Solution manual for introduction to electric circuits

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Solucionario, dorf circuitos electricos 6 edicion

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Solution manual for introduction to electric circuits

  1. 1. Solution Manual to accompanyIntroduction to Electric Circuits, 6e By R. C. Dorf and J. A. Svoboda 1
  2. 2. 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
  3. 3. 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
  4. 4. 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
  5. 5. 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
  6. 6. 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
  7. 7. 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
  8. 8. 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
  9. 9. 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
  10. 10. 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
  11. 11. 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
  12. 12. 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
  13. 13. 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
  14. 14. 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
  15. 15. 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
  16. 16. 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
  17. 17. 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
  18. 18. 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
  19. 19. 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
  20. 20. 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
  21. 21. 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
  22. 22. 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
  23. 23. 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
  24. 24. 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
  25. 25. 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
  26. 26. 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
  27. 27. 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
  28. 28. 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
  29. 29. 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
  30. 30. 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
  31. 31. 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
  32. 32. 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
  33. 33. 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
  34. 34. 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
  35. 35. 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
  36. 36. 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
  37. 37. 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
  38. 38. 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
  39. 39. 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
  40. 40. 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
  41. 41. 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
  42. 42. 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
  43. 43. 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
  44. 44. P3.7-9 3-21
  45. 45. 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
  46. 46. 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
  47. 47. 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
  48. 48. 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
  49. 49. 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
  50. 50. 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
  51. 51. 28
  52. 52. 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
  53. 53. 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
  54. 54. 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
  55. 55. 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
  56. 56. 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
  57. 57. 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
  58. 58. 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
  59. 59. 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
  60. 60. 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
  61. 61. 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
  62. 62. 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
  63. 63. 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

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