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![Chapter 29178
Because Q(0) = Q0, δ = 0 and: tQtQ ωcos)( 0=
The current in the circuit is the
derivative of Q with respect to t: [ ] tQtQ
dt
d
dt
dQ
I ωωω sincos 00 −===
(a) A spreadsheet program was used to plot the following graph showing both the
charge on the capacitor and the current in the circuit as functions of time. L, C,
and Q0 were all arbitrarily set equal to one to obtain these graphs. Note that the
current leads the charge by one-fourth of a cycle or 90°.
-1.2
-0.6
0.0
0.6
1.2
0 2 4 6 8 10
t (s)
Charge
Current
Q(mC)
I(mA)
-1.2
-0.6
0.0
0.6
1.2
(b) The equation for the current is: tQI ωω sin0−= (1)
The sine and cosine functions are
related through the identity: ⎟
⎠
⎞
⎜
⎝
⎛
+=−
2
cossin
π
θθ
Use this identity to rewrite equation
(1): ⎟
⎠
⎞
⎜
⎝
⎛
+=−=
2
cossin 00
π
ωωωω tQtQI
Thus, the current leads the charge by
90°.
Driven RL Circuits
35 •• A coil that has a resistance of 80.0 Ω has an impedance of 200 Ω when
driven at a frequency of 1.00 kHz. What is the inductance of the coil?
Picture the Problem We can solve the expression for the impedance in an LR
circuit for the inductive reactance and then use the definition of XL to find L.](https://image.slidesharecdn.com/ch29ssm-181218172931/85/Ch29-ssm-6-320.jpg)
![Chapter 29184
Picture the Problem We can use Kirchhoff’s loop rule to obtain a differential
equation relating the input, capacitor, and resistor voltages. Because the voltage
drop across the resistor is small compared to the voltage drop across the capacitor,
we can express the voltage drop across the capacitor in terms of the input voltage.
Apply Kirchhoff’s loop rule to the
input side of the filter to obtain:
( ) 0=−− IRVtV C
where VC is the potential difference
across the capacitor.
Substitute for and I to obtain:( )tV
0cos cpeakin =−−
dt
dQ
RVtV ω
Because Q = CVC:
[ ]
dt
dV
CCV
dt
d
dt
dQ C
C ==
Substitute for dQ/dt to obtain:
0cospeak =−−
dt
dV
RCVtV C
Cω
the differential equation describing the
potential difference across the
capacitor.
Because there is no significant
change in the voltage signal during
one time constant:
0=
dt
dVC
⇒ 0=
dt
dV
RC C
Substituting for
dt
dV
RC C
yields:
0cospeakin =− CVtV ω
and
tVVC ωcospeakin=
Consequently, the potential
difference across the resistor is given
by:
[ ]tV
dt
d
RC
dt
dV
RCV C
R ωcospeakin==
49 •• Show that the average power dissipated in the resistor of the high-pass
filter of Problem 44 is given by Pave =
Vin peak
2
2R 1+ ωRC( )−2
⎡
⎣
⎤
⎦
.
Picture the Problem We can express the instantaneous power dissipated in the
resistor and then use the fact that the average value of the square of the cosine
function over one cycle is ½ to establish the given result.
The instantaneous power P(t)
dissipated in the resistor is: R
V
tP
2
out
)( =](https://image.slidesharecdn.com/ch29ssm-181218172931/85/Ch29-ssm-12-320.jpg)
![Alternating-Current Circuits 185
The output voltage is:outV ( )δω −= tVV cosHout
From Problem 44:
( ) 2
peakin
H
1
−
+
=
RC
V
V
ω
Substitute in the expression for P(t)
to obtain: ( )
( )[ ] ( )δω
ω
δω
−
+
=
−=
−
t
RCR
V
t
R
V
tP
2
2
2
peakin
2
2
H
cos
1
cos)(
Because the average value of the
square of the cosine function over
one cycle is ½: ( )[ ]2
2
peakin
ave
12
−
+
=
RCR
V
P
ω
51 •• The circuit shown in Figure 29-36 is an example of a low-pass filter.
(Assume that the output is connected to a load that draws only an insignificant
amount of current.) (a) If the input voltage is given by Vin = Vin peak cos ωt, show
that the output voltage is Vout = VL cos(ωt – δ) where VL = Vin peak 1+ ωRC( )2
.
(b) Discuss the trend of the output voltage in the limiting cases ω → 0 and
ω → ∞.
Picture the Problem In the phasor
diagram for the RC low-pass filter,
appV
r
and CV
r
are the phasors for Vin and
Vout, respectively. The projection of
appV
r
onto the horizontal axis is
Vapp = Vin, the projection of CV
r
onto the
horizontal axis is VC = Vout,
apppeak V
r
=V , and φ is the angle between
CV
r
and the horizontal axis.
δ
CV
r
RV
r
appV
r
ωt
φ
(a) Express :appV tVV ωcospeakinapp =
where ZIV peakpeakin =
and (1)2
C
22
XRZ +=
outV CV= is given by:
φ
φ
cos
cos
peak
peak,out
C
C
XI
VV
=
=](https://image.slidesharecdn.com/ch29ssm-181218172931/85/Ch29-ssm-13-320.jpg)
![Alternating-Current Circuits 193
(b) Noting that the voltage during the
second half of each cycle is now
zero, express the voltage during the
first half cycle of the time interval
TΔ2
1
:
0VV =
Express the square of the voltage
during this half cycle:
2
0
2
VV =
Calculate ( )av
2
V by integrating V2
from t = 0 to t = TΔ2
1
and dividing
by ΔT:
( ) [ ] 2
02
1Δ
0
2
0
Δ
0
2
0
av
2 2
1
2
1
ΔΔ
Vt
T
V
dt
T
V
V
T
T
=== ∫
Substitute to obtain:
V5.8
2
V12
2
02
02
1
rms ====
V
VV
89 ••• A circuit consists of an ac generator, a capacitor and an ideal
inductor⎯all connected in series. The emf of the generator is given by
tωε cospeak . (a) Show that the charge on the capacitor obeys the equation
t
C
Q
dt
Qd
L ωε cospeak2
2
=+ . (b) Show by direct substitution that this equation is
satisfied by tQQ ωcospeak= where
( )2
0
2
peak
peak
ωω
ε
−
−=
L
Q . (c) Show that the current
can be written as ( )δω −= tII cospeak , where
CL XXL
I
−
=
−
=
peak
2
0
2
peak
peak
εε
ωω
ω
and
δ = –90º for ω < ω0 and δ = 90º for ω > ω0, where ω0 is the resonance frequency.
Picture the Problem In Part (a) we can apply Kirchhoff’s loop rule to obtain the
2nd
order differential equation relating the charge on the capacitor to the time. In
Part (b) we’ll assume a solution of the form , differentiate it twice,
and substitute for d
tQQ ωcospeak=
2
Q/dt2
and Q to show that the assumed solution satisfies the
differential equation provided
( )2
0
2
peak
peak
ωω
ε
−
−=
L
Q . In Part (c) we’ll use our
results from (a) and (b) to establish the result for Ipeak given in the problem
statement.
(a) Apply Kirchhoff’s loop rule
to obtain:
0=−−
dt
dI
L
C
Q
ε](https://image.slidesharecdn.com/ch29ssm-181218172931/85/Ch29-ssm-21-320.jpg)
More Related Content
Ch29 ssm
- 1. Chapter 29 Alternating-Current Circuits ConceptualProblems 3 • If the frequency in the circuit shown in Figure 29-27 is doubled, the inductance of the inductor will (a) double, (b) not change, (c) halve, (d) quadruple. Determine the Concept The inductance of an inductor is determined by the details of its construction and is independent of the frequency of the circuit. The inductive reactance, on the other hand, is frequency dependent. )(b is correct. 7 • (a) In a circuit consisting of a generator and a capacitor, are there any time intervals when the capacitor receives energy from the generator? If so, when? Explain your answer. (b) Are there any time intervals when the capacitor supplies power to the generator? If so, when? Explain your answer. Determine the Concept Yes to both questions. (a) While the magnitude of the charge is accumulating on either plate of the capacitor, the capacitor absorbs power from the generator. (b) When the magnitude of the charge is on either plate of the capacitor is decreasing, it supplies power to the generator. 9 • Suppose you increase the rotation rate of the coil in the generator shown in the simple ac circuit in Figure 29-29. Then the rms current (a) increases, (b) does not change, (c) may increase or decrease depending on the magnitude of the original frequency, (d) may increase or decrease depending on the magnitude of the resistance, (e) decreases. Determine the Concept Because the rms current through the resistor is given by ω εε 22 peakrms rms NBA R I === , is directly proportional to ω.rmsI )(a is correct. 11 • Consider a circuit consisting solely of an ideal inductor and an ideal capacitor. How does the maximum energy stored in the capacitor compare to the maximum value stored in the inductor? (a) They are the same and each equal to the total energy stored in the circuit. (b) They are the same and each equal to half of the total energy stored in the circuit. (c) The maximum energy stored in the capacitor is larger than the maximum energy stored in the inductor. (d) The maximum energy stored in the inductor is larger than the maximum energy stored in the capacitor. (e) You cannot compare the maximum energies based on the data given because the ratio of the maximum energies depends on the actual capacitance and inductance values. 173
- 2. Chapter 29174 Determine theConcept The maximum energy stored in the electric field of the capacitor is given by C Q U 2 e 2 1 = and the maximum energy stored in the magnetic field of the inductor is given by 2 m 2 1 LIU = . Because energy is conserved in an LC circuit and oscillates between the inductor and the capacitor, Ue = Um = Utotal. ( )a is correct. 17 • True or false: (a) A transformer is used to change frequency. (b) A transformer is used to change voltage. (c) If a transformer steps up the current, it must step down the voltage. (d) A step-up transformer, steps down the current. (e) The standard household wall-outlet voltage in Europe is 220 V, about twice that used in the United States. If a European traveler wants her hair dryer to work properly in the United States, she should use a transformer that has more windings in its secondary coil than in its primary coil. (f) The standard household wall-outlet voltage in Europe is 220 V, about twice that used in the United States. If an American traveler wants his electric razor to work properly in Europe, he should use a transformer that steps up the current. (a) False. A transformer is a device used to raise or lower the voltage in a circuit. (b) True. A transformer is a device used to raise or lower the voltage in a circuit. (c) True. If energy is to be conserved, the product of the current and voltage must be constant. (d) True. Because the product of current and voltage in the primary and secondary circuits is the same, increasing the current in the secondary results in a lowering (or stepping down) of the voltage. (e) True. Because electrical energy is provided at a higher voltage in Europe, the visitor would want to step-up the voltage in order to make her hair dryer work properly. (f) True. Because electrical energy is provided at a higher voltage in Europe, the visitor would want to step-up the current (and decrease the voltage) in order to make his razor work properly.
- 3. Alternating-Current Circuits 175 AlternatingCurrent in Resistors, Inductors, and Capacitors 19 • A 100-W light bulb is screwed into a standard 120-V-rms socket. Find (a) the rms current, (b) the peak current, and (c) the peak power. Picture the Problem We can use rmsrmsav IP ε= to find ,rmsI rmspeak 2II = to find , andpeakI peakpeakpeak εIP = to find .peakP (a) Relate the average power delivered by the source to the rms voltage across the bulb and the rms current through it: rmsrmsav IP ε= ⇒ rms av rms ε P I = Substitute numerical values and evaluate :rmsI A833.0A8333.0 V120 W100 rms ===I (b) Express in terms of :peakI rmsI rmspeak 2II = Substitute for and evaluate :rmsI peakI ( ) A18.1 A1785.1A8333.02peak = ==I (c) Express the maximum power in terms of the maximum voltage and maximum current: peakpeakpeak εIP = Substitute numerical values and evaluate :peakP ( ) ( ) W200V1202A1785.1peak ==P 21 • What is the reactance of a 1.00-μH inductor at (a) 60 Hz, (b) 600 Hz, and (c) 6.00 kHz? Picture the Problem We can use LXL ω= to find the reactance of the inductor at any frequency. Express the inductive reactance as a function of f: fLLXL πω 2== (a) At f = 60 Hz: ( )( ) mΩ38.0H00.1s602 1 == − μπLX
- 4. Chapter 29176 (b) Atf = 600 Hz: ( )( ) mΩ77.3H00.1s6002 1 == − μπLX (c) At f = 6.00 kHz: ( )( ) mΩ7.37H00.1kHz00.62 == μπLX 25 • A 20-Hz ac generator that produces a peak emf of 10 V is connected to a 20-μF capacitor. Find (a) the peak current and (b) the rms current. Picture the Problem We can use Ipeak = εpeak/XC and XC = 1/ωC to express Ipeak as a function of εpeak, f, and C. Once we’ve evaluate Ipeak, we can use Irms = Ipeak/ 2 to find .rmsI Express in terms of εpeakI peak and XC: CX I peak peak ε = Express the capacitive reactance: fCC XC πω 2 11 == Substitute for XC and simplify to obtain: peakpeak 2 επfCI = (a) Substitute numerical values and evaluate :peakI ( )( )( ) mA25mA1.25 V10F20s202 1 peak == = − μπI (b) Express in terms of :rmsI peakI mA18 2 mA1.25 2 peak rms === I I Undriven Circuits Containing Capacitors, Resistors and Inductors 29 • (a) What is the period of oscillation of an LC circuit consisting of an ideal 2.0-mH inductor and a 20-μF capacitor? (b) A circuit that oscillates consists solely of an 80-μF capacitor and a variable ideal inductor. What inductance is needed in order to tune this circuit to oscillate at 60 Hz? Picture the Problem We can use T = 2π/ω and LC1=ω to relate T (and hence f) to L and C.
- 5. Alternating-Current Circuits 177 (a)Express the period of oscillation of the LC circuit: ω π2 =T For an LC circuit: LC 1 =ω Substitute for ω to obtain: LCT π2= (1) Substitute numerical values and evaluate T: ( )( ) ms3.1F20mH0.22 == μπT (b) Solve equation (1) for L to obtain: CfC T L 222 2 4 1 4 ππ == Substitute numerical values and evaluate L: ( ) ( ) mH88 F80s604 1 212 == − μπ L 33 ••• An inductor and a capacitor are connected, as shown in Figure 29-30. Initially, the switch is open, the left plate of the capacitor has charge Q0. The switch is then closed. (a) Plot both Q versus t and I versus t on the same graph, and explain how it can be seen from these two plots that the current leads the charge by 90º. (b) The expressions for the charge and for the current are given by Equations 29-38 and 29-39, respectively. Use trigonometry and algebra to show that the current leads the charge by 90º. Picture the Problem Let Q represent the instantaneous charge on the capacitor and apply Kirchhoff’s loop rule to obtain the differential equation for the circuit. We can then solve this equation to obtain an expression for the charge on the capacitor as a function of time and, by differentiating this expression with respect to time, an expression for the current as a function of time. We’ll use a spreadsheet program to plot the graphs. Apply Kirchhoff’s loop rule to a clockwise loop just after the switch is closed: 0=+ dt dI L C Q Because :dtdQI = 02 2 =+ C Q dt Qd L or 0 1 2 2 =+ Q LCdt Qd The solution to this equation is: ( )δω −= tQtQ cos)( 0 where LC 1 =ω
- 6. Chapter 29178 Because Q(0)= Q0, δ = 0 and: tQtQ ωcos)( 0= The current in the circuit is the derivative of Q with respect to t: [ ] tQtQ dt d dt dQ I ωωω sincos 00 −=== (a) A spreadsheet program was used to plot the following graph showing both the charge on the capacitor and the current in the circuit as functions of time. L, C, and Q0 were all arbitrarily set equal to one to obtain these graphs. Note that the current leads the charge by one-fourth of a cycle or 90°. -1.2 -0.6 0.0 0.6 1.2 0 2 4 6 8 10 t (s) Charge Current Q(mC) I(mA) -1.2 -0.6 0.0 0.6 1.2 (b) The equation for the current is: tQI ωω sin0−= (1) The sine and cosine functions are related through the identity: ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +=− 2 cossin π θθ Use this identity to rewrite equation (1): ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +=−= 2 cossin 00 π ωωωω tQtQI Thus, the current leads the charge by 90°. Driven RL Circuits 35 •• A coil that has a resistance of 80.0 Ω has an impedance of 200 Ω when driven at a frequency of 1.00 kHz. What is the inductance of the coil? Picture the Problem We can solve the expression for the impedance in an LR circuit for the inductive reactance and then use the definition of XL to find L.
- 7. Alternating-Current Circuits 179 Expressthe impedance of the coil in terms of its resistance and inductive reactance: 22 LXRZ += Solve for XL to obtain: 22 RZXL −= Express XL in terms of L: fLXL π2= Equate these two expressions to obtain: 22 2 RZfL −=π ⇒ f RZ L π2 22 − = Substitute numerical values and evaluate L: ( ) ( ) ( ) mH2.29 kHz00.12 Ω0.80Ω200 22 = − = π L 39 •• A coil that has a resistance R and an inductance L has a power factor equal to 0.866 when driven at a frequency of 60 Hz. What is the coil’s power factor if it is driven at 240 Hz? Picture the Problem We can use the definition of the power factor to find the relationship between XL and R when the coil is driven at a frequency of 60 Hz and then use the definition of XL to relate the inductive reactance at 240 Hz to the inductive reactance at 60 Hz. We can then use the definition of the power factor to determine its value at 240 Hz. Using the definition of the power factor, relate R and XL: 22 cos LXR R Z R + ==δ (1) Square both sides of the equation to obtain: 22 2 2 cos LXR R + =δ Solve for :( )Hz602 LX ( ) ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −= 1 cos 1 Hz60 2 22 δ RXL Substitute for cosδ and simplify to obtain: ( ) ( ) 2 3 1 2 22 1 866.0 1 Hz60 RRXL =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −= Use the definition of XL to obtain: ( ) 222 4 LffXL π= and ( ) 222 4 Lf'f'X L π=
- 8. Chapter 29180 Dividing thesecond of these equations by the first and simplifying yields: ( ) ( ) 2 2 22 22 2 2 4 4 f f' Lf Lf' fX f'X L L == π π or ( ) ( )fX f f' f'X LL 2 2 2 ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = Substitute numerical values to obtain: ( ) ( ) 22 2 2 1 1 2 3 16 3 1 16 Hz60 s60 s240 Hz240 RR XX LL =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = − − Substitute in equation (1) to obtain: ( ) 397.0 19 3 3 16 cos 22 Hz240 = = + = RR R δ 41 •• Figure 29-33 shows a load resistor that has a resistance of RL = 20.0 Ω connected to a high-pass filter consisting of an inductor that has inductance L = 3.20-mH and a resistor that has resistance R = 4.00-Ω. The output of the ideal ac generator is given by ε = (100 V) cos(2πft). Find the rms currents in all three branches of the circuit if the driving frequency is (a) 500 Hz and (b) 2000 Hz. Find the fraction of the total average power supplied by the ac generator that is delivered to the load resistor if the frequency is (c) 500 Hz and (d) 2000 Hz. Picture the Problem 21 VV +=ε , where V1 is the voltage drop across R and V2, is the voltage drop across the parallel combination of L and RL. 21 VVε rrr += is the relation for the phasors. For the parallel combination, LRL III rrr += . Also, V1 is in phase with I and V2 is in phase with . First draw the phasor diagram for the currents in the parallel combination, then add the phasors for the voltages to the diagram. LRI
- 9. Alternating-Current Circuits 181 Thephasor diagram for the currents in the circuit is: δ LRI r LI r I r Adding the voltage phasors to the diagram gives: δ LRI r LI r I r tω 2V r 1V r r ε δ The maximum current in the inductor, I2, peak, is given by: 2 peak,2 peak,2 Z V I = (1) where (2)222 2 −−− += LL XRZ δtan is given by: fL R L R X R RV XV I I LL L L R L πω δ 2 tan Lpeak,2 Lpeak,2 peak, peak, === == Solve for δ to obtain: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = − fL R π δ 2 tan L1 (3) Apply the law of cosines to the triangle formed by the voltage phasors to obtain: δε cos2 peak,2peak,1 2 peak,2 2 peak,1 2 peak VVVV ++= or δcos2 2peakpeak 2 2 2 peak 22 peak 22 peak ZRIIZIRIZI ++= Dividing out the current squared yields: δcos2 2 2 2 22 RZZRZ ++=
- 10. Chapter 29182 Solving forZ yields: δcos2 2 2 2 2 RZZRZ ++= (4) The maximum current in the circuit is given by: peakI Z I peak peak ε = (5) Irms is related to according to: peakI peakrms 2 1 II = (6) (a) Substitute numerical values in equation (3) and evaluate δ : ( )( ) °=⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = − − 31.63 Ω053.10 Ω0.20 tan mH20.3Hz5002 Ω0.20 tan 1 1 π δ Solving equation (2) for Z2 yields: 222 1 −− + = LL XR Z Substitute numerical values and evaluate Z2: ( ) ( ) Ω982.8 Ω053.10Ω0.20 1 22 2 = + = −− Z Substitute numerical values and evaluate Z: ( ) ( ) ( )( ) Ω36.1131.63cosΩ982.8Ω00.42Ω982.8Ω00.4 22 =°++=Z Substitute numerical values in equation (5) and evaluate :peakI A806.8 11.36 V100 peak = Ω =I Substitute for in equation (6) and evaluate : peakI rmsI ( ) A23.6A806.8 2 1 rms ==I The maximum and rms values of V2 are given by: ( )( ) V095.79Ω982.8A806.8 2peakpeak2, == = ZIV and ( ) V929.55V095.79 2 1 2 1 peak,2rms,2 == = VV
- 11. Alternating-Current Circuits 183 Therms values of and are: rms,LRI rms,LI A80.2 Ω20.0 V929.55rms,2 rms,L === L R R V I and A56.5 Ω053.01 V929.55rms,2 rms, === L L X V I (b) Proceed as in (a) with f = 2000 Hz to obtain: Ω= 2.40LX , °= 4.26δ , ,Ω= 9.172Z Ω= 6.21Z , , andA64.4peak =I A28.3rms =I , V0.83max,2 =V , ,V7.58rms,2 =V A94.2rms, =LRI , and A46.1rms, =LI (c) The power delivered by the ac source equals the sum of the power dissipated in the two resistors. The fraction of the total power delivered by the source that is dissipated in load resistor is given by: 1 2 rms, 2 rms 1 11 −− ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ += ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ += + LRR R RR R RI RI P P PP P LLL L Substitute numerical values for f = 500 Hz to obtain: ( ) ( ) ( ) ( ) %2.50502.0 0.20A80.2 00.4A23.6 1 1 2 2 Hz500 ==⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ Ω Ω += + − =fRR R PP P L L (d) Substitute numerical values for f = 2000 Hz to obtain: ( ) ( ) ( ) ( ) %0.80800.0 0.20A94.2 00.4A28.3 1 1 2 2 Hz2000 ==⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ Ω Ω += + − =fRR R PP P L L Filters and Rectifiers 47 •• A slowly varying voltage signal V(t) is applied to the input of the high- pass filter of Problem 44. Slowly varying means that during one time constant (equal to RC) there is no significant change in the voltage signal. Show that under these conditions the output voltage is proportional to the time derivative of V(t). This situation is known as a differentiation circuit.
- 12. Chapter 29184 Picture theProblem We can use Kirchhoff’s loop rule to obtain a differential equation relating the input, capacitor, and resistor voltages. Because the voltage drop across the resistor is small compared to the voltage drop across the capacitor, we can express the voltage drop across the capacitor in terms of the input voltage. Apply Kirchhoff’s loop rule to the input side of the filter to obtain: ( ) 0=−− IRVtV C where VC is the potential difference across the capacitor. Substitute for and I to obtain:( )tV 0cos cpeakin =−− dt dQ RVtV ω Because Q = CVC: [ ] dt dV CCV dt d dt dQ C C == Substitute for dQ/dt to obtain: 0cospeak =−− dt dV RCVtV C Cω the differential equation describing the potential difference across the capacitor. Because there is no significant change in the voltage signal during one time constant: 0= dt dVC ⇒ 0= dt dV RC C Substituting for dt dV RC C yields: 0cospeakin =− CVtV ω and tVVC ωcospeakin= Consequently, the potential difference across the resistor is given by: [ ]tV dt d RC dt dV RCV C R ωcospeakin== 49 •• Show that the average power dissipated in the resistor of the high-pass filter of Problem 44 is given by Pave = Vin peak 2 2R 1+ ωRC( )−2 ⎡ ⎣ ⎤ ⎦ . Picture the Problem We can express the instantaneous power dissipated in the resistor and then use the fact that the average value of the square of the cosine function over one cycle is ½ to establish the given result. The instantaneous power P(t) dissipated in the resistor is: R V tP 2 out )( =
- 13. Alternating-Current Circuits 185 Theoutput voltage is:outV ( )δω −= tVV cosHout From Problem 44: ( ) 2 peakin H 1 − + = RC V V ω Substitute in the expression for P(t) to obtain: ( ) ( )[ ] ( )δω ω δω − + = −= − t RCR V t R V tP 2 2 2 peakin 2 2 H cos 1 cos)( Because the average value of the square of the cosine function over one cycle is ½: ( )[ ]2 2 peakin ave 12 − + = RCR V P ω 51 •• The circuit shown in Figure 29-36 is an example of a low-pass filter. (Assume that the output is connected to a load that draws only an insignificant amount of current.) (a) If the input voltage is given by Vin = Vin peak cos ωt, show that the output voltage is Vout = VL cos(ωt – δ) where VL = Vin peak 1+ ωRC( )2 . (b) Discuss the trend of the output voltage in the limiting cases ω → 0 and ω → ∞. Picture the Problem In the phasor diagram for the RC low-pass filter, appV r and CV r are the phasors for Vin and Vout, respectively. The projection of appV r onto the horizontal axis is Vapp = Vin, the projection of CV r onto the horizontal axis is VC = Vout, apppeak V r =V , and φ is the angle between CV r and the horizontal axis. δ CV r RV r appV r ωt φ (a) Express :appV tVV ωcospeakinapp = where ZIV peakpeakin = and (1)2 C 22 XRZ += outV CV= is given by: φ φ cos cos peak peak,out C C XI VV = =
- 14. Chapter 29186 If wedefine δ as shown in the phasor diagram, then: ( ) ( )δω δω −= −= tX Z V tXIV C C cos cos peakin peakout Solving equation (1) for Z and substituting for XC yields: 2 2 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ += C RZ ω (2) Using equation (2) to substitute for Z and substituting for XC yields: ( )δω ω ω − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = t C C R V V cos 1 1 2 2 peakin out Simplify further to obtain: ( ) ( )δω ω − + = t RC V V cos 1 2 peakin out or ( )δω −= tVV cosLout where ( )2 peakin 1 RC V VL ω+ = (b) Note that, as ω → 0, VL → . This makes sense physically in that, for low frequencies, X peakV C is large and, therefore, a larger peak input voltage will appear across it than appears across it for high frequencies. Note further that, as ω → ∞, VL → 0. This makes sense physically in that, for high frequencies, XC is small and, therefore, a smaller peak voltage will appear across it than appears across it for low frequencies. Remarks: In Figures 29-19 and 29-20, δ is defined as the phase of the voltage drop across the combination relative to the voltage drop across the resistor. 55 ••• The circuit shown in Figure 29-37 is a trap filter. (Assume that the output is connected to a load that draws only an insignificant amount of current.) (a) Show that the trap filter acts to reject signals in a band of frequencies centered at LC1=ω . (b) How does the width of the frequency band rejected depend on the resistance R? Picture the Problem The phasor diagram for the trap filter is shown below. appV r and CL VV rr + are the phasors for Vin and Vout, respectively. The projection of appV r onto the horizontal axis is Vapp = Vin, and the projection of CL VV rr + onto the
- 15. Alternating-Current Circuits 187 horizontalaxis is VL + VC = Vout. Requiring that the impedance of the trap be zero will yield the frequency at which the circuit rejects signals. Defining the bandwidth as trapωωω −=Δ and requiring that RZ =trap will yield an expression for the bandwidth and reveal its dependence on R. δ LV r CV r RV r appV r CL VV rr + ω − δtωt (a) Express :appV tVV ωcospeakapp,app = where ZIVV peakpeakpeakapp, == and ( )222 CL XXRZ −+= (1) outV is given by: ( )δω −= tVV cospeakout,out where trappeakpeakout, ZIV = and CL XXZ −=trap Solving equation (1) for Z yields: ( )22 CL XXRZ −+= (2) Because :Lout CVVV += ( ) ( ) ( )δω δω δω −= −= −= tZ Z V tZI tVV cos cos cos trap peak trappeak peakout,out Using equation (2) to substitute for Z yields: ( )δω − + = tZ ZR V V costrap2 trap 2 peak out Noting that = 0 providedoutV trapZ = 0, set = 0 to obtain:trapZ 0trap =−= CL XXZ Substituting for XL and XC yields: 0 1 =− C L ω ω ⇒ LC 1 =ω
- 16. Chapter 29188 (b) Letthe bandwidth Δω be: trapωωω −=Δ (3) Let the frequency bandwidth be defined by the frequency at which RZ =trap . Then: R C L =− ω ω 1 ⇒ω2 LC −1= ωRC Because LC 1 trap =ω : RCω ω ω =− ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ 1 2 trap For ω ≈ ωtrap: RCtrap trap 2 trap 2 ω ω ωω ≈ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − Solve for :2 trap 2 ωω − ( )( )traptrap 2 trap 2 ωωωωωω +−=− Because ω ≈ ωtrap, :traptrap 2ωωω ≈+ ( )traptrap 2 trap 2 2 ωωωωω −≈− Substitute in equation (3) to obtain: L RRC 22 2 trap trap ==−=Δ ω ωωω Driven RLC Circuits 63 •• Show that the expression 22 rmsav ZRP ε= gives the correct result for a circuit containing only an ideal ac generator and (a) a resistor, (b) a capacitor, and (c) an inductor. In the expression 22 rmsav ZRP ε= , is the average power supplied by the generator, ε avP rms is the root-mean-square of the emf of the generator, R is the resistance, C is the capacitance and L is the inductance. (In Part (a), C = L = 0, in Part (b), R = L = 0 and in Part (c), R = C = 0. Picture the Problem The impedance of an ac circuit is given by ( )22 CL XXRZ −+= . We can evaluate the given expression for first foravP XL = XC = 0 and then for R = 0. (a) For X = 0, Z = R and: RR R Z R P 2 rms 2 2 rms 2 2 rms av εεε === (b) and (c) If R = 0, then: ( ) ( ) 0 0 2 2 rms 2 2 rms av = − == CL XXZ R P εε
- 17. Alternating-Current Circuits 189 Remarks:Recall that there is no energy dissipation in an ideal inductor or capacitor. 65 •• Find (a) the Q factor and (b) the resonance width (in hertz) for the circuit in Problem 64. (c) What is the power factor when ω = 8000 rad/s? Picture the Problem The Q factor of the circuit is given by RLQ 0ω= , the resonance width by QQff πω 200 ==Δ , and the power factor by ZR=δcos . Because Z is frequency dependent, we’ll need to find XC and XL at ω = 8000 rad/s in order to evaluate cosδ. Using their definitions, express the Q factor and the resonance width of the circuit: R L Q 0ω = (1) and QQ f f π ω 2 00 ==Δ (2) (a) Express the resonance frequency for the circuit: LC 1 0 =ω Substituting for ω0 in equation (1) yields: C L RRLC L Q 1 == Substitute numerical values and evaluate Q: 141.14 F2.0 mH10 Ω0.5 1 === μ Q (b) Substitute numerical values in equation (2) and evaluate Δf: ( ) Hz80 1.142 rad/s1007.7 Δ 3 = × = π f (c) The power factor of the circuit is given by: ( ) 2 2 22 1 cos ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −+ = −+ == C LR R XXR R Z R CL ω ω δ Substitute numerical values and evaluate cosδ: ( ) ( )( ) ( )( ) 27.0 F0.2s8000 1 mH10s8000Ω0.5 Ω0.5 cos 2 1 12 = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −+ = − − μ δ
- 18. Chapter 29190 69 ••In the circuit shown in Figure 29-42 the ideal generator produces an rms voltage of 115 V when operated at 60 Hz. What is the rms voltage between points (a) A and B, (b) B and C, (c) C and D, (d) A and C, and (e) B and D? Picture the Problem We can find the rms current in the circuit and then use it to find the potential differences across each of the circuit elements. We can use phasor diagrams and our knowledge of the phase shifts between the voltages across the three circuit elements to find the voltage differences across their combinations. (a) Express the potential difference between points A and B in terms of and XrmsI L: LAB XIV rms= (1) Express in terms of ε and Z:rmsI ( )22 rms CL XXRZ I −+ == εε Evaluate XL and XC to obtain: ( )( ) Ω= == − 648.51 mH137s6022 1 ππfLXL and ( )( ) Ω= == − 10.106 F25s602 1 2 1 1 μππfC XC Substitute numerical values and evaluate :rmsI ( ) ( ) A5556.1 10.106648.5150 V115 22 rms = Ω−Ω+Ω =I Substitute numerical values in equation (1) and evaluate VAB: ( )( ) V80 V344.80Ω648.51A5556.1 = ==ABV (b) Express the potential difference between points B and C in terms of and R:rmsI ( )( ) V78V780.77 Ω50A5556.1rms == == RIVBC (c) Express the potential difference between points C and D in terms of and XrmsI C: ( )( ) kV17.0V05.165 Ω10.106A5556.1rms == == CCD XIV
- 19. Alternating-Current Circuits 191 (d)The voltage across the inductor leads the voltage across the resistor as shown in the phasor diagram to the right: ABV r BCV r ACV r Use the Pythagorean theorem to find VAC: ( ) ( ) kV11.0V58.111 V780.77V0.80 22 22 == += += BCABAC VVV (e) The voltage across the capacitor lags the voltage across the resistor as shown in the phasor diagram to the right: BCV r CDV r BDV r Use the Pythagorean theorem to find VBD: ( ) ( ) kV18.0V46.182 V780.77V05.165 22 22 == += += BCCDBD VVV The Transformer 79 • A rms voltage of 24 V is required for a device whose impedance is 12 Ω. (a) What should the turns ratio of a transformer be, so that the device can be operated from a 120-V line? (b) Suppose the transformer is accidentally connected in reverse with the secondary winding across the 120-V-rms line and the 12-Ω load across the primary. How much rms current will then be in the primary winding? Picture the Problem Let the subscript 1 denote the primary and the subscript 2 the secondary. We can use 2112 NVNV = and 2211 ININ = to find the turns ratio and the primary current when the transformer connections are reversed. (a) Relate the number of primary and secondary turns to the primary and secondary voltages: 2rms1,1rms2, NVNV = (1)
- 20. Chapter 29192 Solve forand evaluate the ratio N2/N1: 5 1 V120 V24 rms,1 rms,2 1 2 === V V N N (b) Relate the current in the primary to the current in the secondary and to the turns ratio: rms,2 1 2 rms,1 I N N I = Express the current in the primary winding in terms of the voltage across it and its impedance: 2 rms,2 rms,2 Z V I = Substitute for I2, rms to obtain: 2 rms,2 1 2 rms,1 Z V N N I = Substitute numerical values and evaluate I1, rms: A50 Ω12 V120 1 5 1 =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =I General Problems 85 •• Figure 29-45 shows the voltage versus time for a square-wave voltage source. If V0 = 12 V, (a) what is the rms voltage of this source? (b) If this alternating waveform is rectified by eliminating the negative voltages, so that only the positive voltages remain, what is the new rms voltage? Picture the Problem The average of any quantity over a time interval ΔT is the integral of the quantity over the interval divided by ΔT. We can use this definition to find both the average of the voltage squared, ( )av 2 V and then use the definition of the rms voltage. (a) From the definition of we have: rmsV ( )av 2 0rms VV = Noting that , evaluate : 2 0 2 0 VV =− rmsV V120 2 0rms === VVV
- 21. Alternating-Current Circuits 193 (b)Noting that the voltage during the second half of each cycle is now zero, express the voltage during the first half cycle of the time interval TΔ2 1 : 0VV = Express the square of the voltage during this half cycle: 2 0 2 VV = Calculate ( )av 2 V by integrating V2 from t = 0 to t = TΔ2 1 and dividing by ΔT: ( ) [ ] 2 02 1Δ 0 2 0 Δ 0 2 0 av 2 2 1 2 1 ΔΔ Vt T V dt T V V T T === ∫ Substitute to obtain: V5.8 2 V12 2 02 02 1 rms ==== V VV 89 ••• A circuit consists of an ac generator, a capacitor and an ideal inductor⎯all connected in series. The emf of the generator is given by tωε cospeak . (a) Show that the charge on the capacitor obeys the equation t C Q dt Qd L ωε cospeak2 2 =+ . (b) Show by direct substitution that this equation is satisfied by tQQ ωcospeak= where ( )2 0 2 peak peak ωω ε − −= L Q . (c) Show that the current can be written as ( )δω −= tII cospeak , where CL XXL I − = − = peak 2 0 2 peak peak εε ωω ω and δ = –90º for ω < ω0 and δ = 90º for ω > ω0, where ω0 is the resonance frequency. Picture the Problem In Part (a) we can apply Kirchhoff’s loop rule to obtain the 2nd order differential equation relating the charge on the capacitor to the time. In Part (b) we’ll assume a solution of the form , differentiate it twice, and substitute for d tQQ ωcospeak= 2 Q/dt2 and Q to show that the assumed solution satisfies the differential equation provided ( )2 0 2 peak peak ωω ε − −= L Q . In Part (c) we’ll use our results from (a) and (b) to establish the result for Ipeak given in the problem statement. (a) Apply Kirchhoff’s loop rule to obtain: 0=−− dt dI L C Q ε
- 22. Chapter 29194 Substitute forε and rearrange the differential equation to obtain: t C Q dt dI L ωε cosmax=+ Because dtdQI = : t C Q dt Qd L ωε cosmax2 2 =+ (b) Assume that the solution is: tQQ ωcospeak= Differentiate the assumed solution twice to obtain: tQ dt dQ ωω sinpeak−= and tQ dt Qd ωω cospeak 2 2 2 −= Substitute for dt dQ and 2 2 dt Qd in the differential equation to obtain: t t C Q tLQ ω ωωω ε cos coscos peak peak peak 2 = +− Factor cosωt from the left-hand side of the equation: t t C Q LQ ω ωω ε cos cos peak peak peak 2 = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +− If this equation is to hold for all values of t it must be true that: peak peak peak 2 εω =+− C Q LQ Solving for yields:peakQ C L Q 12 peak peak +− = ω ε Factor L from the denominator and substitute for 1/LC to obtain: ( )2 0 2 peak 2 peak peak 1 ωω ω ε ε − −= ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +− = L LC L Q
- 23. Alternating-Current Circuits 195 (c)From (a) and (b) we have: ( ) ( )δω ωω ωω ω ωω ε −= = − = −== tI tIt L tQ dt dQ I cos sinsin sin peak peak2 0 2 peak peak where CL XX C L LL I − = − = − = − = peakpeak 2 0 2 peak 2 0 2 peak peak 1 εε εε ω ω ωω ω ωω ω If ω > ω0, XL > XC and the current lags the voltage by 90° (δ = 90°). If ω < ω0, XL < XC and the current leads the voltage by 90°(δ = −90°).
- 24.