° ) = C (12 )( −500 ) sin ( 500t − 45° ) = C ( 6000 ) cos ( 500t + 45° ) 3×10−3 1 1so C= = ×10−6 = µ F 6×10 3 2 2 7-7
  • P7.3-4 1 t 1 t v (t ) = ∫0 i (τ ) dτ + v ( 0 ) = 2 ×10−12 ∫ i (τ ) dτ − 10 −3 C 00 < t < 2 ×10−9 is ( t ) = 0 ⇒ v ( t ) = 1 t ∫ 0 dτ − 10 −3 = −10−3 2 × 10−12 02 × 10−9 < t < 3 × 10−9 is ( t ) = 4 ×10−6 A 1 −12 ∫2ns ( 4 ×10−6 ) dτ − 10−3 = −5 × 10−3 + ( 2 × 106 ) t t ⇒ v (t ) = 2 × 10 In particular, v ( 3 × 10−9 ) = −5 ×10−3 + ( 2 × 106 ) ( 3 × 10−9 ) = 10−33 × 10−9 < t < 5 ×10−9 is ( t ) = −2 × 10−6 A 1 −12 ∫3ns ( −2 × 10−6 ) dτ + 10−3 = 4 × 10−3 − (106 ) t t ⇒ v (t ) = 2 × 10 In particular, v ( 5 × 10−9 ) = 4 ×10−3 − (106 ) ( 5 × 10−9 ) = −10−3 V5 ×10−9 < t is ( t ) = 0 ⇒ v ( t ) = 1 t 2 × 10−12 ∫ 5ns 0 dτ − 10−3 = −10−3 VP7.3-5 (b)  0 0 < t <1  d  4 1< t < 2 i (t ) = C v(t ) =  dt  −4 2 < t < 3 0  3<t (a) 1 t t v ( t ) = ∫ i (τ ) dτ + v ( 0 ) = ∫ i (τ ) dτ C 0 0 t For 0 < t < 1, i(t) = 0 A so v ( t ) = ∫ 0 dτ + 0 = 0 V 0 For 1 < t < 2, i(t) = (4t − 4) A so t ( ) 1=2 t t v ( t ) = ∫ ( 4τ − 4 ) dτ + 0 = 2τ 2 − 4τ 2 − 4t + 2 V 1 ( ) v(2) = 2 22 − 4 ( 2 ) + 2 = 2 V . For 2 < t < 3, i(t) = (−4t + 12) A so t ( ) 1 +2= ( −2 t ) t v ( t ) = ∫ ( −4τ + 12 ) dτ + 2 = −2τ 2 + 12τ 2 + 12 t − 14 V 2 ( ) v(3) = −2 32 + 12 ( 3) − 14 = 4 V t For 3 < t, i(t) = 0 A so v ( t ) = ∫ 0 dτ + 4 = 4 V 0 7-8
  • P7.3-6 (a)  0 0<t <2 d  i (t ) = C v(t ) = 0.1 2 < t < 6 dt 0  6<t (b) 1 t t v ( t ) = ∫ i (τ ) dτ + v ( 0 ) = 2 ∫ i (τ ) dτ C 0 0 t For 0 < t < 2, i(t) = 0 A so v ( t ) = 2 ∫ 0 dτ + 0 = 0 V 0 For 2 < t < 6, i(t) = 0.2 t − 0.4 V so t ( ) 2 =0.2 t t v ( t ) = 2 ∫ ( 0.2τ − 0.4 ) dτ + 0 = 0.2τ 2 − 0.8τ 2 − 0.8 t + 0.8 V 1 ( ) v(6) = 0.2 62 − 0.8 ( 6 ) + 0.8 = 3.2 V . For 6 < t, i(t) = 0.8 A so t v ( t ) = 2 ∫ 0.8 dτ + 3.2 = 1.6 t − 6.4 V 6P7.3-7 1 t v (t ) = v ( 0) + ∫ 0 i(τ ) dτ = 25 + 2.5 ×10 ∫ 0 ( 6×10 ) e dτ 4 t −3 −6τ C t = 25 + 150∫ 0 e −6τ dτ t  1  = 2 5 + 150  − e −6τ  = 50 − 25e−6t V  6 0P7.3-8 v 1 iR = = (1− 2e−2t ) × 10−3 = 25 (1− 2e −2t ) µ A 200×10 3 40 iC = C dv dt ( ) = 10×10−6 ( −2 ) ( −10 e−2t ) = 200 e−2 t µ A i = iR + i C = 200 e−2t + 25 − 50 e −2t = 25 + 150e −2t µA 7-9
  • Section 7-4: Energy Storage in a CapacitorP7.4-1Given  0 t<2  i ( t ) = 0.2 ( t − 2 ) 2 < t < 6  0.8 t >6 The capacitor voltage is given by 1 t t v (t ) = i (τ ) dτ + v ( 0 ) = 2 ∫ i (τ ) dτ + v ( 0 ) 0.5 ∫ 0 0For t < 2 t v ( t ) = 2 ∫ 0 dτ + 0 = 0 0In particular, v ( 2 ) = 0. For 2 < t < 6 v ( t ) = 2 ∫ 2 (τ − 2 ) dτ + 0 = ( 0.2τ 2 − 0.8τ ) = ( 0.2 t 2 − 0.8 t + 0.8 ) V = 0.2 ( t 2 − 4 t + 4 ) V t t 2 2In particular, v ( 6 ) = 3.2 V. For 6 < t t v ( t ) = 2 ∫ 0.8 dτ + 3.2 = 1.6τ 2 + 3.2 = (1.6 t − 6.4 ) V = 1.6 ( t − 4 ) V t 6Now the power and energy are calculated as  0 t<2  p ( t ) = v ( t ) i ( t ) = 0.04 ( t − 2 ) 2 2<t <6  1.28 ( t − 4 ) 6<t and  0 t<2 t   W (t ) = ∫ p (τ ) dτ =  0.01( t − 2 ) 4 2<t<6 0  6<t 0.8 ( t − 4 ) − 0.64 2  7-10
  • 7-11
  • These plots were produced using three MATLAB scripts:capvol.m function v = CapVol(t) if t<2 v = 0; elseif t<6 v = 0.2*t*t - .8*t +.8; else v = 1.6*t - 6.4; endcapcur.m function i = CapCur(t) if t<2 i=0; elseif t<6 i=.2*t - .4; else i =.8; endc7s4p1.m t=0:1:8; for k=1:1:length(t) i(k)=CapCur(k-1); v(k)=CapVol(k-1); p(k)=i(k)*v(k); w(k)=0.5*v(k)*v(k); end plot(t,i,t,v,t,p) text(5,3.6,v(t), V) text(6,1.2,i(t), A) text(6.9,3.4,p(t), W) title(Capacitor Current, Voltage and Power) xlabel(time, s) % plot(t,w) % title(Energy Stored in the Capacitor, J) % xlabel(time, s) 7-12
  • P7.4-2 dv   ic ( 0 ) = 0.2 A ic = C = (10×10−6 ) ( −5 )( −4000 ) e −4000t = 0.2e−4000t A ⇒   ic (10ms ) = 8.5×10 A −19 dt  1 W ( t ) = Cv 2 ( t ) and v ( 0 ) = 5 − 5e0 = 0 ⇒ W ( 0 ) = 0 2 v (10×10 ) = 5 − 5 e −40 = 5 − 21.2 × 10−18 ≅ 5 ⇒ W (10 ) = 1.25×10−4 J −3P7.4–3 dvc i (t ) = C so read off slope of vc (t ) to get i (t ) dt p (t ) = vc (t ) i (t ) so multiply vc (t ) & i(t ) curves to get p (t )P7.4-4 1 t 1 t  π  5 π 5  πvc ( t ) = vc ( 0 ) + ∫0 i dτ = vc ( 0 ) + 2 ∫0 50 cos10t + 6 dτ =  vc ( 0 ) − 2 sin 6  + 2 s in 10t + 6     C    5 π 5  πNow since vc ( t )ave = 0 ⇒ vc ( 0 ) − sin = 0 ⇒ vc ( t ) = sin 10t +  V 2 6 2  6∴ Wmax 1 = C v2 = ( 2×10 ) ( 2.5) = 6.25 µ J −6 2 2 c max 2 π π πFirst non-negative t for max energy occurs when: 10t + = ⇒t = = 0.1047 s 6 2 30 7-13
  • P7.4-5 Max. charge on capacitor = C v = (10×10−6 ) ( 6 ) = 60 µ C ∆q 60×10−6 ∆t = = = 6 sec to charge i 10×10−6 1 1 stored energy = W = C v 2 = (10×10−6 ) ( 6 ) =180 µ J 2 2 2Section 7-5: Series and Parallel capacitorsP7.5-1 2µ F 4 µ F = 6µ F 6µ F⋅3µ F 6µ F in series with 3µ F = = 2µ F 6µ F+3µ F d i (t ) = 2 µ F (6 cos100t ) = (2×10−6 ) (6) (100) (− sin100t ) A = −1.2 sin100t mA dtP7.5-2 4 µ F×4 µ F 4 µ F in series with 4 µ F = = 2 µF 4 µ F+4 µ F 2 µF 2 µF = 4 µF 4 µ F in series with 4 µ F = 2 µ F d i (t ) =(2×10−6 ) (5+ 3 e −250t ) = (2×10−6 ) (0+ 3(−250) e −250t ) A = −1.5 e −250t mA dtP7.5-3 C ⋅C C C in series with C = = C +C 2 C 5 C C = C 2 2 5 C⋅ 5 C 5 C in series with C = 2 = C 2 C+ 5C 7 2 5  d 5  (25×10−3 ) cos 250t =  C  (14sin 250t ) =  C (14)(250) cos 250t  7  dt 7  so 25×10−3 = 2500 C ⇒ C = 10×10−6 = 10 µF 7-14
  • Section 7-6: InductorsP7.6-1 di F in d m a x . v o ltag e a c ro ss c o il: v (t ) = L = 2 0 0 [1 0 0 ( 4 0 0 ) c o s 4 0 0 t ] V dt 8 ×1 0 6 V ∴ v m ax = 8 ×1 0 6 V th u s h a ve a fie ld o f = 4 ×1 0 6 V 2 m m w h ic h e x ce e d s d ie le c tric stre n g th in a ir o f 3 ×1 0 V /m 6 ∴ W e g e t a d isc h a rg e a s th e a ir is io n iz e d .P7.6-2 di v=L + R i = (.1) (4e− t − 4te− t ) + 10(4te− t ) = 0.4 e − t + 39.6t e − t V dtP7.6-3 (a)  0 0 < t <1  4 1< t < 2 d  v(t ) = L i (t ) =  dt  −4 2 < t < 3  0 3<t (b) 1 t t i (t ) = v (τ ) dτ + i ( 0 ) = ∫ v (τ ) dτ L ∫0 0 t For 0 < t < 1, v(t) = 0 V so i ( t ) = ∫ 0 dτ + 0 = 0 A 0 For 1 < t < 2, v(t) = (4t − 4) V so t i ( t ) = ∫ ( 4τ − 4 ) dτ + 0 = ( 2τ 2 − 4τ ) =2 t 2 − 4 t + 2 A t 0 1 i (2) = 4 ( 22 ) − 4 ( 2 ) + 2 = 2 A For 2 < t < 3, v(t) = −4t + 12 V so t i ( t ) = ∫ ( −4τ + 12 ) dτ + 2 = ( −2τ 2 + 12τ ) +2= ( −2 t 2 + 12 t − 14 ) A t 2 2 i (3) = −2 ( 32 ) + 12 ( 3) − 14 = 4 A t For 3 < t, v(t) = 0 V so i ( t ) = ∫ 0 dτ + 4 = 4 A 3 7-15
  • P7.6-4 d v (t ) = (250 × 10 −3 ) (120 × 10 −3 ) sin(500t − 30° ) = (0.25)(0.12)(500) cos(500 t − 30° ) dt = 15 cos(500t − 30° )P7.6-5 1 t iL (t ) = 5 ×10−3 ∫ v (τ ) dτ 0 s − 2 ×10−6 for 0< t < 1 µ s vs (t ) = 4 mV 1 t  4×10−3  ∫0 4 ×10−3 dτ − 2×10−6 =  −6 −6 iL (t ) =  t − 2×10 = 0.8− 2×10 A 5×10−3  5×10−3   4×10−3  6 −3 ( iL (1µs) =  1×10−6 )  − 2×10−6 = − ×10−6 A =−1.2 A  5×10  5 for 1µ s <t < 3 µ s vs (t ) = −1 mV 1 6 1×10−3 6 −3 ∫1µ s ( −1×10−3 ) dτ − ×10−6 =− (t −1×10−6 ) − ×10−6 =( −0.2 t −10−6 ) A t iL (t ) = −3 5×10 5 5×10 5  1×10−3  iL ( 3µ s ) =  − −3 + 3×10−6  − 1×10−6 = −1.6 µA  5×10  for 3µ s < t vs (t ) = 0 so iL (t ) remains − 1.6 µA 7-16
  • P7.6-6 d v(t ) = ( 2 ×10 ) i 3 s (t ) + ( 4 ×10 ) dt −3 is (t ) (in general)  1×10−3  d for 0<t <1 µ s is (t ) = (1)  −6  t = 103 t ⇒ is (t ) = 1×103  1×10  dt v(t ) = (2×103 )(1×103 ) t + 4×10−3 (1×103 ) = ( 2×106 t + 4 ) V d for 1µ s <t < 3µ s is (t ) = 1 mA ⇒ is (t ) = 0 dt v(t ) = (2×103 )(1×10−3 ) + ( 4×10−3 )×0 = 2 V  1×10−3  d 1×10−3 for 3µ s< t < 5µ s is (t ) = 4×10−3 − −6  t ⇒ is (t ) = − −6 =−103  1×10  dt 1×10 v(t ) = ( 2×103 )( 4×10−3 −103 t )+ 4×10−3 ( −103 ) = 4 − ( 2×106 ) t d when 5µ s <t< 7µ s is (t ) = −1×10−3 and is (t ) = 0 dt v(t ) = ( 2×103 )(10−3 ) = − 2 V  1×10−3  d when 7µ s< t < 8µ s is (t ) =  −6  t − 8×10−3 ⇒ is (t ) = 1×103  1×10  dt v(t ) = ( 2 × 103 )(103 t − 8 × 10−3 ) + ( 4 × 10−3 )(103 ) = −12 + ( 2 × 106 ) t d when 8µ s < t , then is (t ) = 0 ⇒ is (t ) = 0 dt v(t )= 0P7.6-8 (a)  0 0<t <2 d  v(t ) = L i (t ) = 0.1 2 < t < 6 dt 0  6<t (b) 1 t t i ( t ) = ∫ v (τ ) dτ + i ( 0 ) = 2 ∫ v (τ ) dτ L 0 0 t For 0 < t < 2, v(t) = 0 V so i ( t ) = 2 ∫ 0 dτ + 0 = 0 A 0 For 2 < t < 6, v(t) = 0.2 t − 0.4 V so t i ( t ) = 2∫ ( 0.2τ − 0.4 ) dτ + 0 = ( 0.2τ 2 − 0.8τ ) =0.2 t 2 − 0.8 t + 0.8 A t 2 2 ( ) i (6) = 0.2 62 − 0.8 ( 6 ) + 0.8 = 3.2 A . For 6 < t, v(t) = 0.8 V so 7-17
  • t i ( t ) = 2 ∫ 0.8 dτ + 3.2 = (1.6 t − 6.4 ) A 6Section 7-7: Energy Storage in an InductorP7.7-1 0 t<0 d  v( t ) =100×10 −3 i ( t ) = 0.4 0≤ t ≤1 dt 0  t>1 0 t <0  p( t ) =v( t ) i( t ) = 1.6t 0≤t ≤1 0 t >1   0 t <0 t  W (t ) = ∫ p (τ ) dτ = 0.8t 2 0<t <1 0  0.8 t >1  7-18
  • P7.7-2  d  p (t ) = v (t ) i (t ) = 5 (4sin 2t )  (4sin 2t )  dt  = 5 (8cos 2t ) (4sin 2t ) = 80 [2 cos 2t sin 2t ] = 80 [sin(2t + 2t ) + sin(2t − 2t )] = 80 sin 4t W t t 80 W(t ) = ∫ p(τ ) dτ = 80∫ sin4τ dτ = − [cos 4τ |t0 ] = 20 (1 − cos 4t ) 0 0 4P7.7-3 1 t 25×10−3 ∫ 0 i(t ) = 6 cos 100τ dτ + 0 6 = −3 [sin 100τ | 0 ] = 2.4sin100 t t (25×10 )(100) p(t ) = v(t ) i(t ) = (6 cos100 t )(2.4 sin100t ) = 7.2 [ 2(cos100 t )(sin100 t ) ] = 7.2 [sin 200 t + sin 0] = 7.2 sin 200 t t t W (t ) = ∫0 p (τ ) dτ = 7.2 ∫0 sin 200τ dτ 7.2 =−  cos 200τ |t0  200   = 0.036[1 − cos 200t ] J = 36 [1 − cos 200t ] mJSection 7-8: Series and Parallel InductorsP7.8-1 6×3 6H 3H = = 2 H and 2 H + 2 H = 4 H 6+3 1 t 6 i (t ) = ∫ 0 6 cos100τ dτ = sin100τ | t0  = 0.015sin100 t A = 15sin100 t mA 4 4×100  P7.8-2 4 mH + 4 mH = 8 mH , 8mH 8mH = (8×10 )×(8×10 ) −3 −3 = 4 mH 8×10−3 +8×10−3 and 4 mH + 4 mH = 8 mH d v(t ) = (8×10−3 ) (5+ 3e −250t ) = (8×10−3 ) (0+ 3(−250) e −250t ) =−6 e −250t V dt 7-19
  • P7.8-3 L⋅ L L L 5 L L = = and L + L + = L L+ L 2 2 2 5  d 25cos 250 t =  L   2  dt ( ) (14×10−3 ) sin 250 t =  5 L (14×10−3 )(250) cos 250 t 2   25 so L = = 2.86 H 5 −3 (14×10 ) (250) 2Section 7-9: Initial Conditions of Switched CircuitsP7.9-1Then i L ( 0+ ) = i L ( 0− ) = 0 and v C ( 0+ ) = v C ( 0− ) = 12 VNext 7-20
  • P7.9-2Then i L ( 0+ ) = i L ( 0− ) = 1 mA and v C ( 0+ ) = v C ( 0− ) = 6 VNextP7.9-3Then i L ( 0+ ) = i L ( 0− ) = 0 and v C ( 0+ ) = v C ( 0− ) = 0 VNext 7-21
  • P7.9-4 at t = 0− KVL: − vc (0− ) + 32 − 15 = 0 ⇒ vc (0− ) = vc (0+ ) =17 V at t = 0+ Apply KCL to supernode shown above: −15 − 9 15 ic ( 0+ ) + − + 0.003 = 0 ⇒ ic ( 0+ ) = 6 mA 4000 5000 Then dvc ic ( 0+ ) 6 ×10−3 V = = −6 = 3000 dt t =0+ C 2 ×10 s 7-22
  • Section 7-10: Operational amplifier Circuits and Linear Differential EquationsP7.10-1P7.10-2 7-23
  • P7.10-3P7.10-4 7-24
  • Verification ProblemsVP7-1 We need to check the values of the inductor current at the ends of the intervals. ? 1 at t = 1 0.025 = − + 0.065 = 0.025 ( Yes!) 25 3 ? 3 − 0.115 at t = 3 − + 0.065 = 25 50 −0.055 = −0.055 ( Yes!) 9 ? at t = 9 − 0.115 = 0.065 50 0.065 = 0.065 ( Yes!) The given equations for the inductor current describe a current that is continuous, as must be the case since the given inductor voltage is bounded.VP7-2 We need to check the values of the inductor current at the ends of the intervals. 1 ? 1 at t = 1 − + 0.025 = − + 0.03 ( Yes!) 200 100 4 ? 4 − 0.03 ( No!) at t = 4 − + 0.03 = 100 100 The equation for the inductor current indicates that this current changes instantaneously at t = 4s. This equation cannot be correct.Design ProblemsDP7-1 d i (t )a) v ( t ) = −6 e −3t is proportional to i(t) so the element is a capacitor. C = = 0.5 F . dt d v (t ) dt d v (t )b) i ( t ) = −6 e −3t is proportional to v(t) so the element is an inductor. L = = 0.5 H . dt d i (t ) dt v (t )c) v(t) is proportional to i(t) so the element is a resistor. R = = 2 Ω. i (t ) 7-25
  • DP7-21.131cos ( 2t + 45° ) = 1.131  cos ( 45° ) cos ( 2t ) − sin ( 45° ) sin ( 2t )    = 0.8 cos 2 t − 0.8 sin 2 tThe first term is proportional to the voltage. Associate it with theresistor. The noticing that t t ∫ v (τ ) dτ = ∫ 4 cos 2 t dτ = 2 sin 2t −∞ −∞ d d v ( t ) = 4 cos 2 t = −8 sin 2 t dt dtassociate the second term with a capacitor to get the minus sign.Then 4 cos 2 t 4 cos 2 t R= = = 5 Ω and i1 (t ) 0.8 cos 2 t i2 (t ) −0.8 sin 2 t C= = = 0.1 F d −8 sin 2 t 4 cos 2 t dt1.131cos ( 2t − 45° ) = 1.131  cos ( −45° ) cos ( 2t ) − sin ( −45° ) sin ( 2t )    = 0.8 cos 2 t + 0.8 sin 2 tThe first term is proportional to the voltage. Associate it with theresistor. Then noticing that t t ∫ v (τ ) dτ = ∫ −∞ −∞ 4 cos 2 t dτ = 2 sin 2t d d v ( t ) = 4 cos 2 t = −8 sin 2 t dt dtassociate the second term with an inductor to get the plus sign. Then 4 cos 2 t 4 cos 2 t R= = = 5 Ω and i1 (t ) 0.8 cos 2 t t L= ∫−∞ 4 cos 2 t dτ = 2 sin 2 t = 2.5 H i2 (t ) 0.8 sin 2 t 7-26
  • DP7-3a) 11.31cos ( 2t + 45° ) = 11.31  cos ( 45° ) cos ( 2t ) − sin ( 45° ) sin ( 2t )    = 8 cos 2 t − 8 sin 2 tThe first term is proportional to the voltage. Associate it with the resistor. The noticing that t t ∫ i (τ ) dτ = ∫ −∞ −∞ 4 cos 2 t dτ = 2 sin 2t d d i ( t ) = 4 cos 2 t = −8 sin 2 t dt dtassociate the second term with an inductor to get the minus sign. Then v (t ) 8 cos 2 t v2 (t ) −8 sin 2 t R= 1 = = 2 Ω and L = = =1H d 4 cos 2 t 4 cos 2 t 4 cos 2 t −8 sin 2 t dtb) 11.31cos ( 2t + 45° ) = 11.31  cos ( −45° ) cos ( 2t ) − sin ( −45° ) sin ( 2t )    = 8 cos 2 t + 8 sin 2 tThe first term is proportional to the voltage. Associate it with the resistor. The noticing that t t ∫ i (τ ) dτ = ∫ 4 cos 2 t dτ = 2 sin 2t −∞ −∞ d d i ( t ) = 4 cos 2 t = −8 sin 2 t dt dtassociate the second term with a capacitor to get the minus sign. Then t R= 1 v (t ) = 8 cos 2 t = 2 Ω and C = ∫−∞ 4 cos 2 t dτ = 2 sin 2 t = 0.25 F 4 cos 2 t 4 cos 2 t v2 (t ) 8 sin 2 t 7-27
  • DP7-4 at t=0− iL ( 0− ) = 0 VB By voltage division: vC ( 0− ) = 4 We require vC ( 0− ) = 3 V so VB = 12 V at t=0+ dvC Now we will check dt t = 0+ First: iL ( 0+ ) = iL ( 0− ) = 0 and vC ( 0+ ) = vC ( 0− ) = 3 V Apply KCL at node a: VB − vC ( 0+ ) iL ( 0 ) + iC ( 0 ) = + + 3 12 − 3 0 + iC ( 0+ ) = ⇒ iC ( 0+ ) = 3 A 3 Finally dvC iC ( 0 + )= 3 V = = 24 dt t =0+ C 0.125 s as required.DP7-5 1 1We require L i L2 = C v C where iL and vC are the steady state inductor current and capacitor 2 2 2 vvoltage. At steady state, i L = C . Then R 2 v  L L 10−2 L  C  = C vC2 ⇒ C= ⇒ R = = = 104 = 10 2 Ω R R2 C 10−6so R = 100 Ω . 7-28
  • Chapter 8 – The Complete Response of RL and RC CircuitExercisesEx. 8.3-1Before the switch closes:After the switch closes: 2Therefore R t = = 8 Ω so τ = 8 ( 0.05 ) = 0.4 s . 0.25Finally, v (t ) = voc + (v (0) − voc ) e− t τ = 2 + e−2.5 t V for t > 0Ex. 8.3-2Before the switch closes:After the switch closes: 8-1
  • 2 6Therefore R t = = 8 Ω so τ = = 0.75 s . 0.25 8 1 1 −1.33 tFinally, i (t ) = isc + (i (0) − isc ) e−t τ = + e A for t > 0 4 12Ex. 8.3-3At steady-state, immediately before t = 0:  10   12  i ( 0) =     = 0.1 A  10 + 40   16+ 40||10 After t = 0, the Norton equivalent of the circuit connected to the inductor is found to be L 20 1 so I sc = 0.3 A, Rt = 40 Ω, τ = = = s Rt 40 2 Finally: i (t ) = (0.1 − 0.3)e −2t + 0.3 = 0.3 − 0.2e −2t A 8-2
  • Ex. 8.3-4At steady-state, immediately before t = 0After t = 0, we have: so Voc = 12 V, Rt = 200 Ω, τ = Rt C = (200)(20 ⋅10 −6 ) = 4 ms Finally: v(t ) = (12 − 12) e −t 4 + 12 = 12 VEx. 8.3-5Immediately before t = 0, i (0) = 0.After t = 0, replace the circuit connected to the inductor by its Norton equivalent to calculate theinductor current: L 25 5 I sc = 0.2 A, Rt = 45 Ω, τ = = = Rth 45 9 So i (t ) = 0.2 (1 − e−1.8t ) ANow that we have the inductor current, we can calculate v(t): d v(t ) = 40 i (t ) + 25 i (t ) dt = 8(1 − e −1.8t ) + 5(1.8)e −1.8t = 8 + e −1.8t V for t > 0 8-3
  • Ex. 8.3-6At steady-state, immediately before t = 0 so i(0) = 0.5 A.After t > 0: Replace the circuit connected to the inductor by its Norton equivalent to get I sc = 93.75 mA, Rt = 640 Ω, L .1 1 τ = = = s Rt 640 6400 i (t ) = 406.25 e −6400t + 93.75 mAFinally: d v (t ) = 400 i (t ) + 0.1 i (t ) = 400 (.40625e −6400t + .09375) + 0.1 (−6400) (0.40625e −6400t ) dt = 37.5 − 97.5e −6400t VEx. 8.4-1 τ = ( 2×103 )(1×10−6 ) = 2 ×10−3 s vc (t ) = 5 + (1.5−5 ) e −500 t V vc (0.001) = 5 − 3.5e − 0.5 = 2.88 V So vc (t ) will be equal to vT at t = 1 ms if v = 2.88 V. T 8-4
  • Ex. 8.4-2 iL (0) = 1 mA, I sc = 10 mA  −500  t L  ⇒ iL ( t ) = 10 − 9 e L mA Rt =500 Ω, τ =  500  500 − t vR (t ) = 300 iL ( t ) = 3 − 2.7 e L VWe require that vR ( t ) = 1.5 V at t = 10 ms = 0.01 s . That is 500 − (0.01) 5 1.5 = 3 − 2.7e L ⇒ L= = 8.5 H 0.588Ex. 8.6-10 < t < t1 v(t ) = v(∞)+Ae− t / RC where v(∞) = (1 A)(1 Ω ) = 1 V v(t ) = 1 + A e− t /(1)(.1) = 1 + A e−10 t Now v(0− ) = v(0+ ) = 0 = 1 + A ⇒ A = −1 ∴ v(t ) =1−e−10t Vt > t1 − t −.5 t1 = 0.5 s, v(t ) = v(t1 ) e (1)(.1) = v(0.5) e−10(t −.5) Now v(0.5) 1 − e −10(0.5) = 0.993 V ∴ v(t ) = 0.993 e−10(t −0.5) V 8-5
  • Ex. 8 6-2t < 0 no sources ∴ v(0− ) = v(0+ ) = 00 < t <t1 − t 2 × 1 0 5 (1 0 − 7 ) − t / RC v (t ) = v (∞ ) + A e = v (∞ ) + A e where for t = ∞ (steady-state) ∴capacitor becomes an open ⇒ v(∞) = 10 V v(t ) = 10 + Ae −50t Now v(0) = 0 = 10 + A ⇒ A = −10 ∴ v(t ) = 10(1− e −50t ) Vt > t1 , t1 = .1 s v(t ) = v(.1) e−50(t −.1) where v(.1) = 10(1− e −50(.1) ) = 9.93 V ∴ v(t ) = 9.93e−50( t −.1) VEx. 8.6-3For t < 0 i = 0.For 0 < t < 0.2 s KCL: −5 + v / 2 + i = 0   di di  + 10i = 50 also: v = 0.2  dt dt  −10 t ∴ i (t ) = 5+ Ae i (0) = 0 = 5+ A ⇒ A = −5 so we have i (t ) = 5 (1−e−10t ) AFor t > 0.2 s i (0.2) = 4.32 A ∴ i(t ) = 4.32e−10(t −.2) A 8-6
  • Ex. 8.7-1 vs ( t ) = 10sin 20t V  d v( t )  KVL: −10sin 20t +10 .01  + v( t ) = 0  dt  d v( t ) ⇒ +10 v( t ) = 100sin 20t dt Natural response: vn (t ) = Ae−t τ where τ = Rt C ∴ vn (t ) = Ae−10t Forced response: try v f (t ) = B1 cos 20t + B2 sin 20 t Plugging v f (t ) into the differential equation and equating like terms yields: B1 = − 4 & B2 = 2 Complete response: v(t ) = vn (t ) + v f (t ) v(t ) = Ae −10t − 4 cos 20t + 2 sin 20t Now v(0− ) = v(0+ ) = 0 = A − 4 ∴ A = 4 ∴ v(t ) = 4 e −10t − 4 cos 20t + 2sin 20 t VEx. 8.7-2 is ( t ) = 10 e −5t for t > 0 KCL at top node: −10e −5t + i ( t ) + v( t ) /10 = 0 di ( t ) di ( t ) Now v ( t ) = 0.1 ⇒ +100 i ( t ) = 1000 e −5t dt dtNatural response: in (t ) = Ae −t τ where τ = L R t ∴ in (t ) = Ae −100t Forced response: try i f (t ) = Be −5t & plug into the differential equation −5 Be −5t + 100 Be −5t = 1000e −5t ⇒ B = 10.53 −100 t −5t Complete response: i (t ) = Ae + 10.53e − + Now i (0 ) = i (0 ) = 0 = A + 10.53 ⇒ A = −10.53 ∴ i (t ) = 10.53 (e −5t − e −100t ) AEx. 8.7-3When the switch is closed, the inductor current is iL = vs / R = vs . When the switch opens, theinductor current is forced to change instantaneously. The energy stored in the inductorinstantaneously dissipates in the spark. To prevent the spark, add a resistor (say 1 kΩ) across theswitch terminals. 8-7
  • ProblemsSection 8.3: The Response of a First Order Circuit to a Constant InputP8.3-1 Here is the circuit before t = 0, when the switch is open and the circuit is at steady state. The open switch is modeled as an open circuit. A capacitor in a steady-state dc circuit acts like an open circuit, so an open circuit replaces the capacitor. The voltage across that open circuit is the initial capacitor voltage, v (0). By voltage division 6 v (0) = (12 ) = 4 V 6+6+6 Next, consider the circuit after the switch closes. The closed switch is modeled as a short circuit. We need to find the Thevenin equivalent of the part of the circuit connected to the capacitor. Here’s the circuit used to calculate the open circuit voltage, Voc. 6 Voc = (12 ) = 6 V 6+6 Here is the circuit that is used to determine Rt. A short circuit has replaced the closed switch. Independent sources are set to zero when calculating Rt, so the voltage source has been replaced by a short circuit. Rt = ( 6 )( 6 ) = 3 Ω 6+6 Then τ = R t C = 3 ( 0.25 ) = 0.75 sFinally, v ( t ) = Voc + ( v ( 0 ) − Voc ) e−t / τ = 6 − 2 e−1.33 t V for t > 0 8-8
  • P8.3-2 Here is the circuit before t = 0, when the switch is closed and the circuit is at steady state. The closed switch is modeled as a short circuit. An inductor in a steady-state dc circuit acts like an short circuit, so a short circuit replaces the inductor. The current in that short circuit is the initial inductor current, i(0). 12 i ( 0) = =2 A 6 Next, consider the circuit after the switch opens. The open switch is modeled as an open circuit. We need to find the Norton equivalent of the part of the circuit connected to the inductor. Here’s the circuit used to calculate the short circuit current, Isc. 12 I sc = =1 A 6+6 Here is the circuit that is used to determine Rt. An open circuit has replaced the open switch. Independent sources are set to zero when calculating Rt, so the voltage source has been replaced by an short circuit. Rt = ( 6 + 6 )( 6 ) = 4 Ω ( 6 + 6) + 6 Then L 8 τ= = =2 s Rt 4Finally, i ( t ) = I sc + ( i ( 0 ) − I sc ) e −t / τ = 1 + e −0.5 t A for t > 0 8-9
  • P8.3-3Before the switch closes:After the switch closes: −6Therefore R t = = 3 Ω so τ = 3 ( 0.05 ) = 0.15 s . −2Finally, v (t ) = voc + (v (0) − voc ) e− t τ = −6 + 18 e−6.67 t V for t > 0P8.3-4Before the switch closes:After the switch closes: 8-10
  • −6 6Therefore R t = = 3 Ω so τ = = 2 s . −2 3 t − 10Finally, i (t ) = isc + (i (0) − isc ) e τ = −2 + e−0.5 t A for t > 0 3P8.3-5Before the switch opens, v o ( t ) = 5 V ⇒ v o ( 0 ) = 5 V . After the switch opens the part of thecircuit connected to the capacitor can be replaced by its Thevenin equivalent circuit to get:Therefore τ = ( 20 × 103 )( 4 × 10−6 ) = 0.08 s . t −Next, v C (t ) = voc + (v (0) − voc ) e τ = 10 − 5 e−12.5 t V for t > 0Finally, v 0 (t ) = vC (t ) = 10 − 5 e −12.5 t V for t > 0 8-11
  • P8.3-6Before the switch opens, v o ( t ) = 5 V ⇒ v o ( 0 ) = 5 V . After the switch opens the part of thecircuit connected to the capacitor can be replaced by its Norton equivalent circuit to get: 5Therefore τ = = 0.25 ms . 20 × 103 t −Next, i L (t ) = isc + (i L (0) − isc ) e τ = 0.5 − 0.25 e−4000 t mA for t > 0 dFinally, vo ( t ) = 5 i L ( t ) = 5 e −4000 t V for t > 0 dtP8.3-7At t = 0− (steady-state)Since the input to this circuit is constant, theinductor will act like a short circuit when thecircuit is at steady-state:for t > 0 iL ( t ) = iL ( 0 ) e − ( R L ) t = 6 e−20t A 8-12
  • P8.3-8Before the switch opens, the circuit will be at steady state. Because the only input to this circuitis the constant voltage of the voltage source, all of the element currents and voltages, includingthe capacitor voltage, will have constant values. Opening the switch disturbs the circuit.Eventually the disturbance dies out and the circuit is again at steady state. All the elementcurrents and voltages will again have constant values, but probably different constant values thanthey had before the switch opened. Here is the circuit before t = 0, when the switch is closed and the circuit is at steady state. The closed switch is modeled as a short circuit. The combination of resistor and a short circuit connected is equivalent to a short circuit. Consequently, a short circuit replaces the switch and the resistor R. A capacitor in a steady-state dc circuit acts like an open circuit, so an open circuit replaces the capacitor. The voltage across that open circuit is the capacitor voltage, vo(t). Because the circuit is at steady state, the value of the capacitor voltage will be constant.This constant is the value of the capacitor voltage just before the switch opens. In the absence ofunbounded currents, the voltage of a capacitor must be continuous. The value of the capacitorvoltage immediately after the switch opens is equal to the value immediately before the switchopens. This value is called the initial condition of the capacitor and has been labeled as vo(0).There is no current in the horizontal resistor due to the open circuit. Consequently, vo(0) is equalto the voltage across the vertical resistor, which is equal to the voltage source voltage. Therefore vo ( 0 ) = VsThe value of vo(0) can also be obtained by setting t = 0 in the equation for vo(t). Doing so gives vo ( 0 ) = 2 + 8 e0 = 10 VConsequently, Vs = 10 V 8-13
  • Next, consider the circuit after the switch opens. Eventually (certainly as t →∞) the circuit will again be at steady state. Here is the circuit at t = ∞, when the switch is open and the circuit is at steady state. The open switch is modeled as an open circuit. A capacitor in a steady-state dc circuit acts like an open circuit, so an open circuit replaces the capacitor. The voltage across that open circuit is the steady-state capacitor voltage, vo(∞). There is no current in the horizontal resistor and vo(∞) is equal to the voltage across the vertical resistor. Using voltage division, 10 vo ( ∞ ) = (10 ) R + 10The value of vo(∞) can also be obtained by setting t = ∞ in the equation for vo(t). Doing so gives vo ( ∞ ) = 2 + 8 e −∞ = 2 VConsequently, 10 2= (10 ) ⇒ 2 R + 20 = 100 ⇒ R = 40 Ω R + 10 −t τ Finally, the exponential part of vo(t) is known to be of the form e where τ = R t C andRt is the Thevenin resistance of the part of the circuit connected to the capacitor. Here is the circuit that is used to determine Rt. An open circuit has replaced the open switch. Independent sources are set to zero when calculating Rt, so the voltage source has been replaced by a short circuit. R t = 10 + ( 40 )(10 ) = 18 Ω 40 + 10 so τ = R t C = 18 C From the equation for vo(t) t −0.5 t = − ⇒ τ =2s τ Consequently, 2 = 18 C ⇒ C = 0.111 = 111 mF 8-14
  • P8.3-9:Before the switch closes, the circuit will be at steady state. Because the only input to this circuitis the constant voltage of the voltage source, all of the element currents and voltages, includingthe inductor current, will have constant values. Closing the switch disturbs the circuit by shortingout the resistor R1. Eventually the disturbance dies out and the circuit is again at steady state. Allthe element currents and voltages will again have constant values, but probably different constantvalues than they had before the switch closed.The inductor current is equal to the current in the 3 Ω resistor. Consequently, − 0.35 t vo (t ) 6 − 3 e − 0.35 t i (t ) = = = 2− e A when t > 0 3 3In the absence of unbounded voltages, the current in any inductor is continuous. Consequently,the value of the inductor current immediately before t = 0 is equal to the value immediately aftert = 0. Here is the circuit before t = 0, when the switch is open and the circuit is at steady state. The open switch is modeled as an open circuit. An inductor in a steady-state dc circuit acts like a short circuit, so a short circuit replaces the inductor. The current in that short circuit is the steady state inductor current, i(0). Apply KVL to the loop to get R1 i ( 0 ) + R 2 i ( 0 ) + 3 i ( 0 ) − 24 = 0 24 ⇒ i (0) = R1 + R 2 + 3The value of i(0) can also be obtained by setting t = 0 in the equation for i(t). Do so gives i ( 0 ) = 2 − e0 = 1 AConsequently, 24 1= ⇒ R1 + R 2 = 21 R1 + R 2 + 3 Next, consider the circuit after the switch closes. Here is the circuit at t = ∞, when the switch is closed and the circuit is at steady state. The closed switch is modeled as a short circuit. The combination of resistor and a short circuit connected is equivalent to a short circuit. Consequently, a short circuit replaces the switch and the resistor R1. 8-15
  • An inductor in a steady-state dc circuit acts like a short circuit, so a short circuit replaces theinductor. The current in that short circuit is the steady state inductor current, i(∞). Apply KVL tothe loop to get 24 R 2 i ( ∞ ) + 3 i ( ∞ ) − 24 = 0 ⇒ i ( ∞ ) = R2 + 3The value of i(∞) can also be obtained by setting t = ∞ in the equation for i(t). Doing so gives i ( ∞ ) = 2 − e −∞ = 2 AConsequently 24 2= ⇒ R2 = 9 Ω R2 + 3Then R1 = 12 Ω −t τ L Finally, the exponential part of i(t) is known to be of the form e where τ = and RtRt is the Thevenin resistance of the part of the circuit that is connected to the inductor. Here is shows the circuit that is used to determine Rt. A short circuit has replaced combination of resistor R1 and the closed switch. Independent sources are set to zero when calculating Rt, so the voltage source has been replaced by an short circuit. R t = R 2 + 3 = 9 + 3 = 12 Ω so L L τ= = R t 12From the equation for i(t) t −0.35 t = − ⇒ τ = 2.857 s τConsequently, L 2.857 = ⇒ L = 34.28 H 12 8-16
  • P8.3-10First, use source transformations to obtain the equivalent circuitfor t < 0: for t > 0: 1 L 1So iL ( 0 ) = 2 A, I sc = 0, Rt = 3 + 9 = 12 Ω, τ = = 2 = s Rt 12 24and iL ( t ) = 2e−24t t >0Finally v ( t ) = 9 iL ( t ) = 18 e−24t t >0 8-17
  • Section 8-4: Sequential SwitchingP8.4-1Replace the part of the circuit connected to the capacitor by its Thevenin equivalent circuit toget:Before the switch closes at t = 0 the circuit is at steady state so v(0) = 10 V. For 0 < t < 1.5s, voc =5 V and Rt = 4 Ω so τ = 4 × 0.05 = 0.2 s . Therefore v (t ) = voc + (v (0) − voc ) e−t τ = 5 + 5e−5 t V for 0 < t < 1.5 sAt t =1.5 s, v (1.5) = 5 + 5e ( −0.05 1.5) = 5 V . For 1.5 s < t, voc = 10 V and Rt = 8 Ω soτ = 8 × 0.05 = 0.4 s . Therefore −( t −1.5) τ −2.5 ( t −1.5 ) v (t ) = voc + (v (1.5) − voc ) e = 10 − 5 e V for 1.5 s < tFinally   5+5 e V −5 t for 0 < t < 1.5 s v (t ) =  −2.5 ( t −1.5) 10 − 5 e  V for 1.5 s < tP8.4-2Replace the part of the circuit connected to the inductor by its Norton equivalent circuit to get:Before the switch closes at t = 0 the circuit is at steady state so i(0) = 3 A. For 0 < t < 1.5s, isc = 2 12A and Rt = 6 Ω so τ = = 2 s . Therefore 6 i (t ) = isc + (i (0) − isc ) e−t τ = 2 + e−0.5 t A for 0 < t < 1.5 s 8-18
  • −0.5 (1.5) 12At t =1.5 s, i (1.5) = 2 + e = 2.47 A . For 1.5 s < t, isc = 3 A and Rt = 8 Ω so τ = = 1.5 s . 8Therefore −( t −1.5 ) τ −0.667 ( t −1.5 ) i (t ) = isc + (i (1.5) − isc ) e = 3 − 0.53 e V for 1.5 s < tFinally   2 + e −0.5 t A for 0 < t < 1.5 s i (t ) =  −0.667 ( t −1.5 ) for 1.5 s < t 3 − 0.53 e  AP8.4-3At t = 0−: KVL : − 52 + 18 i + (12 8) i = 0 ⇒ i (0− ) =104 39 A  6  + ∴ iL = i   = 2 A = iL (0 )  6+ 2 For 0 < t < 0.051 s iL (t ) = iL (0) e −t τ where τ = L R t R t = 6 12 + 2 = 6 Ω iL (t ) = 2 e −6t A  6 +12  −6 t ∴ i (t ) = iL (t )   =6e A  6 For t > 0.051 s i L ( t ) = i L (0.051) e − ( R L ) ( t − 0.051) i L (0.051) = 2 e − 6 (.051) = 1.473 A i L ( t ) = 1.473 e − 14 ( t − 0.051) A i ( t ) = i L ( t ) = 1.473 e − 14 ( t − 0.051) A 8-19
  • P8.4-4At t = 0-: Assume that the circuit has reached steady state so that the voltage across the 100 µFcapacitor is 3 V. The charge stored by the capacitor is q ( 0− ) = (100 × 10−6 ) ( 3) = 300 × 10−6 C0 < t < 10ms: With R negligibly small, the circuit reaches steady state almost immediately (i.e. att = 0+). The voltage across the parallel capacitors is determined by considering chargeconservation: q ( 0+ ) = (100 µ F) v ( 0+ ) + (400 µ F) v ( 0+ ) q ( 0+ ) q ( 0− ) 300 × 10−6 v (0 + ) = 100 ×10 −6 = = + 400 ×10−6 500 × 10−6 500 × 10−6 ( ) v 0+ = 0.6 V10 ms < t < l s: Combine 100 µF & 400 µF in parallel to obtain v(t ) = v ( 0+ ) e − (t −.01) RC 3 −4 ) = 0.6e − (t −.01) (10 ) (5 x10 v(t ) = 0.6 e−2(t −.01) VP8.4-5For t < 0:Find the Thevenin equivalent of thecircuit connected to the inductor (aftert >0). First, the open circuit voltage: 40 i1 = =1 A 20 + 20 voc = 20 i1 − 5 i1 = 15 V 8-20
  • Next, the short circuit current: 20 i 1 = 5 i 1 ⇒ i 1 = 0 40 i sc + 0 = ⇒ i sc = 2 A 20Then voc 15 Rt = = = 7.5 Ω i sc 2Replace the circuit connected to theinductor by its Norton equivalentcircuit. First L 15 × 10−3 1 τ= = = Rt 7.5 500Next i ( t ) = 2 − 2 e − 500 t A t >0After t = 0, the steady state inductorcurrent is 2 A. 99% of 2 is 1.98.1.98 = 2 − 2 e − 500 t ⇒ t = 9.2 msP8.4-6 v ( 0 ) = 5 V , v ( ∞ ) = 0 and τ = 105 ×10−6 = 0.1 s ∴ v ( t ) = 5 e −10 t V for t > 0 2.5 = 5 e −10 t1 t 1 = 0.0693 s v (t1 ) 2.5 i (t1 ) = = = 25 µ A 100 ×10 3 100 × 103 8-21
  • Section 8-5: Stability of First Order CircuitsP8.5-1This circuit will be stable if the Thèvenin equivalent resistance of the circuit connected to theinductor is positive. The Thèvenin equivalent resistance of the circuit connected to the inductoris calculated as 100  v T (400− R) 100 i (t ) = iT  100+ 400  ⇒ Rt = =  iT 100+ 400 vT = 400 i (t ) − R i (t )  The circuit is stable when R < 400 Ω.P8.5-2The Thèvenin equivalent resistance of the circuit connected to the inductor is calculated as Ohm’s law: v ( t ) = 1000 iT ( t ) KVL: A v ( t ) + vT ( t ) − v ( t ) − 4000 iT ( t ) = 0 ∴ vT ( t ) = (1 − A ) 1000 iT ( t ) + 4000 iT ( t ) vT ( t ) Rt = = ( 5 − A ) ×1000 iT ( t ) The circuit is stable when A < 5 V/V.P8.5-3The Thèvenin equivalent resistance of the circuit connected to the inductor is calculated as v (t ) Ohm’s law: i ( t ) = − T 6000 v (t ) KCL: i ( t ) + B i ( t ) + i T ( t ) = T 3000  v ( t )  vT ( t ) ∴ i T ( t ) = − ( B + 1)  − T +  6000  3000 = ( B + 3) vT ( t ) 6000 vT ( t ) 6000 Rt = = iT ( t ) B + 3 The circuit is stable when B > −3 A/A. 8-22
  • P8.5-4The Thèvenin equivalent resistance of the circuit connected to the inductor is calculated as v(t ) = (1000 )( 4000 ) i (t ) = 800 i (t ) T T 1000 + 4000 vT (t ) = v(t ) − Av(t ) = (1 − A ) v(t ) vT (t ) Rt = = 800 (1 − A) iT (t ) The circuit is stable when A < 1 V/V. 8-23
  • Section 8-6: The Unit Step ResponseP8.6-1The value of the input is one constant, 8 V, before time t = 0 and a different constant, −7 V, aftertime t = 0. The response of the first order circuit to the change in the value of the input will be vo ( t ) = A + B e − a t for t > 0where the values of the three constants A, B and a are to be determined. The values of A and B are determined from the steady state responses of this circuitbefore and after the input changes value. Capacitors act like open circuits when the input is constant and the circuit is at steady state. Consequently, the capacitor is replaced by an open circuit. The value of the capacitor voltage at time t = 0, will be equal to the steady state capacitor voltage before the input changes. At time t = 0 the output voltage is vo ( 0 ) = A + B e ( ) = A + B −a 0 The steady-state circuit for t < 0. Consequently, the capacitor voltage is labeled as A + B. Analysis of the circuit gives A+ B = 8 V Capacitors act like open circuits when the input is constant and the circuit is at steady state. Consequently, the capacitor is replaced by an open circuit. The value of the capacitor voltage at time t = ∞, will be equal to the steady state capacitor voltage after the input changes. At time t = ∞ the output voltage is vo ( ∞ ) = A + B e − a (∞) =A The steady-state circuit for t > 0. Consequently, the capacitor voltage is labeled as A. Analysis of the circuit gives A = -7 V Therefore B = 15 V 8-24
  • The value of the constant a is determined from the time constant, τ, which is in turncalculated from the values of the capacitance C and of the Thevenin resistance, Rt, of the circuitconnected to the capacitor. 1 =τ = Rt C a Here is the circuit used to calculate Rt. Rt = 6 Ω Therefore 1 1 a= = 2.5 ( 6 ) ( 66.7 ×10 ) −3 s (The time constant is τ = ( 6 ) ( 66.7 × 10−3 ) = 0.4 s .)Putting it all together:  8 V for t ≤ 0 vo ( t ) =  − 2.5 t −7 + 15 e V for t ≥ 0P8.6-2The value of the input is one constant, 3 V, before time t = 0 and a different constant, 6 V, aftertime t = 0. The response of the first order circuit to the change in the value of the input will be vo ( t ) = A + B e − a t for t > 0where the values of the three constants A, B and a are to be determined. The values of A and B are determined from the steady state responses of this circuitbefore and after the input changes value. Capacitors act like open circuits when the input is constant and the circuit is at steady state. Consequently, the capacitor is replaced by an open circuit. The value of the capacitor voltage at time t = 0, will be equal to the steady state capacitor voltage before the input changes. At time t = 0 the output voltage is The steady-state circuit for t < 0. vo ( 0 ) = A + B e ( ) = A + B −a 0 Consequently, the capacitor voltage is labeled as A + B. Analysis of the circuit gives 8-25
  • 6 A+ B = ( 3) = 2 V 3+ 6 Capacitors act like open circuits when the input is constant and the circuit is at steady state. Consequently, the capacitor is replaced by an open circuit. The value of the capacitor voltage at time t = ∞, will be equal to the steady state capacitor voltage after the input changes. At time t = ∞ the output voltage is vo ( ∞ ) = A + B e − a (∞) The steady-state circuit for t > 0. =A Consequently, the capacitor voltage is labeled as A. Analysis of the circuit gives 6 A= (6) = 4 V 3+ 6 Therefore B = −2 V The value of the constant a is determined from the time constant, τ, which is in turncalculated from the values of the capacitance C and of the Thevenin resistance, Rt, of the circuitconnected to the capacitor. 1 =τ = Rt C a Here is the circuit used to calculate Rt. Rt = ( 3)( 6 ) = 2 Ω 3+ 6 Therefore 1 1 a= =1 ( 2 )(.5 ) s (The time constant is τ = ( 2 )( 0.5 ) = 1 s .)Putting it all together:  2 V for t ≤ 0 vo ( t ) =  − t  4 − 2 e V for t ≥ 0 8-26
  • P8.6-3The value of the input is one constant, −7 V, before time t = 0 and a different constant, 6 V, aftertime t = 0. The response of the first order circuit to the change in the value of the input will be vo ( t ) = A + B e − a t for t > 0where the values of the three constants A, B and a are to be determined. The values of A and B are determined from the steady state responses of this circuitbefore and after the input changes value. Inductors act like short circuits when the input is constant and the circuit is at steady state. Consequently, the inductor is replaced by a short circuit. The value of the inductor current at time t = 0, will be equal to the steady state inductor current before the input changes. At time t = 0 the output current is The steady-state circuit for t < 0. io ( 0 ) = A + B e − a ( 0) = A+ B Consequently, the inductor current is labeled as A + B. Analysis of the circuit gives −7 A+ B = = −1.4 A 5 Inductors act like short circuits when the input is constant and the circuit is at steady state. Consequently, the inductor is replaced by a short circuit. The value of the inductor current at time t = ∞, will be equal to the steady state inductor current after the input changes. At time t = ∞ the output current is The steady-state circuit for t > 0. io ( ∞ ) = A + B e − a (∞) =A Consequently, the inductor current is labeled as A. Analysis of the circuit gives 6 A= = 1.2 A 5 Therefore B = −2.6 V 8-27
  • The value of the constant a is determined from the time constant, τ, which is in turncalculated from the values of the inductance L and of the Thevenin resistance, Rt, of the circuitconnected to the inductor. 1 L =τ = a Rt Here is the circuit used to calculate Rt. Rt = ( 5 ) ( 4 ) = 2.22 Ω 5+ 4 Therefore 2.22 1 a= = 1.85 1.2 s 1.2 (The time constant is τ = = 0.54 s .) 2.22Putting it all together:  −1.4 A for t ≤ 0 io ( t ) =  − 1.85 t 1.2 − 2.6 e A for t ≥ 0P8.6-4 v (t ) = 4u (t ) − u (t − 1) − u (t − 2) + u (t − 4) − u (t − 6)P8.6-5 Assume that the circuit is at steady state at t = 1−. Then 0 t <1  v ( t ) = 4 − 4 e − (t −1) V for 1 ≤ t ≤ 2 vs ( t ) = 4 1<t < 2 0 t >2  so v ( 2 ) = 4 − 4 e − (2−1) = 2.53 V τ = R C = ( 5 ×105 )( 2 × 10−6 ) = 1 s and v ( t ) = 2.53 e − (t − 2) V for t ≥ 2 0 t ≤1  ∴ v(t ) = 4− 4e − ( t −1) 1≤t ≤ 2 2.53e − (t − 2) t ≥2  8-28
  • P8.6-6The capacitor voltage is v(0−) = 10 V immediately before the switch opens at t = 0.For 0 < t < 0.5 s the switch is open: 1 1 v ( 0 ) = 10 V, v ( ∞ ) = 0 V, τ = 3 × = s 6 2 so v ( t ) = 10 e − 2 t V In particular, v ( 0.5 ) = 10 e − 2 ( 0.5) = 3.679 VFor t > 0.5 s the switch is closed: v ( 0 ) = 3.679 V, v ( ∞ ) = 10 V, Rt = 6 || 3 = 2 Ω, 1 1 τ = 2× = s 6 3 so v ( t ) = 10 + ( 3.679 − 10 ) e − 3 ( t − 0.5) V − 3 ( t − 0.5 ) = 10 − 6.321 e VP8.6-7Assume that the circuit is at steady state before t = 0. Then the initial inductor current isi(0−) = 0 A.For 0 < t < 1 ms: The steady state inductor current will be 30 i ( ∞ ) = lim i ( t ) = ( 40 ) = 24 A t →∞ 30 + 20 The time constant will be 50 × 10−3 1 τ= = 10−3 = s 30 + 20 1000 The inductor current is i ( t ) = 24 (1 − e −1000 t ) A In particular, i ( 0.001) = 24 (1 − e −1 ) = 15.2 AFor t > 1 ms Now the initial current is i(0.001) = 15.2 A and the steady state current is 0 A. As before, the time constant is 1 ms. The inductor current is i ( t ) = 15.2 e −1000 ( t − 0.001) A 8-29
  • The output voltage is 480 (1 − e −1000 t ) V t < 1 ms v ( t ) = 20 i ( t ) =  −1000 ( t − 0.001) 303 e  V t > 1 msP8.6-8For t < 0, the circuit is:After t = 0, replace the part of the circuit connected to the capacitor by its Thevenin equivalentcircuit to get: vc ( t ) = −15 + ( −6 − ( −15 ) ) e − t / ( 4000×0.00005 ) = −15 + 9 e − 5 t V 8-30
  • Section 8-7 The Response of an RL or RC Circuit to a Nonconstant SourceP8.7-1Assume that the circuit is at steady state before t = 0: KVL : 12ix + 3(3 ix ) + 38.5 = 0 ⇒ ix = −1.83 AThen vc (0− ) = −12 ix = 22 V = vc (0+ )After t = 0:KVL : 12i (t ) − 8e −5t + v ( t ) = 0 x c dv ( t ) 1 dvc ( t )KCL : −ix ( t )−2ix ( t ) + (1 36) c = 0 ⇒ ix ( t ) = dt 108 dt  1 dvc ( t ) ∴ 12   − 8e −5t + v (t ) = 0 108 dt  cdv (t ) c + 9v (t ) = 72e−5t ⇒ v (t ) = Ae −9t dt c cnTry v (t ) = Be −5t & substitute into the differential equation ⇒ B = 18 cf∴ v (t ) = Ae −9t + 18 e−5t c v (0) = 22 = A + 18 ⇒ A = 4 c∴ v (t ) = 4e−9t + 18e−5t V c 8-31
  • P8.7-2Assume that the circuit is at steady state before t = 0: 12 iL (0+ ) = iL (0− ) = = 3A 4After t = 0: v( t ) −12 v( t )KCL : + iL ( t ) + = 6 e −2t 4 2 di ( t )also : v ( t ) = (2 / 5) L dt 3 di ( t ) iL ( t ) + (2 / 5) L  = 3 + 6 e −2t 4 dt diL ( t ) 10 + iL ( t ) = 10 + 20 e −2t dt 3 − (10 / 3)t∴ in (t ) = Ae , try i f (t ) = B + Ce−2t , substitute into the differential equation,and then equating like terms ⇒ B =3, C =15 ⇒ i f (t ) =3+15 e−2t∴iL (t ) =in (t ) + i f (t ) = Ae −(10 / 3)t + 3+15e−2t , iL (0) = 3 = A + 3 + 15 ⇒ A = −15 ∴ iL (t ) = −15e − (10 / 3) t + 3 + 15e−2t diLFinally, v(t ) =( 2 / 5 ) = 20 e − (10 / 3) t −12 e −2t V dtP8.7-3Assume that the circuit is at steady state before t = 0:  6 Current division: iL (0− ) = −5   = −1 mA  6 + 24  8-32
  • After t = 0: diL ( t )KVL: − 25sin 4000 t + 24iL ( t ) + .008 =0 dt di L ( t ) 25 +3000i L ( t ) = sin4000t dt .008in (t ) = Ae −3000t , try i f (t ) = B cos 4000t + C sin 4000t , substitute into the differential equationand equate like terms ⇒ B = −1/ 2, C = 3 / 8 ⇒ i f (t ) = −0.5cos 4000 t + 0.375 sin 4000 tiL (t ) = in (t ) + i f (t ) = Ae −3000t − 0.5cos 4000 t + 0.375 sin 4000 tiL (0+ ) = iL (0− ) =−10−3 = A− 0.5 ⇒ A ≅ 0.5∴ iL (t ) = 0.5 e −3000t − 0.5cos 4000 t + 0.375 sin 4000 t mAbut v(t ) = 24iL (t ) = 12 e −3000t − 12 cos 4000t + 9sin 4000t VP8.7-4Assume that the circuit is at steady state before t = 0:Replace the circuit connected to the capacitor by its Thevenin equivalent (after t=0) to get:  dvc ( t )  dvc ( t ) 30 dt  c ( )KVL: − 10 cos 2t + 15  1 +v t =0 ⇒ + 2vc ( t ) = 20 cos 2t   dtvn (t ) = Ae−2t , Try v f (t ) = B cos 2t + C sin 2t & substitute into the differential equation to getB = C = 5 ⇒ v f (t ) = 5cos 2t + 5sin 2t. ∴ vc (t ) = vn (t ) + v f (t ) = Ae−2t + 5cos 2t + 5sin 2tNow vc (0) = 0 = A + 5 ⇒ A = −5 ⇒ vc (t ) = −5e−2t + 5cos 2t + 5sin 2t V 8-33
  • P8.7-5Assume that the circuit is at steady state before t = 0. There are no sources in the circuit so i(0) =0 A. After t = 0, we have: di ( t ) KVL : − 10sin100t + i ( t ) + 5 + v (t ) = 0 dt v( t ) Ohms law : i ( t ) = 8 dv( t ) ∴ +18 v( t ) = 160sin100t dt∴ vn (t ) = Ae−18t , try v f (t ) = B cos100t + C sin100t , substitute into the differential equationand equate like terms ⇒ B = −1.55 & C = 0.279 ⇒ v f (t ) = −1.55cos100t + 0.279sin100t∴ v(t ) = vn (t ) + v f (t ) = Ae−18t −1.55 cos100 t + 0.279 sin100 tv(0) = 8 i (0) = 0 ⇒ v (0) = 0 = A−1.55 ⇒ A = 1.55so v(t ) = 1.55e−18t −1.55cos100t + 0.279 sin100t VP8.7-6Assume that the circuit is at steady statebefore t = 0. vo ( t ) = −vc ( t ) vC (0+ ) = vC (0− ) = −10 VAfter t = 0, we have v (t ) 8 e−5 t i (t ) = s = = 0.533 e−5 t mA 15000 15000The circuit is represented by the differential dv ( t ) vC ( t )equation: i ( t ) = C C + . Then dt R dvc ( t ) dvc ( t ) ( 0.533 ×10 ) e −3 −5 t = ( 0.25 ×10−6 ) dt + (10−3 ) vc ( t ) ⇒ dt + 4000 vc ( t ) = 4000 e−5tThen vn ( t ) = Ae−4000t . Try v f ( t ) = Be−5t . Substitute into the differential equation to get ( d B e−5t ) + 4000 ( B e ) = 4000 e −5t −5t ⇒ B= 4000 = −1.00125 ≅ −1 dt −3995 8-34
  • vC (t ) = v f ( t ) + vn ( t ) = e −5t + Ae −4000t vC (0) = −10 = 1 + A ⇒ A = −11 ⇒ vC (t ) = 1 e−2t − 11 e−4000t VFinally vo (t ) = − vC (t ) = 11e−4000t −1e −5t V , t ≥ 0P8.7-7From the graph iL (t ) = 1 t mA . Use KVL to get 4 diL (t ) diL (t ) (1) iL (t ) + 0.4 = v1 (t ) ⇒ + 2.5 iL (t ) = 2.5 v1 (t ) dt dtThen di  1  t + 2.5  1 t  = 2.5 v1 (t ) ⇒ v1 = 0.1+ 0.25t V dt  4  4 P8.7-8Assume that the circuit is at steady state beforet = 0. 2 v (0+ ) = v (0− ) = 30 = 10 V 4+ 2After t = 0 we have 5 d v( t )  1 d v( t )  KVL : + v (t ) + 4  −i  = 30 2 dt  2 dt   1 d v( t )  2 i ( t ) + 4 i ( t ) −  + 30 = e −3t  2 dt The circuit is represented by the differential equation d v( t ) 6 6 2 + v (t ) = (10 + e −3t ) dt 19 19 3 −( 6 /19 ) tTake vn ( t ) = Ae . Try , v f ( t ) = B + Ce −3t , substitute into the differential equation to get 8-35
  • 6 60 4 −3t −3Ce −3t + ( B + Ce −3t ) = + e 19 19 19Equate coefficients to get 4 4 B = 10 , C = − ⇒ v f ( t ) = e −3t + Ae− (6 /19) t 51 51Then 4 −3t v ( t ) = vn ( t ) + v f ( t ) = 10 − e + Ae− (6 /19) t 51Finally 4 4 vc (0+ ) = 10 V, ⇒ 10 = 10 − +A ⇒ A= 51 51 4 − (6 /19) t −3t ∴ vc (t ) = 10 + (e −e ) V 51P8.7-9We are given v(0) = 0. From part b of thefigure: 5t 0 ≤ t ≤ 2 s vs ( t ) =  10 t > 2sFind the Thevenin equivalent of the part of thecircuit that is connected to the capacitor:The open circuit voltage: The short circuit current: (ix=0 because of the short across the right 2 Ω resistor)Replace the part of the circuit connected to thecapacitor by its Thevenin equivalent: KVL: dv( t ) 2 + v ( t ) − vs ( t ) = 0 dt dv( t ) v ( t ) vs ( t ) + = dt 2 2 vn ( t ) = Ae−0.5 t 8-36
  • For 0 < t < 2 s, vs ( t ) = 5 t . Try v f ( t ) = B + C t . Substituting into the differential equation andequating coefficients gives B = −10 and C =5. Therefore v ( t ) = 5t − 10 + A e − t / 2 . Using v(0) = 0,we determine that A =10. Consequently, v ( t ) = 5t + 10(e−t / 2 − 1) .At t = 2 s, v( 2 ) = 10e−1 = 3.68 .Next, for t > 2 s, vs ( t ) = 10 V . Try v f ( t ) = B . Substituting into the differential equation andequating coefficients gives B = 10. Therefore v ( t ) = 10 + Ae . Using v ( 2 ) = 3.68 , we − (t −2) / 2determine that A = −6.32. Consequently, v ( t ) = 10 − 6.32 e − (t − 2) / 2 .P8.7-10  d v (t )  KVL: − kt + Rs C C  + vC ( t ) = 0  dt  d vC ( t ) 1 k ⇒ + vC ( t ) = t dt Rs C Rs Cvc ( t ) = vn ( t ) + v f ( t ) , where vc ( t ) = Ae− t / Rs C . Try v f ( t ) = B0 + B1 t 1 k& plug into D.E. ⇒ B1 + [ B0 + B1t ] = t thus B0 = − kRs C , B1 = k . Rs C Rs CNow we have vc (t ) = Ae − t / Rs C + k (t − Rs C ). Use vc (0) = 0 to get 0 = A − kRs C ⇒ A = kRs C.∴ vc (t ) = k[t − Rs C (1− e −t / Rs C )]. Plugging in k =1000 , Rs = 625 kΩ & C = 2000 pF getvc (t ) = 1000[t − 1.25 × 10−3 (1 − e −800 t )]v(t) and vC(t) track well on a millisecond time scale. 8-37
  • Spice ProblemsSP 8-1 8-38
  • SP 8-2 8-39
  • SP 8-3 v(t ) = A + B e −t / τ for t > 0 7.2 = v(0) = A + B e 0 ⇒ 7.2 = A + B   −∞  ⇒ B = −0.8 V 8.0 = v(∞) = A + B e ⇒ A = 8.0 V  0.05  8 − 7.7728  7.7728 = v(0.05) = 8 − 0.8 e −0.05 / τ ⇒ − = ln   = −1.25878 τ  0.8  0.05 ⇒ τ= = 39.72 ms 1.25878Therefore v(t ) = 8 − 0.8 e −t / 0.03972 V for t > 0 8-40
  • SP 8-4 i (t ) = A + B e −t / τ for t > 0 0 = i (0) = A + B e 0 ⇒ 0 = A+ B   −3  ⇒ B = −4 × 10 A 4 × 10−3 = i (∞) = A + B e −∞ ⇒ A = 4 × 10−3 A  2.4514 ×10 = v(5 ×10 ) = ( 4 ×10 −3 −6 −3 ) − ( 4 ×10 ) e −3 ( ) − 5×10−6 / τ 5 × 10−6  ( 4 − 2.4514 ) × 10−3  ⇒ − = ln   = −0.94894 τ  4 × 10−3  5 × 10−6 ⇒ τ= = 5.269 µ s 0.94894Therefore −6 i (t ) = 4 − 4 e −t / 5.269×10 mA for t > 0 8-41
  • Verification ProblemsVP 8-1First look at the circuit. The initial capacitor voltage is vc(0) = 8 V. The steady-state capacitorvoltage is vc = 4 V.We expect an exponential transition from 8 volts to 4 volts. That’s consistent with the plot.Next, let’s check the shape of the exponential transition. The Thevenin resistance of the part ofthe circuit connected to the capacitor is R t = ( 2000 )( 4000 ) = 4 kΩ so the time constant is 2000 + 4000 3  4  2τ = R t C =  × 103  ( 0.5 × 10−6 ) = ms . Thus the capacitor voltage is 3  3 vc (t ) = 4 e− t 0.67 +4 Vwhere t has units of ms. To check the point labeled on the plot, let t1 = 1.33 ms. Then  1.33  −  vc (t1 ) = 4 e  .67  + 4 = 4.541 ~ 4.5398 VSo the plot is correct.VP 8-2The initial and steady-state inductor currents shown on the plot agree with the values obtainedfrom the circuit.Next, let’s check the shape of the exponential transition. The Thevenin resistance of the part ofthe circuit connected to the inductor is R t = ( 2000 )( 4000 ) = 4 kΩ so the time constant is 2000 + 4000 3 L 5 15τ= = = ms . Thus inductor current is R t 4 ×103 4 3 iL (t ) − 2 e− t 3.75 + 5 mAwhere t has units of ms. To check the point labeled on the plot, let t1 = 3.75 ms. Then  3.75  −  iL (t1 ) = −2 e  3.75  + 5 = 4.264 mA ≠ 4.7294 mAso the plot does not correspond to this circuit. 8-42
  • VP 8-3 Notice that the steady-state inductor current does not depend on the inductance, L. The initialand steady-state inductor currents shown on the plot agree with the values obtained from thecircuit.After t = 0 L So I sc = 5 mA and τ = 1333The inductor current is given by iL (t ) = −2e −1333t L + 5 mA , where t has units of seconds and Lhas units of Henries. Let t 1 = 3.75 ms, then 4.836 = iL (t1 ) = −2 e −(1333)⋅(0.00375) L + 5 = −2e−5 L + 5so 4.836−5 = e −5 L −2and −5 L= =2 H  4.836−5  ln    −2 is the required inductance.VP 8-4First consider the circuit. When t < 0 and the circuit is at steady-state:For t > 0So R2 R1 R2 RRC Voc = ( A + B) , Rt = and τ = 1 2 R1 + R2 R1 + R2 R1 + R2 8-43
  • Next, consider the plot. The initial capacitor voltage is (vc (0)=) –2 and the steady-state capacitorvoltage is (Voc =) 4 V, so vC (t ) = − 6e −t τ + 4At t 1 = 1.333 ms 3.1874 = vC (t1 ) = − 6 e −0.001333 τ + 4so −0.001333 τ = = 0.67 ms  −4+ 3.1874  ln    −6 Combining the information obtained from the circuit with the information obtained from the plotgives R2 R2 R1 R2C A = −2, ( A + B ) = 4, = 0.67 ms R1 + R2 R1 + R2 R1 + R2There are many ways that A, B, R , R , and C can be chosen to satisfy these equations. Here is 1 2one convenient way. Pick R = 3000 and R = 6000. Then 1 2 2A = −2 ⇒ A = −3 3 2( A+ B) = 4 ⇒ B −3 = 6 ⇒ B = 9 3 2 1 2000 ⋅ C = ms ⇒ µF = C 3 3Design ProblemsDP 8-1 R3Steady-state response when the switch is open: 6= 12 ⇒ R1 + R 2 = R 3 . R1 + R 2 + R 3 R3 R3Steady-state response when the switch is open: 10 = 12 ⇒ R1 = . R1 + R 3 5 R310 ms = 5 τ = ( R 1 || R 3 ) C = C 6Let C = 1 µF. Then R 3 = 60 kΩ, R 1 = 30 kΩ and R 2 = 30 kΩ. 8-44
  • DP 8-2 12steady state response when the switch is open : 0.001 = R + R ⇒ R + R = 12 kΩ . 1 2 1 2 12steady state response when the switch is open: 0.004 = ⇒ R1 = 3 kΩ . R1Therefore, R 2 = 9 kΩ.  L  L10 ms = 5 τ = 5  = ⇒ L = 240 H  R1 + R 2  2400  DP 8-3Rt = 50 kΩ when the switch is open and Rt = 49 kΩ ≈ 50 kΩ when the switch is closed so useRt = 50 kΩ. 10−6(a) ∆t = 5 Rt C ⇒ C = = 4 pF 5 50×103 ( )(b) ∆ t = 5 ( 50×103 )( 2×10−6 ) = 0.5 sDP 8-4Rt = 50 kΩ when the switch is open and Rt = 49 kΩ ≈ 50 kΩ when the switch is closed so useRt = 50 kΩ. −∆t τ ∆tWhen the switch is open: 5 e = (1 − k ) 5 ⇒ ln (1− k ) = − ⇒ ∆ t = −τ ln (1− k ) τ −∆t τWhen the switch is open: 5 − 5 e = k 5 ⇒ ∆ t = −τ ln (1− k ) 10−6(a) C = = 6.67 pF − ln (1−.95 ) ( 50×103 )(b) ∆ t = − ln(1 − .95) ( 50×103 )( 2×10−6 ) = 0.3 sDP 8-5 20 R1 i (0) = × 40 R1 R1 + 40 40 + 40 + R1 8-45
  • Fort > 0: −t L 10−2 i (t ) = i (0) e τ where τ = = R t 40+ R 2At t < 200 µ s we need i ( t ) > 60 mA and i ( t ) <180 mAFirst lets find a value of R 2 to cause i (0) < 180 mA. 1 tTry R 2 = 40 Ω . Then i (0) = A = 166.7 mA so i (t ) = 0.1667 e− τ . 6Next, we find a value of R 2 to cause i (0.0002) > 60 mA. 10−2 1Try R 2 = 10Ω, then τ = = 0.2 ms = s. 50 5000i (0.0002) = 166.7×10−3 e −5000×0.0002 = 166.7×10−3 e −1 = 61.3 mADP 8-6The current waveform will look like this: We only need to consider the rise time: −t Vs −t A τ iL (t ) = (1 − e ) =τ (1 − e ) R+2 R+2 where L 0.2 1 τ = = = s Rt 3 15 A ∴ iL (t ) = (1 − e −15t ) 3Now find A so that iL R fuse ≥ 10 W during 0.25 ≤ t ≤ 0.75 s 2 A2∴ we want [iL (0.25)]R fuse = 10 W ⇒ 2 (1−e −15(.25) ) 2 (1) =10 ⇒ A = 9.715 V 9 8-46
  • Chapter 9 - Complete Response of Circuits with Two Energy Storage ElementsExercises Apply KVL to right mesh:Ex. 9.3-1 di ( t ) 2 + v ( t ) + 1( i ( t ) −is ( t ) ) = 0 dt di ( t ) ⇒ v ( t ) = −2 − ( i ( t )−is ( t ) ) dtThe capacitor current and voltage are related by 1 dv ( t ) 1 d  di ( t )  1 dis ( t ) 1 di ( t ) d 2i ( t ) i (t ) = =  −2 −i ( t )+ is ( t )  = − − 2 dt 2 dt  dt  2 dt 2 dt dt 2 d 2i ( t ) 1 di ( t ) 1 dis ( t ) ∴ 2 + + i( t ) = dt 2 dt 2 dt The inductor voltage is related to the inductorEx. 9.3-2 current by di ( t ) v (t ) = 1 dt Apply KCL at the top node: v (t ) 1 dv ( t ) is ( t ) = + i (t ) + 1 2 dt dUsing the operator s = we have dt v (t ) = s i (t )   v (t ) 1 1  ⇒ is ( t ) = v ( t ) + + sv ( t ) is ( t ) = v ( t ) + i ( t ) + sv ( t )  s 2 2 Therefore d 2v ( t ) dv ( t ) d i (t ) 2s is ( t ) = 2 s v ( t ) + 2 v ( t ) + s 2 v ( t ) ⇒ 2 +2 + 2 v (t ) = 2 s dt dt dt 9-1
  • Ex. 9.3-3 d Using the operator s = , apply KVL to the dt left mesh: i1 ( t ) + s ( i1 ( t ) −i 2 ( t ) ) = vs ( t ) Apply KVL to the right mesh: 1 2 i 2 ( t ) + 2 i 2 ( t ) + s ( i 2 ( t ) −i1 ( t ) ) = 0 s 1 2 i1 ( t ) = 2 i 2 ( t ) + 2 i 2 ( t ) + i 2 ( t ) s sCombining these equations gives: d 2i2 ( t ) di2 ( t ) d 2 vs ( t ) 3s i2 ( t ) + 4si2 ( t ) + 2i2 ( t ) = s vs ( t ) 2 2 or 3 +4 + 2i2 ( t ) = dt 2 dt dt 2Ex. 9.4-1 d Using the operator s = , apply KCL at the dt top node: v (t ) 1 i s (t ) = + i (t ) + s v (t ) 4 4 Apply KVL to the right-most mesh: v (t ) − ( s i (t ) + 6 i (t )) = 0 Combining these equations gives: s 2 i ( t ) + 7 s i ( t ) + 10 i ( t ) = 4 i s ( t )The characteristic equation is: s 2 + 7 s + 10 = 0 .The natural frequencies are: s = −2 and s = −5 . Assume zero initial conditions . Write meshEx. 9.4-2 d equations using the operator s = : dt 1 s i1 ( t ) − i 2 ( t )  + 7 + 10 i1 ( t ) − 10 = 0 2   and 1 v ( t ) − 7 − s i1 ( t ) − i 2 ( t )  = 0   2 9-2
  • i2 (t )Now 0.005 s v ( t ) = i 2 ( t ) ⇒ v ( t ) = 200 so the second mesh equation becomes: s i2 (t ) 1 200 − 7 − s  i1 ( t ) − i 2 ( t )  = 0   s 2Writing the mesh equation in matrix form:  s 1  10 + 2 2 − s   i1 ( t )   3    =  −1 s 1 200  i 2 ( t )  7    s+  2  2 s Obtain the characteristic equation by calculating a determinant: s 1 10 + − s 2 2 = s 2 + 20s + 400 = 0 ⇒ s1,2 = −10 ± j 17.3 1 1 200 − s s+ 2 2 sEx. 9.5-1After t = 0 , we have a parallel RLC circuit with 1 1 7 1 1α = = = and ω o = 2 = =6 2 RC 2(6)(1/ 42) 2 LC (7)(1/ 42) 2 7 7∴ s 1 , s 2 = −α ± α −ω = − ± 2 2 o   − 6 = −1, − 6 2  2 dv ( t )∴ vn (t ) = A1e −t + A2 e−6t . We need vn (0) and n to evaluate A1 & A2 . dt t =0At t = 0+ we have: dv ( t ) 10 iC ( 0+ ) = −10 A ⇒ = = 420 V s dt t =0+ 1 42 Then vn (0+ ) = 0 = A1 + A2   dvn  A1 = −84 , A2 = 84 =−420= − A1 − 6 A2  dt t =0+  9-3
  • Finally ∴ vn (t ) = −84e −t + 84e −6t VEx. 9.5-2 1 1 s2 + s+ = 0 ⇒ s 2 + 40s + 100 = 0 RC LC Therefore s 1 , s 2 = −2.68 , −37.3vn ( t ) = A1 e −2.68t + A2 e−37.3t , v(0) = 0 = A1 + A2 + v(0+ ) + 1 dv(0+ )KCL at t = 0 yields + i (0 ) + = 0 so 1 40 dtdv(0+ ) = − 40 v(0+ ) − 40 i (0+ ) = − 40(0) − 40(1) = − 2.7 A1 − 37.3 A2 dtTherefore: A1 = −1.16 , A2 = 1.16 ⇒ v(t ) = vn (t ) = −1.16e −2.68t + 1.16e−37.3tEx. 9.6-1 1 1 1 1For parallel RLC circuits: α = = −3 = 50, ω o = 2 = = 2500 2 RC 2(10)(10 ) LC (0.4)(10−3 )The roots of the characteristic equations are:∴ s1,2 = −50 ± (50) 2 − 2500 = −50, −50The natural response is vn (t ) = A1 e−50t + A2 t e−50t .At t = 0+ we have: + −v(0+ ) ic (0 ) = = − .8V 10Ω dv( t ) i (0+ ) ∴ = c = −800 V / s dt + c t =0 9-4
  • So vn (0+ ) = 8 = A1 ⇒ vn (t ) = 8e−50t + A2 t e−50t dv(0+ ) = −800 = − 400 + A2 ⇒ A2 = − 400 dt ∴ vn (t ) = 8 e −50t − 400 t e−50t VEx. 9.7-1 1 1 α= = = 8000 2 RC 2(62.5)(10−6 ) 1 1 ω o2 = = −6 = 1 08 LC (.01)(10 )∴ s = − α ± α 2 −ω o = − 8000 ± 2 (8000) 2 −108 = − 8000 ± j 6000∴ vn (t ) = e −8000t [ A1 cos 6000 t + A2 sin 6000 t ]at t = 0+ 10 0.08 + + ic (0+ ) = 0 62.5 ⇒ ic (0+ ) = −.24 A dv(0+ ) ic (0+ ) ∴ = = −2.4 ×10+5 V/s dt C + dvn (0+ )vn (0 ) = 10 = A1 and = −2.4 × 105 = 6000 A2 − 8000 A1 ⇒ A2 = −26.7 dt∴ vn (t ) = e−8000t [10 cos 6000 t − 26.7 sin 6000 t ] VEx. 9.8-1 d 2v ( t ) dv ( t )The differential equation is 2 +5 + 6 v ( t ) = vs ( t ) so the characteristic equation is dt dts 2 + 5 s + 6 = 0 . The roots are s 1 , s 2 = −2, − 3 . d 2v ( t ) dv ( t )(a) 2 +5 + 6 v ( t ) = 8 . Try v f ( t ) = B . Substituting into the differential equation gives dt dt6 B = 8 ∴ v f (t ) = 8 / 6 V . 9-5
  • d 2v ( t ) dv ( t )(b) 2 +5 + 6 v ( t ) = 3 e − 4 t . Try v f ( t ) = B e− 4 t . Substituting into the differential dt dt 3 3equation gives (−4) 2 B + 5(−4) B + 6 B = 3 ⇒ B = . ∴ v f ( t ) = e−4t . 2 2 d v (t ) 2 dv ( t )(c) 2 +5 + 6 v ( t ) = 2 e − 2 t . Try v f ( t ) = B t e− 2 t because –2 is a natural frequency. dt dtSubstituting into the differential equation gives (4t − 4) B + 5 B (1 − 2t ) + 6 Bt = 2 ⇒ B = 2. ∴ v f ( t ) = 2 t e −2t .Ex. 9.8-2d 2i ( t ) di ( t ) +9 + 20 i ( t ) = 36 + 12 t . Try i f ( t ) = A + B t . Substituting into the differential dt 2 dtequation gives 0 + 9 B + 20( A + Bt ) = 36 + 12t ⇒ B = 0.6 and A = 1.53. ∴ i f ( t ) = 1.53 + 0.6 t AEx. 9.9-1When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and theinductor acts like a short circuit. Under these conditions R2 vC ( ∞ ) = 1 R1 + R 2Next, represent the circuit by a 2nd order differential equation: vC ( t ) dKCL at the top node of R2 gives: +C vC ( t ) = iL ( t ) R2 dt dKVL around the outside loop gives: vs ( t ) = L iL ( t ) + R1 iL ( t ) + vC ( t ) dtUse the substitution method to get 9-6
  • d  vC ( t ) d   v (t ) d  vs ( t ) = L  + C vC ( t )  + R1  C + C vC ( t )  + vC ( t ) dt  R 2  dt    R2  dt   d2  L d  R1  vC ( t ) +  + R1 C  vC ( t ) +  1 +  R2  C ( ) = LC v t dt 2  R2  dt     (a) C = 1 F, L = 0.25 H, R1 = R2 = 1.309 ΩUse the steady state response as the forced response: R2 1 v f = vC ( ∞ ) = 1= V R1 + R 2 2The characteristic equation is  R1   1  1+  R1   R2  2 s + 2 + s+ = s + 6 s + 8 = ( s + 2 )( s + 4 ) = 0  R 2 C L   LC          so the natural response is vn = A1 e −2 t + A2 e −4 t VThe complete response is 1 vc ( t ) = + A1 e −2 t + A2 e−4 t V 2 vC ( t ) d iL ( t ) = + vC ( t ) = −1.236 A1 e −2 t − 3.236 A2 e−4 t + 0.3819 1.309 dt +At t = 0 ( ) 0 = vc 0+ = A1 + A2 + 0.5 0 = iL 0 ( ) = −1.236 A − 3.236 A + 1 2 + 0.3819Solving these equations gives A1 = −1 and A2 = 0.5, so 1 −2 t 1 −4 t vc ( t ) = −e + e V 2 2(b) C = 1 F, L = 1 H, R1 = 3 Ω, R2 = 1 ΩUse the steady state response as the forced response: R2 1 v f = vC ( ∞ ) = 1= V R1 + R 2 4The characteristic equation is 9-7
  •  R1   1  1+  R1   R2  2 = s + 4s + 4 = ( s + 2 ) = 0 2 s + 2 + s+  R 2 C L   LC          so the natural response is vn = ( A1 + A2 t ) e −2 t VThe complete response is 1 vc ( t ) = + ( A1 + A2 t ) e −2 t V 4 iL ( t ) = vC ( t ) + d dt 1 vC ( t ) = + 4 (( A 2 ) − A1 ) − A2 t e −2 tAt t = 0+ 1 ( ) 0 = vc 0+ = A1 + 4 1 0 = iL 0+ =( ) 4 + A2 − A1Solving these equations gives A1 = −0.25 and A2 = −0.5, so 1  1 1  −2 t vc ( t ) = −  + t e V 4 4 2 (c) C = 0.125 F, L = 0.5 H, R1 = 1 Ω, R2 = 4 ΩUse the steady state response as the forced response: R2 4 v f = vC ( ∞ ) = 1= V R1 + R 2 5The characteristic equation is  R1   1  1+  R1   R2  2 s + 2 + s+ = s + 4s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 ) = 0  R 2 C L   LC          so the natural response is vn = e −2 t ( A1 cos 4 t + A2 sin 4 t ) VThe complete response is vc ( t ) = 0.8 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V vC ( t ) 1d A 2 −2 t A1 iL ( t ) = + vC ( t ) = 0.2 + e cos 4 t − e−2 t sin 4 t 4 8 dt 2 2 9-8
  • At t = 0+ ( ) 0 = vc 0+ = 0.8 + A1 A2 ( ) 0 = iL 0+ = 0.2 + 2Solving these equations gives A1 = -0.8 and A2 = -0.4, so vc ( t ) = 0.8 − e −2 t ( 0.8cos 4 t + 0.4sin 4 t ) VEx 9.9-2When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and theinductor acts like a short circuit. Under these conditions R2 1 R2 vC ( ∞ ) = 1, iL ( ∞ ) = and vo ( ∞ ) = 1 R1 + R 2 R1 + R 2 R1 + R 2Next, represent the circuit by a 2nd order differential equation: dKVL around the right-hand mesh gives: vC ( t ) = L iL ( t ) + R 2 iL ( t ) dt vs ( t ) − vC ( t ) dKCL at the top node of the capacitor gives: − C vC ( t ) = iL ( t ) R1 dtUse the substitution method to get d  d   d  vs ( t ) = R1 C  L iL ( t ) + R 2 iL ( t )  +  L iL ( t ) + R 2 iL ( t )  + R1 iL ( t ) dt  dt   dt  d2 d = R1 LC 2 iL ( t ) + ( L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t ) dt dt vo ( t )Using iL ( t ) = gives R2 R1 d2  L d  R1 + R 2  vs ( t ) = LC 2 vo ( t ) +  + R1 C  vo ( t ) +   R2  o ( ) v t R2 dt  R2  dt      9-9
  • (a) C = 1 F, L = 0.25 H, R1 = R2 = 1.309 ΩUse the steady state response as the forced response: R2 1 v f = vo ( ∞ ) = 1= V R1 + R 2 2The characteristic equation is  R2   1  1+  R2   R1  2 s + 2 + s+ = s + 6 s + 8 = ( s + 2 )( s + 4 ) = 0  R1 C L   LC          so the natural response is vn = A1 e −2 t + A2 e −4 t VThe complete response is 1 vo ( t ) = + A1 e −2 t + A2 e −4 t V 2 v (t ) 1 A1 −2 t A2 −4 t iL ( t ) = o = + e + e V 1.309 2.618 1.309 1.309 1 d 1 vC ( t ) = 1.309 iL ( t ) + iL ( t ) = + 0.6180 A1 e −2 t + 0.2361 A2 e −4 t 4 dt 2At t = 0+ 1 A1 A2 ( ) 0 = iL 0+ = + + 2.618 1.309 1.309 1 ( ) 0 = vC 0+ = 2 + 0.6180 A1 + 0.2361 A2Solving these equations gives A1 = −1 and A2 = 0.5, so 1 −2 t 1 −4 t vo ( t ) = −e + e V 2 2(b) C = 1 F, L = 1 H, R1 = 1 Ω, R2 = 3 ΩUse the steady-state response as the forced response: R2 3 v f = vo ( ∞ ) = 1= V R1 + R 2 4The characteristic equation is  R2   1  1+  R2   R1  2 = s + 4s + 4 = ( s + 2 ) = 0 2 s2 +  + s+  R1 C L   LC           9-10
  • so the natural response is vn = ( A1 + A2 t ) e −2 t VThe complete response is 3 vo ( t ) = + ( A1 + A2 t ) e −2 t V 4 vo ( t ) 1  A1 A2  −2 t iL ( t ) = = + + t e V 3 4  3 3  d 3   A1 A2  A2  −2 t vC ( t ) = 3 iL ( t ) + iL ( t ) = +   + + t e dt 4  3  3  3  At t = 0+ A1 1 0 = iL 0+ = ( ) 3 + 4 3 A1 A2 0 = vC 0+ ( ) = + + 4 3 3Solving these equations gives A1 = -0.75 and A2 = -1.5, so 3  3 3  −2 t vo ( t ) = −  + t e V 4 4 2 (c) C = 0.125 F, L = 0.5 H, R1 = 4 Ω, R2 = 1 ΩUse the steady state response as the forced response: R2 1 v f = vo ( ∞ ) = 1= V R1 + R 2 5The characteristic equation is  R2   1  1+  R2   R1  2 s + 2 + s+ = s + 4s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 ) = 0  R1 C L   LC          so the natural response is vn = e −2 t ( A1 cos 4 t + A2 sin 4 t ) VThe complete response is vo ( t ) = 0.2 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V vo ( t ) iL ( t ) = = 0.2 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V 1 9-11
  • 1 d vC ( t ) = iL ( t ) + iL ( t ) = 0.2 + 2 A2 e−2 t cos 4 t − 2 A1 e−2 t sin 4 t 2 dtAt t = 0+ ( ) 0 = iL 0+ = 0.2 + A1 0 = vC ( 0 ) = 0.2 + 2 A + 2Solving these equations gives A1 = -0.8 and A2 = -0.4, so vc ( t ) = 0.2 − e −2 t ( 0.2 cos 4 t + 0.1sin 4 t ) VEx. 9.10-1At t = 0+ no initial stored energy ⇒ v1 (0+ ) = v2 (0+ ) = i (0+ ) = 0 3 di (0+ ) di (0+ )KVL : − 0 + +0=0 ⇒ =0 10 dt dt 0 + dv1 (0+ )KCL at A : + i1 (0 ) + 0 = 0 ⇒ = 0 1 dt 5 dv2 (0+ ) dv2 (0+ )KCL at B : − 0 + i2 (0+ ) − 10 = 0 ⇒ i2 (0+ ) = = 10 ⇒ = 12 V s 6 dt dtFor t > 0: v1 1 d v1 KCL at A : + +i = 0 1 12 dt 5 d v2 KCL at B : − i + = 10 6 dt 3 di KVL: − v1 + + v2 = 0 10 dt Eliminating i yields 1 d v1 5 d v2 v1 + + − 10 = 0 12 dt 6 dt 3  5 d 2 v2  −v1 +   + v2 = 0 10  6 dt 2 Next 9-12
  • 1 d 2 v2 d v1 d 2 v2 1 d 3 v2 v1 = v2 + ⇒ = + 4 dt 2 dt dt 2 4 dt 3Now, eliminating v1 1 d 2 v2 1  d v2 1 d 3 v2  5 d v2 v2 + +  + + = 10 4 dt 2 12  dt 4 dt 3  6 dtFinally, the circuit is represented by the differential equation: d 3 v2 d2 v dv 3 + 12 2 2 + 44 2 + 48v2 = 480 dt dt dtThe characteristic equation is s 3 + 12s 2 + 44s + 48 = 0 . It’s roots are s1,2,3 = −2, −4, −6 . Thenatural response is vn = A1e −2t + A2 e −4t + A3e −6tTry v f = B as the forced response. Substitute into the differential equation and equatecoefficients to get B = 10. Then v2 (t ) = vn (t ) + v f (t ) = A1e −2t + A2 e −4t + A3e−6t + 10 dv2 (0+ ) d 2 v2 (0+ )We have seen that v2 (0+ ) = 0 and = 12 V/s . Also = 4[v1 (0+ ) − v2 (0+ )] = 0 . dt dt 2Then v2 (0+ ) = 0 = A1 + A2 + A3 + 10 dv2 (0+ ) = 12 = −2 A1 − 4 A2 − 6 A3 dt d 2 v2 (0+ ) = 0 = 4 A1 + 16 A2 + 36 A3 dt 2Solving these equations yields A1 = −15, A2 = 6, A3 = −1 so v2 (t ) = ( −15 e −2t + 6 e −4t − e −6t + 10 ) VEx. 9.11-1 1 1 10s2 + s+ =0 In our case L = 0.1, C = 0.1 so we have s 2 + s + 100 = 0 RC LC R 9-13
  • a) R = 0.4 Ω ⇒ s 2 + 25s + 100 = 0 s1,2 = −5, − 20b) R = 1 Ω ⇒ s 2 + 10 s + 100 = 0 s1,2 = − 5 ± j 5 3 9-14
  • ProblemsSection 9-3: Differential Equations for Circuits with Two Energy Storage ElementsP9.3-1 v (t ) dv ( t ) KCL: iL ( t ) = +C R2 dt diL ( t ) KVL: vs ( t ) = R1iL ( t ) + L + v (t ) dt  v (t ) dv ( t )  L dv ( t ) d 2v ( t ) vs ( t ) = R1  +C + + LC + v (t )  R2 dt  R2 dt dt 2 R   L  dv ( t ) d 2v ( t ) vs ( t ) =  1 + 1 v ( t ) +  R1C +  + [ LC ]  R2   R2  dt dt 2In this circuit R1 = 2 Ω, R 2 = 100 Ω, L = 1 mH, C = 10 µ F so dv ( t ) −8 d v ( t ) 2 vs ( t ) = 1.02v ( t ) + .00003 + 10 dt dt 2 dv ( t ) d 2 v ( t ) 10 vs ( t ) = 1.02 × 10 v ( t ) + 3000 8 8 + dt dt 2P9.3-2 d Using the operator s = we have dt v (t ) KCL: is ( t ) = + iL ( t ) + Csv ( t ) R1 KVL: v ( t ) = R2iL ( t ) + LsiL ( t ) is ( t )Solving by usingCramers rule for iL ( t ) : iL ( t ) = R2 Ls + + R2Cs + LCs 2 + 1 R1 R1  R2  L  1 +  iL ( t ) +  + R2C  siL ( t ) + [ LC ] s iL ( t ) = is ( t ) 2  R1   R1  9-15
  • In this circuit R1 = 100 Ω, R 2 = 10 Ω, L = 1 mH, C = 10 µ F so 1.1iL ( t ) + .00011siL ( t ) + 10−8 s 2iL ( t ) = is ( t ) diL ( t ) d 2iL ( t ) 1.1× 10 iL ( t ) + 11000 8 + 2 = 108 is ( t ) dt dtP9.3-3After the switch closes, a source KCL:transformation gives: dvc ( t ) vs ( t ) + vc ( t ) iL ( t ) + C + =0 dt R2 KVL: diL ( t ) R1is ( t ) + R1iL ( t ) + L − vc ( t ) − vs ( t ) = 0 dt di ( t ) vc ( t ) = R1is ( t ) + R1iL ( t ) + L L − vs ( t ) dt Differentiating d vc ( t ) d is ( t ) d iL ( t ) d 2iL ( t ) d vs ( t ) = R1 + R1 +L − dt dt dt dt 2 dtThen  d is ( t ) d iL ( t ) d 2iL ( t ) d vs ( t )  vs ( t )iL ( t ) + C  R1 + R1 +L − +  dt dt dt 2 dt  R2 1  di ( t )  +  R1is ( t ) + R1iL ( t ) + L L − vs ( t )  = 0 R2  dt Solving for i L ( t ) : d 2i L ( t )  R1 1  di L ( t )  R1 1  − R1 R1 di s ( t ) 1 dv s ( t ) 2 + +  + +  i L (t ) = i s (t ) − + dt  L R 2C  dt    LR 2C LC    LCR 2 L dt L dt 9-16
  • Section 9-4: Solution of the Second Order Differential Equation - The Natural ResponseP9.4-1From Problem P 9.3-2 the characteristic equation is: −11000± (11000) 2 − 4(1.1×108 ) 1.1×108 +11000 s + s 2 = 0 ⇒ s1 , s2 = = −5500± j8930 2P9.4-2 diL ( t ) KVL: 40 ( is ( t ) − iL ( t ) ) = (100 × 10−3 ) + vc ( t ) dt The current in the inductor is equal to the current in the capacitor so 1  dv ( t ) iL ( t ) =  ×10−3  c 3  dt  dv ( t )  100  d vC ( t ) 2  40 vC ( t ) = 40 is ( t ) −  × 10−3  C −  × 10−6  2  3  dt  3  dt d 2 vC ( t ) dv ( t ) 2 + 400 C + 30000 vC ( t ) = 40 is ( t ) dt dt s + 400s + 30000 = 0 ⇒ ( s + 100)( s + 300) = 0 ⇒ s1 = −100, s2 = −300 2P9.4-3 v ( t ) − vs ( t ) dv ( t ) KCL: + iL ( t ) + (10 × 10−6 ) =0 1 dt di ( t ) KVL: v ( t ) = 2iL ( t ) + (1× 10−3 ) L dt diL ( t ) diL ( t ) d 2iL ( t ) 0 = 2 iL ( t ) + (1× 10−3 ) − vs ( t ) + iL ( t ) + (10 × 10−6 ) ( 2 ) + (10 × 10−6 )(10−3 ) dt dt dt diL ( t ) d 2iL ( t ) d 2iL ( t ) diL ( t )vs ( t ) = 3iL ( t ) + 0.00102 + 10−8 2 ⇒ + 102000 + 3 × 108 iL ( t ) = 108 vs ( t ) dt dt dt dt s 2 + 102000s + 3 ×108 = 0 ⇒ s1 = 3031, s2 = −98969 9-17
  • Section 9.5: Natural Response of the Unforced Parallel RLC CircuitP9.5-1 dv( 0 ) dThe initial conditions are v ( 0 ) = 6 V, = −3000 V/s . Using the operator s = , the node dt dt v( t ) v( t ) − vs ( t )  L equation is Csv ( t ) + + = 0 or  LCs 2 + s + 1 v ( t ) = vs ( t ) R sL  R  1 1The characteristic equation is: s 2 + s+ = 0 ⇒ s 2 + 500 s + 40, 000 = 0 RC LCThe natural frequencies are: s1,2 = −250 ± 2502 − 40, 000 = −100, −400The natural response is of the form v ( t ) = Ae−100t + Be−400t . We will use the initial conditionsto evaluate the constants A and B. v ( 0) = 6 = A + B   dv ( 0 )  ⇒ A = −2 and B = 8 = −3000 = −100 A − 400 B  dt Therefore, the natural response is v ( t ) = −2e−100t + 8e−400t t >0P9.5-2The initial conditions are v ( 0 ) = 2 V, i ( 0 ) = 0 . 1 1The characteristic equation is: s 2 + s+ = 0 ⇒ s 2 + 4s + 3 = 0 RC LCThe natural frequencies are: s1 , s2 = −1, − 3The natural response is of the form v ( t ) = Ae − t + Be −3t . We will use the initial conditions toevaluate the constants A and B. dv ( t )Differentiating the natural response gives = − Ae − t − 3Be −3t . At t = 0 this becomes dt dv ( 0 ) dv ( t ) v ( t ) dv ( t ) v (t ) i (t ) = − A − 3B . Applying KCL gives C + + i ( t ) = 0 or = − − . dt dt R dt RC C dv ( 0 ) v ( 0) i ( 0)At t = 0 this becomes = − − . Consequently dt RC C v ( 0) i ( 0) 2 −1A − 3B = − − =− −0 = − 8 RC C 14 9-18
  • Also, v ( 0 ) = 2 = A + B . Therefore A = −1 and B = 3. The natural response is v ( t ) = −e − t + 3e −3t VP9.5-3 di1 ( t ) di ( t ) KVL : i1 + 5 −3 2 =0 (1) dt dt di ( t ) di ( t ) KVL : − 3 1 +3 2 + 2i2 ( t ) = 0 ( 2) dt dt dUsing the operator s = , the KVL equations are dt (1+5s )i1 + ( −3s )i2 = 0   3s  ⇒ (1 + 5s ) i1 − ( 3s ) i1 = 0 ⇒ (1 + 5s )( 3s + 2 ) − ( 3s )  i1 = 0 2 ( −3s ) i1 + ( 3s + 2 ) i2 = 0  3s + 2   1The characteristic equation is (1+5s )( 3s + 2 ) − 9 s 2 = 6 s 2 + 13s + 2 = 0 ⇒ s1,2 = − , −2 6 −t −tThe currents are i1 ( t ) = Ae 6 + Be−2t and i2 ( t ) = Ce 6 + De −2t , where the constants A, B, Cand D must be evaluated using the initial conditions. Using the given initial values of thecurrents gives i1 ( 0 ) = 11 = A + B and i 2 ( 0 ) = 11 = C + DLet t = 0 in the KCL equations (1) and (2) to get di1 ( 0 ) 33 A di2 ( 0 ) 143 C =− = − − 2 B and =− =− −D dt 2 6 dt 6 6So A = 3, B = 8, C = −1 and D = 12. Finally, i1 (t ) = 3e− t / 6 + 8e−2t A and i 2 (t ) = −e −t / 6 + 12e −2t A 9-19
  • Section 9.6: Natural Response of the Critically Damped Unforced Parallel RLC CircuitP9.6-1After t = 0 dic ( t ) Using KVL: 100 ic ( t ) + 0.025 + vc ( t ) = 0 dt The capacitor current and voltage are related by: dvc ( t ) ic ( t ) = 10−5 dt d 2 vc ( t ) dvc ( t ) ∴ 2 + 4000 + 4 × 106 vc ( t ) = 0 dt dtThe characteristic equation is: s 2 + 4000 s + 4 × 106 = 0The natural frequencies are: s1,2 = − 2000, − 2000The natural response is of the form: vc ( t ) = A1e−2000t + A2 t e−2000tBefore t = 0 the circuit is at steady state (The capacitor current is continuous at t = 0 in this circuit because it is equal to the inductor current.) vc ( 0+ ) = 3 = A1 dvc ( 0+ ) = 0 = −2000 A1 + A2 ⇒ A2 = 6000 dt ∴ vc ( t ) = ( 3 + 6000t ) e−2000t V for t ≥ 0P9.6-2After t = 0 Using KCL: t 1 dvc ( t ) ∫ vc (τ ) dτ + vc ( t ) + =0 −∞ 4 dt d 2 vc ( t ) dv ( t ) ⇒ +4 c + 4vc ( t ) = 0 dt dtThe characteristic equation is: s 2 + 4 s + 4 = 0 9-20
  • The natural frequencies are: s1,2 = −2, −2The natural response is of the form: vc ( t ) = A1e−2t + A2 t e−2 tBefore t = 0 the circuit is at steady state i L ( 0+ ) = i L ( 0 − ) and vC ( 0+ ) = vC ( 0− ) i C ( 0 + ) = −i L ( 0 + ) = − 2 A dv ( 0+ ) i C ( 0+ ) = = −8 V dt 14 dvc ( 0+ ) vc ( 0 + )=0= A1 and = −8 = A2 ⇒ vc ( t ) = −8 t e−2 t V dtP9.6-3Assume that the circuit is at steady state before t = 0. The initial conditions are vc ( 0− ) = 104 V & iL ( 0− ) = 0 AAfter t = 0 diL ( t ) KVL: − vc ( t ) + .01 + 106 iL ( t ) = 0 (1) dt KCL: dvC ( t )  d 2iL ( t ) di ( t )  iL ( t ) = −C = −C .01 2 + 106 L  ( 2) dt  dt dt  d 2iL ( t ) di ( t ) ∴ 0.01 C 2 + 106 C L + iL ( t ) = 0 dt dtThe characteristic equation is: ( 0.01 C ) s 2 + (106 C ) s + 1 = 0 (10 C ) 2 −106 C ± 6 − 4 ( 0.01C )The natural frequencies are: s1,2 = 2 ( 0.01C )For critically-damped response: 1012 C 2 − .04C = 0 ⇒ C = 0.04 pF so s1,2 = −5 ×107 , −5 × 107 . 9-21
  • The natural response is of the form: iL ( t ) = A1e −5×10 t + A2 t e −5×10 7 7 t diL +Now from (1) ⇒ dt ( 0 ) = 100 vc ( 0+ ) − 106 iL ( 0+ )  = 106 A   s di ( 0 )So iL ( 0 ) = 0 = A1 and L = A2 ∴ iL ( t ) = 106 t e−5×10 t A 7 = 106 dtNow v ( t ) = 106 iL ( t ) = 1012 t e−5×10 t V 7P9.6-4 1 1The characteristic equation can be shown to be: s 2 + s+ = s 2 + 500 s + 62.5 × 103 = 0 RC LCThe natural frequencies are: s1,2 = −250, − 250The natural response is of the form: v ( t ) = Ae −250t + B t e−250t dv ( 0 )v ( 0 ) = 6 = A and = −3000 = −250 A + B ⇒ B = −1500 dt∴ v ( t ) = 6e−250t − 1500 t e−250t VP9.6-5After t=0, using KVL yields:di ( t ) + Ri ( t ) + 2 + 4 ∫0 i (τ ) dτ = 6 (1) t dt v( t )Take the derivative with respect to t: d 2i ( t ) di ( t ) +R + 4i ( t ) = 0 dt 2 dtThe characteristic equation is s 2 + Rs + 4 = 0 ( s + 2) 2Let R = 4 for critical damping ⇒ =0So the natural response is i ( t ) = A t e −2t + B e −2t di ( 0 )i ( 0 ) = 0 ⇒ B = 0 and = 4 − R ( i( 0 ) ) = 4 − R ( 0 ) = 4 = A dt∴ i ( t ) = 4 t e −2t A 9-22
  • Section 9-7: Natural Response of an Underdamped Unforced Parallel RLC CircuitP9.7-1After t = 0 KCL: vc ( t ) dv ( t ) + iL ( t ) + 5 ×10−6 c = 0 (1) 250 dt KVL: diL ( t ) vc ( t ) = 0.8 ( 2) dtd 2vc ( t ) dv ( t ) + 800 c + 2.5 ×105 v ( t ) = 0 ⇒ s 2 + 800 s + 250, 000 = 0, s1,2 = −400 ± j 300 dt 2 dt cThe natural response is of the form v c (t ) = e −400t  A1 cos 300t + A 2sin 300t   Before t = 0 the circuit is at steady state: −6 iL ( 0+ ) = iL ( 0− ) = A 500 vc ( 0+ ) = vc ( 0− ) = 250 −6 ( 500 )+6 = 3 V From equation (1) : dvc ( 0+ ) = − 2 × 105 iL ( 0+ ) − 800vc ( 0+ ) = 0 dt vc ( 0+ ) = 3 = A1 dvc ( 0+ ) = 0 = −400 A1 + 300 A2 ⇒ A2 = 4 dt ∴ vc ( t ) = e −400t [3cos 300t + 4sin 300t ] V 9-23
  • P9.7-2Before t = 0 v ( 0+ ) = v ( 0− ) = 0 V i ( 0+ ) = i ( 0− ) = 2 AAfter t = 0 KCL : v (t ) 1 d v (t ) t + + 2 ∫ v (τ ) dτ + i ( 0 ) = 0 1 4 dt 0 dUsing the operator s = we have dt 1  2 v (t ) +   s v (t ) +   v (t ) + i (0) = 0 ⇒  4 s (s 2 + 4s + 8) v ( t ) = 0The characteristic equation and natural frequencies are: s 2 + 4s + 8 = 0 ⇒ s = −2 ± j 2The natural response is of the form: v ( t ) = e −2 t  B1 cos 2t + B 2 sin 2t    dv ( 0 ) v ( 0 ) = 0 = B1 and = 4  −i ( 0 ) − v ( 0 )  = −4 [ 2] = −8 = 2 B2 or B2 = −4   dtso v ( t ) = −4e −2t sin 2t VP9.7-3After t = 0 1 dvc ( t ) vc ( t ) KCL : + + iL ( t ) = 0 (1) 4 dt 2 4 diL ( t ) KVL : vc ( t ) = + 8 iL ( t ) ( 2) dt d 2i L ( t ) di L ( t )Characteristic Equation: +4 + 5 i L ( t ) = 0 ⇒ s 2 + 4 s + 5 = 0 ⇒ s1,2 = −2 ± i dt 2 dtNatural Response: i L ( t ) = e −2t  A1 cos t + A2 sin t    9-24
  • Before t = 0 vc ( 0− )  48   4 8 + 2  ⇒ vc ( 0 ) = vc ( 0 ) = 8 V + − = 7  2   8 iL ( 0+ ) = iL ( 0− ) = − = −4 A 2 diL ( 0+ ) vc ( 0+ ) 8 A = − 2iL ( 0+ ) = − 2 ( −4 ) = 10 dt 4 4 s iL ( 0+ ) = −4 = A1 diL ( 0+ ) = 10 = − 2 A1 + A2 ⇒ A2 = 2 dt∴ iL ( t ) = e −2t [ −4 cos t + 2sin t ] AP9.7-4The plot shows an underdamped response, i.e. v ( t ) = e − α t [ k1 cos ω t + k2 sin ω t ] + k3 .Examining the plot shows v ( ∞ ) = 0 ⇒ k3 = 0, v ( 0 ) = 0 ⇒ k1 = 0 .Therefore, v ( t ) = k2 e − α t sin ω t .Again examining the plot we see that the maximum voltage is approximately 260 mV the time isapproximately 5 ms and that the minimum voltage is approximately −200 mV the time isapproximately 7.5 ms. The time between adjacent maximums is approximately 5 ms so 2πω≈ = 1257 rad/s . Then 5 ×10−3 −α ( 0.005 ) 0.26 = k2 e sin (1257 (.005 ) ) (1) −α ( 0.0075 ) −0.2 = k2 e sin (1257 (.0075 ) ) ( 2)To find α we divide (1) by (2) to get − 1.3 = e α ( 0.0025 )  sin( 6.29 rad )  ⇒ e0.0025α = 1.95 ⇒ α = 267    sin ( 9.43 rad ) From (1) we get k2 = 544 . Then v ( t ) = 544e −267t sin1257t V 9-25
  • P9.7-5After t = 0 The characteristic equation is: 1 1 s2 + s+ = 0 or s 2 + 2 s + 5 = 0 RC LC The natural frequencies are: s1,2 = −1 ± j 2 The natural response is of the form: v(t ) = e − t  B1 cos 2t + B 2 sin 2t    v ( 0+ )v(0 ) = 2 = B1 . From KCL, ic ( 0 + ) = − 5 − i ( 0 ) = − 5 − 10 = − 1 V + + 2 1 2 s so dv ( 0 )+  1 3 = 10 − = − B + 2 B ⇒ B = −   1 2 2 dt  2 2 3Finally, v ( t ) = 2e − t cos 2t − e− t sin 2t V t≥0 2 9-26
  • Section 9-8: Forced Response of an RLC CircuitP9.8-1After t = 0 v (t ) dv ( t ) KCL : is ( t ) = + iL ( t ) + C R dt diL ( t ) KVL : v ( t ) = L dt L diL ( t ) d 2iL ( t ) is ( t ) = + iL ( t ) + LC R dt dt 2 d 2iL ( t ) 1 diL ( t ) 1 1 2 + + iL ( t ) = is ( t ) dt RC dt LC LC d 2iL ( t ) diL ( t ) 2 + ( 650 ) + (105 ) iL ( t ) = (105 ) is ( t ) dt dt(a) Try a forced response of the form i f ( t ) = A . Substituting into the differential equations gives 1 10+0+ A = ⇒ A = 1 . Therefore i f ( t ) = 1 A . (.01) (1×10 ) (.01) (1×10−3 ) −3(b) Try a forced response of the form i f ( t ) = A t + B . Substituting into the differential equations 65 1gives 0 + A + ( A t + B) = 0.5 t . Therefore A = 0.5 and (100 ) ( 0.001) ( 0.01)( 0.001)B = −3.25 × 10 −3 . Finally i f ( t ) = 5 t − 3.25 × 10−3 A .(c) Try a forced response of the form i f ( t ) = A e−250t . It doesn’t work so try a forced response ofthe form i f ( t ) = B t e−250t . Substituting into the differential equation gives ( −250 )2 B e−250t − 500 B e−250t  + 650 ( −250 ) B t e−250t + B e−250t  + 105 B t e−250t = 2 e−250t .    Equating coefficients gives( 250 ) B + 650 ( −250 ) B + 105 B = 0 ⇒ ( 250 ) + 650 ( −250 ) + 105  B = 0 ⇒ [ 0] B = 0 2 2  and −500 B + 650 B = 2 ⇒ B = 0.0133Finally i f ( t ) = 0.0133 t e −250t A . 9-27
  • P9.8-2After t = 0 d 2v ( t ) R dv ( t ) 1 v (t ) + + v (t ) = s dt L dt LC LC d 2v ( t ) dv ( t ) + 70 + 12000v ( t ) = 12000 vs ( t ) dt dt(a) Try a forced response of the form v f ( t ) = A . Substituting into the differential equations gives0 + 0 + 12000 A = 24000 ⇒ A = 2 . Therefore v f ( t ) = 2 V .(b) Try a forced response of the form v f ( t ) = A + B t . Substituting into the differential equations −70 Agives 70 A + 12000 A t + 12000 B = 2400 t . Therefore A = 0.2 and B = = −1.167 × 10 −3 . 12000Finally v f ( t ) = ( −1.167 × 10−3 ) t + 0.2 V .(c) Try a forced response of the form v f ( t ) = A e −30t . Substituting into the differential equations 12000gives 900 Ae −30t − 2100 Ae −30 t + 12000 Ae −30 t = 12000 e −30t . Therefore A = = 1.11 . Finally 10800 v f ( t ) = 1.11e−250 t V . 9-28
  • Section 9-9: Complete Response of an RLC CircuitP9.9-1First, find the steady state response for t < 0, when the switch is open. Both inputs are constantso the capacitor will act like an open circuit at steady state, and the inductor will act like a shortcircuit. After a source transformation at the left of the circuit: 11 − 4 i L ( 0) = = 2.33 mA 3000 and v C ( 0) = 4 VAfter the switch closesApply KCL at node a: vC d +C vC + iL = 0 R dtApply KVL to the right mesh: d dL i L + Vs − vC = 0 ⇒ vC = L i L + Vs dt dtAfter some algebra: d2 1 d 1 V d2 d i + 2 L iL + iL = − s ⇒ i + ( 500 ) i L + (1.6 × 105 ) i L = −320 2 L dt R C dt LC R LC dt dtThe characteristic equation is s 2 + 500 s + 1.6 × 105 = 0 ⇒ s1,2 = −250 ± j 312 rad/s 9-29
  • After the switch closes the steady-state inductor current is iL(∞) = -2 mA so i L ( t ) = −0.002 + e −250 t ( A1 cos 312 t + A2 sin 312 t ) d v C ( t ) = 6.25 i L (t ) + 4 dt = 6.25 e −250 t  −250 ( A1 cos 312 t − A2 sin 312 t ) − 312 ( A1 sin 312 t − A2 cos 312 t )  + 4   = 6.25 e −250 t ( 312 A2 − 250 A1 ) cos 312 t + ( 250 A2 + 312 A1 ) sin 312 t  + 4  Let t = 0 and use the initial conditions: iL ( 0+ ) = 0.00233 = −0.002 + A1 ⇒ 0.00433 = A1 250 250 vC ( 0+ ) = 4 = 6.25 ( 312 A2 − 250 A1 ) + 4 ⇒ A2 = A2 = ( 0.00433) = 0.00347 312 312Then i L ( t ) = −0.002 + e −250 t ( 0.00433cos 312 t + 0.00345sin 312 t ) = −0.002 + 0.00555 e −250 t cos ( 312 t − 36.68° ) A v C ( t ) = 4 + 13.9 e −250 t sin ( 312 t ) V vC (t ) i (t ) = = 2 + 6.95 e −250 t sin ( 312 t ) mA 2000 (checked using LNAP on 7/22/03)P9.9-2First, find the steady state response for t < 0. The input is constant so the capacitor will act likean open circuit at steady state, and the inductor will act like a short circuit. −1 i ( 0) = = −0.2 A 1+ 4 and 4 v (0) = ( −1) = −0.8 V 1+ 4 9-30
  • For t > 0Apply KCL at node a: v − Vs d +C v+i = 0 R1 dtApply KVL to the right mesh: d dR2 i + L i − v = 0 ⇒ v = R2 i + L iL dt dtAfter some algebra: d2 L + R1 R 2C d R1 + R 2 Vs d2 d 2 i+ i+ i= ⇒ 2 i +5 i +5i =1 dt R1 L C dt R1 L C R1 L C dt dt d2 dThe forced response will be a constant, if = B so 1 = 2 B + 5 B + 5 B ⇒ B = 0.2 A . dt dtTo find the natural response, consider the characteristic equation: 0 = s 2 + 5 s + 5 = ( s + 3.62 )( s + 1.38 )The natural response is in = A1 e −3.62 t + A2 e−1.38 tso i ( t ) = A1 e −3.62 t + A2 e−1.38 t + 0.2Then  d  v ( t ) =  4 i ( t ) + 4 i ( t )  = −10.48 A1 e−3.62 t − 1.52 A2 e−1.38 t + 0.8  dt At t=0+ −0.2 = i ( 0 + ) = A1 + A2 + 0.2 −0.8 = v ( 0 + ) = −10.48 A1 − 1.52 A2 + 0.8so A1 = 0.246 and A2 = −0.646. Finally i ( t ) = 0.2 + 0.246 e −3.62 t − 0.646 e−1.38 t A 9-31
  • P9.9-3First, find the steady state response for t < 0. The input is constant so the capacitors will act likean open circuits at steady state. 1000 v1 ( 0 ) = (10 ) = 5 V 1000 + 1000 and v2 ( 0 ) = 0 VFor t > 0,Node equations: v1 − 10  1  d v −v +  × 10−6  v1 + 1 2 = 0 1000  6  dt 1000 1  d ⇒ 2 v1 +  × 10−3  v1 − 10 = v2 6  dt v1 − v2  1  d =  × 10−6  v2 1000  16  dt 1  d ⇒ v1 − v2 =  × 10−3  v2  16  dtAfter some algebra: d2 d v + ( 2.8 × 104 ) v1 + ( 9.6 ×107 ) v1 = 9.6 × 108 2 1 dt dtThe forced response will be a constant, vf = B so d2 d 2 B + ( 2.8 × 104 ) B + ( 9.6 × 107 ) B = 9.6 × 108 ⇒ B = 10 V . dt dtTo find the natural response, consider the characteristic equation: s 2 + ( 2.8 × 104 ) s + ( 9.6 × 107 ) = 0 ⇒ s1,2 = −4 × 103 , −2.4 × 104The natural response is 3 4t vn = A1 e −4×10 t + A2 e−2.4×10so 3 4t v1 ( t ) = A1 e−4×10 t + A2 e−2.4×10 + 10At t = 0 9-32
  • −4×103 ( 0 ) −2.4×104 ( 0 ) 5 = v1 ( 0 ) = A1 e + A2 e + 10 = A1 + A2 + 10 (1)Next 1  d d 2 v1 +  ×10−3  v1 − 10 = v2 ⇒ v1 = −12000v1 + 6000 v2 + 6 × 104 6  dt dtAt t = 0 d v1 ( 0 ) = −12000v1 ( 0 ) + 6000 v2 ( 0 ) + 6 × 104 = −12000 ( 5 ) + 6000 ( 0 ) + 6 × 104 = 0 dtso d 3 4 dt ( ) ( v1 ( t ) = A1 −4 × 103 e −4×10 t + A2 −2.4 × 104 e−2.4×10 t )At t = 0+ d −4×103 ( 0 ) −2.4×104 ( 0 )0= dt ( v1 ( 0 ) = A1 −4 × 103 e ) ( + A2 −2.4 × 104 e ) ( = A1 −4 × 103 + A2 −2.4 × 104 ) ( )so A1 = −6 and A2 = 1. Finally 4t 3t v1 ( t ) = 10 + e −2.4 ×10 − 6 e−4 ×10 V for t > 0P9.9-4For t > 0 KCL at top node:  diL ( t )  1 dv ( t )  0.5 − 5 cos t  + iL ( t ) + =0 (1)  dt  12 dt KVL for right mesh: diL ( t ) 1 dv ( t ) 0.5 = + v (t ) ( 2) dt 12 dtTaking the derivative of these equations gives: d 2iL ( t ) diL ( t ) 1 d 2 v ( t ) d of (1) ⇒ 0.5 + + = −5 sin t (3) dt dt 2 dt 12 dt 2 d of ( 2 ) ⇒ 0.5 d iL ( t ) = 1 d v ( t ) + dv ( t ) 2 2 dt ( 4) dt 2 12 dt 2 dt 9-33
  • d 2iL ( t ) di ( t )Solving for 2 in ( 4 ) and L in ( 2 ) & plugging into ( 3) gives dt dtd 2v ( t ) dv ( t ) +7 + 12v ( t ) = −30sin t dt 2 dtThe characteristic equation is: s 2 + 7s+12 = 0 .The natural frequencies are s1,2 = −3, −4 .The natural response is of the form vn (t ) = A1e −3t + A2 e −4t . Try a forced response of the formv f ( t ) = B1 cos t + B 2 sin t . Substituting the forced response into the differential equation and 21 33equating like terms gives B1 = and B 2 = − . 17 17 21 33 v ( t ) = vn (t ) + v f ( t ) = A1e−3t + A2 e−4t + cos t − sin t 17 17We will use the initial conditions to evaluate A1 and A2. We are given iL ( 0 ) = 0 and v ( 0 ) = 1 V .Apply KVL to the outside loop to get 1 iC ( t ) + iL ( t )  + 1( iC ( t ) ) + v ( t ) − 5cos t = 0  At t = 0+ 5cos ( 0 ) + iL ( 0 ) − v ( 0 ) 5 + 0 − 1 iC ( 0 ) = = =2 A 2 2 dv ( 0 ) iC ( 0 ) 2 = = = 24 V/s dt 1 12 1 12 21  v(0+ ) = 1 = A1 + A2 +  A1 = 25 17   ⇒ 429 + dv(0 ) 33  A2 = − = 24 = −3 A1 − 4 A2 − 17 dt 17  Finally, 429e −4t − 21cos t + 33sin t ∴ v(t ) = 25e −3t − V 17 9-34
  • P9.9-5Use superposition. Find the response to inputs 2u(t) and –2u(t-2) and then add the two responses.First, consider the input 2u(t):For 0< t < 2 s d Using the operator s = we have dt KVL: vc ( t ) + siL ( t ) + 4 iL ( t ) − 2  = 0   (1) KCL: 1 3 iL ( t ) = s vc ( t ) ⇒ vc ( t ) = iL ( t ) (2) 3 sPlugging (2) into (1) yields the characteristic equation: ( s 2 + 4 s + 3) = 0 . The natural frequenciesare s1,2 = −1 , −3 . The inductor current can be expressed as iL (t ) = in (t ) + i f (t ) = ( A1 e −t + A2 e−3t ) + 0 = A1 e− t + A2 e −3t .Assume that the circuit is at steady state before t = 0. Then vc (0+ ) = 0 and iL (0+ ) = 0 . diL (0+ )Using KVL we see that (1) = 4  2 − iL (0+ )  − vC (0+ ) = 8 A/s . Then   dt iL (0) = 0 = A1 + A2   diL (0)  A1 = 4 , A2 = −4 . = 8 = − A1 − 3 A2  dt Therefore iL (t ) = 4e − t − 4e−3t A . The response to 2u(t) is  0 t<0 v1 (t ) = 8 − 4 iL (t ) =  −t −3t 8 − 16 e + 16 e V t > 0 . = 8 − 16 e − t + 16 e −3t  u ( t ) V  The response to –2u(t-2) can be obtained from the response to 2u(t) by first replacing t by t-2everywhere is appears and the multiplying by –1. Therefore, the response to –2u(t-2) is v2 (t ) =  −8 + 16e − (t − 2) − 16 e−3(t − 2)  u ( t - 2 ) V .  By superposition, v(t ) = v1 (t ) + v2 (t ) . Therefore v(t ) = 8 − 16e −t + 16e−3t  u (t ) +  −8 + 16e− (t − 2) − 16 e −3(t − 2)  u (t − 2) V     9-35
  • P9.9-6First, find the steady state response for t < 0, when the switch is closed. The input is constant sothe capacitor will act like an open circuit at steady state, and the inductor will act like a shortcircuit. 5 i ( 0) = − = −1.25 A 4 and v (0) = 5 VAfter the switch opens Apply KCL at node a: v d + 0.125 v = i 2 dt Apply KVL to the right mesh: d −10 cos t + v + 4 i+4i =0 dtAfter some algebra: d2 d 2 v + 5 v + 6 v = 20 cos t dt dtThe characteristic equation is s 2 + 5 s + 6 = 0 ⇒ s1,2 = −2, − 3Try vf = A cos t + B sin t d2 d 2 ( A cos t + B sin t ) + 5 ( A cos t + B sin t ) + 6 ( A cos t + B sin t ) = 20 cos t dt dt 9-36
  • ( − A cos t − B sin t ) + 5 ( − A sin t + B cos t ) + 6 ( A cos t + B sin t ) = 20 cos t ( − A + 5 B + 6 A) cos t + ( − B − 5 A + 6 B ) sin t = 20 cos tEquating the coefficients of the sine and cosine terms yields A =2 and B =2. Then vf = 2 cos t + 2 sin t v ( t ) = 2 cos t + 2 sin t + A1 e −2 t + A2 e −3 tNext v (t ) d d + 0.125 v ( t ) = i ( t ) ⇒ v (t ) = 8 i (t ) − 4 v (t ) 2 dt dt d  5 V v ( 0 ) = 8 i ( 0 ) − 4 v ( 0 ) = 8  −  − 4 ( 5 ) = −30 dt  4 sLet t = 0 and use the initial conditions: 5 = v ( 0 ) = 2 cos 0 + 2 sin 0 + A1 e −0 + A2 e −0 = 2 + A1 + A2 d v ( t ) = −2 sin t + 2 cos t − 2 A1 e −2 t − 3 A2 e−3 t dt d −30 = v ( 0 ) = −2 sin 0 + 2 cos 0 − 2 A1 e −0 − 3 A2 e −0 = 2 − 2 A1 − 3 A2 dtSo A1 = −23 and A2 = 26 and v ( t ) = 2 cos t + 2 sin t − 23 e −2 t + 26 e −3 t V for t > 0 9-37
  • P9.9-7First, find the steady-state response for t < 0. The input is constant so the capacitor will act likean open circuit at steady state, and the inductor will act like a short circuit. i ( 0) = 0 A and v (0) = 0 VAfter t = 0 d Apply KCL at node a: C v=i dt Apply KVL to the right mesh: d 8 i + v + 2 i + 4 (2 + i) = 0 dt d 12 i + v + 2 i = −8 dt d 2 d  1  4After some algebra: 2 v + ( 6) v +  v = − dt dt 2C CThe forced response will be a constant, vf = B so d2 d  1  4 2 B + ( 6) B +  B = − ⇒ B = −8 V dt dt 2C C(a) d2 d When C = 1/18 F the differential equation is 2 v + ( 6 ) v + ( 9 ) v = −72 . dt dt The characteristic equation is s + 6 s + 9 = 0 ⇒ s1,2 = −3, −3 2 Then v ( t ) = ( A1 + A2 t ) e −3t − 8 . Using the initial conditions: 9-38
  • 0 = v ( 0 ) = ( A1 + A2 ( 0 ) ) e0 − 8 ⇒ A1 = 8 d 0 = i ( 0) = C v ( 0 ) = C  −3 ( A1 + A2 ( 0 ) ) e0 + A2 e0  ⇒   A2 = 3 A1 = 24 dt So v ( t ) = ( 8 + 24 t ) e −3t − 8 V for t > 0(b) d2 d When C = 1/10 F the differential equation is 2 v + ( 6 ) v + ( 5 ) v = −40 dt dt The characteristic equation is s + 6 s + 5 = 0 ⇒ s1,2 = −1, −5 2 Then v ( t ) = A1 e − t + A2 e −5 t − 8 . Using the initial conditions: 0 = v ( 0 ) = A1 e0 + A2 e0 − 8 ⇒ A1 + A2 = 8   d  ⇒ A1 = 10 and A2 = −2 0 = v ( 0 ) = − A1 e − 5 A2 e ⇒ − A1 − 5 A2 = 0  0 0 dt  So v ( t ) = 10 e − t − 2 e −5 t − 8 V for t > 0(c) d2 d When C = 1/20 F the differential equation is 2 v + ( 6 ) v + (10 ) v = −80 dt dt The characteristic equation is s + 6 s + 10 = 0 ⇒ s1,2 = −3 ± j 2 Then v ( t ) = e −3 t ( A1 cos t + A2 sin t ) − 8 . Using the initial conditions: 0 = v ( 0 ) = e0 ( A1 cos 0 + A2 sin 0 ) − 8 ⇒ A1 = 8 d 0= v ( 0 ) = −3 e0 ( A1 cos 0 + A2 sin 0 ) + e0 ( − A1 sin 0 + A2 cos 0 ) ⇒ A2 = 3 A1 = 24 dt So v ( t ) = e −3 t ( 8cos t + 24 sin t ) − 8 V for t > 0 9-39
  • P9.9-8The circuit will be at steady state for t<0:so iL(0+) = iL(0−) = 0.5 Aand vC(0+) = vC(0−) = 2 V.For t>0: Apply KCL at node b to get: 1 1 d 1 1 d = i (t ) + v (t ) ⇒ i (t ) = − v (t ) 4 L 4 dt C L 4 4 dt C Apply KVL at the right-most mesh to get: d 1 d  L( 4i t) + 2 i (t ) = 8  vc ( t )  + v ( t ) dt L  4 dt  cUse the substitution method to get 1 1 d  d 1 1 d  1 d  4 − vC ( t )  + 2  − vC ( t )  = 8  vc ( t )  + v ( t )  4 4 dt  dt  4 4 dt   4 dt  cor d2 d 2 C ( ) v t + 6 v (t ) + 2 v (t ) 2= dt dt C C d2 dThe forced response will be a constant, vC = B so 2 = B + 6 B + 2B ⇒ B = 1 V . dt 2 dtTo find the natural response, consider the characteristic equation: 0 = s 2 + 6 s + 2 = ( s + 5.65 )( s + 0.35 )The natural response is −5.65 t −0.35 t vn = A1 e + A2 eso −5.65 t −0.35 t vC ( t ) = A1 e + A2 e +1Then 1 1 d 1 −5.65 t −0.35 t iL ( t ) = + vC ( t ) = + 1.41A1 e + 0.0875 A2 e 4 4 dt 4 9-40
  • At t=0+ 2 = vC ( 0 + ) = A1 + A + 1 2 1 1 = iL ( 0+ ) = + 1.41A1 + 0.0875 A 2 4 2so A1 = 0.123 and A2 = 0.877. Finally −5.65 t −0.35 t vC ( t ) = 0.123 e + 0.877 e +1 VP9.9-9After t = 0 The inductor current and voltage are related be di ( t ) v (t ) = L (1) dt Apply KCL at the top node to get dv( t ) v( t ) C + i (t ) + =5 (2) dt 2 dUsing the operator s = , and substituting (1) into (2) yields ( s 2 + 4s + 29 ) i ( t ) = 5 . dtThe characteristic equation is s 2 + 4 s + 29 = 0 . The characteristic roots are s1,2 = − 2 ± j 5 .The natural response is of the form in ( t ) = e −2t [ A cos 5t + B sin 5t ] .Try a forced response of the form i f ( t ) = A . Substituting into the differential equation givesA = 5 . Therefore i f ( t ) = 5 A .The complete response is i(t ) = 5 + e−2t [ A cos 5t + B sin 5t ] where the constants A and B areyet to be evaluated using the initial condition: i (0) = 0 = A + 5 ⇒ A = −5 di (0) di (0) 2A 0 = v ( 0) = L ⇒ = 0 = −2 A + 5 B ⇒ B = = −2 dt dt 5Finally, i (t ) = 5 + e −2t [ −5cos 5t − 2sin 5t ] A .P9.9-10 9-41
  • Assume that the circuit is at steady before t = 0. 2 i (0+ ) = i (0− ) = ×9 = 6 A 2+1 1 v(0+ ) = v(0− ) = × 9× 1.5 = 4.5 V 2+1After t = 0: Apply KCL at the top node of the current source to get dv( t ) v( t ) i ( t ) + 0.5 + = is ( t ) (1) dt 1.5 Apply KVL and KCL to get  dv( t ) v( t )  5di( t ) v ( t ) + 0.5 +  0.5 = + i (t ) (2)  dt 1.5  dtSolving for i(t) in (1) and plugging into (2) yields d 2 v( t ) 49 dv( t ) 4 2 di ( t ) 2 + + v ( t ) = is ( t ) + 2 s where is ( t ) = 9 + 3e −2t A dt 30 dt 5 5 dt d 49 4Using the operator s = , the characteristic equation is s 2 + s + = 0 and the characteristic dt 30 5roots are s1,2 =−.817 ± j.365 . The natural response has the form vn (t ) = e−0.817t  A1 cos (0.365 t ) + A2 sin (0.365 t )    −2 tTry a forced response of the form v f (t ) = B0 + B1e . Substituting into the differential equationsgives B0 = 4.5 and B1 = −7.04 . The complete response has the form v(t ) = e−.817 t  A1 cos(0.365 t ) + A2 sin (0.365 t )  + 4.5 − 7.04 e −2t  Next, consider the initial conditions: v (0) = 4.5 = A1 + 4.5 − 7.04 ⇒ A1 = 7.04 9-42
  • d v(0) 4 4 = 2 is (0) - 2 i (0) − v(0) = 2(9 + 3) − 2(6) − (4.5) = 6 dt 3 3 d v( 0 ) 6= = −0.817 A1 + 0.365 A2 + 14.08 ⇒ A2 = −6.38 dtSo the voltage is given by v(t ) = e−0.817 t 7.04 cos(0.365 t ) + A2 sin (0.365 t )  + 4.5 − 7.04 e−2t  Next the current given by v(t ) d v(t ) i (t ) = is (t ) − − 0.5 1.5 dtFinally i (t ) = e −0.817 t [ 2.37 cos(0.365t ) + 7.14sin(0.365t ) ] + 6 + 0.65e −2t AP9.9-11First, find the steady state response for t < 0. The input is constant so the capacitor will act likean open circuit at steady state, and the inductor will act like a short circuit. va ( 0− ) = −4 i ( 0− ) ( ) i ( 0− ) = 2va ( 0− ) = 2 −4 i ( 0− ) ⇒ i ( 0− ) = 0 A and v ( 0− ) = 10 V 9-43
  • For t > 0Apply KCL at node 2: va d + K va + C v=0 R dtKCL at node 1 and Ohm’s Law: va = − R iso d 1+ K R v= i dt C dApply KVL to the outside loop: L i + R i + v − Vs = 0 dtAfter some algebra: d2 R d 1+ K R 1+ K R d2 d 2 v+ v+ v= Vs ⇒ 2 v + 40 v + 144 v = 2304 dt L dt LC LC dt dtThe forced response will be a constant, vf = B so d2 d 2 B + ( 40 ) B + (144 ) B = 2304 ⇒ B = 16 V dt dtThe characteristic equation is s 2 + 40 s + 144 = 0 ⇒ s1,2 = −4, −36 .Then v ( t ) = A1 e− 4 t + A2 e−36 t + 16 .Using the initial conditions: 10 = v ( 0+ ) = A1 e0 + A2 e0 + 16 ⇒ A1 + A2 = −6   d  ⇒ A1 = −6.75 and A2 = 0.75 0= v ( 0+ ) = −4 A1 e0 − 36 A2 e0 ⇒ − 4 A1 − 36 A2 = 0  dt So v ( t ) = 0.75 e −36 t − 6.75 e−4 t + 16 V for t > 0 (checked using LNAP on 7/22/03) 9-44
  • Section 9-10: State Variable Approach to Circuit AnalysisP9.10-1At t = 0− the circuit is source free ∴ iL (0) = 0 and v(0) = 0.After t = 0 Apply KCL at the top node to get 1 dv( t ) iL ( t ) + =4 (1) 5 dt Apply KVL to the right mesh to get diL ( t ) v( t ) −(1) − 6 iL ( t ) = 0 (2) dt d 2 v( t ) dv( t )Solving for i1 ( t ) in (1) and plugging into (2) ⇒ 2 + 6 + 5v ( t ) = 120 . dt dtThe characteristic equation is s 2 + 6 s + 5= 0 . The natural frequencies are s1,2 =−1, −5 . The naturalresponse has the form vn (t ) = A1 e−t + A2 e−5t . Try v f ( t ) = B as the forced response.Substituting into the differential equation gives B = 24 so v f ( t ) = 24 V. The complete responsehas the form v (t ) = A1 e−t + A2 e−5t + 24 . dv(0)Now consider the initial conditions. From (1) = 20 − 5 iL (0) = 20 V . Then dt sv(0) = 0 = A1 + A2 + 24  dv(0)  ⇒ A1= −25, A2 = 1 = 20 = − A1 −5 A2  dt Finally v (t ) = − 25e −t + e −5t + 24 V . 9-45
  • P9.10-2Before t = 0 there are no sources in the circuit so iL (0) = 0 and v(0) = 0 .After t = 0 we have: Apply KCL at the top node to get 1 dv ( t ) iL ( t ) = 4 − (1) 10 dt Apply KVL to the left mesh to get diL ( t ) v( t ) − − 6iL ( t ) = 0 (2) dt Substituting iL ( t ) from (1) into (2) gives d 2 v( t ) dv( t ) 2 +6 + 10v( t ) = 240 dt dtThe characteristic equation is s 2 + 6 s + 10 = 0 . The natural frequencies are s1,2 = −3 ± j . Thenatural response has the form vn (t ) = e −3t  A1 cos t + A2 sin t  . Try v f ( t ) = B as the forced  response. Substituting into the differential equation gives B = 24 so v f ( t ) = 24 V. The completeresponse has the form v(t ) = e −3t  A1 cos t + A2 sin t  + 24 .   dv(0)Now consider the initial conditions. From (1) = 40 − 10 iL (0) = 40 V . Then dt s v (0) = 0 = A1 + 24 ⇒ A1 = −24 dv(0) = 40 = −3 A1 + A2 = 72 + A2 ⇒ A2 = −32 dtFinally, v (t ) = e −3t [ −24 cos t −32sin t ] + 24 V 9-46
  • P9.10-3Assume that the circuit is at steady state before t = 0 so iL (0) = −3 A and v(0) = 0 V .After t = 0 we have dv( t ) v( t ) KCL: i ( t ) + C + + 6=0 dt R di ( t ) KVL: v( t ) = L dt d  di( t )  1  di ( t )  i (t ) + C L  + L + 6=0 dt  dt  R  dt  d 2i ( t ) 1 di ( t ) 1 −6 d 2i ( t ) di ( t ) 2 + + i (t ) = ⇒ 2 + 100 + 250i ( t ) = −1500 dt R C dt LC LC dt dtThe characteristic equation is s 2 + 100 s + 250 = 0 . The natural frequencies ares1,2 = −2.57, − 97.4 . The natural response has the form in (t ) = A1 e−2.57 t + A2 e−97.4t . Tryi ( t ) = B as the forced response. Substituting into the differential equation gives B = −6 so fi f ( t ) = −6 A . The complete response has the form i(t ) = A1 e−2.57 t + A2 e−97.4t − 6 .Now consider the initial conditions: i (0) = A1 + A2 − 6 = −3   A1 = 3 .081 di (0)  = 0 = − 2.57 A1 −97.4 A2  A2 =−0.081 dt Finally: −2.57 t −97.4 t i (t ) = 3. 081 e −.081e −6 A di ( t ) −2.57 t v(t ) = .2 = −1.58e +1.58e−97.4t V dt 9-47
  • P9.10-4 Apply KCL to the supernode corresponding to the dependent voltage source to get dv ( t ) vx ( t ) ix ( t ) − 2ix ( t ) − 0.01 + =0 dt 2 Apply KCL at node 1 to get vx ( t ) i ( t ) − 2ix ( t ) + =0 2(Encircled numbers are node numbers.) Apply KVL to the top-right mesh to get di ( t ) vx ( t ) + v ( t ) − 0.1 =0 dtApply KVL to the outside loop to get ix ( t ) = −2 vx ( t ) − v ( t ) .Eliminate ix ( t ) to get 5 dv ( t ) vx ( t ) + v ( t ) − 0.01 =0 2 dt 9 i ( t ) + vx ( t ) + 2 v ( t ) = 0 2 d i (t ) vx ( t ) = −v ( t ) + 0.01 dtThen eliminate vx ( t ) to get dv ( t ) di ( t ) −1.5 v ( t ) − 0.01 + 0.25 =0 dt dt di ( t ) −2.5 v ( t ) + i ( t ) + 0.45 =0 dt dUsing the operator s = we have dt (−1.5 − .01s )v ( t ) + (.25s ) i ( t ) = 0 (−2.5)v( t ) + (1+.45s ) i ( y ) = 0The characteristic equation is s 2 +13.33 s + 333.33 = 0 . The natural frequenciesare s1 , s2 = −6.67 ± j 17 . The natural response has the form vn (t ) = [ A cos17 t + B sin17 t ] e−6.67 t .The forced response is v f ( t ) = 0 . The complete response has the formv(t ) = [ A cos17 t + B sin 17 t ] e−6.67 t . 9-48
  • The given initial conditions are i (0) = 0 and v (0) = 10 V. Then dv(0) v(0) =10 = A and =−111= − 6.67 A +17 B ⇒ B =−2.6 dtFinally i (t ) = [3.27 sin 17 t ] e−6.67 t A . (Checked using LNAP on 7/22/03)P9.10-5 v(0) 10Assume that the circuit is at steady state before t = 0 so v(0) = 10 V and iL (0) = = A. 3 3The switch is open when 0 < t < 0.5 s For this series RLC circuit we have: R 1 α= = 3 and ω 02 = = 12 2L LC −α ± α 2 −ω02 = s1,2 = −3 ± j 3The natural response has the form vn (t ) = e −3t ( A cos1.73 t + B sin1.72 t ) . There is no source sov f ( t ) =0 . The complete response has the form v (t ) = e −3t ( A cos1.73 t + B sin1.72 t ) .Next v(0) = 10 = A    A= 10 dv(0) i ( 0 ) 10 3  ⇒  =− =− = − 20 = −3 A +1.73 B   B =5.77 dt C 16 so v(t ) = e−3t (10 cos1.73 t + 5.77 sin1.73 t ) V i(t ) = e−3t ( 3.33 cos1.73 t − 5.77 sin1.73 t ) AIn particular,  1.73 1.73  v (0.5) = e −1.5  10 cos + 5.77 sin  = 0.2231× ( 6.4864 + 4.3915 ) = 2.43 V  2 2 and  1.73 1.73  i (0.5) = e −1.5  3.33cos −5.77 sin  = 0.2231× ( 2.1600 − 4.3915 ) = −0.50 A  2 2  9-49
  • The switch is closed when t > 0.5 s Apply KCL at the top node: v( t ) −30 1 dv( t ) + iL ( t ) + =0 6 6 dt 1 dv( t )  ⇒ iL ( t ) = 5 −  v( t ) +  6 dt  d iL ( t ) 1  dv( t ) d v( t )  2 ⇒ =−  +  dt 6  dt dt 2  Apply KVL to the right mesh: 1 diL ( t ) v( t ) = 3 iL ( t ) + 2 dtThe circuit is represented by the differential equation d 2v ( t ) dv ( t ) 2 +7 + 18 v ( t ) = 180 dt dtThe characteristic equation is 0 = s 2 + 7 s + 18 . The natural frequencies are s1,2 = −3.7 ± j 2.4 .The natural response has the form vn (t ) = e −3.5 t ( A cos 2.4 t + B sin 2.4 t ) . The forced response isv f ( t ) = 10 V . The complete response has the form v (t ) = e −3.5 t ( A cos 2.4 t + B sin 2.4 t ) + 10 .Next v (0.5) = e −3.5×0.5 ( A cos1.2 + B sin1.2 ) = 0.063 A + 0.162 B dv ( 0.5 ) = e−3.5×0.5 ( −3.5 A + 2.4 B ) cos1.2 − ( 3.5B + 2.4 A ) sin1.2    dt = e−3.5×0.5 ( −3.5cos1.2 − 2.4sin1.2 ) A + e−3.5×0.5 ( 2.4 cos1.2 − 3.5sin1.2 ) B = −0.6091A − 0.4158BUsing the initial conditions yields 2.43=v(0.5) = 0.063 A+ 0.162 B   A=−20.65 dv(0.5) i ( 0.5 ) −1 2  ⇒ =− =− =3=−0.6091A− 0.4158B  B = 23.03 dt C 16 Finally v(t ) = e −3.5 t ( −20.65 cos 2.4 t + 23.03 sin 2.4 t ) + 10In summary   e −3t (10 cos1.73 t + 5.77 sin1.73 t ) V 0<t <0.5 v(t ) =  −3.5 t e ( −20.65 cos 2.4 t + 23.03 sin 2.4 t )+10 V 0.5<t  9-50
  • Section 9-11: Roots in the Complex PlaneP9.11-1After t = 0 d i1 2 ×10−3 −6 i1 + i 2 + dt =0 2000 d i1 d i2 2 ×10−3 = 3000 i 2 + 2 ×10−3 dt dt d Using the operator s = yields dt  2000 + 2 × 10−3 s 2000   i1   6   −3   =   −3 2 × 10 s 3000 + 2 × 10 s  i 2   0  s 2 + 3.5 × 106 s + 1.5 ×1012 = 0 ⇒ s1,2 = −5 × 105 , −3 ×106P9.11-2From P9.7-1 1 1 1 1s2 + s+ = 0 ⇒ s2 + s+ =0 RC LC ( 250 ) ( 5 ×10 ) ( 0.8) ( 5 ×10−6 ) −6 ⇒ s 2 + 800 s + 250000 = 0 s1,2 = 400 ± j 300P9.11-3 1 dv( t ) v( t ) KCL: i L ( t ) = × 10−6 + 4 dt 4000 di ( t ) KVL: vs ( t ) = 4 L + v( t ) dt d 1 dv( t ) v( t )  d 2v ( t ) dv ( t ) v ( t ) = 4  ×10−6 +  + v ( t ) = 10−6 + 10−3 + v (t ) s dt  4 dt 4000  dt 2 dt 9-51
  • Characteristic equation: s 2 + 103 s + 106 = 0Characteristic roots: s1,2 = −500 ± j 866P9.11-4Before t = 0 the voltage source voltage is 0 V so vb (0+) = vb (0−) = 0 V andi (0+) = i (0−) = 0 A . Apply KCL at node a to get va (0+ ) −36 v (0+ ) − vb (0+ ) − i (0+) + a = 0 ⇒ va (0+ ) + 2 va (0+ ) = 36 ⇒ va (0) = 12 V 12 6After t = 0 the node equations are: va ( t ) − vs ( t ) 1 t v ( t ) − va ( t ) − + ∫ ( vb (τ ) − va (τ ) ) dτ + b =0 12 L 0 6 d vb ( t ) vb ( t ) − va ( t ) 1 t C + + ∫ ( vb (τ ) − va (τ ) ) dτ = 0 dt 6 L 0 dUsing the operator s = we have dt 1 1 1  1 1 v (t )  + +  va ( t ) +  − −  vb ( t ) = s  12 6 s  6 s 12  1 1 1 1 1  − −  va ( t ) +  s + +  vb ( t ) = 0  6 s  18 6 s Using Cramer’s rule 9-52
  • ( s 2 +5s + 6) vb ( t ) = ( s + 6) vs ( t ) =( s + 6) ( 36 )The characteristic equation is s 2 + 5s + 6 = 0 . The naturalfrequencies are s1,2 = −2, −3 . The natural response has theform vn (t ) = A1 e −2 t + A2 e −3 t . Try v f ( t ) = B as the forcedresponse. Substituting into the differential equation givesB = 36 so v f ( t ) = 36 V. The complete response has the formvb (t ) = A1 e −2 t + A2 e −3t + 36 .Next vb (0+ ) = 36 + A1 + A2 dvb + (0 ) = −2 A1 − 3 A2 dtApply KCL at node a to get 1 dvb ( t ) vb ( t ) − va ( t ) + + i (t ) = 0 18 dt 6At t = 0+ 1 d vb ( 0 ) = v ( 0 )−v ( 0 ) − i (0+ ) = 126−0 − 0 = 2 + + + 1 a b (−2 A1 − 3 A2 ) = 18 18 dt 6So 0 = vb (0+ ) =36+ A1 + A 2   1  ⇒ A1 = −72, A2 = 36 ( −2 A 1 − 3 A 2 ) = 2  18 Finally vb ( t ) = 36 − 72e −2 t + 36e −3t V for t ≥ 0 9-53
  • PSpice ProblemsSP 9-1Make three copies of the circuit: one for each set of parameter values. (Cut and paste, but be sureto edit the labels of the parts so, for example, there is only one R1.) 9-54
  • V(C1:2), V(C2:2) and V(C3:2) are the capacitor voltages, listed from top to bottom. 9-55
  • SP 9-2Make three copies of the circuit: one for each set of parameter values. (Cut and paste, but be sureto edit the labels of the parts so, for example, there is only one R1.) 9-56
  • V(R2:2), V(R4:2) and V(R6:2) are the output voltages, listed from top to bottom. 9-57
  • SP 9-3 9-58
  • SP 9-4 9-59
  • Verification ProblemsVP 9-1This problem is similar to the verification example in this chapter. First, check the steady-stateinductor current v ( t ) 25 i (t ) = s = = 250 mA 100 100This agrees with the value of 250.035 mA shown on the plot. Next, the plot shows anunderdamped response. That requires 12 ×10−3 = L < 4 R 2C = 4(100) 2 (2 × 10−6 ) = 8 × 10−2This inequality is satisfied, which also agrees with the plot.The damped resonant frequency is given by 2 1  1  2 1  1  ω = −  = − −6  = 5.95 × 103 d LC  2 RC  ( 2 ×10 )(12 ×10 )  −6 −3 2(100)(2 × 10 ) The plot indicates a maxima at 550.6µs and a minima at 1078.7µs. The period of the dampedoscillation is T d = 2 (1078.7 µ s − 550.6 µ s) = 1056.2 µ sFinally, check that 2π 2π 5.95 ×103 = ω d = = −6 = 5.949 × 103 Td 1056.2 ×10The value of ω d determined from the plot agrees with the value obtained from the circuit.The plot is correct.VP 9-2This problem is similar to the verification example in this chapter. First, check the steady-stateinductor current. v ( t ) 15 i (t ) = s = = 150 mA 100 100 9-60
  • This agrees with the value of 149.952 mA shown on the plot.Next, the plot shows an underdamped response. This requires 8 ×10−3 = L < 4 R 2C = 4 (100)2 (0.2 × 10−6 ) = 8 × 10−3This inequality is not satisfied. The values in the circuit would produce a critically damped, notunderdamped, response.This plot is not correct.Design ProblemsDP 9-1When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and theinductor acts like a short circuit. Under these conditions R2 vC ( ∞ ) = 1 R1 + R 2 1The specifications require that vC ( ∞ ) = so 2 1 R2 = ⇒ R1 = R 2 2 R1 + R 2Next, represent the circuit by a 2nd order differential equation: vC ( t ) dKCL at the top node of R2 gives: +C vC ( t ) = iL ( t ) R2 dt dKVL around the outside loop gives: vs ( t ) = L iL ( t ) + R1 iL ( t ) + vC ( t ) dtUse the substitution method to get 9-61
  • d  vC ( t ) d   v (t ) d  vs ( t ) = L  + C vC ( t )  + R1  C + C vC ( t )  + vC ( t ) dt  R 2  dt    R2  dt   d2  L d  R1  vC ( t ) +  + R1 C  vC ( t ) +  1 +  R2  C ( ) = LC v t dt 2  R2  dt     The characteristic equation is  R1   1  1+  R1  R2 s2 +  + s+  = s 2 + 6 s + 8 = ( s + 2 )( s + 4 ) = 0  R 2 C L   LC         Equating coefficients of like powers of s: R1 1+ 1 R1 R2 + = 6 and =8 R2 C L LCUsing R1 = R 2 = R gives 1 R 1 + =6 ⇒ =4 RC L LC 1These equations do not have a unique solution. Try C = 1 F. Then L = H and 4 1 3 1 + 4 R = 6 ⇒ R 2 − R + = 0 ⇒ R = 1.309 Ω or R = 0.191 Ω R 2 4Pick R = 1.309 Ω. Then 1 vc ( t ) = + A1 e −2 t + A2 e−4 t V 2 vC ( t ) d iL ( t ) = + vC ( t ) = −1.236 A1 e−2 t − 3.236 A2 e−4 t + 0.3819 1.309 dtAt t = 0+ ( ) 0 = vc 0+ = A1 + A2 + 0.5 0 = iL 0 ( ) = −1.236 A − 3.236 A + 1 2 + 0.3819Solving these equations gives A1 = -1 and A2 = 0.5, so 1 −2 t 1 −4 t vc ( t ) = −e + e V 2 2 9-62
  • DP 9-2When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and theinductor acts like a short circuit. Under these conditions R2 vC ( ∞ ) = 1 R1 + R 2 1The specifications require that vC ( ∞ ) = so 4 1 R2 = ⇒ 3 R 2 = R1 4 R1 + R 2Next, represent the circuit by a 2nd order differential equation: vC ( t ) dKCL at the top node of R2 gives: +C vC ( t ) = iL ( t ) R2 dt dKVL around the outside loop gives: vs ( t ) = L iL ( t ) + R1 iL ( t ) + vC ( t ) dtUse the substitution method to get d  vC ( t ) d   v (t ) d  vs ( t ) = L  + C vC ( t )  + R1  C + C vC ( t )  + vC ( t ) dt  R 2  dt    R2  dt   d2  L d  R1  2 C ( )  + R1 C  vC ( t ) +  1 +  R2  C ( ) = LC v t + v t dt  R2  dt     The characteristic equation is  R1   1  1+  R1   R2  2 = s + 4s + 4 = ( s + 2 ) = 0 2 s + 2 + s+  R 2 C L   LC         Equating coefficients of like powers of s: R 1+ 1 1 R1 R2 + = 4 and =4 R2 C L LC 9-63
  • Using R 2 = R and R1 = 3R gives 1 3R 1 + =4 ⇒ =1 RC L LCThese equations do not have a unique solution. Try C = 1 F. Then L = 1 H and 1 4 1 1 + 3 R = 4 ⇒ R 2 − R + = 0 ⇒ R = 1 Ω or R = Ω R 3 3 3Pick R = 1 Ω. Then R1 = 3 Ω and R 2 = 1 Ω . 1 vc ( t ) = + ( A1 + A2 t ) e −2 t V 4 iL ( t ) = vC ( t ) + d dt 1 vC ( t ) = + 4 (( A 2 ) − A1 ) − A2 t e −2 tAt t = 0+ 1 ( ) 0 = vc 0+ = A1 + 4 1 0 = iL 0+ =( ) 4 + A2 − A1Solving these equations gives A1 = -0.25 and A2 = -0.5, so 1  1 1  −2 t vc ( t ) = −  + t e V 4 4 2 DP 9-3When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and theinductor acts like a short circuit. Under these conditions R2 vC ( ∞ ) = 1 R1 + R 2 4The specifications require that vC ( ∞ ) = so 5 9-64
  • 4 R2 = ⇒ 4 R1 = R 2 5 R1 + R 2Next, represent the circuit by a 2nd order differential equation: vC ( t ) dKCL at the top node of R2 gives: +C vC ( t ) = iL ( t ) R2 dt dKVL around the outside loop gives: vs ( t ) = L iL ( t ) + R1 iL ( t ) + vC ( t ) dtUse the substitution method to get d  vC ( t ) d   v (t ) d  vs ( t ) = L  + C vC ( t )  + R1  C + C vC ( t )  + vC ( t ) dt  R 2  dt    R2  dt   d2  L d  R1  2 C ( )  + R1 C  vC ( t ) +  1 +  R2  C ( ) = LC v t + v t dt  R2  dt     The characteristic equation is  R1   1  1+  R1  R2 s2 +  + s+  = s 2 + 4s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 ) = 0  R 2 C L   LC         Equating coefficients of like powers of s: R1 1+ 1 R1 R2 + = 4 and = 20 R2 C L LCUsing R1 = R and R 2 = 4 R gives 1 R 1 + = 4 and = 16 4R C L LC 1 1These equations do not have a unique solution. Try C = F . Then L = H and 8 2 2 + 2 R = 4 ⇒ R2 − 2R + 1 = 0 ⇒ R = 1 Ω RThen R1 = 1 Ω and R 2 = 4 Ω . Next vc ( t ) = 0.8 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V 9-65
  • vC ( t ) 1d A 2 −2 t A1 iL ( t ) = + vC ( t ) = 0.2 + e cos 4 t − e−2 t sin 4 t 4 8 dt 2 2 +At t = 0 ( ) 0 = vc 0+ = 0.8 + A1 A2 ( ) 0 = iL 0+ = 0.2 + 2Solving these equations gives A1 = −0.8 and A2 = −0.4, so vc ( t ) = 0.8 − e −2 t ( 0.8cos 4 t + 0.4sin 4 t ) VDP 9-4When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and theinductor acts like a short circuit. Under these conditions R2 vC ( ∞ ) = 1 R1 + R 2 1The specifications require that vC ( ∞ ) = so 2 1 R2 = ⇒ R1 = R 2 2 R1 + R 2Next, represent the circuit by a 2nd order differential equation: vC ( t ) dKCL at the top node of R2 gives: +C vC ( t ) = iL ( t ) R2 dt dKVL around the outside loop gives: vs ( t ) = L iL ( t ) + R1 iL ( t ) + vC ( t ) dtUse the substitution method to get 9-66
  • d  vC ( t ) d   v (t ) d  vs ( t ) = L  + C vC ( t )  + R1  C + C vC ( t )  + vC ( t ) dt  R 2  dt    R2  dt   d2  L d  R1  vC ( t ) +  + R1 C  vC ( t ) +  1 +  R2  C ( ) = LC v t dt 2  R2  dt     The characteristic equation is  R1   1  1+  R1  R2 s2 +  + s+  = s 2 + 4s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 ) = 0  R 2 C L   LC         Equating coefficients of like powers of s: R1 1+ 1 R1 R2 + = 4 and = 20 R2 C L LCUsing R1 = R 2 = R gives 1 R 1 + = 4 and = 10 RC L LC 1Substituting L = into the first equation gives 10 C 4 1 0.4 ± 0.42 − 4 ( 0.1) ( RC ) − ( RC ) + = 0 ⇒ RC = 2 10 10 2Since RC cannot have a complex value, the specification cannot be satisfied.DP 9-5When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and theinductor acts like a short circuit. Under these conditions 9-67
  • R2 1 R2 vC ( ∞ ) = 1, iL ( ∞ ) = and vo ( ∞ ) = 1 R1 + R 2 R1 + R 2 R1 + R 2 1The specifications require that vo ( ∞ ) = so 2 1 R2 = ⇒ R1 = R 2 2 R1 + R 2Next, represent the circuit by a 2nd order differential equation: dKVL around the right-hand mesh gives: vC ( t ) = L iL ( t ) + R 2 iL ( t ) dt vs ( t ) − vC ( t ) dKCL at the top node of the capacitor gives: − C vC ( t ) = iL ( t ) R1 dtUse the substitution method to get d  d   d  vs ( t ) = R1 C  L iL ( t ) + R 2 iL ( t )  +  L iL ( t ) + R 2 iL ( t )  + R1 iL ( t ) dt  dt   dt  d2 d 2 L( ) ( = R1 LC i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t ) dt dt vo ( t )Using iL ( t ) = gives R2 R1 d2  L d  R1 + R 2  vs ( t ) = 2 o( )  + R1 C  vo ( t ) +   R2  o ( ) LC v t + v t R2 dt  R2  dt     The characteristic equation is  R2   1  1+  R2   R1  = s 2 + 6 s + 8 = ( s + 2 )( s + 4 ) = 0 s2 +  + s+  R1 C L   LC         Equating coefficients of like powers of s: R2 1+ 1 R2 R1 + = 6 and =8 R1 C L LCUsing R1 = R 2 = R gives 1 R 1 + =6 ⇒ =4 RC L LC 1These equations do not have a unique solution. Try C = 1 F. Then L = H and 4 9-68
  • 1 3 1 + 4 R = 6 ⇒ R 2 − R + = 0 ⇒ R = 1.309 Ω or R = 0.191 Ω R 2 4Pick R = 1.309 Ω. Then 1 vo ( t ) = + A1 e −2 t + A2 e −4 t V 2 v (t ) 1 A1 −2 t A2 −4 t iL ( t ) = o = + e + e V 1.309 2.618 1.309 1.309 1 d 1 vC ( t ) = 1.309 iL ( t ) + iL ( t ) = + 0.6167 A1 e−2 t + 0.2361 A2 e−4 t 4 dt 2At t = 0+ 1 A1 A2 ( ) 0 = iL 0+ = + + 2.618 1.309 1.309 1 ( ) 0 = vC 0+ = 2 + 0.6167 A1 + 0.2361 A2Solving these equations gives A1 = -1 and A2 = 0.5, so 1 −2 t 1 −4 t vo ( t ) = −e + e V 2 2DP 9-6When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and theinductor acts like a short circuit. Under these conditions R2 1 R2 vC ( ∞ ) = 1, iL ( ∞ ) = and vo ( ∞ ) = 1 R1 + R 2 R1 + R 2 R1 + R 2 3The specifications require that vo ( ∞ ) = so 4 3 R2 = ⇒ 3R1 = R 2 4 R1 + R 2 9-69
  • Next, represent the circuit by a 2nd order differential equation: dKVL around the right-hand mesh gives: vC ( t ) = L iL ( t ) + R 2 iL ( t ) dt vs ( t ) − vC ( t ) dKCL at the top node of the capacitor gives: − C vC ( t ) = iL ( t ) R1 dtUse the substitution method to get d  d   d  vs ( t ) = R1 C  L iL ( t ) + R 2 iL ( t )  +  L iL ( t ) + R 2 iL ( t )  + R1 iL ( t ) dt  dt   dt  d2 d 2 L( ) ( = R1 LC i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t ) dt dt vo ( t )Using iL ( t ) = gives R2 R1 d2  L d  R1 + R 2  vs ( t ) = vo ( t ) +  + R1 C  vo ( t ) +   R2  o ( ) LC v t R2 dt 2  R2  dt     The characteristic equation is  R2   1  1+  R2   R1  = s 2 + 4 s + 4 = ( s + 2 )2 = 0 s2 +  + s+  R1 C L   LC         Equating coefficients of like powers of s: R2 1+ 1 R2 R1 + = 4 and =4 R1 C L LCUsing R1 = R and R 2 = 3R gives 1 3R 1 + = 4 and =1 RC L LCThese equations do not have a unique solution. Try C = 1 F. Then L = 1 H and 1 4 1 1 + 3 R = 4 ⇒ R 2 − R + = 0 ⇒ R = 1 Ω or R = Ω R 3 3 3Pick R = 1 Ω. Then R1 = 1 Ω and R 2 = 3 Ω . 3 vo ( t ) = + ( A1 + A2 t ) e −2 t V 4 9-70
  • vo ( t ) 1  A1 A2  −2 t iL ( t ) = = + + t e V 3 4  3 3  d 3   A1 A2  A2  −2 t vC ( t ) = 3 iL ( t ) + iL ( t ) = +   + + t e dt 4  3  3  3  At t = 0+ A1 1 0 = iL ( 0 + ) = + 3 4 3 A1 A2 0 = vC ( 0 + ) = + + 4 3 3Solving these equations gives A1 = −0.75 and A2 = −1.5, so 3  3 3  −2 t vo ( t ) = −  + t e V 4 4 2 DP 9-7When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and theinductor acts like a short circuit. Under these conditions R2 1 R2 vC ( ∞ ) = 1, iL ( ∞ ) = and vo ( ∞ ) = 1 R1 + R 2 R1 + R 2 R1 + R 2 1The specifications require that vo ( ∞ ) = so 5 1 R2 = ⇒ R1 = 4 R 2 5 R1 + R 2Next, represent the circuit by a 2nd order differential equation: dKVL around the right-hand mesh gives: vC ( t ) = L iL ( t ) + R 2 iL ( t ) dt vs ( t ) − vC ( t ) dKCL at the top node of the capacitor gives: − C vC ( t ) = iL ( t ) R1 dt 9-71
  • Use the substitution method to get d  d   d  vs ( t ) = R1 C  L iL ( t ) + R 2 iL ( t )  +  L iL ( t ) + R 2 iL ( t )  + R1 iL ( t ) dt  dt   dt  d2 d 2 L( ) ( = R1 LC i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t ) dt dt vo ( t )Using iL ( t ) = gives R2 R1 d2  L d  R1 + R 2  vs ( t ) = 2 o( )  + R1 C  vo ( t ) +   R2  o ( ) LC v t + v t R2 dt  R2  dt     The characteristic equation is  R2   1  1+  R2   R1  = s 2 + 4s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 ) = 0 s2 +  + s+  R1 C L   LC         Equating coefficients of like powers of s: R2 1+ 1 R2 R1 + = 4 and = 20 R1 C L LCUsing R 2 = R and R1 = 4 R gives 1 R 1 + = 4 and = 16 4R C L LC 1 1These equations do not have a unique solution. Try C = F . Then L = H and 8 2 2 + 2 R = 4 ⇒ R2 − 2R + 2 = 0 ⇒ R = 1 Ω RThen R1 = 4 Ω and R 2 = 1 Ω . Next vo ( t ) = 0.2 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V vo ( t ) iL ( t ) = = 0.2 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V 1 1 d vC ( t ) = iL ( t ) + iL ( t ) = 0.2 + 2 A2 e−2 t cos 4 t − 2 A1 e−2 t sin 4 t 2 dtAt t = 0+ 9-72
  • ( ) 0 = iL 0+ = 0.2 + A1 0 = vC ( 0 ) = 0.2 + 2 A + 2Solving these equations gives A1 = −0.8 and A2 = −0.4, so vc ( t ) = 0.2 − e −2 t ( 0.2 cos 4 t + 0.1sin 4 t ) VDP 9-8When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and theinductor acts like a short circuit. Under these conditions R2 1 R2 vC ( ∞ ) = 1, iL ( ∞ ) = and vo ( ∞ ) = 1 R1 + R 2 R1 + R 2 R1 + R 2 1The specifications require that vC ( ∞ ) = so 2 1 R2 = ⇒ R1 = R 2 2 R1 + R 2Next, represent the circuit by a 2nd order differential equation: dKVL around the right-hand mesh gives: vC ( t ) = L iL ( t ) + R 2 iL ( t ) dt vs ( t ) − vC ( t ) dKCL at the top node of the capacitor gives: − C vC ( t ) = iL ( t ) R1 dtUse the substitution method to get d  d   d  vs ( t ) = R1 C  L iL ( t ) + R 2 iL ( t )  +  L iL ( t ) + R 2 iL ( t )  + R1 iL ( t ) dt  dt   dt  d2 d 2 L( ) ( = R1 LC i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t ) dt dt vo ( t )Using iL ( t ) = gives R2 9-73
  • R1 d2  L d  R1 + R 2  vs ( t ) = 2 o( )  + R1 C  vo ( t ) +   R2  o ( ) LC v t + v t R2 dt  R2  dt     The characteristic equation is  R2   1  1+  R2   R1  = s 2 + 4s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 ) = 0 s2 +  + s+  R1 C L   LC         Equating coefficients of like powers of s: R2 1+ 1 R2 R1 + = 4 and = 20 R1 C L LCUsing R1 = R 2 = R gives 1 R 1 + =4 ⇒ = 10 RC L LC 1Substituting L = into the first equation gives 10 C 4 1 0.4 ± 0.42 − 4 ( 0.1) ( RC ) − ( RC ) + = 0 ⇒ RC = 2 10 10 2Since RC cannot have a complex value, the specification cannot be satisfied. 9-74
  • DP 9-9Let’s simulate the three copies of the circuit simultaneously. Each copy uses a different value ofthe inductance.The PSpice transient response shows that when L = 1 H the inductor current has its maximum atapproximately t=0.5 s.Consequently, we choose L = 1 H. 9-75
  • Chapter 10 – Sinusoidal Steady-State AnalysisExercisesEx. 10.3-1 (a) T = 2π / ω = 2π / 4 (b) v leads i by 30 − (−70) = 100°Ex. 10.3-2 v ( t ) = 3cos 4 t + 4sin 4 t = ( ( 3 )) = 5 cos(4 t −53° ) (3) 2 + (4) 2 cos 4 t − tan −1 4Ex. 10.3-3    12    i ( t ) = −5 cos 5t + 12 sin 5t = (−5) 2 + (12) 2 cos  5 t − 180 + tan −1     = 13 cos (5 t −112.6°)    −5   Ex. 10.4-1 v( t ) d d v( t ) I KCL: is ( t ) = + C v (t ) ⇒ v (t ) + = m cos ω t R dt dt RC CTry v f ( t ) = A cos ω t + B sin ω t & plug into above differential equation to get 1 I −ω A sin ω t + ω B cos ω t + ( A cos ω t + B sin ω t ) = m cos ω t RC CEquating sin ω t & cos ω t terms yields R Im ω R2 C Im A = and B = 1+ω 2 R 2 C 2 1+ω 2 R 2 C 2Therefore R Im ω R2 C Im R Im v f (t ) = cos ω t + sin ω t = cos ω t − tan −1 (ω RC )    1+ω 2 R 2 C 2 1+ω 2 R 2 C 2 1+ ω R C 2 2 2 10-1
  • Ex. 10.4-2 10 10∠0 10 KVL : − 10 + j 3 I + 2 I = 0 ⇒ I = = = ∠ − 56.3 A 2+ j 3 13 ∠56.3 13Therefore 10 i (t ) = cos (3 t − 56.3 ) A 13Ex. 10.5-1 10 = 4.24 e− j 45 = 3− j 3 2.36 e j 45Ex. 10.5-2 j 32 32e j 90 32 j (90-111) = = e = 3.75 e − j 21 −3+ j 8 8.54 e j111 8.54Ex. 10.6-1(a) i = 4 cos(ω t − 80 ) = Re{4 e j ω t e − j 80° } ⇒ I = 4 e − j 80° = 4∠ − 80° A(b) i = 10 cos(ω t + 20° ) = Re{10 e j ω t e j 20° } ⇒ I = 10e j 20° = 10∠20°(c) i = 8sin (ω t − 20 ) = 8cos (ω t − 110 ) = 8 Re{e j ω t e − j110° } ⇒ I = 8e − j110° = 8∠ − 110° AEx. 10.6-2(a) V = 10∠ − 140° = 10 e − j140° V ⇒ v(t ) = Re{10 e − j140° e j ω t } = 10 cos (ω t − 140°) V(b) V = 80 + j 75 = 109.7∠43.2° = 109.7 e j 43.2° ⇒ v (t ) = Re{109.7 e j 43.2° e jω t } = 109.7 cos(ω t + 43.2°) V 10-2
  • Ex. 10.6-3 d 0.01 v + v = 10 cos 100 t dt ( 0.01)( j 100 )V + V =10 10 V= =7.071 ∠− 45° 1+ j v = 7.071 cos 100 t VEx. 10.6-4 { vs = 40 cos100t = Re 4 e j100 t } di (t ) 1 t KVL: i (t ) + 10 × 10 −3 dt + 5×10 −3 ∫ −∞ i (t ) dt = vSAssume i (t ) = Ae j100 t where A is complex number to be determined. Plugging intothe differential equation yields 4 Ae j100 t + j Ae j100 t + (− j 2 A)e j100 t = 4 e j100 t ⇒ A= = 2 2 e j 45° 1− jIn the time domain: { } { } i (t ) = Re 2 2 e j100 t e j 45° = Re 2 2 e j (100 t -45°) = 2 2 cos (100 t + 45° ) AEx. 10.7-1(a) v = R i = 10 (5 cos100 t ) = 50 cos 100 t(b) di v = L = 0.01[5(−100) sin100 t ] = −5sin100 t = 5cos (100 t + 90° ) V dt(c) 1 v = ∫ i dt = 103 ∫ 5cos100 t dt = 50sin100 t = 50 cos (100 t − 90°) V CEx. 10.7-2 dv i=C = 10 × 10−6 [100(−500) sin (500 t + 30°)] dt = − 0.5sin (500 t +30°) = 0.5sin (500 t + 210°) = 0.5cos(500t +120°) A 10-3
  • Ex. 10.7-3From Figure E10.7-3 we get i (t ) = I m sin ω t = I m cos(ω t − 90°) A ⇔ I = I m ∠ − 90°A v(t ) = Vm cos ω t ⇔ V = Vm ∠0° VThe voltage leads the current by 90° so the element is an inductor: V V ∠0° V Z eq = = m = m ∠90° Ω I I m ∠−90° ImAlso Vm Vm Z eq = j ω L = ω L ∠90° ⇒ ω L = ⇒ L = Im ω ImEx. 10.8-1 1 2.4 j 2.4 ZR = 8 Ω, ZC = = = = − j 2.4 Ω, ZL1 = j 5 (2) = j 10 Ω, 1 j j× j j5 12 ZL2 = j 5 (4) = j 20 Ω and VS = 5 ∠-90° V.Ex. 10.8-2 1 4 j4 ZR = 8 Ω, ZC = = = = − j 4 Ω, ZL1 = j 3 (2) = j 6 Ω, 1 j j× j j3 12 ZL2 = j 3 (4) = j 12 Ω and IS = 4 ∠15° V. 10-4
  • Ex 10.9-1 j10 V1 (ω ) = 5 e − j 90 = 3.9 e − j 51 8 + j10 j 20 V 2 (ω ) = 5 e − j 90 =