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# Chapter 29 Alternating-Current Circuits

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```Chapter 29
Alternating-Current Circuits
Conceptual Problems
1
•
A coil in an ac generator rotates at 60 Hz. How much time elapses
between successive peak emf values of the coil?
Determine the Concept Successive peaks are one-half period apart. Hence the
1
1
1
=
= 8.33 ms .
elapsed time between the peaks is T =
2
2 f 2 60 s -1
(
)
2
•
If the rms voltage in an ac circuit is doubled, the peak voltage is
(a) doubled, (b) halved, (c) increased by a factor of 2 , (d) not changed.
Picture the Problem We can use the relationship between V and Vpeak to decide
the effect of doubling the rms voltage on the peak voltage.
Express the initial rms voltage in
terms of the peak voltage:
Vrms =
Express the doubled rms voltage in
terms of the new peak voltage V' peak :
2Vrms =
Divide the second of these equations
by the first and simplify to obtain:
Vpeak
2
V' peak
2
V' peak
V' peak
2Vrms
= 2 ⇒ 2=
Vpeak
Vpeak
Vrms
2
Solving for V' peak yields:
V' peak = 2Vpeak ⇒ (a) is correct.
3
•
[SSM] 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.
4
•
If the frequency in the circuit shown in Figure 29-27 is doubled, the
inductive reactance of the inductor will (a) double, (b) not change, (c) halve,
(d) quadruple.
2739
2740 Chapter 29
Determine the Concept The inductive reactance of an inductor varies with the
frequency according to X L = ωL. Hence, doubling ω will double XL. (a) is
correct.
5
•
If the frequency in the circuit in Figure 29-28 is doubled, the
capacitive reactance of the circuit will (a) double, (b) not change, (c) halve,
(d) quadruple.
Determine the Concept The capacitive reactance of an capacitor varies with the
frequency according to X C = 1 ωC . Hence, doubling ω will halve XC. (c) is
correct.
6
•
(a) In a circuit consisting solely of a ac generator and an ideal
inductor, are there any time intervals when the inductor receives energy from the
generator? If so, when? Explain your answer. (b) Are there any time intervals
when the inductor supplies energy back to the generator? If so when? Explain
your answer.
Determine the Concept Yes to both questions. (a) While the magnitude of the
current in the inductor is increasing, the inductor absorbs power from the
generator. (b) When the magnitude of the current in the inductor decreases, the
inductor supplies power to the generator.
7
•
[SSM] (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.
8
•
(a) Show that the SI unit of inductance multiplied by the SI unit of
capacitance is equivalent to seconds squared. (b) Show that the SI unit of
inductance divided by the SI unit of resistance is equivalent to seconds.
Determine the Concept
(a) Substitute the SI units of
inductance and capacitance and
simplify to obtain:
V ⋅s C s
⋅ = ⋅ C = s2
A V C
s
Alternating-Current Circuits
(b) Substitute the SI units of
inductance divided by resistance and
simplify to obtain:
2741
V ⋅s V ⋅s
A = A = s
V
Ω
A
9
•
[SSM] 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
ε peak NBA
ε
ω , I rms is directly proportional to ω. (a) is correct.
I rms = rms =
=
R
2
2
10 •
If the inductance value is tripled in a circuit consisting solely of a
variable inductor and a variable capacitor, how would you have to change the
capacitance so that the natural frequency of the circuit is unchanged? (a) triple the
capacitance, (b) decrease the capacitance to one-third of its original value, (c) You
should not change the capacitance.(d) You cannot determine how to change the
capacitance from the data given.
Determine the Concept The natural frequency of an LC circuit is given by
f 0 = 1 2π LC .
Express the natural frequencies of
the circuit before and after the
inductance is tripled:
Divide the second of the these
equations by the first and simplify to
obtain:
Because the natural frequency is
unchanged:
When the inductance is tripled:
f0 =
1
2π LC
and f 0' =
1
2π L'C'
1
f
= 2π L'C' =
1
f0
2π LC
'
0
LC
L'C'
1=
LC
LC
L
⇒
= 1 ⇒ C' = C
L'C'
L'C'
L'
C' =
L
1
C = C⇒
3L
3
(b )
is correct.
11 •
[SSM] 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
2742 Chapter 29
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.
Determine the Concept The maximum energy stored in the electric field of the
1 Q2
and the maximum energy stored in the magnetic
capacitor is given by U e =
2 C
1
field of the inductor is given by U m = LI 2 . Because energy is conserved in an
2
LC circuit and oscillates between the inductor and the capacitor, Ue = Um = Utotal.
(a ) is correct.
12
•
True or false:
(a) A driven series RLC circuit that has a high Q factor has a narrow resonance
curve.
(b) A circuit consists solely of a resistor, an inductor and a capacitor, all
connected in series. If the resistance of the resistor is doubled, the natural
frequency of the circuit remains the same.
(c) At resonance, the impedance of a driven series RLC combination equals the
resistance R.
(d) At resonance, the current in a driven series RLC circuit is in phase with the
voltage applied to the combination.
(a) True. The Q factor and the width of the resonance curve at half power are
related according to Q = ω0 Δω ; i.e., they are inversely proportional to each
other.
(b) True. The natural frequency of the circuit depends only on the inductance L of
the inductor and the capacitance C of the capacitor and is given by ω = 1 LC .
(c) True. The impedance of an RLC circuit is given by Z = R 2 + ( X L − X C ) . At
2
resonance XL = XC and so Z = R.
(d) True. The phase angle δ is related to XL and XC according to
⎛ X − XC ⎞
δ = tan −1 ⎜ L
⎟ . At resonance XL = XC and so δ = 0.
R
⎝
⎠
Alternating-Current Circuits
13
•
2743
True or false (all questions related to a driven series RLC circuit):
(a) Near resonance, the power factor of a driven series RLC circuit is close to
zero.
(b) The power factor of a driven series RLC circuit does not depend on the value
of the resistance.
(c) The resonance frequency of a driven series RLC circuit does not depend on the
value of the resistance.
(d) At resonance, the peak current of a driven series RLC circuit does not depend
on the capacitance or the inductance.
(e) For frequencies below the resonant frequency, the capacitive reactance of a
driven series RLC circuit is larger than the inductive reactance.
(f) For frequencies below the resonant frequency of a driven series RLC circuit,
the phase of the current leads (is ahead of) the phase of the applied voltage.
(a) False. Near resonance, the power factor, given by cos δ =
R
(X L − X C )
2
+R
2
,
is close to 1.
(b) False. The power factor is given by cos δ =
R
( X L − X C )2 + R 2
.
(c) True. The resonance frequency for a driven series RLC circuit depends only on
L and C and is given by ω res = 1 LC
(d) True. At resonance X L − X C = 0 and so Z = R and the peak current is given by
I peak = Vapp, peak R .
(e) True. Because the capacitive reactance varies inversely with the driving
frequency and the inductive reactance varies directly with the driving frequency,
at frequencies well below the resonance frequency the capacitive reactance is
larger than the inductive reactance.
(f) True. For frequencies below the resonant frequency, the circuit is more
capacitive than inductive and the phase constant φ is negative. This means that the
current leads the applied voltage.
14 •
You may have noticed that sometimes two radio stations can be heard
when your receiver is tuned to a specific frequency. This situation often occurs
when you are driving and are between two cities. Explain how this situation can
occur.
2744 Chapter 29
Determine the Concept Because the power curves received by your radio from
two stations have width, you could have two frequencies overlapping as a result
of receiving signals from both stations.
15
•
True or false (all questions related to a driven series RLC circuit):
(a) At frequencies much higher than or much lower than the resonant frequency
of a driven series RLC circuit, the power factor is close to zero.
(b) The larger the resonance width of a driven series RLC circuit is, the larger the
Q factor for the circuit is.
(c) The larger the resistance of a driven series RLC circuit is, the larger the
resonance width for the circuit is.
(a) True. Because the power factor is given by cos δ =
R
2
, for
1 ⎞
⎛
2
⎜ ωL −
⎟ +R
ωC ⎠
⎝
values of ω that are much higher or much lower than the resonant frequency, the
term in parentheses becomes very large and cosδ approaches zero.
(b) False. When the resonance curve is reasonably narrow, the Q factor can be
approximated by Q = ω 0 Δω . Hence a large value for Q corresponds to a narrow
resonance curve.
(c) True. See Figure 29-21.
16 •
An ideal transformer has N1 turns on its primary and N2 turns on its
secondary. The average power delivered to a load resistance R connected across
the secondary is P2 when the primary rms voltage is V1. The rms current in the
primary windings can then be expressed as (a) P2/V1, (b) (N1/N2)(P2/V1),
(c) (N2/N1)(P2/V1), (d) (N2/N1)2(P2/V1).
Picture the Problem Let the subscript 1 denote the primary and the subscript 2
the secondary. Assuming no loss of power in the transformer, we can equate the
power in the primary circuit to the power in the secondary circuit and solve for
the rms current in the primary windings.
Assuming no loss of power in the
transformer:
P1 = P2
Substitute for P1 and P2 to obtain:
I1, rmsV1, rms = I 2, rmsV2, rms
Alternating-Current Circuits
Solving for I1,rms and simplifying
yields:
I1, rms =
(a )
17
•
[SSM]
I 2, rmsV2, rms
V1, rms
=
2745
P2
V1, rms
is correct.
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.
Estimation and Approximation
18 ••
The impedances of motors, transformers, and electromagnets include
both resistance and inductive reactance. Suppose that phase of the current to a
large industrial plant lags the phase of the applied voltage by 25&deg; when the plant
2746 Chapter 29
is under full operation and using 2.3 MW of power. The power is supplied to the
plant from a substation 4.5 km from the plant; the 60 Hz rms line voltage at the
plant is 40 kV. The resistance of the transmission line from the substation to the
plant is 5.2 Ω. The cost per kilowatt-hour to the company that owns the plant is
\$0.14, and the plant pays only for the actual energy used. (a) Estimate the
resistance and inductive reactance of the plant’s total load. (b) Estimate the rms
current in the power lines and the rms voltage at the substation. (c) How much
power is lost in transmission? (d) Suppose that the phase that the current lags the
phase of the applied voltage is reduced to 18&ordm; by adding a bank of capacitors in
series with the load. How much money would be saved by the electric utility
during one month of operation, assuming the plant operates at full capacity for 16
h each day? (e) What must be the capacitance of this bank of capacitors to achieve
this change in phase angle?
Picture the Problem We can find the resistance and inductive reactance of the
plant’s total load from the impedance of the load and the phase constant. The
current in the power lines can be found from the total impedance of the load the
potential difference across it and the rms voltage at the substation by applying
Kirchhoff’s loop rule to the substation-transmission wires-load circuit. The power
2
Rtrans . We can find the cost
lost in transmission can be found from Ptrans = I rms
savings by finding the difference in the power lost in transmission when the phase
angle is reduced to 18&deg;. Finally, we can find the capacitance that is required to
reduce the phase angle to 18&deg; by first finding the capacitive reactance using the
definition of tanδ and then applying the definition of capacitive reactance to find
C.
Rtrans = 5.2 Ω
εsubstation
∼
f = 60 Hz
Z
(a) Relate the resistance and
inductive reactance of the plant’s
total load to Z and δ:
R = Z cos δ
and
X L = Z sin δ
Express Z in terms of the rms current
I rms in the power lines and the rms
Z=
voltage ε rms at the plant:
ε rms
I rms
ε rms= 40 kV
δ = 25&deg;
Alternating-Current Circuits
Express the power delivered to the
Pav = ε rms I rms cos δ
plant in terms of εrms, I rms , and δ and
and
solve for I rms :
I rms =
Substitute to obtain:
Substitute numerical values and
evaluate Z:
Substitute numerical values and
evaluate R and XL:
Z=
Pav
ε rms cos δ
(1)
2
ε rms
cos δ
Pav
2
(
40 kV ) cos 25&deg;
Z=
= 630 Ω
2.3 MW
R = (630 Ω ) cos 25&deg; = 571 Ω
= 0.57 kΩ
and
X L = (630 Ω )sin 25&deg; = 266 Ω
= 0.27 kΩ
(b) Use equation (1) to find the
current in the power lines:
I rms =
2.3 MW
= 63.4 A
(40 kV )cos 25&deg;
= 63 A
Apply Kirchhoff’s loop rule to the
circuit:
ε sub − I rms Rtrans − I rms Z tot = 0
Solve for εsub:
ε sub = I rms (Rtrans + Z tot )
Substitute numerical values and
ε sub = (63.4 A )(5.2 Ω + 630 Ω)
evaluate εsub:
(c) The power lost in transmission is:
= 40.3 kV
2
Ptrans = I rms
Rtrans = (63.4 A ) (5.2 Ω )
2
= 20.9 kW = 21kW
(d) Express the cost savings ΔC in
terms of the difference in energy
consumption (P25&deg; − P18&deg;)Δt and the
per-unit cost u of the energy:
ΔC = (P25&deg; − P18&deg; )Δtu
2747
2748 Chapter 29
Express the power lost in
transmission when δ = 18&deg;:
P18&deg; = I182 &deg; Rtrans
Find the current in the transmission
lines when δ = 18&deg;:
I18&deg; =
Evaluate P18&deg; :
P18&deg; = (60.5 A ) (5.2 Ω ) = 19.0 kW
2.3 MW
= 60.5 A
(40 kV ) cos18&deg;
2
Substitute numerical values and evaluate ΔC:
d ⎞ ⎛ \$0.14 ⎞
⎛ h ⎞⎛
ΔC = (20.9 kW − 19.0 kW )⎜16 ⎟ ⎜ 30
⎟⎜
⎟ = \$128
⎝ d ⎠ ⎝ month ⎠ ⎝ kW ⋅ h ⎠
(e) The required capacitance is given
by:
Relate the new phase angle δ to the
inductive reactance XL, the reactance
due to the added capacitance XC, and
the resistance of the load R:
Substituting for XC yields:
C=
1
2πfX C
tan δ =
C=
XL − XC
⇒ X C = X L − R tan δ
R
1
2πf ( X L − R tan δ )
Substitute numerical values and evaluate C:
C=
1
= 33 μF
2π 60 s (266 Ω − (571Ω ) tan 18&deg;)
(
-1
)
Alternating Current in Resistors, Inductors, and Capacitors
19 •
[SSM] 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 Pav = ε rms I rms to find I rms , I peak = 2I rms to find
I peak , and Ppeak = I peakε peak to find Ppeak .
(a) Relate the average power
delivered by the source to the rms
voltage across the bulb and the rms
current through it:
Pav = ε rms I rms ⇒ I rms =
Pav
ε rms
Alternating-Current Circuits
2749
100 W
= 0.8333 A = 0.833 A
120 V
Substitute numerical values and
evaluate I rms :
I rms =
(b) Express I peak in terms of I rms :
I peak = 2I rms
Substitute for I rms and evaluate I peak :
I peak = 2 (0.8333 A ) = 1.1785 A
= 1.18 A
(c) Express the maximum power in
terms of the maximum voltage and
maximum current:
Substitute numerical values and
evaluate Ppeak :
Ppeak = I peakε peak
Ppeak = (1.1785 A ) 2 (120 V ) = 200 W
20 •
A circuit breaker is rated for a current of 15 A rms at a voltage of
120 V rms. (a) What is the largest value of the peak current that the breaker can
carry? (b) What is the maximum average power that can be supplied by this
circuit?
Picture the Problem We can I peak = 2I rms to find the largest peak current the
breaker can carry and Pav = I rmsVrms to find the average power supplied by this
circuit.
(a) Express I peak in terms of I rms :
I peak = 2 I rms = 2 (15 A ) = 21 A
(b) Relate the average power to
the rms current and voltage:
Pav = I rmsVrms = (15 A )(120 V ) = 1.8 kW
21 •
[SSM] 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 X L = ωL to find the reactance of the inductor at
any frequency.
Express the inductive reactance
as a function of f:
X L = ωL = 2πfL
(a) At f = 60 Hz:
X L = 2π 60 s −1 (1.00 mH ) = 0.38 Ω
(
)
2750 Chapter 29
(b) At f = 600 Hz:
X L = 2π (600 s −1 )(1.00 mH ) = 3.77 Ω
(c) At f = 6.00 kHz:
X L = 2π (6.00 kHz )(1.00 mH ) = 37.7 Ω
22 •
An inductor has a reactance of 100 Ω at 80 Hz. (a) What is its
inductance? (b) What is its reactance at 160 Hz?
Picture the Problem We can use X L = ωL to find the inductance of the inductor
at any frequency.
(a) Relate the reactance of the
inductor to its inductance:
X L = ωL = 2πfL ⇒ L =
Solve for and evaluate L:
L=
(b) At 160 Hz:
X L = 2π (160 s −1 )(0.199 H ) = 0.20 kΩ
XL
2πf
100 Ω
= 0.199 H = 0.20 H
2π 80 s −1
(
)
23 •
At what frequency would the reactance of a 10-μF capacitor equal the
reactance of a 1.0-μH inductor?
Picture the Problem We can equate the reactances of the capacitor and the
inductor and then solve for the frequency.
Express the reactance of the
inductor:
Express the reactance of the
capacitor:
Equate these reactances to obtain:
Substitute numerical values and
evaluate f:
X L = ωL = 2πfL
XC =
1
1
=
ωC 2πfC
2πfL =
f =
1
2π
1
1
⇒ f =
2πfC
2π
1
LC
1
= 1.6 kHz
(10 μF)(1.0 mH )
24 •
What is the reactance of a 1.00-nF capacitor at (a) 60.0 Hz,
(b) 6.00 kHz, and (c) 6.00 MHz?
Picture the Problem We can use X C = 1 ωC to find the reactance of the
capacitor at any frequency.
Alternating-Current Circuits
Express the capacitive reactance as
a function of f:
XC =
(a) At f = 60.0 Hz:
XC =
2751
1
1
=
ωC 2πfC
1
(
2π 60.0 s
−1
)(1.00 nF) =
2.65 MΩ
(b) At f = 6.00 kHz:
XC =
1
= 26.5 kΩ
2π (6.00 kHz )(1.00 nF)
(c) At f = 6.00 MHz:
XC =
1
= 26.5 Ω
2π (6.00 MHz )(1.00 nF)
25 •
[SSM] 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 I rms .
Express I peak in terms of εpeak
and XC:
Express the capacitive reactance:
Substitute for XC and simplify to
obtain:
(a) Substitute numerical values and
evaluate I peak :
(b) Express I rms in terms of I peak :
I peak =
XC =
ε peak
XC
1
1
=
ωC 2πfC
I peak = 2πfCε peak
(
)
I peak = 2π 20 s −1 (20 μF)(10 V )
= 25.1 mA = 25 mA
I rms =
I peak
2
=
25.1 mA
= 18 mA
2
26 •
At what frequency is the reactance of a 10-μF capacitor (a) 1.00 Ω,
(b) 100 Ω, and (c) 10.0 mΩ?
2752 Chapter 29
Picture the Problem We can use X C = 1 ωC = 1 2πfC to relate the reactance of
the capacitor to the frequency.
1
1
1
=
⇒f =
ωC 2πfC
2πCX C
The reactance of the capacitor is
given by:
XC =
(a) Find f when XC = 1.00 Ω:
f =
1
= 16 kHz
2π (10 μF)(1.00 Ω )
(b) Find f when XC = 100 Ω:
f =
1
= 0.16 kHz
2π (10 μF)(100 Ω )
(c) Find f when XC = 10.0 mΩ:
f =
1
2π (10 μF)(10.0 mΩ )
= 1.6 MHz
27 ••
A circuit consists of two ideal ac generators and a 25-Ω resistor, all
connected in series. The potential difference across the terminals of one of the
generators is given by V1 = (5.0 V) cos(ωt – α), and the potential difference
across the terminals of the other generator is given by V2 = (5.0 V) cos(ωt + α),
where α = π/6. (a) Use Kirchhoff’s loop rule and a trigonometric identity to find
the peak current in the circuit. (b) Use a phasor diagram to find the peak current in
the circuit. (c) Find the current in the resistor if α = π/4 and the amplitude of V2 is
increased from 5.0 V to 7.0 V.
Picture the Problem We can use the trigonometric identity
cosθ + cos φ = 2 cos 12 (θ + φ ) cos 12 (θ − φ )
to find the sum of the phasors V1 and V2 and then use this sum to express I as a
function of time. In (b) we’ll use a phasor diagram to obtain the same result and in
(c) we’ll use the phasor diagram appropriate to the given voltages to express the
current as a function of time.
(a) Applying Kirchhoff’s loop rule to
the circuit yields:
V1 + V2 − IR = 0
Solve for I to obtain:
I=
V1 + V2
R
Alternating-Current Circuits
2753
Use the trigonometric identity cos θ + cos φ = 2 cos 12 (θ + φ ) cos 12 (θ − φ )
to find V1 + V2:
V1 + V2 = (5.0 V )[cos(ωt − α ) + cos(ωt + α )] = (5 V )[2 cos 12 (2ωt )cos 12 (− 2α )]
= (10 V )cos
π
6
cos ωt = (8.66 V )cos ωt
Substitute for V1 + V2 and R to obtain:
I=
(8.66 V ) cos ωt = (0.346 A ) cos ωt
25 Ω
= (0.35 A ) cos ωt
where
I peak = 0.35 A
(b) Express the magnitude of the
current in R:
I =
r
V
R
r
V2
The phasor diagram for the voltages
is shown to the right.
r
V
30&deg;
30&deg;
r
V1
r
Use vector addition to find V :
r
r
V = 2 V1 cos 30&deg; = 2(5.0 V ) cos 30&deg;
= 8.66 V
r
Substitute for V and R to obtain:
I =
8.66 V
= 0.346 A
25 Ω
and I = (0.35 A ) cos ωt
where
I peak = 0.35 A
2754 Chapter 29
r
V2
(c) The phasor diagram is shown to
the right. Note that the phase angle
r
r
between V1 and V2 is now 90&deg;.
α
r
V
o
90 − α
δ
r
V1
Use the Pythagorean theorem to
r
find V :
Express I as a function of t:
r
r2 r 2
V = V1 + V2 =
(5.0 V )2 + (7.0 V )2
= 8.60 V
r
V
cos(ωt + δ )
R
where
δ = 45&deg; − (90&deg; − α ) = α − 45&deg;
I=
⎛ 7 .0 V ⎞
⎟⎟ − 45&deg;
= tan −1 ⎜⎜
⎝ 5.0 V ⎠
= 9.462&deg; = 0.165 rad
Substitute numerical values and
evaluate I:
I=
=
8.60 V
cos(ωt + 0.165 rad )
25 Ω
(0.34 A )cos(ωt + 0.17 rad )
Undriven Circuits Containing Capacitors, Resistors and
Inductors
28 •
(a) Show that 1 LC has units of inverse seconds by substituting SI
units for inductance and capacitance into the expression. (b) Show that ω0L/R (the
expression for the Q-factor) is dimensionless by substituting SI units for angular
frequency, inductance, and resistance into the expression.
Picture the Problem We can substitute the units of the various physical
quantitities in 1 / LC and Q = ω0 L R to establish their units.
Alternating-Current Circuits
(a) Substitute the units for L and C in
the expression 1 LC and simplify
to obtain:
(b) Substitute the units for ω0, L, and
R in the expression Q = ω0 L R and
simplify to obtain:
1
=
H⋅F
1
(Ω ⋅ s )⎛⎜ s ⎞⎟
=
1
s
2
2755
= s −1
⎝Ω⎠
1 V ⋅s 1 V ⋅s
⋅
⋅
s A = s A = 1 ⇒ no units
V
V
A
A
29 •
[SSM] (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 ω = 1
f) to L and C.
2π
(a) Express the period of oscillation
of the LC circuit:
T=
For an LC circuit:
ω=
Substitute for ω to obtain:
T = 2π LC
Substitute numerical values and
evaluate T:
T = 2π
(b) Solve equation (1) for L to
obtain:
L=
Substitute numerical values and
evaluate L:
L=
LC to relate T (and hence
ω
1
LC
(1)
(2.0 mH)(20 μF) =
1.3 ms
T2
1
=
2
2 2
4π C 4π f C
(
1
4π 2 60 s −1
) (80 μF)
2
= 88 mH
An LC circuit has capacitance C0 and inductance L. A second LC
30 •
circuit has capacitance 12 C0 and inductance 2L, and a third LC circuit has
capacitance 2C0 and inductance 12 L. (a) Show that each circuit oscillates with the
same frequency. (b) In which circuit would the peak current be greatest if the
peak voltage across the capacitor in each circuit was the same?
2756 Chapter 29
Picture the Problem We can use the expression f 0 = 1 2π LC for the resonance
frequency of an LC circuit to show that each circuit oscillates with the same
frequency. In (b) we can use I peak = ωQ0 , where Q0 is the charge of the capacitor
at time zero, and the definition of capacitance Q0 = CV to express Ipeak in terms of
ω, C and V.
Express the resonance frequency
for an LC circuit:
(a) Express the product of L and C0
for each circuit:
f0 =
1
2π LC
Circuit 1: L1C1 = L1C0 ,
Circuit 2: L2C2 = (2 L1 )( 12 C0 ) = L1C1 ,
and
Circuit 3: L3C3 = ( 12 L1 )(2C0 ) = L1C1
Because L1C1 = L2 C 2 = L3C3 , the resonance frequencies of the three circuits are
the same.
(b) Express I peak in terms of the
I peak = ωQ0
charge stored in the capacitor:
Express Q0 in terms of the
capacitance of the capacitor and the
potential difference across the
capacitor:
Q0 = CV
Substituting for Q0 yields:
I peak = ωCV
or, for ω and V constant, I peak ∝ C .
Hence, the circuit with capacitance 2C0
has the greatest peak current.
31 ••
A 5.0-μF capacitor is charged to 30 V and is then connected across an
ideal 10-mH inductor. (a) How much energy is stored in the system? (b) What is
the frequency of oscillation of the circuit? (c) What is the peak current in the
circuit?
Picture the Problem We can use U = 12 CV 2 to find the energy stored in the
electric field of the capacitor, ω0 = 2πf 0 = 1
Q0 = CV to find I peak .
LC to find f0, and I peak = ωQ0 and
Alternating-Current Circuits
(a) Express the energy stored in the
system as a function of C and V:
U = 12 CV 2
Substitute numerical values and
evaluate U:
U=
(b) Express the resonance frequency
of the circuit in terms of L and C:
ω0 = 2πf 0 =
Substitute numerical values and
evaluate f0:
f0 =
1
2
(5.0 μF)(30 V )2 =
2.3 mJ
1
1
⇒ f0 =
LC
2π LC
1
2π
2757
(10 mH)(5.0 μF)
= 712 Hz
= 0.71 kHz
(c) Express I peak in terms of the
I peak = ωQ0
charge stored in the capacitor:
Express Q0 in terms of the
capacitance of the capacitor and the
potential difference across the
capacitor:
Q0 = CV
Substituting for Q0 yields:
I peak = ωCV
Substitute numerical values and
evaluate I peak :
I peak = 2π 712 s −1 (5.0 μF)(30 V )
(
)
= 0.67 A
32 ••
A coil with internal resistance can be modeled as a resistor and an
ideal inductor in series. Assume that the coil has an internal resistance of 1.00 Ω
and an inductance of 400 mH. A 2.00-μF capacitor is charged to 24.0 V and is
then connected across coil. (a) What is the initial voltage across the coil? (b) How
much energy is dissipated in the circuit before the oscillations die out? (c) What is
the frequency of oscillation the circuit? (Assume the internal resistance is
sufficiently small that has no impact on the frequency of the circuit.) (d) What is
the quality factor of the circuit?
2758 Chapter 29
Picture the Problem In Part (a) we can apply Kirchhoff’s loop rule to find the
initial voltage across the coil. (b) The total energy lost via joule heating is the total
energy initially stored in the capacitor. (c) The natural frequency of the circuit is
given by f 0 = 1 2π LC . In Part (d) we can use its definition to find the quality
factor of the circuit.
(a) Application of Kirchhoff’s loop rule leads us to conclude that the initial
voltage across the coil is 24.0 V .
(b) Because the ideal inductor can
not dissipate energy as heat, all of
the energy initially stored in the
capacitor will be dissipated as joule
heat in the resistor:
1
1
2
U = CV 2 = (2.00 μF)(24.0 V )
2
2
(c) The natural frequency of the
circuit is:
f0 =
= 0.576 mJ
1
2π LC
=
1
2π (400 mH )(2.00 μF)
= 178 Hz
(d) The quality factor of the circuit is
given by:
Substituting for ω0 and simplifying
yields:
Substitute numerical values and
evaluate Q:
Q=
Q=
Q=
ω0 L
R
1
L
LC = 1 L
R
R C
1
400 mH
= 447
1.00 Ω 2.00 μF
33 ••• [SSM] 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 90&ordm;. (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 90&ordm;.
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
Alternating-Current Circuits
2759
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:
Q
dI
+L
=0
C
dt
Because I = dQ dt :
L
d 2Q Q
d 2Q 1
+
Q=0
+
=
0
or
dt 2 LC
dt 2 C
Q(t ) = Q0 cos(ωt − δ )
The solution to this equation is:
1
LC
where ω =
Because Q(0) = Q0, δ = 0 and:
Q(t ) = Q0 cos ωt
The current in the circuit is the
derivative of Q with respect to t:
I=
dQ d
= [Q0 cos ωt ] = −ωQ0 sin ωt
dt
dt
(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&deg;.
1.2
1.2
Charge
0.6
0.0
0.0
-0.6
-0.6
I (mA)
Q (mC)
Current
0.6
-1.2
-1.2
0
2
4
6
8
10
t (s)
(b) The equation for the current is:
I = −ωQ0 sin ωt
The sine and cosine functions are
related through the identity:
π⎞
⎛
− sin θ = cos⎜θ + ⎟
2⎠
⎝
(1)
2760 Chapter 29
Use this identity to rewrite equation
(1):
π⎞
⎛
I = −ωQ0 sin ωt = ωQ0 cos⎜ ωt + ⎟
2⎠
⎝
Thus, the current leads the charge by
90&deg;.
Driven RL Circuits
34 ••
A circuit consists of a resistor, an ideal 1.4-H inductor and an ideal
60-Hz generator, all connected in series. The rms voltage across the resistor is
30 V and the rms voltage across the inductor is 40 V. (a) What is the resistance of
the resistor? (b) What is the peak emf of the generator?
Picture the Problem We can express the ratio of VR to VL and solve this
expression for the resistance R of the circuit. In (b) we can use the fact that, in an
LR circuit, VL leads VR by 90&deg; to find the ac input voltage.
(a) Express the potential
differences across R and L in terms
of the common current through
these components:
VL = IX L = IωL
and
VR = IR
Divide the second of these
equations by the first to obtain:
⎛V ⎞
VR
IR
R
=
=
⇒ R = ⎜⎜ R ⎟⎟ωL
VL IωL ωL
⎝ VL ⎠
Substitute numerical values and
evaluate R:
⎛ 30 V ⎞
⎟⎟2π 60 s −1 (1.4 H ) = 0.40 kΩ
R = ⎜⎜
⎝ 40 V ⎠
(b) Because VR leads VL by 90&deg; in
an LR circuit:
Vpeak = 2Vrms = 2 VR2 + VL2
Substitute numerical values and
evaluate Vpeak :
Vpeak = 2
(
)
(30 V )2 + (40 V )2 =
71 V
35 ••
[SSM] 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.
Alternating-Current Circuits
Express the impedance of the coil in
terms of its resistance and inductive
reactance:
Z = R 2 + X L2
Solve for XL to obtain:
X L = Z 2 − R2
Express XL in terms of L:
X L = 2πfL
Equate these two expressions to
obtain:
2πfL = Z 2 − R 2 ⇒ L =
Substitute numerical values and
evaluate L:
L=
2761
Z 2 − R2
2πf
(200 Ω )2 − (80.0 Ω )2
2π (1.00 kHz )
= 29.2 mH
36 ••
A two conductor transmission line simultaneously carries a
superposition of two voltage signals, so the potential difference between the two
conductors is given by V = V1 + V2, where V1 = (10.0 V) cos(ω1t) and
V2 = (10.0 V) cos(ω2t), where ω1 = 100 rad/s and ω2 = 10 000 rad/s. A 1.00 H
inductor and a 1.00 kΩ shunt resistor are inserted into the transmission line as
shown in Figure 29-31. (Assume that the output is connected to a load that draws
only an insignificant amount of current.) (a) What is the voltage (Vout) at the
output of the transmission line? (b) What is the ratio of the low-frequency
amplitude to the high-frequency amplitude at the output?
Picture the Problem We can express the two output voltage signals as the
product of the current from each source and R = 1.00 kΩ. We can find the
currents due to each source using the given voltage signals and the definition of
the impedance for each of them.
(a) Express the voltage signals
observed at the output side of the
transmission line in terms of the
potential difference across the
resistor:
V1, out = I1 R
and
V2, out = I 2 R
2762 Chapter 29
Evaluate I1 and I2:
I1 =
V1
=
Z1
(10.0 V )cos100t
= (9.95 mA )cos100t
(1.00 kΩ )2 + [(100 s -1 )(1.00 H )] 2
I2 =
V2
=
Z2
(10.0 V )cos10 4 t
= (0.995 mA )cos10 4 t
2
(1.00 kΩ )2 + [(10 4 s −1 )(1.00 H )]
and
Substitute for I1 and I2 to obtain:
V1, out = (1.00 kΩ )(9.95 mA ) cos100t
=
(9.95 V )cos100t
where ω1 = 100 rad/s and
ω2 = 10 000 rad/s.
and
V2, out = (1.00 kΩ )(0.995 mA ) cos10 4 t
=
(b) Express the ratio of V1,out to
V2,out:
V1, out
V2, out
=
(0.995 V )cos104 t
9.95 V
= 10 : 1
0.995 V
37 ••
A coil is connected to a 120-V rms, 60-Hz line. The average power
supplied to the coil is 60 W, and the rms current is 1.5 A. Find (a) the power
factor, (b) the resistance of the coil, and (c) the inductance of the coil. (d) Does
the current lag or lead the voltage? Explain your answer. (e) Support your answer
to Part (d) by determining the phase angle.
Picture the Problem The average power supplied to coil is related to the power
2
R to find R. Because the
factor by Pav = ε rms I rms cos δ . In (b) we can use Pav = I rms
inductance L is related to the resistance R and the phase angle δ according to
X L = ωL = R tan δ , we can use this relationship to find the resistance of the coil.
Finally, we can decide whether the current leads or lags the voltage by noting that
the circuit is inductive.
(a) Express the average power
supplied to the coil in terms of the
power factor of the circuit:
Pav = ε rms I rms cos δ ⇒ cos δ =
Substitute numerical values and
evaluate cosδ:
cos δ =
Pav
ε rms I rms
60 W
= 0.333 = 0.33
(120 V )(1.5 A )
Alternating-Current Circuits
(b) Express the power supplied by
the source in terms of the resistance
of the coil:
Pav
2
I rms
2
Pav = I rms
R⇒R =
60 W
= 26.7 Ω = 27 Ω
(1.5 A )2
Substitute numerical values and
evaluate R:
R=
(c) Relate the inductive reactance to
the resistance and phase angle:
X L = ωL = R tan δ
Solving for L yields:
L=
2763
R tan δ
ω
=
[
]
R tan cos −1 (0.333)
2πf
(26.7 Ω ) tan (70.5&deg;) =
Substitute numerical values and
evaluate L:
L=
(d) Evaluate XL:
X L = (26.7 Ω) tan(70.5&deg;) = 75.4 Ω
(
2π 60 s −1
)
0.20 H
Because the circuit is inductive, the current lags the voltage.
δ = cos −1 (0.333) = 71&deg;
(e) From Part (a):
38
••
A 36-mH inductor that has a resistance of 40 Ω is connected to an
ideal ac voltage source whose output is given by ε = (345 V) cos(150πt), where t
is in seconds. Determine (a) the peak current in the circuit, (b) the peak and rms
voltages across the inductor, (c) the average power dissipation, and
(d) the peak and average magnetic energy stored in the inductor.
Picture
the
Problem
(a)
We
can
use
I peak = ε peak
R 2 + (ωL ) and
2
VL , peak = I peak X L = ωLI peak to find the peak current in the circuit and the peak
voltage across the inductor. (b) Once we’ve found VL, peak we can find VL , rms using
VL , rms = VL , peak
2
R to find the average power
2 . (c) We can use Pav = 12 I rms
2
dissipation, and (d) U L , peak = 12 LI peak
to find the peak and average magnetic energy
stored in the inductor. The average energy stored in the magnetic field of the
inductor can be found using U L ,av = ∫ Pav dt .
2764 Chapter 29
(a) Apply Kirchhoff’s loop rule to
the circuit to obtain:
Substitute numerical values and
evaluate I:
ε − IZ = 0 ⇒ I = ε
=
Z
ε
R 2 + (ωL )
2
(345 V )cos(150πt )
(40 Ω )2 + [(150π s −1 )(36 mH )]2
= (7.94 A ) cos(150πt )
I=
and I peak = 7.9 A .
(b) Because ε = (345 V) cos(150πt) :
Find VL ,rms from VL, peak :
(c) Relate the average power
dissipation to Ipeak and R:
V L , peak = 345 V
VL , rms =
VL , peak
2
=
345 V
= 244 V
2
2
Pav = I
2
rms
⎛I ⎞
2
R = ⎜⎜ peak ⎟⎟ R = 12 I peak
R
⎝ 2 ⎠
Substitute numerical values and
evaluate Pav :
Pav = 12 (7.94 A ) (40 Ω ) = 1.3 kW
(d) The maximum energy stored in
the magnetic field of the inductor is:
2
U L , peak = 12 LI peak
=
The definition of U L,av is:
U(t) is given by:
Substitute for U(t) to obtain:
2
1
2
(36 mH)(7.94 A )2
= 1.1 J
T
U L ,av =
1
U (t )dt
T ∫0
U (t ) =
1
2
L[I (t )]
2
L
2T
T
U L ,av =
1 2
⎡1 2 ⎤
⎢⎣ 2 I peak ⎥⎦T = 4 LI peak
∫ [I (t )] dt
2
0
Evaluating the integral yields:
U L ,av =
L
2T
Substitute numerical values and
evaluate U L,av :
U L ,av =
1
(36 mH )(7.94 A )2 = 0.57 J
4
Alternating-Current Circuits
2765
39 ••
[SSM] 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 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:
cos δ =
Square both sides of the equation
to obtain:
cos2 δ =
R
=
Z
R
(1)
R + X L2
2
R2
R 2 + X L2
Solve for X L2 (60 Hz) :
⎛ 1
⎞
− 1⎟
X L2 (60 Hz ) = R 2 ⎜
2
⎝ cos δ
⎠
Substitute for cosδ and simplify to
obtain:
⎛
⎞
1
X L2 (60 Hz ) = R 2 ⎜⎜
− 1⎟⎟ = 13 R 2
2
⎝ (0.866 )
⎠
Use the definition of XL to obtain:
X L2 ( f ) = 4πf 2 L2 and X L2 ( f' ) = 4πf' 2 L2
Dividing the second of these
equations by the first and simplifying
yields:
X L2 ( f' ) 4πf' 2 L2 f' 2
=
= 2
X L2 ( f ) 4πf 2 L2
f
or
2
⎛ f' ⎞
X ( f' ) = ⎜⎜ ⎟⎟ X L2 ( f )
⎝ f ⎠
2
L
Substitute numerical values to
obtain:
2
⎛ 240 s −1 ⎞ 2
⎟ X L (60 Hz )
(
)
X 240 Hz = ⎜⎜
−1 ⎟
60
s
⎠
⎝
⎛ 1 ⎞ 16
= 16⎜ R 2 ⎟ = R 2
⎝3 ⎠ 3
2
L
2766 Chapter 29
Substitute in equation (1) to obtain:
(cos δ )240 Hz =
R
R2 +
16 2
R
3
=
3
19
= 0.397
40
••
A resistor and an inductor are connected in parallel across an ideal ac
voltage source whose output is given by ε = εpeakcosωt as shown in Figure 29-32.
Show that (a) the current in the resistor is given by IR = εpeak/R cos ωt, (b) the
current in the inductor is given by IL = εpeak/XL cos(ωt – 90&ordm;), and (c) the current
in the voltage source is given by I = IR + IL = Ipeak cos(ωt – δ), where
Ipeak = εmax/Z.
Picture the Problem Because the resistor and the inductor are connected in
parallel, the voltage drops across them are equal. Also, the total current is the sum
of the current through the resistor and the current through the inductor. Because
these two currents are not in phase, we’ll need to use phasors to calculate their sum.
The amplitudes of the applied voltage and the currents are equal to the magnitude
r
r
r
r
of the phasors. That is ε = ε peak , I = I peak , I R = I R , peak , and I L = I L , peak .
(a) The ac source applies a voltage
given by ε = ε peak cos ωt . Thus, the
ε peak cos ωt = I R R
voltage drop across both the load
resistor and the inductor is:
The current in the resistor is in phase
with the applied voltage:
Because I R , peak =
ε peak
R
:
(b) The current in the inductor lags the
applied voltage by 90&deg;:
Because I L , peak =
ε peak
XL
:
(c) The net current I is the sum of the
currents through the parallel
branches:
I R = I R , peak cos ωt
IR =
ε peak
R
cos ωt
I L = I L , peak cos(ωt − 90&deg;)
IL =
ε peak
XL
I = IR + IL
cos(ωt − 90&deg;)
Alternating-Current Circuits
r
Draw the phasor diagram for the
circuit. The projections of the phasors
onto the horizontal axis are the
instantaneous values. The current in
the resistor is in phase with the applied
voltage, and the current in the inductor
lags the applied voltage by 90&deg;. The
net current phasor is the sum of the
r r r
branch current phasors I = I L + I R .
(
Z is the impedance of the combination:
From the phasor diagram we have:
ε
r
IR
r
I
δ
ωt −δ
ωt
90&deg;− ω t
r
IL
)
The peak current through the parallel
combination is equal to ε peak Z , where
2767
I = I peak cos(ωt − δ ) ,
where I peak =
ε peak
Z
2
I peak
= I R2 , peak + I L2, peak
⎛ε ⎞ ⎛ε ⎞
= ⎜⎜ peak ⎟⎟ + ⎜⎜ peak ⎟⎟
⎝ R ⎠ ⎝ XL ⎠
2
2
2
⎛ 1
1 ⎞ ε peak
= ε ⎜⎜ 2 + 2 ⎟⎟ = 2
XL ⎠
Z
⎝R
1
1
1
where 2 = 2 + 2
Z
R
XL
.
2
peak
Solving for Ipeak yields:
From the phasor diagram:
I peak =
ε peak
Z
where Z −2 = R −2 + X L−2
I = I peak cos(ωt − δ
where
tan δ =
)
ε peak
I L , peak
I R , peak
=
XL
ε peak
=
R
XL
R
41 ••
[SSM] 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.
2768 Chapter 29
Picture the Problem ε = V1 + V2 , where V1 is the voltage drop across R and V2 is
r r r
the voltage drop across the parallel combination of L and RL. ε = V1 + V 2 is the
r r
r
relation for the phasors. For the parallel combination I = I RL + I L . Also, V1 is in
phase with I and V2 is in phase with I RL . First draw the phasor diagram for the
currents in the parallel combination, then add the phasors for the voltages to the
diagram.
r
I RL
The phasor diagram for the currents
in the circuit is:
r
I
δ
r
IL
r
Adding the voltage phasors to the
diagram gives:
ε
r
V2
r
I RL
δ
r
I
ωt
δ
r
V1
r
IL
The maximum current in the
inductor, I2, peak, is given by:
I 2, peak =
V2, peak
(1)
Z2
where Z 2−2 = RL−2 + X L−2
tan δ is given by:
tan δ =
=
Solve for δ to obtain:
I L , peak
I R , peak
=
(2)
V2, peak X L
V2, peak RL
RL
R
R
= L = L
X L ωL 2πfL
⎛ RL ⎞
⎟⎟
⎝ 2πfL ⎠
δ = tan −1 ⎜⎜
(3)
Alternating-Current Circuits
Apply the law of cosines to the triangle formed by the voltage phasors to obtain:
2
ε peak
= V1,2peak + V22, peak + 2V1, peakV2, peak cos δ
or
2
2
2
I peak
Z 2 = I peak
R 2 + I peak
Z 22 + 2 I peak RI peak Z 2 cos δ
Dividing out the current squared
yields:
Z 2 = R 2 + Z 22 + 2 RZ 2 cos δ
Solving for Z yields:
Z = R 2 + Z 22 + 2 RZ 2 cos δ
The maximum current I peak in the
circuit is given by:
Irms is related to I peak according
to:
(a) Substitute numerical values in
equation (3) and evaluate δ :
I peak =
I rms =
ε peak
(4)
(5)
Z
1
I peak
2
(6)
⎛
⎞
20.0 Ω
⎟⎟
⎝ 2π (500 Hz )(3.20 mH ) ⎠
δ = tan −1 ⎜⎜
⎛ 20.0 Ω ⎞
⎟⎟ = 63.31&deg;
= tan −1 ⎜⎜
⎝ 10.053 Ω ⎠
Solving equation (2) for Z2 yields:
Substitute numerical values and
evaluate Z2:
Z2 =
Z2 =
1
−2
L
R + X L−2
1
(20.0 Ω )−2 + (10.053 Ω )−2
= 8.982 Ω
Substitute numerical values and evaluate Z:
Z=
(4.00 Ω )2 + (8.982 Ω)2 + 2(4.00 Ω )(8.982 Ω )cos 63.31&deg; = 11.36 Ω
Substitute numerical values in
equation (5) and evaluate I peak :
I peak =
100 V
= 8.806 A
11.36 Ω
2769
2770 Chapter 29
Substitute for I peak in equation
(6) and evaluate I rms :
The maximum and rms values of
V2 are given by:
1
(8.806 A ) = 6.23 A
2
I rms =
V2, peak = I peak Z 2
= (8.806 A )(8.982 Ω ) = 79.095 V
and
1
V2, peak
2
1
(79.095 V ) = 55.929 V
=
2
V2,rms =
The rms values of I RL ,rms and
I RL ,rms =
I L ,rms are:
V2,rms
RL
=
55.929 V
= 2.80 A
20.0 Ω
and
I L ,rms =
V2,rms
XL
=
55.929 V
= 5.53 A
10.053 Ω
X L = 40.2 Ω , δ = 26.4&deg; , Z 2 = 17.9 Ω ,
(b) Proceed as in (a) with
f = 2000 Hz to obtain:
Z = 21.6 Ω , I peak = 4.64 A , and
I rms = 3.28 A ,
V2,max = 83.0V , V2, rms = 58.7 V ,
I RL ,rms = 2.94 A , and I L,rms = 1.46 A
(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
⎛
⎛
P ⎞
I2 R ⎞
= ⎜1 + R ⎟ = ⎜1 + 2 rms ⎟
⎜ I R ,rms RL ⎟
PRL + PR ⎜⎝ PRL ⎟⎠
L
⎝
⎠
PRL
−1
Substitute numerical values for f = 500 Hz to obtain:
−1
PRL
PRL + PR
f =500 Hz
⎛ (6.23 A )2 (4.00 Ω ) ⎞
⎟ = 0.502 = 50.2%
= ⎜⎜1 +
2
⎟
(
)
(
)
2
.
80
A
20
.
0
Ω
⎝
⎠
Alternating-Current Circuits
2771
(d) Substitute numerical values for f = 2000 Hz to obtain:
−1
PRL
PRL + PR
42
••
f = 2000 Hz
⎛ (3.28 A )2 (4.00 Ω ) ⎞
⎟ = 0.800 = 80.0%
= ⎜⎜1 +
2
⎟
(
)
(
)
2
.
94
A
20
.
0
Ω
⎝
⎠
An ideal ac voltage source whose emf ε1 is given by (20 V) cos(2πft)
and an ideal battery whose emf ε2 is 16 V are connected to a combination of two
resistors and an inductor (Figure 29-34), where R1 = 10 Ω, R2 = 8.0 Ω, and
L = 6.0 mH. Find the average power delivered to each resistor if (a) the driving
frequency is 100 Hz, (b) the driving frequency is 200 Hz, and (c) the driving
frequency is 800 Hz.
Picture the Problem We can treat the ac and dc components separately. For the
dc component, L acts like a short circuit. Let ε1, peak denote the peak value of the
voltage supplied by the ac voltage source. We can use P = ε 22 R to find the
power dissipated in the resistors by the current from the ideal battery. We’ll apply
Kirchhoff’s loop rule to the loop including L, R1, and R2 to derive an expression
for the average power delivered to each resistor by the ac voltage source.
(a) The total power delivered to R1
and R2 is:
The dc power delivered to the
resistors whose resistances are R1
and R2 is:
Express the average ac power
delivered to R1:
Apply Kirchhoff’s loop rule to a
clockwise loop that includes R1,
L, and R2:
Solving for I2 yields:
P1 = P1, dc + P1, ac
(1)
and
P2 = P2, dc + P2, ac
(2)
P1,dc =
P1, ac =
ε 22 and P
2 ,dc
R1
=
ε 22
R2
ε12, rms ε12, peak
R1
=
2R1
R1 I1 − Z 2 I 2 = 0
I2 =
R1
R
I1 = 1
Z2
Z2
ε1, peak ε1, peak
R1
=
Z2
2772 Chapter 29
P2, ac
Substituting in equations (1) and (2)
yields:
P1 =
and
P2 =
Substitute numerical values and
evaluate P1:
⎛ε
⎞
R2 = ⎜⎜ 1, peak ⎟⎟ R2
= I
⎝ Z2 ⎠
ε2 R
= 1, peak2 2
2Z 2
2
Express the average ac power
delivered to R2:
1
2
2
2 , rms
1
2
ε 22 + ε12, peak
R1
2R1
ε 22 + ε12, peak R2
R2
2Z 22
2
2
(
(
16 V )
20 V )
P1 =
+
=
10 Ω
2(10 Ω )
46 W
Substitute numerical values and evaluate P2:
P2 =
(16 V )2
8 .0 Ω
+
(20 V )2 (8.0 Ω )
2
2
2 (8.0 Ω ) + (2π {100 s −1 }{6.0 mH})
[
(b) Proceed as in (a) to evaluate P1
and P2 with f = 200 Hz:
(c) Proceed as in (a) to evaluate P1
and P2 with f = 800 Hz:
]=
52 W
P1 = 25.6 W + 20.0 W = 46 W
P2 = 32.0 W + 13.2 W = 45 W
P1 = 25.6 W + 20.0 W = 46 W
P2 = 32.0 W + 1.64 W = 34 W
An ac circuit contains a resistor and an ideal inductor connected in
43 ••
series. The voltage rms drop across this series combination is 100-V and the rms
voltage drop across the inductor alone is 80 V. What is the rms voltage drop
across the resistor?
Picture the Problem We can use the phasor diagram for an RL circuit to find the
voltage across the resistor.
Alternating-Current Circuits
2773
r
VL
The phasor diagram for the voltages in
the circuit is shown to the right:
ε
rms
r
VR
Use the Pythagorean theorem to
express VR:
VR =
2
ε rms
− VL2
Substitute numerical values and
evaluate VR:
VR =
(100 V ) 2 − (80 V ) 2
= 60 V
Filters and Rectifiers
44 ••
The circuit shown in Figure 29-35 is called an RC high-pass filter
because it transmits input voltage signals that have high frequencies with greater
amplitude than it transmits input voltage signals that have low frequencies. If the
input voltage is given by Vin = Vin peak cos ωt, show that the output voltage is
Vout = VH cos(ωt – δ) where VH = Vin peak
1 + (ωRC ) . (Assume that the output is
−2
connected to a load that draws only an insignificant amount of current.) Show that
this result justifies calling this circuit a high-pass filter.
Picture the Problem The phasor
diagram for the RC high-pass filter is
r
r
shown to the right. Vapp and VR are the
phasors for Vin and Vout, respectively.
Note that tan δ = − X C R. That δ is
r
negative follows from the fact that Vapp
r
lags VR by δ . The projection of
r
Vapp onto the horizontal axis is
r
Vapp = Vin, and the projection of VR
onto the horizontal axis is VR = Vout.
Express Vapp :
r
VR
ωt − δ
δ
r
Vapp
ωt
r
VC
Vapp = Vapp,peak cos ωt
where Vapp,peak = Vpeak = I peak Z
and Z 2 = R 2 + X C2
Because δ &lt; 0:
ωt + δ = ωt − δ
(1)
2774 Chapter 29
VR is given by:
VR = VR , peak cos(ωt − δ )
where VR , peak = VH = I peak R
Solving equation (1) for Z and
substituting for XC yields:
⎛ 1 ⎞
Z = R +⎜
⎟
⎝ ωC ⎠
Because Vout = VR :
Vout = VR , peak cos(ωt − δ )
Simplify further to obtain:
(2)
= I in peak R cos(ωt − δ )
=
Using equation (2) to substitute for Z
yields:
2
2
Vout =
Vout =
Vin peak
Z
R cos(ωt − δ )
Vin peak
⎛ 1 ⎞
R2 + ⎜
⎟
⎝ ωC ⎠
2
Vin peak
1 + (ωRC )
−2
R cos(ωt − δ )
cos(ωt − δ )
or
Vout = VH cos(ωt − δ )
where
VH =
As ω → ∞:
VH →
Vin peak
1 + (ωRC )
−2
Vin peak
= Vin peak showing that
1 + (0 )
the result is consistent with the highpass name for this circuit.
2
45 ••
(a) Find an expression for the phase constant δ in Problem 44 in terms
of ω, R and C. (b) What is the value of δ in the limit that ω → 0? (c) What is the
value of δ in the limit that ω → ∞? (d) Explain your answers to Parts (b) and (c).
Alternating-Current Circuits
r
VR
Picture the Problem The phasor
diagram for the RC high-pass filter is
r
r
shown below. Vapp and VR are the
r
Vapp
phasors for Vin and Vout, respectively.
r
The projection of Vapp onto the
δ
horizontal axis is Vapp = Vin, and the
r
projection of VR onto the horizontal
axis is VR = Vout.
r
r
(a) Because Vapp lags VR by δ .
Use the definition of XC to obtain:
Solving for δ yields:
2775
ωt
r
VC
tan δ = −
VC
IX
X
=− C =− C
VR
IR
R
1
1
tan δ = − ωC = −
R
ωRC
⎡
1 ⎤
δ = tan −1 ⎢−
⎣ ωRC ⎥⎦
(b) As ω → 0:
δ → − 90&deg;
(c) As ω → ∞:
δ→ 0
r
r
(d) For very low driving frequencies, X C &gt;&gt; R and so VC effectively lags Vin by
r
90&deg;. For very high driving frequencies, X C &lt;&lt; R and so VR is effectively in phase
r
with Vin .
46 ••
Assume that in Problem 44, R = 20 kΩ and C = 15 nF. (a) At what
frequency is V H = 12 V in peak ? This particular frequency is known as the 3 dB
frequency, or f3dB for the circuit. (b) Using a spreadsheet program, make a graph
of log10(VH) versus log10(f), where f is the frequency. Make sure that the scale
extends from at least 10% of the 3-dB frequency to ten times the 3-dB frequency.
(c) Make a graph of δ versus log10(f) for the same range of frequencies as in Part
(b). What is the value of the phase constant when the frequency is equal to the
3-dB frequency?
Picture the Problem We can use the results obtained in Problems 44 and 45 to
find f3 dB and to plot graphs of log(Vout) versus log(f) and δ versus log(f).
2776 Chapter 29
(a) Use the result of Problem 44 to
express the ratio Vout/Vin peak:
Vin peak
1 + (ωRC )
Vout
=
Vin peak
Vin peak
1
When Vout = Vin peak / 2 :
1 + (ωRC )
Square both sides of the
equation and solve for ωRC to
obtain:
Substitute numerical values and
evaluate f3 dB:
ωRC = 1 ⇒ ω =
f 3 dB =
B
(b) From Problem 44 we have:
In Problem 45 it was shown that:
Rewrite these expressions in terms
of f3 dB to obtain:
−2
=
Vout =
−2
=
1
1 + (ωRC )
1
2
1
1
⇒ f 3 dB =
RC
2πRC
1
= 0.53 kHz
2π (20 kΩ )(15 nF)
Vin peak
1 + (ωRC )
−2
1 ⎤
⎡
⎣ ωRC ⎥⎦
δ = tan −1 ⎢−
Vout =
Vin peak
⎛ 1 ⎞
⎟⎟
1 + ⎜⎜
⎝ 2πfRC ⎠
2
Vpeak
=
⎛f ⎞
1 + ⎜⎜ 3 dB ⎟⎟
⎝ f ⎠
and
⎡
⎡ f ⎤
1 ⎤
= tan −1 ⎢− 3 dB ⎥
⎥
⎣ 2πfRC ⎦
⎣ f ⎦
δ = tan −1 ⎢−
A spreadsheet program to generate the data for a graph of Vout versus f and δ
versus f is shown below. The formulas used to calculate the quantities in the
columns are as follows:
Cell
B1
B2
B3
B4
A8
−2
Formula/Content
2.00E+03
1.50E−08
1
531
53
Algebraic Form
R
C
Vin peak
f3 dB
0.1f3 dB
2
Alternating-Current Circuits
C8
\$B\$3/SQRT(1+(1(\$B\$4/A8))^2)
Vin peak
2
D8
E8
LOG(C8)
ATAN(−\$B\$4/A8)
F8
E8*180/PI()
A
⎛f ⎞
1 + ⎜⎜ 3 dB ⎟⎟
⎝ f ⎠
log(Vout)
⎡ f ⎤
tan −1 ⎢− 3 dB ⎥
⎣ f ⎦
δ in degrees
1
2
3
4
5
6
7
8
9
10
11
R=
C=
Vin peak=
f3 dB=
B
2.00E+04
1.50E−08
1
531
C
ohms
F
V
Hz
D
E
F
f
53
63
73
83
log(f)
1.72
1.80
1.86
1.92
Vout
0.099
0.118
0.136
0.155
log(Vout)
−1.003
−0.928
−0.865
−0.811
55
56
57
523
533
543
2.72
2.73
2.73
0.702
0.709
0.715
−0.154
−0.150
−0.146
−0.793
−0.783
−0.774
−45.4
−44.9
−44.3
531
532
533
534
5283
5293
5303
5313
3.72
3.72
3.72
3.73
0.995
0.995
0.995
0.995
−0.002
−0.002
−0.002
−0.002
−0.100
−0.100
−0.100
−0.100
−5.7
−5.7
−5.7
−5.7
delta(rad) delta(deg)
−1.471
−84.3
−1.453
−83.2
−1.434
−82.2
−1.416
−81.1
The following graph of log(Vout) versus log(f) was plotted for Vin peak = 1 V.
0.0
log(V out)
-0.2
-0.4
-0.6
-0.8
-1.0
1.5
2.0
2.5
3.0
log(f )
3.5
4.0
2777
2778 Chapter 29
A graph of δ (in degrees) as a function of log(f) follows.
0
-10
-20
delta (deg)
-30
-40
-50
-60
-70
-80
-90
1.5
2.0
2.5
3.0
3.5
4.0
log(f )
Referring to the spreadsheet program, we see that when f = f3 dB, δ ≈ − 44.9&deg;.
This result is in good agreement with its calculated value of −45.0&deg;.
[SSM] A slowly varying voltage signal V(t) is applied to the input of
47 ••
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.
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:
Substitute for V (t ) and I to obtain:
V (t ) − VC − IR = 0
where VC is the potential difference
across the capacitor.
Vin peak cos ωt − Vc − R
dQ
=0
dt
Because Q = CVC:
dQ d
dV
= [CVC ] = C C
dt
dt
dt
Substitute for dQ/dt to obtain:
dVC
=0
dt
the differential equation describing the
potential difference across the
capacitor.
Vpeak cos ωt − VC − RC
Alternating-Current Circuits
Because there is no significant
change in the voltage signal during
one time constant:
Substituting for RC
2779
dVC
dV
= 0 ⇒ RC C = 0
dt
dt
Vin peak cos ωt − VC = 0
and
VC = Vin peak cos ωt
dVC
yields:
dt
Consequently, the potential
difference across the resistor is given
by:
V R = RC
[
dVC
d
= RC Vin peak cos ωt
dt
dt
]
48 ••
We can describe the output from the high-pass filter from Problem 44
using a decibel scale: β = (20 dB) log10 (VH Vin peak ) , where β is the output in
decibels. Show that for V H =
VH =
1
2
1
2
V in peak , β = 3.0 dB. The frequency at which
V in peak is known as f3dB (the 3-dB frequency). Show that for f &lt;&lt; f3dB, the
output β drops by 6 dB if the frequency f is halved.
Picture the Problem We can use the expression for VH from Problem 44 and the
definition of β given in the problem to show that every time the frequency is
halved, the output drops by 6 dB.
From Problem 44:
VH =
Vin peak
1 + (ωRC )
−2
or
VH
1
=
−2
Vin peak
1 + (ωRC )
Express this ratio in terms of f and
f3 dB and simplify to obtain:
VH
=
Vpeak
For f &lt;&lt; f3dB:
VH
≈
Vpeak
B
From the definition of β we have:
1
⎛f ⎞
1 + ⎜⎜ 3 dB ⎟⎟
⎝ f ⎠
2
=
f
⎛
f2 ⎞
f 32dB ⎜⎜1 + 2 ⎟⎟
f 3 dB ⎠
⎝
⎛ VH ⎞
⎟
⎟
V
⎝ peak ⎠
β = 20 log10 ⎜⎜
f
⎛
f2 ⎞
f 32dB ⎜⎜1 + 2 ⎟⎟
f 3 dB ⎠
⎝
=
f
f 3 dB
2780 Chapter 29
Substitute for VH/Vpeak to obtain:
Doubling the frequency yields:
The change in decibel level is:
⎛ f ⎞
⎟⎟
⎝ f 3 dB ⎠
β = 20 log10 ⎜⎜
⎛
f ⎞
⎟⎟
⎝ f 3 dB ⎠
β' = 20 log10 ⎜⎜
1
2
Δβ = β ' − β
⎛ 1 f ⎞
⎛ f ⎞
⎟⎟
= 20 log10 ⎜⎜ 2 ⎟⎟ − 20 log10 ⎜⎜
⎝ f 3 dB ⎠
⎝ f 3 dB ⎠
= 20 log10 ( 12 ) ≈ − 6 dB
Show that the average power dissipated in the resistor of the
V in2 peak
high-pass filter of Problem 44 is given by Pave =
.
−2
2R ⎡1+ (ω RC ) ⎤
⎣
⎦
49
••
[SSM]
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 &frac12; to establish the given result.
The instantaneous power P(t)
dissipated in the resistor is:
The output voltage Vout is:
From Problem 44:
Substitute in the expression for P(t)
to obtain:
2
Vout
P(t ) =
R
Vout = VH cos(ωt − δ )
VH =
Vin peak
1 + (ωRC )
−2
VH2
P (t ) =
cos 2 (ωt − δ )
R
Vin2 peak
=
cos 2 (ωt − δ )
−2
R 1 + (ωRC )
[
Because the average value of the
square of the cosine function over
one cycle is &frac12;:
Pave =
]
[
Vin2 peak
2 R 1 + (ωRC )
−2
]
Alternating-Current Circuits
2781
50 ••
One application of the high-pass filter of Problem 44 is a noise filter
for electronic circuits (a filter that blocks out low-frequency noise). Using a
resistance value of 20 kΩ, find a value for the capacitance for the high-pass filter
that attenuates a 60-Hz input voltage signal by a factor of 10. That is, so
V H = 101 V in peak .
Picture the Problem We can solve the expression for VH from Problem 44 for the
required capacitance of the capacitor.
From Problem 44:
VH =
We require that:
VH
Vin peak
Vin peak
1 + (ωRC )
=
−2
1
1 + (ωRC )
−2
=
1
10
or
1 + (ωRC )
Solving for C yields:
C=
Substitute numerical values and
evaluate C:
C=
−2
= 10
1
1
=
99ωR 2π 99 Rf
1
= 13 nF
2π 99 (20 kΩ )(60 Hz )
51 ••
[SSM] The circuit shown in Figure 29-36 is an example of a lowpass 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
V L = V in peak
1+ (ω RC ) . (b) Discuss the trend of the output voltage in the
2
limiting cases ω → 0 and ω → ∞.
Picture
diagram
r
Vapp and
the Problem In the phasor
for the RC low-pass filter,
r
VC are the phasors for Vin and
Vout, respectively. The projection of
r
Vapp onto the horizontal axis is Vapp =
r
Vin, the projection of VC onto the
horizontal axis is VC = Vout,
r
Vpeak = Vapp , and φ is the angle between
r
VC and the horizontal axis.
r
VR
r
Vapp
ωt
φ δ
r
VC
2782 Chapter 29
(a) Express Vapp :
Vapp = Vin peak cos ωt
where Vin peak = I peak Z
and Z 2 = R 2 + X C2
Vout = VC is given by:
(1)
Vout = VC , peak cos φ
= I peak X C cos φ
If we define δ as shown in the phasor
diagram, then:
Vout = I peak X C cos(ωt − δ )
=
Solving equation (1) for Z and
substituting for XC yields:
Using equation (2) to substitute for Z
and substituting for XC yields:
Simplify further to obtain:
Vin peak
Z
X C cos(ωt − δ )
⎛ 1 ⎞
Z = R +⎜
⎟
⎝ ωC ⎠
2
2
Vout =
Vout =
(2)
1
cos(ωt − δ )
2 ωC
⎛ 1 ⎞
R2 + ⎜
⎟
⎝ ωC ⎠
Vin peak
Vin peak
1 + (ωRC )
2
cos(ωt − δ )
or
Vout = VL cos(ωt − δ )
where
VL =
Vin peak
1 + (ωRC )
2
(b) Note that, as ω → 0, VL → Vpeak . This makes sense physically in that, for low
frequencies, XC 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.
Alternating-Current Circuits
2783
52 ••
(a) Find an expression for the phase angle δ for the low-pass filter of
Problem 51 in terms of ω, R and C. (b) Find the value of δ in the limit that ω → 0
and in the limit that ω → ∞. Explain your answer.
Picture the Problem The phasor
diagram for the RC low-pass filter is
r
r
shown to the right. Vapp and VC are the
r
VR
r
Vapp
phasors for Vin and Vout, respectively.
r
The projection of Vapp onto the
ωt
δ
horizontal axis is Vapp = Vin and the
r
projection of VC onto the horizontal
r
axis is VC = Vout . Vpeak = Vapp .
(a) From the phasor diagram we
have:
r
VC
tan δ =
I peak R
VR
R
=
=
VC I peak X C X C
Use the definition of XC to obtain:
tan δ =
R
= ωRC
1
ωC
Solving for δ yields:
δ = tan −1 (ωRC )
(b) As ω → 0, δ → 0&deg; . This behavior makes sense physically in that, at
low frequencies, XC is very large compared to R and, as a consequence, VC
is in phase with Vin .
As ω → ∞, δ → 90&deg; . This behavior makes sense physically in that, at
high frequencies, XC is very small compared to R and, as a consequence, VC
is out of phase with Vin .
Remarks: See the spreadsheet solution in the following problem for
additional evidence that our answer for Part (b) is correct.
53 ••
Using a spreadsheet program, make a graph of VL versus input
frequency f and a graph of phase angle δ versus input frequency for the low-pass
filter of Problems 51 and 52. Use a resistance value of 10 kΩ and a capacitance
value of 5.0 nF.
2784 Chapter 29
Picture the Problem We can use the expressions for VL and δ derived in
Problems 51 and 52 to plot the graphs of VL versus f and δ versus f for the lowpass filter of Problem 51. We’ll simplify the spreadsheet program by expressing
both VL and δ as functions of f3 dB.
From Problems 51 and 52 we have:
VL =
Vin peak
1 + (ωRC )
2
and
δ = tan −1 (ωRC )
Rewrite each of these expressions in
terms of f to obtain:
VL =
Vin peak
1 + (2πfRC )
2
and
δ = tan −1 (2πfRC )
A spreadsheet program to generate the data for graphs of VL versus f and δ versus
f for the low-pass filter is shown below. Note that Vin peak has been arbitrarily set
equal to 1 V. The formulas used to calculate the quantities in the columns are as
follows:
Cell
B1
B2
B3
B8
Formula/Content
2.00E+03
5.00E−09
1
\$B\$3/SQRT(1+((2*PI()*A8*
1000*\$B\$1*\$B\$2)^2))
Algebraic Form
R
C
Vin peak
Vin peak
C8
ATAN(2*PI()*A8*1000*\$B\$1*\$B\$2)
D8
C8*180/PI()
tan −1 (2πfRC )
δ in degrees
A
1
2
3
4
5
6
7
8
9
10
B
C
R= 1.00E+04 ohms
C= 5.00E−09 F
Vin peak= 1
V
f(kHz)
0
1
2
3
Vout
1.000
0.954
0.847
0.728
1 + (2πfRC )
D
δ(rad)
δ (deg)
0.000
0.304
0.561
0.756
0.0
17.4
32.1
43.3
2
Alternating-Current Circuits
54
55
56
57
47
48
49
50
0.068
0.066
0.065
0.064
1.503
1.505
1.506
1.507
2785
86.1
86.2
86.3
86.4
A graph of VL as a function of f follows:
1.0
0.8
V L ( V)
0.6
0.4
0.2
0.0
0
10
20
30
40
50
40
50
f ( kHz)
A graph of δ as a function of f follows:
90
delta (deg)
60
30
0
0
10
20
30
f (kHz)
54 ••• A rapidly varying voltage signal V(t) is applied to the input of the lowpass filter of Problem 51. Rapidly varying means that during one time constant
(equal to RC) there are significant changes in the voltage signal. Show that under
these conditions the output voltage is proportional to the integral of V(t) with
respect to time. This situation is known as an integration circuit.
Picture the Problem We can use Kirchhoff’s loop rule to obtain a differential
equation relating the input, capacitor, and resistor voltages. We’ll then assume a
solution to this equation that is a linear combination of sine and cosine terms with
coefficients that we can find by substitution in the differential equation. The
solution to these simultaneous equations will yield the amplitude of the output
voltage.
2786 Chapter 29
Apply Kirchhoff’s loop rule to the
input side of the filter to obtain:
V (t ) − IR − VC = 0
where VC is the potential difference
across the capacitor.
Substitute for V(t) and I to
obtain:
Vin peak cos ωt − R
Because Q = CVC:
dQ d
dV
= [CVC ] = C C
dt
dt
dt
Substitute for dQ/dt to obtain:
dVC
− VC = 0
dt
the differential equation describing the
potential difference across the
capacitor.
VC is given by:
The fact that V(t) varies rapidly
means that ω &gt;&gt; 1 and so:
Vin peak cos ωt − RC
VC = IX c =
I
ωC
VC ≈ 0
and
Vpeak cos ωt − RC
Separating the variables in this
differential equation and solving
for VC yields:
dQ
− VC = 0
dt
VC =
dVC
=0
dt
1
Vpeak cos ωtdt
RC ∫
55 ••• [SSM] 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 ω = 1/ LC . (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.
r
r
r
Vapp and VL + VC are the phasors for Vin and Vout, respectively. The projection of
r
r
r
Vapp onto the horizontal axis is Vapp = Vin, and the projection of VL + VC onto the
horizontal axis 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 Δω = ω − ωtrap and requiring that Z trap = R will yield an expression
for the bandwidth and reveal its dependence on R.
Alternating-Current Circuits
r
Vapp
r r
VL + VC
r
VL
r
VR
δ
ωt ω t − δ
r
VC
(a) Express Vapp :
Vapp = Vapp, peak cos ωt
where Vapp, peak = Vpeak = I peak Z
and Z 2 = R 2 + ( X L − X C )
2
Vout is given by:
(1)
Vout = Vout, peak cos(ωt − δ )
where Vout, peak = I peak Z trap
and Z trap = X L − X C
Solving equation (1) for Z yields:
Z = R2 + (X L − X C )
Because Vout = VL + VC :
Vout = Vout, peak cos(ωt − δ )
2
(2)
= I peak Z trap cos(ωt − δ )
=
Vpeak
Z
Z trap cos(ωt − δ )
Using equation (2) to substitute for Z
yields:
Vout =
Noting that Vout = 0 provided
Z trap = X L − X C = 0
Vpeak
2
R 2 + Z trap
Z trap cos(ωt − δ )
Z trap = 0, set Z trap = 0 to obtain:
Substituting for XL and XC yields:
(b) Let the bandwidth Δω be:
ωL −
1
= 0 ⇒ω =
ωC
Δω = ω − ω trap
1
LC
(3)
2787
2788 Chapter 29
Let the frequency bandwidth be
defined by the frequency at which
Z trap = R . Then:
Because ωtrap =
1
:
LC
For ω ≈ ωtrap:
ωL −
1
2
= R ⇒ ω LC − 1= ωRC
ωC
⎛ ω
⎜
⎜ω
⎝ trap
2
⎞
⎟ − 1 = ωRC
⎟
⎠
2
⎛ ω 2 − ω trap
⎞
⎜
⎟ ≈ ω trap RC
⎜ ω trap ⎟
⎝
⎠
2
Solve for ω 2 − ω trap
:
2
ω 2 − ω trap
= (ω − ω trap )(ω + ω trap )
Because ω ≈ ωtrap, ω + ω trap ≈ 2ω trap :
2
ω 2 − ω trap
≈ 2ω trap (ω − ω trap )
Substitute in equation (3) to obtain:
Δω = ω − ωtrap =
2
RCωtrap
2
=
R
2L
56 ••• A half-wave rectifier for transforming an ac voltage into a dc voltage
is shown in Figure 29-38. The diode in the figure can be thought of as a one-way
valve for current. It allows current to pass in the forward direction (the direction
of the arrowhead) only when Vin is at a higher electric potential than Vout by
0.60 V (that is, whenever Vin −Vout ≥ +0.60 V ). The resistance of the diode is
effectively infinite when Vin − Vout is less than +0.60 V. Plot two cycles of both
input and output voltages as a function of time, on the same graph, assuming the
input voltage is given by Vin = Vin peak cos ωt.
Picture the Problem For voltages greater than 0.60 V, the output voltage will
mirror the input voltage minus a 0.60 V drop. But when the voltage swings below
0.60 V, the output voltage will be 0. A spreadsheet program was used to plot the
following graph. The angular frequency and the peak voltage were both arbitrarily
set equal to one.
Alternating-Current Circuits
1.0
2789
1.0
V_in
0.5
0.0
0.0
-0.5
-0.5
Vout (V)
V in (V)
V_out
0.5
-1.0
-1.0
0
2
4
6
8
10
12
t (s)
57 ••• The output of the rectifier of Problem 56, can be further filtered by
putting its output through a low-pass filter as shown in Figure 29-39a. The
resulting output is a dc voltage with a small ac component (ripple) shown in
Figure 29-39b. If the input frequency is 60 Hz and the load resistance is 1.00 kΩ,
find the value for the capacitance so that the output voltage varies by less than 50
percent of the mean value over one cycle.
Picture the Problem We can use the decay of the potential difference across the
capacitor to relate the time constant for the RC circuit to the frequency of the
input signal. Expanding the exponential factor in the expression for VC will allow
us to find the approximate value for C that will limit the variation in the output
voltage by less than 50 percent (or any other percentage).
The voltage across the capacitor is
given by:
VC = Vin e −t RC
Expand the exponential factor to
obtain:
e −t RC ≈ 1 −
For a decay of less than 50 percent:
Because the voltage goes positive
every cycle, t = 1/60 s and:
1−
1
t
RC
1
2
t ≤ 0. 5 ⇒ C ≤ t
RC
R
C≤
2 ⎛ 1 ⎞
⎜ s ⎟ = 33 μF
1.00 kΩ ⎝ 60 ⎠
Driven LC Circuits
58
ε
••
The generator voltage in Figure 29-40, is given by
= (100 V) cos (2πft). (a) For each branch, what is the peak current and what is
the phase of the current relative to the phase of the generator voltage? (b) At the
resonance frequency there is no current in the generator. What is the angular
frequency at resonance? (c) At the resonance frequency, find the current in the
inductor and what is the current in the capacitor. Express your results as
2790 Chapter 29
functions of time. (d) Draw a phasor diagram showing the phasors for the applied
voltage, the generator current, the capacitor current, and the inductor current for
the case where frequency is higher than the resonance frequency.
Picture the Problem We know that the current leads the voltage across and
capacitor and lags the voltage across an inductor. We can use I L , peak = ε peak X L
and I C , peak = ε peak X C to find the amplitudes of these currents. The current in the
generator will vanish under resonance conditions, i.e., when I L = I C . To find the
currents in the inductor and capacitor at resonance, we can use the common
potential difference across them and their reactances together with our knowledge
of the phase relationships mentioned above.
(a) Express the amplitudes of the
currents through the inductor and
the capacitor:
I L , peak =
and
I C , peak =
ε peak ε peak
=
XL
ε peak
XC
=
2πfL
ε peak
1
2πfC
= 2πfCε peak
Substitute numerical values to obtain:
I L , peak =
and
100 V
25.0 V/H
=
, lagging ε by 90&deg;
(4.00 H )2πf
2πf
I C , peak = (25.0 μF)(100 V )ω =
(b) Express the condition that
I = 0:
Solve for ω to obtain:
Substitute numerical values and
evaluate ω:
(2.50 mV ⋅ F) 2πf , leading ε by 90&deg;
I L = I C or
ω=
ω=
ε
ωL
=
ε
1
ωC
= ωCε
1
LC
1
(4.00 H )(25.0 μF)
= 100 rad/s
Alternating-Current Circuits
2791
⎛ 25.0 V/H ⎞ ⎛
π⎞
⎟ cos⎜ ω t − ⎟
I L = ⎜⎜
−1 ⎟
2⎠
⎝ 100 s ⎠ ⎝
(c) Express the current in the
inductor at ω = ω0:
=
(250 mA)cos⎛⎜ ω t − π ⎞⎟
2⎠
⎝
where ω = 100 rad/s.
π⎞
⎛
I C = (2.50 mV ⋅ F) 100 s −1 cos⎜ ωt + ⎟
2⎠
⎝
(
Express the current in the capacitor
at ω = ω0:
)
π⎞
⎛
= − (250 mA )cos⎜ ωt + ⎟
2⎠
⎝
where ω = 100 rad/s.
r
IC
(d) The phasor diagram for the case
where the inductive reactance is
larger than the capacitive reactance
is shown to the right.
r
ε
max
r
I
r
IL
59 ••
A circuit consist of an ideal ac generator, a capacitor and an ideal
inductor, all connected in series. The charge on the capacitor is given by
Q = (15 μC) cos(ωt + π4 ), where ω = 1250 rad/s. (a) Find the current in the circuit
as a function of time. (b) Find the capacitance if the inductance is 28 mH.
(c) Write expressions for the electrical energy Ue, the magnetic energy Um, and
the total energy U as functions of time.
Picture the Problem We can differentiate Q with respect to time to find I as a
function of time. In (b) we can find C by using ω = 1 LC . The energy stored in
the magnetic field of the inductor is given by U m = 12 LI 2 and the energy stored in
the electric field of the capacitor by U e =
1
2
Q2
.
C
2792 Chapter 29
(a) Differentiate the charge with respect to time to obtain the current:
I (t ) =
dQ d ⎡
π ⎞⎤
π⎞
⎛
⎛
= ⎢(15 μC )cos⎜ ωt + ⎟⎥ = −(15 μC ) 1250 s −1 sin ⎜ ωt + ⎟
4 ⎠⎦
4⎠
dt dt ⎣
⎝
⎝
(
)
π⎞
π⎞
⎛
⎛
= −(18.75 mA )sin ⎜ ωt + ⎟ = − (19 mA )sin ⎜ ωt + ⎟
4⎠
4⎠
⎝
⎝
where ω = 1250 rad/s
(b) Relate C to L and ω:
ω=
Solve for C to obtain:
C=
Substitute numerical values and
evaluate C:
C=
1
LC
1
ω 2L
1
(1250 s ) (28 mH)
−1 2
= 22.86 μF
= 23 μF
(c) Express and evaluate the magnetic energy U m (t ) :
U m (t ) = 12 LI 2 =
1
2
(28 mH )(18.75 mA )2 sin 2 ⎛⎜ ωt + π ⎞⎟ = (4.9 μJ )sin 2 ⎛⎜ ωt + π ⎞⎟
4⎠
⎝
⎝
4⎠
where ω = 1250 rad/s
Q2
to find the electrical
C
energy stored in the capacitor as a
function of time:
Use U e =
1
2
U e (t ) =
1
2
(15 μF)2
π⎞
⎛
cos 2 ⎜ ωt + ⎟
22.86 μF
4⎠
⎝
π⎞
⎛
= (4.92 μJ )cos 2 ⎜ ωt + ⎟
4⎠
⎝
=
(4.9 μJ )cos 2 ⎛⎜ ωt + π ⎞⎟
⎝
where ω = 1250 rad/s
4⎠
The total energy stored in the electric and magnetic fields is the sum of U m (t ) and
U e (t ) :
π⎞
π⎞
⎛
⎛
U = (4.92 μJ )sin 2 ⎜ ωt + ⎟ + (4.92 μJ )cos 2 ⎜ ωt + ⎟ = 4.9 μJ
4⎠
4⎠
⎝
⎝
where ω = 1250 rad/s.
Alternating-Current Circuits
2793
60 ••• One method for determining the compressibility of a dielectric
material uses a driven LC circuit that has a parallel-plate capacitor. The dielectric
is inserted between the plates and the change in resonance frequency is
determined as the capacitor plates are subjected to a compressive stress. In one
such arrangement, the resonance frequency is 120 MHz when a dielectric of
thickness 0.100 cm and dielectric constant κ = 6.80 is placed between the plates.
Under a compressive stress of 800 atm, the resonance frequency decreases to
116 MHz. Find the Young’s modulus of the dielectric material. (Assume that the
dielectric constant does not change with pressure.)
Picture the Problem We can use the definition of the capacitance of a dielectricfilled capacitor and the expression for the resonance frequency of an LC circuit to
derive an expression for the fractional change in the thickness of the dielectric in
terms of the resonance frequency and the frequency of the circuit when the
dielectric is under compression. We can then use this expression for Δt/t to
calculate the value of Young’s modulus for the dielectric material.
stress ΔP
=
strain Δt t
Use its definition to express Young’s
modulus of the dielectric material:
Y=
Letting t be the initial thickness of
the dielectric, express the initial
capacitance of the capacitor:
C0 =
Express the capacitance of the
capacitor when it is under
compression:
Cc =
Express the resonance frequency of
the capacitor before the dielectric is
compressed:
ω0 =
1
=
C0 L
1
κ ∈0 AL
t
When the dielectric is compressed:
ωc =
1
=
Cc L
1
κ ∈0 AL
t − Δt
Express the ratio of ωc to ω0 and
simplify to obtain:
κ ∈0 A
t
κ ∈0 A
t − Δt
κ ∈0 AL
ωc
Δt
t
=
= 1−
ω0
t
κ ∈0 AL
t − Δt
(1)
2794 Chapter 29
Expand the radical binomially to
obtain:
12
ωc ⎛ Δt ⎞
= ⎜1 − ⎟
ω0 ⎝
t ⎠
≈ 1−
Δt
2t
provided Δt &lt;&lt; t.
Solve for Δt/t:
⎛ ω ⎞
Δt
= 2⎜⎜1 − c ⎟⎟
t
⎝ ω0 ⎠
Substitute in equation (1) to obtain:
Y=
Substitute numerical values and
evaluate Y:
Y=
ΔP
⎛ ω ⎞
2⎜⎜1 − c ⎟⎟
⎝ ω0 ⎠
(800 atm )(101.325 kPa/atm )
⎛ 116 MHz ⎞
⎟⎟
2⎜⎜1 −
⎝ 120 MHz ⎠
= 1.22 &times; 109 N/m 2
61 ••• Figure 29-41 shows an inductor in series with a parallel plate
capacitor. The capacitor has a width w of 20 cm and a gap of 2.0 mm. A dielectric
that has a dielectric constant of 4.8 can be slid in and out of the gap. The inductor
has an inductance of 2.0 mH. When half the dielectric is between the capacitor
plates (when x = 12 w ), the resonant frequency of this combination is 90 MHz.
(a) What is the capacitance of the capacitor without the dielectric?
(b) Find the resonance frequency as a function of x for 0 ≤ x ≤ w .
Picture the Problem We can model this capacitor as the equivalent of two
capacitors connected in parallel. Let C1 be the capacitance of the dielectric-filled
capacitor and C2 be the capacitance of the air-filled capacitor. We’ll derive
expressions for the capacitances of the parallel capacitors and add these
expressions to obtain C(x). We can then use the given resonance frequency when
x = w/2 and the given value for L to evaluate C0. In Part (b) we can use our result
for C(x) and the relationship between f, L, and C(x) at resonance to express f(x).
κ ∈0 A1 ∈0 A2
(a) Express the equivalent
capacitance of the two capacitors in
parallel:
C (x ) = C1 + C2 =
Express A2 in terms of the total area
of a capacitor plate A, w, and the
distance x:
A2 x
x
= ⇒ A2 = A
A w
w
d
+
d
(1)
Alternating-Current Circuits
Express A1 in terms of A and A2:
x⎞
⎛
A1 = A − A2 = A⎜1 − ⎟
⎝ w⎠
Substitute in equation (1) and
simplify to obtain:
C (x ) =
κ ∈0 A ⎛
x⎞ ∈ A x
⎜1 − ⎟ + 0
d ⎝ w⎠
d w
∈ A⎡ ⎛
x⎞ x⎤
= 0 ⎢κ ⎜1 − ⎟ + ⎥
d ⎣ ⎝ w ⎠ w⎦
⎡ κ −1 ⎤
= κC0 ⎢1 −
x
κw ⎥⎦
⎣
∈ A
where C0 = 0
d
Find C(w/2):
⎛ w⎞
⎡ κ −1 w ⎤
C ⎜ ⎟ = κC0 ⎢1 −
κw 2 ⎥⎦
⎝2⎠
⎣
⎡ κ − 1⎤
= κC0 ⎢1 −
2κ ⎥⎦
⎣
κ +1
= C0
2
Express the resonance frequency
of the circuit in terms of L and
C(x):
f (x ) =
Evaluate f(w/2):
⎛ w⎞
f⎜ ⎟=
⎝2⎠
1
2π LC ( x )
=
Solve for C0 to obtain:
C0 =
(2)
1
2π LC0
1
2π
κ +1
2
2
(κ + 1)LC0
1
⎛ w⎞
2π 2 f 2 ⎜ ⎟ L(κ + 1)
⎝2⎠
Substitute numerical values and evaluate C0:
C0 =
1
= 5.4 fF
2 ⎛ 20 cm ⎞
2
2π (90 MHz ) ⎜
⎟ (2.0 mH )(4.8 + 1)
⎝ 2 ⎠
2795
2796 Chapter 29
(b) Substitute for C(x) in equation (2)
to obtain:
f (x ) =
1
⎡ κ −1 ⎤
2π LκC0 ⎢1 −
x
κw ⎥⎦
⎣
Substitute numerical values and evaluate f(x):
1
f (x ) =
2π
(2.0 mH )(4.8)(5.39 &times;10
−16
⎡
⎤
4.8 − 1
x⎥
F ⎢1 −
⎣ 4.8(0.20 m ) ⎦
)
70 MHz
=
(
)
1 − 4.0 m −1 x
Driven RLC Circuits
62 •
A circuit consists of an ideal ac generator, a 20-μF capacitor and an
80-Ω resistor, all connected in series. The output of the generator has a peak emf
of 20-V, and the armature of the generator rotates at 400 rad/s. Find (a) the power
factor, (b) the rms current, and (c) the average power supplied by the generator.
Picture the Problem The diagram shows the relationship between δ, XL, XC, and
R. We can use this reference triangle to express the power factor for the given
circuit. In (b) we can find the rms current from the rms potential difference and
the impedance of the circuit. We can find the average power delivered by the
source from the rms current and the resistance of the resistor.
)
2
2
Z=
R
XL
+(
C
−X
X L − XC
δ
R
(a) The power factor is defined to
be:
cos δ =
With no inductance in the circuit:
XL = 0
and
cos δ =
R
=
Z
R
R2 + (X L − X C )
2
R
R +X
2
2
C
=
R
R2 +
1
ω C2
2
Alternating-Current Circuits
Substitute numerical values and
evaluate cosδ:
2797
80 Ω
cos δ =
(80 Ω )2 +
1
(400 s ) (20 μF)
−1 2
2
= 0.54
(b) Express the rms current in the
circuit:
I rms =
ε rms
Z
ε max
=
2
R + X C2
2
ε max
=
2 R2 +
Substitute numerical values and
evaluate I rms :
1
ω C2
2
20 V
I rms =
2
(80 Ω )2 +
1
(400 s ) (20 μF)
−1 2
2
= 95.3 mA = 95 mA
(c) The average power delivered
by the generator is given by:
2
Pav = I rms
R
Substitute numerical values and
evaluate Pav :
Pav = (95.3 mA) (80 Ω ) = 0.73 W
2
2
63 ••
[SSM] Show that the expression Pav = Rε rms
Z 2 gives the correct
result for a circuit containing only an ideal ac generator and (a) a resistor, (b) a
2
capacitor, and (c) an inductor. In the expression Pav = Rε rms
Z 2 , Pav is the
average power supplied by the generator, εrms 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
Z = R 2 + ( X L − X C ) . We can evaluate the given expression for Pav first for
2
XL = XC = 0 and then for R = 0.
(a) For X = 0, Z = R and:
Pav =
2
2
Rε rms
Rε rms
=
=
Z2
R2
2
ε rms
R
2798 Chapter 29
(b) and (c) If R = 0, then:
2
2
(
0)ε rms
Rε rms
=
= 0
Pav =
Z2
( X L − X C )2
Remarks: Recall that there is no energy dissipation in an ideal inductor or
capacitor.
64 ••
A series RLC circuit that has an inductance of 10 mH, a capacitance of
2.0 μF, and a resistance of 5.0 Ω is driven by an ideal ac voltage source that has a
peak emf of 100 V. Find (a) the resonant frequency and (b) the root-mean-square
current at resonance. When the frequency is 8000 rad/s, find (c) the capacitive and
inductive reactances, (d) the impedance, (e) the root-mean-square current, and
(f) the phase angle.
Picture the Problem We can use ω0 = 1
LC to find the resonant frequency of
the circuit, I rms = ε rms R to find the rms current at resonance, the definitions of XC
and XL to find these reactances at ω = 8000 rad/s, the definitions of Z and Irms to
find the impedance and rms current at ω = 8000 rad/s, and the definition of the
phase angle to find δ.
(a) Express the resonant frequency
ω0 in terms of L and C:
ω0 =
Substitute numerical values and
evaluate ω0:
ω0 =
1
LC
1
(10 mH)(2.0 μF)
= 7.1 &times; 10 3 rad/s
(b) Relate the rms current at
resonance to εrms and the impedance
of the circuit at resonance:
(c) Express and evaluate XC and
XL at ω = 8000 rad/s:
I rms =
ε rms = ε max
2R
R
100 V
=
2 (5.0 Ω )
= 14 A
XC =
1
1
=
−1
ωC 8000 s (2.0 μF)
(
)
= 62.50 Ω = 63 Ω
and
X L = ωL = 8000 s −1 (10 mH) = 80 Ω
(
)
Alternating-Current Circuits
(d) Express the impedance in terms
of the reactances, substitute the
results from (c), and evaluate Z:
Z = R 2 + (X L − X C )
2799
2
(5.0 Ω )2 + (80 Ω − 62.5 Ω )2
=
= 18.2 Ω = 18 Ω
(e) Relate the rms current at
ω = 8000 rad/s to εrms and the
I rms =
impedance of the circuit at this
frequency:
(f) δ is given by:
Substitute numerical values and
evaluate δ:
ε rms = ε max
2Z
Z
=
100 V
2 (18.2 Ω )
= 3.9 A
⎛ X L − XC ⎞
⎟
R
⎝
⎠
δ = tan −1 ⎜
⎛ 80 Ω − 62.5 Ω ⎞
⎟⎟ = 74&deg;
5.0 Ω
⎝
⎠
δ = tan −1 ⎜⎜
65 ••
[SSM] 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 Q = ω0 L R , the
resonance width by Δf = f 0 Q = ω0 2πQ , and the power factor by cos δ = R Z .
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:
Q=
ω0 L
(1)
R
and
Δf =
(a) Express the resonance frequency
for the circuit:
ω0 =
Substituting for ω0 in equation (1)
yields:
Q=
Substitute numerical values and
evaluate Q:
Q=
ω
f0
= 0
Q 2πQ
(2)
1
LC
L
LC R
=
1 L
R C
1
10 mH
= 14.1 = 14
5.0 Ω 2.0 μF
2800 Chapter 29
7.07 &times; 103 rad/s
Δf =
= 80 Hz
2π (14.1)
(b) Substitute numerical values in
equation (2) and evaluate Δf:
(c) The power factor of the circuit is given by:
cos δ =
R
=
Z
R
R 2 + (X L − X C )
2
R
=
1 ⎞
⎛
R + ⎜ ωL −
⎟
ωC ⎠
⎝
2
2
Substitute numerical values and evaluate cosδ:
5 .0 Ω
cos δ =
(5.0 Ω )2
⎛
⎞
1
⎟⎟
+ ⎜⎜ 8000 s −1 (10 mH ) −
−1
(
)
8000
s
2
.
0
F
μ
⎝
⎠
(
)
(
2
= 0.27
)
66 ••
FM radio stations typically operate at frequencies separated by
0.20 MHz. Thus, when your radio is tuned to a station operating at a frequency of
100.1 MHz, the resonance width of the receiver circuit should be much smaller
than 0.20 MHz, so that you do not receive a signal from stations operating at
adjacent frequencies. Assume your receiving circuit has a resonance width of
0.050 MHz. When tuned in to this particular station, what is the Q factor of your
circuit?
Picture the Problem We can use its definition, Q = f 0 Δf to find the Q factor for
the circuit.
The Q factor for the circuit is given
by:
Q=
f0
Δf
Substitute numerical values and
evaluate Q:
Q=
100.1 MHz
≈ 2.0 &times; 103
0.050 MHz
67 ••
A coil is connected to a 60-Hz ac generator with a peak emf equal to
100 V. At this frequency, the coil has an impedance of 10 Ω and a reactance of
8.0 Ω. (a) What is the peak current in the coil? (b) What is the phase angle
between the current and the applied voltage? (c) A capacitor is put in series with
the coil and the generator. What capacitance is required so that the current is in
phase with the generator emf? (d) What is the peak voltage measured across this
capacitor?
Alternating-Current Circuits
2801
Picture the Problem We can use I peak = ε peak Z to find the current in the coil and
the definition of the phase angle to evaluate δ. We can equate XL and XC to find
the capacitance required so that the current and the voltage are in phase. Finally,
we can find the voltage measured across the capacitor by using VC = IX C .
(a) Express the current in the coil
in terms of the potential difference
across it and its impedance:
Substitute numerical values and
evaluate I peak :
(b) The phase angle δ is given by:
Substitute numerical values and
evaluate δ:
I peak =
I peak =
ε peak
Z
100 V
= 10 A
10 Ω
⎛R⎞
−1 ⎛ X ⎞
⎟ = sin ⎜ L ⎟
⎝Z⎠
⎝ Z ⎠
δ = cos −1 ⎜
⎛ 8.0 Ω ⎞
⎟⎟ = 53&deg;
⎝ 10 Ω ⎠
δ = sin −1 ⎜⎜
(c) Express the condition on the
reactances that must be satisfied if
the current and voltage are to be in
phase:
X L = XC =
Substitute numerical values and
evaluate C:
C=
1
1
1
⇒C =
=
ωC
ωX L 2πfX L
1
(
2π 60 s
−1
)(8.0 Ω ) = 332 μF
= 0.33 mF
(d) Express the potential difference
across the capacitor:
Relate the peak current in the circuit
to the impedance of the circuit when
XL = XC:
Substitute for I to obtain:
Relate the impedance of the circuit to
the resistance of the coil:
VC = I peak X C
I peak =
VC =
Vpeak
R
Vpeak X C
R
=
Vpeak
2πfCR
Z = R2 + X 2 ⇒ R = Z 2 − X 2
2802 Chapter 29
Substituting for R yields:
VC =
Vpeak
2πfC Z 2 − X 2
Substitute numerical values and evaluate VC:
VC =
100 V
2π (60 s −1 )(332 μF) (10 Ω ) − (8.0 Ω )
2
2
= 0.13 kV
68 ••
An ideal 0.25-H inductor and a capacitor are connected in series with
an ideal 60-Hz generator. An digital voltmeter is used to measure the rms voltages
across the inductor and capacitor independently. The voltmeter reading across the
capacitor is 75 V and that across the inductor is 50 V. (a) Find the capacitance
and the rms current in the circuit. (b) What is the rms voltage across the series
combination of the capacitor and the inductor?
Picture the Problem We can find C using VC = I rms X C and I rms from the
potential difference across the inductor. In the absence of resistance in the circuit,
the measured rms voltage across both the capacitor and inductor is V = VL − VC .
(a) Relate the capacitance C to the
potential difference across the
capacitor:
VC = I rms X C =
Use the potential difference across
the inductor to express the rms
current in the circuit:
I rms =
Substitute for I rms to obtain:
C=
Substitute numerical values and
evaluate C:
C=
VL
V
= L
X L 2πfL
VL
(2πf )2 LVC
50 V
[2π (60 s )] (0.25 H )(75 V )
= 19 μF
(b) Express the measured rms
voltage V across both the capacitor
and the inductor when R = 0:
I rms
I
⇒ C = rms
2πfC
2πfVC
V = VL − VC
−1
2
Alternating-Current Circuits
2803
Substitute numerical values and
V = 50 V − 75 V = 25 V
evaluate V:
69 ••
[SSM] 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
I rms and XL:
VAB = I rms X L
ε
(1)
ε
Express I rms in terms of ε and Z:
I rms =
Evaluate XL and XC to obtain:
X L = 2πfL = 2π (60 s −1 )(137 mH)
Z
=
R2 + (X L − X C )
2
= 51.648 Ω
and
1
1
=
−1
2πfC 2π 60 s (25 μF)
= 106.10 Ω
XC =
Substitute numerical values and
evaluate I rms :
I rms =
(
)
115 V
(50 Ω)2 + (51.648 Ω − 106.10 Ω)2
= 1.5556 A
Substitute numerical values in
equation (1) and evaluate VAB:
(b) Express the potential difference
between points B and C in terms of
I rms and R:
V AB = (1.5556 A )(51.648 Ω ) = 80.344 V
= 80 V
VBC = I rms R = (1.5556 A )(50 Ω )
= 77.780 V = 78 V
2804 Chapter 29
(c) Express the potential difference
between points C and D in terms of
I rms and XC:
(d) The voltage across the inductor
leads the voltage across the resistor
as shown in the phasor diagram to
the right:
VCD = I rms X C = (1.5556 A )(106.10 Ω )
= 165.05 V = 0.17 kV
r
V AC
r
V AB
r
V BC
Use the Pythagorean theorem to find
VAC:
2
2
+ VBC
V AC = V AB
=
(80.0 V )2 + (77.780 V )2
= 111.58 V = 0.11 kV
r
V BC
(e) The voltage across the capacitor
lags the voltage across the resistor as
shown in the phasor diagram to the
right:
r
VCD
Use the Pythagorean theorem to
find VBD:
r
V BD
2
2
VBD = VCD
+ V BC
=
(165.05 V )2 + (77.780 V )2
= 182.46 V = 0.18 kV
70 ••
When an RLC series circuit is connected to a 120-V rms, 60-Hz line,
the rms current in the circuit is 11 A and this current leads the line voltage by 45&ordm;.
(a) Find the average power supplied to the circuit. (b) What is the resistance in the
circuit? (c) If the inductance in the circuit is 50 mH, find the capacitance in the
circuit. (d) Without changing the inductance, by how much should you change the
capacitance to make the power factor equal to 1? (e) Without changing the
capacitance, by how much should you change the inductance to make the power
factor equal to 1?
Picture the Problem We can use Pav = ε rms I rms cos δ to find the power supplied
2
R to find the resistance. In Part (c) we can relate the
to the circuit and Pav = I rms
capacitive reactance to the impedance, inductive reactance, and resistance of the
circuit and solve for the capacitance C. We can use the condition on XL and XC at
Alternating-Current Circuits
2805
resonance to find the capacitance or inductance you would need to add to the
circuit to make the power factor equal to 1.
(a) Express the power supplied to
the circuit in terms of εrms, I rms ,
Pav = ε rms I rms cos δ
and the power factor cosδ:
Substitute numerical values and
evaluate Pav :
(b) Relate the power dissipated in
the circuit to the resistance of the
resistor:
Pav = (120 V )(11A )cos 45&deg; = 933 W
= 0.93 kW
2
Pav = I rms
R ⇒ R=
Pav
2
I rms
Substitute numerical values and
evaluate R:
R=
933 W
= 7.71Ω = 7.7 Ω
(11A )2
(c) Express the capacitance of the
capacitor in terms of its reactance:
C=
1
1
=
ωX C 2πfX C
Relate the capacitive reactance to the
impedance, inductive reactance, and
resistance of the circuit:
Z 2 = R2 + (X L − X C )
Express the impedance of the circuit
2
in terms of the rms emf ε and the
rms current I rms :
Z2 =
Equating these two expressions
yields:
ε2
Solve for X L − X C :
Note that because I leads ε, the
circuit is capacitive and XC &gt; XL.
Hence:
(1)
I
2
rms
ε2
2
I rms
= R 2 + (X L − X C )
XL − XC =
ε2
2
I rms
2
− R2
X L − X C = −( X L − X C )
and
2806 Chapter 29
ε2
XC = X L +
I
− R2
ε2
= 2πfL +
Substitute numerical values and
evaluate XC:
2
rms
I
2
rms
− R2
X C = 2π (60 s −1 )(50 mH )
(120 V )2 − (7.71Ω )2
(11 A )2
+
= 18.8 Ω + 7.7 Ω = 26.6 Ω
Substitute in equation (1) and
evaluate C:
C=
2π (60 s
1
−1
)(26.6 Ω) = 99.9 μF
= 0.10 mF
(d) Let Cpf = 1 represent the
ΔC = C pf = 1 − C = C pf = 1 − 99.9 μF
capacitance required to make
cosδ = 1. The necessary change in
capacitance is given by:
Relate Cpf = 1 to XL:
XL = XC =
Solving for Cpf = 1 yields:
C pf = 1 =
Substitute for Cpf = 1 in the expression
ΔC =
for ΔC to obtain:
Substitute numerical values and
evaluate ΔC:
ΔC =
1
2πfC pf = 1
1
2πfX L
1
− 99.9 μF
2πfX L
(
2π 60 s
1
−1
)(18.8 Ω ) − 99.9 μF
= 41 μF
(e) Let Lpf = 1 represent the inductance
required to make cosδ = 1. The
necessary change in inductance is
given by:
ΔL = Lpf = 1 − L = Lpf = 1 − 50 mH
Alternating-Current Circuits
2807
X L = X C = 2πfLpf = 1
Relate Lpf = 1 to XC:
Solving for Lpf = 1 yields:
Lpf = 1 =
Substitute for Lpf = 1 in the expression
for ΔL to obtain:
Substitute numerical values and
evaluate ΔL:
XC
2πf
ΔL =
XC
− 50 mH
2πf
ΔL =
26.6 Ω
− 50 mH = 20 mH
2π 60 s −1
(
)
71 ••
Plot the circuit impedance versus the angular frequency for each of the
following circuits. (a) A driven series LR circuit, (b) a driven series RC circuit,
and (c) a driven series RLC circuit.
Picture the Problem The impedance for the three circuits as functions of the
angular frequency is shown in the three figures below. Also shown in each figure
(dashed line) is the asymptotic approach for large angular frequencies.
(a)
(b)
(c)
72 ••
In a driven series RLC circuit, the ideal generator has a peak emf equal
to 200 V, the resistance is 60.0 Ω and the capacitance is 8.00 μF. The inductance
can be varied from 8.00 mH to 40.0 mH by the insertion of an iron core in the
solenoid. The angular frequency of the generator is 2500 rad/s. If the capacitor
voltage is not to exceed 150 V, find (a) the peak current and (b) the range of
inductances that is safe to use.
Picture the Problem We can find the maximum current in the circuit from the
maximum voltage across the capacitor and the reactance of the capacitor. To find
the range of inductance that is safe to use we can express Z2 for the circuit in
2
2
terms of ε peak
and I peak
and solve the resulting quadratic equation for L.
(a) Express the peak current in terms
of the maximum potential difference
across the capacitor and its
I peak =
VC , peak
XC
= ωCVC , peak
2808 Chapter 29
reactance:
Substitute numerical values and
evaluate I peak :
(b) Relate the maximum current in
the circuit to the emf of the source
and the impedance of the circuit:
Express Z2 in terms of R, XL, and
XC:
Substitute to obtain:
I peak = (2500 rad/s )(8.00 μF)(150 V )
= 3.00 A
I peak =
Z
⇒Z2 =
2
ε peak
2
I peak
Z 2 = R2 + (X L − X C )
2
2
ε peak
I
Evaluating XC yields:
ε peak
2
peak
= R 2 + (X L − X C )
2
1
1
=
ωC (2500 rad/s )(8.00 μF)
= 50.0 Ω
XC =
Substitute numerical values to obtain:
(200 V )2 = (60.0 Ω )2 + ((2500 rad/s)L − 50.0 Ω )2
(3.00 A )2
Solving for L yields:
Denoting the solutions as L+ and L−,
find the values for the inductance:
The ranges for L are:
L=
50.0 Ω &plusmn; 844 Ω 2
2500 s −1
L+ = 31.6 mH and L− = 8.38 mH
8.00 mH &lt; L &lt; 8.38 mH
and
31.6 mH &lt; L &lt; 40.0 mH
73 ••
A certain electrical device draws an rms current of 10 A at an average
power of 720 W when connected to a 120-V rms, 60-Hz power line.
(a) What is the impedance of the device? (b) What series combination of
resistance and reactance would have the same impedance as this device? (c) If the
current leads the emf, is the reactance inductive or capacitive?
Alternating-Current Circuits
2809
Picture the Problem We can find the impedance of the circuit from the applied
2
R to find
emf and the current drawn by the device. In Part (b) we can use Pav = I rms
R and the definition of the impedance of a series RLC circuit to find X = XL − XC.
(a) Express the impedance of the
device in terms of the current it
draws and the emf provided by
the power line:
Z=
Substitute numerical values and
evaluate Z:
Z=
(b) Use the relationship between
the average power supplied to the
device and the rms current it
draws to express R:
ε rms
I rms
120 V
= 12 Ω
10 A
2
Pav = I rms
R⇒R =
Pav
2
I rms
720 W
= 7.20 Ω = 7.2 Ω
(10 A )2
Substitute numerical values and
evaluate R:
R=
The impedance of a series RLC
circuit is given by:
Z = R2 + (X L − X C )
2
or
2
Z 2 = R2 + (X L − X C )
Solving for X L − X C yields:
X = X L − X C = Z 2 − R2
Substitute numerical values and
evaluate X:
X=
(12 Ω )2 − (7.20 Ω )2
= 10 Ω
(c) If the current leads the emf, the reactance is capacitive.
A method for measuring inductance is to connect the inductor in series
74 ••
with a known capacitance, a known resistance, an ac ammeter, and a variablefrequency signal generator. The frequency of the signal generator is varied and
the emf is kept constant until the current is maximum. (a) If the capacitance is
10 μF, the peak emf is 10 V, the resistance is 100 Ω, and the rms current in the
circuit is maximum when the driving frequency is 5000 rad/s, what is the value of
the inductance? (b) What is the maximum rms current?
Picture the Problem (a) We can use the fact that when the current is a maximum,
2810 Chapter 29
XL = XC, to find the inductance of the circuit. In Part (b), we can find Irms, max from
εpeak and the impedance of the circuit at resonance.
(a) Relate XL and XC at resonance:
X L = X C or ω0 L =
Solve for L to obtain:
L=
Substitute numerical values and
evaluate L:
L=
(b) Noting that, at resonance,
X = 0, express Irms, max in terms of the
applied emf and the impedance of
the circuit at resonance:
I rms, max =
75
••
1
ω0C
1
ω02C
1
(5000 s ) (10 μF)
−1 2
=
= 4.0 mH
ε rms = ε max
2Z
Z
10 V
2 (100 Ω )
= 71 mA
A resistor and a capacitor are connected in parallel across an ac
generator (Figure 29-43) that has an emf given by ε = εpeak cos ωt. (a) Show that
the current in the resistor is given by IR = (εpeak/R) cos ωt. (b) Show that the
current in the capacitor branch is given by IC = (εpeak/XC) cos(ωt + 90&ordm;). (c) Show
that the total current is given by I = Ipeak cos(ωt + δ), where tan δ = R/XC and
Ipeak = εpeak/Z.
Picture the Problem Because the resistor and the capacitor are connected in
parallel, the voltage drops across them are equal. Also, the total current is the sum
of the current through the resistor and the current through the capacitor. Because
these two currents are not in phase, we use phasors to calculate their sum. The
amplitudes of the applied voltage and the currents are equal to the magnitude of the
r
r
r
r
phasors. That is ε = ε peak , I = I peak , I R = I R , peak , and I C = I C , peak .
(a) The ac source applies a voltage
given by ε = ε peak cos ωt . Thus, the
voltage drop across both the load
resistor and the capacitor is:
ε peak cos ωt = I R R
The current in the resistor is in phase
with the applied voltage:
I R = I R , peak cos ωt
Alternating-Current Circuits
Because I R , peak =
ε peak
:
R
(b) The current in the capacitor leads
the applied voltage by 90&deg;:
Because I C , peak =
ε peak
XC
(c) The net current I is the sum of the
currents through the parallel branches:
Draw the phasor diagram for the
circuit. The projections of the phasors
onto the horizontal axis are the
instantaneous values. The current in
the resistor is in phase with the applied
voltage, and the current in the capacitor
leads the applied voltage by 90&deg;. The
net current phasor is the sum of the
r r r
branch current phasors I = I C + I R .
cos ωt
I C = I C, peak cos(ωt + 90&deg;)
ε peak
XC
cos(ωt + 90&deg;)
I = I R + IC
r
I
r
ε
r
IC
max
r
IR
δ
)
The peak current through the parallel
combination is equal to ε peak Z , where
Z is the impedance of the combination:
From the phasor diagram we have:
R
IC =
:
(
ε peak
IR =
2811
ωt
I = I peak cos(ωt − δ ),
where I peak =
ε peak
Z
2
I peak
= I R2 , peak + I C2 , peak
⎛ε ⎞ ⎛ε ⎞
= ⎜⎜ peak ⎟⎟ + ⎜⎜ peak ⎟⎟
⎝ R ⎠ ⎝ XC ⎠
2
2
2
⎛ 1
1 ⎞ ε peak
= ε ⎜⎜ 2 + 2 ⎟⎟ = 2
XC ⎠ Z
⎝R
1
1
1
where 2 = 2 + 2
Z
R
XC
2
peak
Solving for Ipeak yields:
From the phasor diagram:
I peak =
ε peak
Z
where Z −2 = R −2 + X C−2
I = I peak cos(ωt + δ )
where
2812 Chapter 29
ε peak
IC
R
X
= C =
I R ε peak
XC
R
76 ••• Figure 29-44 shows a plot of average power Pav versus generator
frequency ω for a series RLC circuit driven by an ac generator. The average
power Pav is given by Equation 29-56. The full width at half-maximum, Δω, is the
width of the resonance curve between the two points, where Pav is one-half its
maximum value. Show that for a sharply peaked resonance, Δω ≈ R/L and that
Q ≈ ω0/Δω in this case (Equation 29-58). Hint: The half-power points occur when
the denominator of Equation 29-56 is equal to twice the value it has at resonance;
that is, when L2 ω 2 − ω02 + ω 2 R 2 ≈ 2ω02 R 2 . Let ω1 and ω2 be the solutions of this
tan δ =
(
)
equation. Then, show that Δω = ω2 – ω1 ≈ R/L.
Picture the Problem We can use the condition determining the half-power points
to obtain a quadratic equation that we can solve for the frequencies corresponding
to the half-power points. Letting ω1 be the half-power frequency that is less than
ω0 and ω2 be the half-power frequency that is greater than ω0 will lead us to the
result that Δω = ω2 −ω1 ≈ R/L. We can then use the definition of Q to complete
the proof that Q ≈ ω0 /Δω.
Equation 29-56 is:
The half-power points occur when
the denominator of Equation 29-56 is
twice the value near resonance; that
is, when:
For a sharply peaked resonance,
ω + ω0 ≈ 2ω0 . Hence:
Let ω1 be a solution to this equation.
Noting that, for a sharply peaked
resonance, ω1 ≈ ω0 , it follows that:
Pav =
(
(
2
2
Vapp,
rms Rω
L2 ω 2 − ω 02
)
2
+ ω 2R2
)
2
L2 ω 2 − ω02 + ω 2 R 2 ≈ 2ω02 R 2
or
2
L2 [(ω − ω0 )(ω + ω0 )] + ω 2 R 2 ≈ 2ω02 R 2
L2 [(ω − ω0 )(2ω0 )] + ω 2 R 2 ≈ 2ω02 R 2
2
or
2
4ω02 L2 (ω − ω0 ) + ω 2 R 2 ≈ 2ω02 R 2
4ω02 L2 (ω1 − ω0 ) + ω02 R 2 ≈ 2ω02 R 2
2
or, simplifying,
2
(ω1 − ω0 )2 ≈ R 2
4L
Alternating-Current Circuits
Solving for ω1 yields:
R
2L
where we’ve used the minus sign
because ω1 &lt; ω0.
ω1 ≈ ω0 −
2813
2814 Chapter 29
Similarly for ω2:
R
2L
where we’ve used the plus sign because
ω2 &gt; ω0.
ω 2 ≈ ω0 +
Evaluating Δω = ω2 − ω1 yields:
Δω ≈ ω 0 +
R ⎛
R ⎞
R
− ⎜ ω0 −
⎟=
2L ⎝
2L ⎠
L
R ω0
=
L Q
From the definition of Q:
Substitute in the expression for Δω to
obtain:
Δω ≈
ω0
Q
⇒Q ≈
ω0
Δω
2
77
•••
dQ 1
dQ
+ Q =0
2 +R
dt
dt C
cos ω ' t , where τ = 2 L R ,
Show by direct substitution that L
(Equation 29-43b) is satisfied by Q = Q0e− t τ
ω ' = 1 (LC ) − 1 τ 2 , and Q0 is the charge on the capacitor at t = 0.
Picture the Problem We’ll differentiate Q = Q0e− t τ cos ω ' t twice and substitute
this function and both its derivatives in the differential equation of the circuit.
Rewriting the resulting equation in the form Acosω′t + Bsinω′t = 0 will reveal
that B vanishes. Requiring that Acosω′t = 0 hold for all values of t will lead to the
result that ω ' = 1 (LC ) − 1 τ 2 .
Equation 29-43b is:
L
d 2Q
dQ 1
+R
+ Q=0
2
dt C
dt
Q = Q0 e −t τ cos ω ' t
Assume a solution of the form:
Differentiate Q(t) twice to obtain:
[
]
d −t τ
dQ
= Q0
e cos ω ' t
dt
dt
d
d
⎛
⎞
= Q0 ⎜ e −t τ cos ω ' t + cos ω ' t e −t τ ⎟
dt
dt
⎝
⎠
1
⎛
⎞
= Q0e −t τ ⎜ − ω ' sin ω ' t − cos ω ' t ⎟
τ
⎝
⎠
Alternating-Current Circuits
2815
and
d 2Q
d ⎡
1
⎛
⎞⎤
= Q0 ⎢e − t τ ⎜ − ω ' sin ω ' t − cos ω ' t ⎟⎥
2
dt
dt ⎣
τ
⎝
⎠⎦
⎡⎛ 1
2ω'
⎤
⎞
sin ω't ⎥
= Q0e − t τ ⎢⎜ 2 − ω' 2 ⎟ cos ω't +
τ
⎦
⎠
⎣⎝ τ
Substitute these derivatives in the differential equation and simplify to obtain:
⎡⎛ 1
2ω'
1
⎞
⎤
⎛
⎞
LQ0e −t τ ⎢⎜ 2 − ω' 2 ⎟ cos ω't +
sin ω't ⎥ + RQ0e −t τ ⎜ − ω ' sin ω ' t − cos ω ' t ⎟
τ
τ
⎠
⎦
⎝
⎠
⎣⎝ τ
1
+ Q0e −t τ cos ω ' t = 0
C
Because Q0 and e −t τ are never zero, divide them out of the equation and simplify
to obtain:
2 Lω'
R
1
⎛1
⎞
L⎜ 2 − ω' 2 ⎟ cos ω't +
sin ω't − ω ' R sin ω ' t − cos ω ' t + cos ω ' t = 0
τ
τ
C
⎝τ
⎠
Rewriting this equation in the form Acosω′t + Bsinω′t = 0 yields:
(Rω' − Rω' )sin ω't + ⎡⎢ L⎛⎜ 12 − ω' 2 ⎞⎟ + 1 − R ⎤⎥ cos ω't = 0
⎣ ⎝τ
⎠ C
τ⎦
or
⎡ ⎛1
1 R⎤
2⎞
⎢ L⎜ τ 2 − ω' ⎟ + C − τ ⎥ cos ω't = 0
⎠
⎣ ⎝
⎦
If this equation is to hold for all
values of t, its coefficient must
vanish:
Solving for ω′ yields:
⎛1
⎞ 1 R
L⎜ 2 − ω' 2 ⎟ + − = 0
⎝τ
⎠ C τ
ω' =
1
⎛ 1 ⎞
−⎜
⎟
LC
⎝ 2L ⎠
2
,
the condition that must be satisfied if
Q = Q0e−t τ cos ω't is the solution to
Equation 29-43b.
78 ••• One method for measuring the magnetic susceptibility of a sample
uses an LC circuit consisting of an air-core solenoid and a capacitor. The resonant
frequency of the circuit without the sample is determined and then measured
2816 Chapter 29
again with the sample inserted in the solenoid. Suppose you have a solenoid that
is 4.00 cm long, 3.00 mm in diameter, and has 400 turns of fine wire. You have a
sample that is inserted in the solenoid and completely fills the air space. Neglect
end effects. (a) Calculate the inductance of your empty solenoid. (b) What value
for the capacitance of the capacitor should you choose that the resonance
frequency of the circuit without a sample is exactly 6.0000 MHz? (c) When a
sample is inserted in the solenoid, you determine that the resonance frequency
drops to 5.9989 MHz. Use your data to determine the sample’s susceptibility.
Picture the Problem We can use L = μ 0 n 2 Al to determine the inductance of the
empty solenoid and the resonance condition to find the capacitance of the samplefree circuit when the resonance frequency of the circuit is 6.0000 MHz. By
expressing L as a function of f0 and then evaluating df0/dL and approximating the
derivative with Δf0/ΔL , we can evaluate χ from its definition.
(a) Express the inductance of an aircore solenoid:
L = μ 0 n 2 Al
Substitute numerical values and evaluate L:
2
⎛ 400 ⎞ π
⎟⎟
(3.00 cm )2 (4.00 cm ) = 3.553 mH = 3.55 mH
L = 4π &times; 10 N/A ⎜⎜
⎝ 4.00 cm ⎠ 4
(
−7
2
)
1
2πf 0C
(b) Express the condition for
resonance in the LC circuit:
X L = X C ⇒ 2πf 0 L =
Solving for C yields:
C=
1
4π f 02 L
C=
1
2
4π (6.0000 MHz ) (3.553 mH )
Substitute numerical values and
evaluate C:
(1)
2
2
= 1.9803 &times;10 −13 F = 0.198 pF
(c) Express the sample’s
susceptibility in terms of L and
ΔL:
χ=
Solve equation (1) for f0:
f0 =
ΔL
L
(2)
1
2π LC
Alternating-Current Circuits
Differentiate f0 with respect to L:
df 0
1
d −1 2
1
L =−
L−3 2
=
dL 2π C dL
4π C
f
1
=−
=− 0
2L
4πL LC
Approximate df0/dL by Δf0/ΔL:
Δf
ΔL
Δf 0
f
= − 0 or 0 = −
f0
2L
ΔL
2L
Substitute in equation (2) to obtain:
χ = −2
Substitute numerical values and
evaluate χ:
χ = −2⎜⎜
2817
Δf 0
f0
⎛ 5.9989 MHz − 6.0000 MHz ⎞
⎟⎟
6.0000
MHz
0
⎝
⎠
= 3.7 &times; 10− 4
The Transformer
79 •
[SSM] 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 V2 N1 = V1 N 2 and N1 I1 = N 2 I 2 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:
V2, rms N1 = V1, rms N 2
Solve for and evaluate the ratio
N2/N1:
1
N 2 V2, rms 24 V
=
=
=
5
N1 V1, rms 120 V
(b) Relate the current in the
primary to the current in the
secondary and to the turns ratio:
I1, rms =
N2
I 2, rms
N1
(1)
2818 Chapter 29
Express the current in the primary
winding in terms of the voltage
across it and its impedance:
Substitute for I2, rms to obtain:
Substitute numerical values and
evaluate I1, rms:
I 2 , rms =
V2 , rms
I1, rms =
N 2 V2, rms
N1 Z 2
Z2
⎛ 5 ⎞⎛ 120 V ⎞
⎟⎟ = 50 A
I1 = ⎜ ⎟⎜⎜
⎝ 1 ⎠⎝ 12 Ω ⎠
80 •
A transformer has 400 turns in the primary and 8 turns in the
secondary. (a) Is this a step-up or a step-down transformer? (b) If the primary is
connected to a 120 V rms voltage source, what is the open-circuit rms voltage
across the secondary? (c) If the primary rms current is 0.100 A, what is the
secondary rms current, assuming negligible magnetization current and no power
loss?
Picture the Problem Let the subscript 1 denote the primary and the subscript 2
the secondary. We can decide whether the transformer is a step-up or step-down
transformer by examining the ratio of the number of turns in the secondary to the
number of terms in the primary. We can relate the open-circuit rms voltage in the
secondary to the primary rms voltage and the turns ratio.
(a) Because there are fewer turns in the secondary than in the primary it is a stepdown transformer.
(b) Relate the open-circuit rms
voltage V2, rms in the secondary to the
rms voltage V1, rms in the primary:
V2, rms =
N2
V1, rms
N1
Substitute numerical values and
evaluate V2, rms :
V2, rms =
8
(120 V ) = 2.40 V
400
(c) Because there are no power
losses:
V1, rms I1, rms = V2, rms I 2, rms
and
I 2, rms =
Substitute numerical values and
evaluate I2, rms:
I 2, rms =
V1, rms
V2, rms
I1, rms
120 V
(0.100 A ) = 5.00 A
2.40 V
Alternating-Current Circuits
2819
81 •
The primary of a step-down transformer has 250 turns and is
connected to a 120-V rms line. The secondary is to supply 20 A rms at 9.0 V rms.
Find (a) the rms current in the primary and (b) the number of turns in the
secondary, assuming 100 percent efficiency.
Picture the Problem Let the subscript 1 denote the primary and the subscript 2
the secondary. We can use I1, rmsV1, rms = I 2, rmsV2, rms to find the current in the primary
and V2, rms N1 = V1, rms N 2 to find the number of turns in the secondary.
(a) Because we have 100 percent
efficiency:
I1, rmsV1, rms = I 2, rmsV2, rms
and
I1, rms = I 2, rms
V2, rms
V1, rms
9.0 V
= 1.5 A
120 V
Substitute numerical values and
evaluate I1, rms:
I1, rms = (20 A )
(b) Relate the number of primary
and secondary turns to the
primary and secondary voltages:
V2, rms N1 = V1, rms N 2 ⇒ N 2 =
Substitute numerical values and
evaluate N2/N1:
N2 =
V2, rms
V1, rms
N1
9.0 V
(250) ≈ 19
120 V
82 ••
An audio oscillator (ac source) that has an internal resistance of
2000 Ω and an open-circuit rms output voltage of 12.0 V is to be used to drive a
loudspeaker coil that has a resistance of 8.00 Ω. (a) What should be the ratio of
primary to secondary turns of a transformer, so that maximum average power is
transferred to the speaker? (b) Suppose a second identical speaker is connected in
parallel with the first speaker. How much average power is then supplied to the
two speakers combined?
Picture the Problem Note: In a simple circuit maximum power transfer from
source to load requires that the load resistance equals the internal resistance of the
source. We can use Ohm’s law and the relationship between the primary and
secondary currents and the primary and secondary voltages and the turns ratio of
the transformer to derive an expression for the turns ratio as a function of the
effective resistance of the circuit and the resistance of the speaker(s).
(a) Express the effective loudspeaker
resistance at the primary of the
transformer:
Reff =
V1, rms
I1, rms
2820 Chapter 29
Relate V1, rms to V2, rms, N1, and N2:
V1, rms = V2, rms
N1
N2
Express I1, rms in terms of I2, rms, N1,
and N2:
I1, rms = I 2, rms
N2
N1
Substitute for V1, rms and I1, rms and
simplify to obtain:
V2, rms
Reff =
I 2, rms
Solve for N1/N2:
N1
=
N2
N1
2
N 2 ⎛⎜ V2, rms ⎞⎟ ⎛ N1 ⎞
⎜
⎟
=
N 2 ⎜⎝ I 2, rms ⎟⎠ ⎜⎝ N 2 ⎟⎠
N1
I 2, rms Reff
V2, rms
=
Reff
R2
(1)
Evaluate N1/N2 for Reff = Rcoil:
N1
2000 Ω
=
= 15.811 = 15.8
8.00 Ω
N2
(b) Express the power delivered to
the two speakers connected in
parallel:
Psp = I12, rms Reff
Find the equivalent resistance Rsp of
1
1
1
⇒ Rsp = 4.00 Ω
=
+
Rsp 8.00 Ω 8.00 Ω
the two 8.00-Ω speakers in parallel:
Solve equation (1) for Reff to obtain:
Reff
⎛N ⎞
= R2 ⎜⎜ 1 ⎟⎟
⎝ N2 ⎠
(2)
2
Substitute numerical values and
evaluate Reff :
Reff = (4.00 Ω )(15.811) = 1000 Ω
Find the current supplied by the
source:
I1, rms =
2
12.0 V
Vrms
=
Rtot 2000 Ω + 1000 Ω
= 4.00 mA
Substitute numerical values in
equation (2) and evaluate the power
delivered to the parallel speakers:
Psp = (4.00 mA ) (1000 Ω ) = 16.0 mW
2
Alternating-Current Circuits
2821
General Problems
83 •
The distribution circuit of a residential power line is operated at
2000 V rms. This voltage must be reduced to 240 V rms for use within residences.
If the secondary side of the transformer has 400 turns, how many turns are in the
primary?
Picture the Problem We can relate the input and output voltages to the number
of turns in the primary and secondary using V2, rms N1 = V1, rms N 2 .
Relate the output voltages V2, rms to
the input voltage V1, rms and the
number of turns in the primary N1
and secondary N2:
V2, rms =
V
N2
V1, rms ⇒ N1 = N 2 1, rms
V2, rms
N1
⎛ 2000 V ⎞
⎟⎟ = 3.33 &times; 103
N1 = (400)⎜⎜
⎝ 240 V ⎠
Substitute numerical values and
evaluate N1:
84 ••
A resistor that has a resistance R carries a current given by
(5.0 A) sin 2πft + (7.0 A) sin 4πft, where f = 60 Hz. (a) What is the rms current in
the resistor? (b) If R =12 Ω, what is the average power delivered to the resistor?
(c) What is the rms voltage across the resistor?
(I ) to relate the rms
current to the current carried by the resistor and find (I ) by integrating I2.
Picture the Problem We can use its definition, I rms =
2
av
2
av
(a) Express the rms current in terms
of the I 2 av :
I rms =
( )
(I )
2
av
Evaluate I2:
I 2 = [(5.0 A ) sin 2πft + (7.0 A ) sin 4πft ]
(
)
(
2
)
(
)
= 25 A 2 sin 2 2πft + 70 A 2 sin 2πft sin 4πft + 49 A 2 sin 2 4πft
( )
Find I 2
av
by integrating I2 from t = 0 to t = T = 2π/ω and dividing by T:
(I )
2
av
=
ω
2π
2π ω
∫ {(25 A )sin
2
0
(
)
2
(
)
2πft + 70 A 2 sin 2πft sin 4πft
}
+ 49 A 2 sin 2 4πft dt
2822 Chapter 29
Use the trigonometric identity
sin 2 x = 12 (1 − cos 2 x ) to simplify and
(I )
2
av
= 12.5 A 2 + 0 + 24.5 A 2 = 37.0 A 2
evaluate the 1st and 3rd integrals and
recognize that the middle term is of
the form sinxsin2x to obtain:
( )
Substitute for I 2
av
and evaluate Irms:
I rms = 37.0 A 2 = 6.1 A
(b) Relate the power dissipated in
the resistor to its resistance and the
rms current in it:
2
P = I rms
R
Substitute numerical values and
evaluate P:
P = (6.08 A ) (12 Ω ) = 0.44 kW
(c) Express the rms voltage across
the resistor in terms of R and I rms :
Vrms = I rms R
Substitute numerical values and
evaluate Vrms :
Vrms = (6.08 A )(12 Ω ) = 73 V
2
85 ••
[SSM] Figure 29-45 shows the voltage versus time for a squarewave 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, V 2 av and then use the definition
( )
of the rms voltage.
(a) From the definition of Vrms we
have:
Noting that − V02 = V02 , evaluate
Vrms :
Vrms =
(V )
2
0 av
Vrms = V02 = V0 = 12 V
Alternating-Current Circuits
(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
1
2 ΔT :
V = V0
Express the square of the voltage
during this half cycle:
V 2 = V02
( )
Calculate V 2
av
from t = 0 to t =
by integrating V2
1
2
ΔT and dividing
(V )
2
av
V2
= 0
ΔT
1
ΔT
2
by ΔT:
Substitute to obtain:
Vrms =
1
2
∫ dt =
0
V02 =
V0
2
2823
V02 12 ΔT 1 2
[t ] 0 = 2 V0
ΔT
=
12 V
2
= 8.5 V
86 ••
What are the average values and rms values of current for the two
current waveforms shown in Figure 29-46?
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 current I av , and the average of the current squared,
(I )
2
av
From the definition of I av and
I rms we have:
Waveform (a) Express the
current during the first half cycle
of time interval ΔT:
Evaluate I av, a :
I av =
1
ΔT
∫
ΔT
0
Idt and I rms =
(I )
2
av
4A
t
ΔT
where I is in A when t and T are in
seconds.
Ia =
I av, a =
=
1
ΔT
ΔT
4.0 A
4.0 A
∫0 ΔT tdt = (ΔT )2
4.0 A
(ΔT )2
ΔT
⎡t ⎤
⎢ ⎥ = 2.0 A
⎣ 2 ⎦0
2
ΔT
∫ tdt
0
2824 Chapter 29
Express the square of the current
during this half cycle:
Noting that the average value of the
squared current is the same for each
time interval ΔT, calculate I a2 av by
( )
integrating I from t = 0 to t = ΔT
2
a
and dividing by ΔT:
Substitute in the expression for
I rms, a to obtain:
Waveform (b) Noting that the
current during the second half of
each cycle is zero, express the
current during the first half cycle of
the time interval 12 ΔT :
2
(
4.0 A ) 2
I =
t
(ΔT )2
2
ΔT
(I a2 )av = Δ1T ∫ (4.0 A2) t 2 dt
0 (ΔT )
ΔT
2
(
4.0 A ) ⎡ t 3 ⎤
16 2
=
⎥ = A
3 ⎢
(ΔT ) ⎣ 3 ⎦ 0 3
2
a
I rms, a =
16 2
A = 2 .3 A
3
I b = 4.0 A
Evaluate I av, b :
1
I av, b
ΔT
4.0 A 2
4.0 A 12 ΔT
dt =
[t ] 0
=
∫
ΔT 0
ΔT
= 2.0 A
Express the square of the current
during this half cycle:
( )
Calculate I b2
av
from t = 0 to t =
by integrating I b2
1
2
ΔT and dividing
by ΔT:
I b2 = (4.0 A )
2
(I )
2
b av
2
(
4 .0 A )
=
ΔT
Substitute in the expression for I rms, b
∫ dt
0
(4.0 A ) [t ] ΔT
0
2
=
1
Δ
2
ΔT
1
2
= 8 .0 A 2
I rms, b = 8.0 A 2 = 2.8 A
to obtain:
87
••
In the circuit shown in Figure 29-47, ε1 = (20 V) cos 2πft, where
f = 180 Hz; ε2 = 18 V, and R = 36 Ω. Find the maximum, minimum, average, and
rms values of the current in the resistor.
Picture the Problem We can apply Kirchhoff’s loop rule to express the current in
the circuit in terms of the emfs of the sources and the resistance of the resistor.
Alternating-Current Circuits
2825
We can then find Imax and I min by considering the conditions under which the
time-dependent factor in I will be a maximum or a minimum. Finally, we can use
I rms =
(I )
2
av
to derive an expression for Irms that we can use to determine its
value.
ε 1, peak cos ωt + ε 2 − IR = 0
Apply Kirchhoff’s loop rule to obtain:
Solving for I yields:
I=
ε 1, peak
R
cos ωt +
ε2
R
or
I = A1 cos ωt + A2
ε 1, peak
where A1 =
R
and A2 =
ε2
R
⎛ 20 V ⎞
18 V
⎟⎟ cos 2π 180 s −1 t +
I = ⎜⎜
36 Ω
⎝ 36 Ω ⎠
= (0.556 A ) cos 1131s −1 t + 0.50 A
( (
Substitute numerical values to obtain:
(
))
)
The current is a maximum
when cos 1131s −1 t = 1 . Hence :
I max = 0.50 A + 0.556 A = 1.06 A
Evaluate I min :
I min = 0.50 A − 0.556 A = − 0.06 A
Because the average value of
cosωt = 0:
I av = 0.50 A
The rms current is the square root of
the average of the squared current:
I rms =
(
[I ]
2
av
)
[I ]
2
(1)
av
is given by:
[I ] = [( A cos ωt + A ) ]
= [A cos ωt + 2 A A cos ωt + A ]
= [A cos ωt ] + [2 A A cos ωt ] + [A ]
= A [cos ωt ] + 2 A A [cos ωt ] + A
[I ] = A + A
ωt ] = and
2
2
av
1
2
1
2
2
1
2
2
1
[
Because cos 2
[cos ωt ]av = 0 :
av
1
2
2
av
1
2
2 av
2
1
av
2
2
av
1
av
2
2
av
2
2 av
av
1
2
2
1
2
2
2
2
2826 Chapter 29
Substituting in equation (1) yields:
Substitute for A1 and A2 to obtain:
Substitute numerical values and
evaluate Irms:
I rms =
1
2
A12 + A22
I rms =
1
2
⎛ ε1 ⎞ ⎛ ε 2 ⎞
⎜ ⎟ +⎜ ⎟
⎝R⎠ ⎝ R⎠
I rms =
1
2
⎛ 20 V ⎞ ⎛ 18 V ⎞
⎜
⎟ +⎜
⎟
⎝ 36 Ω ⎠ ⎝ 36 Ω ⎠
2
2
2
2
= 0.64 A
88
••
Repeat Problem 87 if the resistor is replaced by a 2.0-μF capacitor.
Picture the Problem We can apply Kirchhoff’s loop rule to obtain an expression
for charge on the capacitor as a function of time. Differentiating this expression
with respect to time will give us the current in the circuit. We can then find Imax
and Imin by considering the conditions under which the time-dependent factor in I
(I )
will be a maximum or a minimum. Finally, we can use I rms =
2
av
expression for Irms that we can use to determine its value.
Apply Kirchhoff’s loop rule to
obtain:
Solving for q(t) yields:
ε1 peak cos ωt + ε 2 − q(t ) = 0
C
q(t ) = C (ε1 peak cos ωt + ε 2 )
= A1 cos ωt + A2
where
A1 = Cε 1 peak and A2 = Cε 2
Differentiate this expression with
respect to t to obtain the current
as a function of time:
dq d
= ( A1 cos ωt + A2 )
dt dt
= −ωA1 sin ωt
I=
Substituting numerical values yields:
(
)
I = −2π (180 Hz )(2.0 μF)sin (2π (180 Hz )t ) = (− 2.26 mA )sin 1131 s −1 t
The current is a minimum when
sin 1131s −1 t = 1 . Hence:
I min = − 2.3 mA
The current is a maximum when
sin 1131s −1 t = −1 . Hence:
I max = 2.3 mA
(
(
)
)
to derive an
Alternating-Current Circuits
2827
2828 Chapter 29
Because the dc source sees the
capacitor as an open circuit and the
average value of the sine function
over a period is zero:
I av = 0
The rms current is the square root of
the average of the squared current:
I rms =
[I ]
2
av
[I ]
2
(1)
av
is given by:
[I ] = [( A cos ωt + A ) ]
= [A cos ωt + 2 A A cos ωt + A ]
= [A cos ωt ] + [2 A A cos ωt ] + [A ]
= A [cos ωt ] + 2 A A [cos ωt ] + A
[I ] = A + A
ωt ] = and
2
2
av
1
2
1
2
2
1
2
2
1
[
Because cos 2
[cos ωt ]av = 0 :
av
2
av
1
2
2 av
2
1
av
2
2
2 av
av
2
1
2
av
1
2
2
av
2
2
1
2
av
2
1
2
2
Substituting in equation (1) yields:
I rms =
1
2
A12 + A22
Substitute for A1 and A2 to obtain:
I rms =
1
2
(Cε1 )2 + (Cε 2 )2
=C
Substitute numerical values and
evaluate Irms:
1
2
(ε1 )2 + (ε 2 )2
I rms = (2.0 μF)
1
2
(20 V )2 + (18 V )2
= 46 μA
89 ••• [SSM] 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
ε peak cos ωt . (a) Show that the charge on the capacitor obeys the equation
d 2Q Q
L 2 + = ε peak cos ωt . (b) Show by direct substitution that this equation is
dt
C
satisfied by Q = Qpeak cos ωt where Qpeak = −
ε peak
. (c) Show that the current
L(ω 2 − ω02 )
ωε peak
ε peak
can be written as I = I peak cos(ωt − δ ) , where I peak =
and
=
2
2
XL − XC
L ω − ω0
δ = –90&ordm; for ω &lt; ω0 and δ = 90&ordm; for ω &gt; ω0, where ω0 is the resonance frequency.
Alternating-Current Circuits
2829
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 Q = Qpeak cos ωt , differentiate it twice,
and substitute for d2Q/dt2 and Q to show that the assumed solution satisfies the
differential equation provided Qpeak = −
ε peak
L(ω 2 − ω02 )
. 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:
ε − Q − L dI
Substitute for ε and rearrange the
differential equation to obtain:
L
Because I = dQ dt :
C
dt
=0
dI Q
+ = ε max cos ωt
dt C
d 2Q Q
L 2 + = ε max cos ωt
dt
C
(b) Assume that the solution is:
Q = Qpeak cos ωt
Differentiate the assumed
solution twice to obtain:
dQ
= −ωQpeak sin ωt
dt
and
d 2Q
= −ω 2 Qpeak cos ωt
2
dt
d 2Q
dQ
and
in the
Substitute for
dt 2
dt
differential equation to obtain:
− ω 2 LQpeak cos ωt +
Factor cosωt from the left-hand
side of the equation:
Q ⎞
⎛
⎜⎜ − ω 2 LQpeak + peak ⎟⎟ cos ωt
C ⎠
⎝
= ε peak cos ωt
If this equation is to hold for all
values of t it must be true that:
Qpeak
cos ωt
C
= ε peak cos ωt
− ω 2 LQpeak +
Qpeak
C
= ε peak
2830 Chapter 29
Solving for Qpeak yields:
Factor L from the denominator
and substitute for 1/LC to obtain:
Qpeak =
Qpeak =
ε peak
− ω 2L +
ε peak
1 ⎞
⎛
L⎜ − ω 2 +
⎟
LC ⎠
⎝
= −
(c) From (a) and (b) we have:
I=
=
1
C
(
ε peak
)
L ω 2 − ω 02
dQ
= −ωQpeak sin ωt
dt
ωε peak
sin ωt = I peak sin ωt
L (ω 2 − ω 02 )
= I peak cos(ωt − δ )
where
I peak =
=
ωε peak
Lω −ω
2
ε peak
ωL −
1
ωC
2
0
=
=
ε peak
L
ω
ω 2 − ω02
ε peak
XL − XC
If ω &gt; ω0, XL &gt; XC and the current lags the voltage by 90&deg; (δ = 90&deg;).
If ω &lt; ω0, XL &lt; XC and the current leads the voltage by 90&deg;(δ = −90&deg;).
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