Formatted
Chapter 29
Alternating-Current Circuits
Conceptual Problems
3
•
If the frequency in the circuit shown in Figure 29-27 is doubled, the
inductance of the inductor will (a) double, (b) not change, (c) halve,
(d) quadruple.
Determine the Concept The inductance of an inductor is determined by the
details of its construction and is independent of the frequency of the circuit. The
inductive reactance, on the other hand, is frequency dependent. (b) is correct.
7
•
(a) In a circuit consisting of a generator and a capacitor, are there any
time intervals when the capacitor receives energy from the generator? If so,
when? Explain your answer. (b) Are there any time intervals when the capacitor
supplies power to the generator? If so, when? Explain your answer.
Determine the Concept Yes to both questions. (a) While the magnitude of the
charge is accumulating on either plate of the capacitor, the capacitor absorbs
power from the generator. (b) When the magnitude of the charge is on either plate
of the capacitor is decreasing, it supplies power to the generator.
9
•
Suppose you increase the rotation rate of the coil in the generator
shown in the simple ac circuit in Figure 29-29. Then the rms current (a) increases,
(b) does not change, (c) may increase or decrease depending on the magnitude of
the original frequency, (d) may increase or decrease depending on the magnitude
of the resistance, (e) decreases.
Determine the Concept Because the rms current through the resistor is given by
ε peak NBA
ε
I rms = rms =
=
ω , I rms is directly proportional to ω. (a) is correct.
R
2
2
11 •
Consider a circuit consisting solely of an ideal inductor and an ideal
capacitor. How does the maximum energy stored in the capacitor compare to the
maximum value stored in the inductor? (a) They are the same and each equal to
the total energy stored in the circuit. (b) They are the same and each equal to half
of the total energy stored in the circuit. (c) The maximum energy stored in the
capacitor is larger than the maximum energy stored in the inductor. (d) The
maximum energy stored in the inductor is larger than the maximum energy stored
in the capacitor. (e) You cannot compare the maximum energies based on the data
given because the ratio of the maximum energies depends on the actual
capacitance and inductance values.
Deleted: 171
173
Chapter 29
174
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.
True or false:
17
•
(a)
(b)
(c)
(d)
(e)
A transformer is used to change frequency.
A transformer is used to change voltage.
If a transformer steps up the current, it must step down the voltage.
A step-up transformer, steps down the current.
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.
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.
(f)
(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.
Alternating-Current Circuits
175
Alternating Current in Resistors, Inductors, and Capacitors
19 •
A 100-W light bulb is screwed into a standard 120-V-rms socket. Find
(a) the rms current, (b) the peak current, and (c) the peak power.
Picture the Problem We can use 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 =
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
Pav
ε rms
100 W
= 0.8333 A = 0.833 A
120 V
= 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
21 •
What is the reactance of a 1.00-μH inductor at (a) 60 Hz, (b) 600 Hz,
and (c) 6.00 kHz?
Picture the Problem We can use 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 μH ) = 0.38 mΩ
(
)
Deleted: (a) At f = 60 Hz:
Deleted:
¶
(
)
X L = 2π 60 s −1 (1.00
176
Chapter 29
(b) At f = 600 Hz:
(c) At f = 6.00 kHz:
(
)
X L = 2π 600 s −1 (1.00 μH ) = 3.77 mΩ
X L = 2π (6.00 kHz )(1.00 μH ) = 37.7 mΩ
Deleted: (b) At f = 600 Hz:
Deleted:
(
)
X L = 2π 600 s −1 (1.0
¶
Deleted: (c) At f = 6.00 kHz:
25 •
A 20-Hz ac generator that produces a peak emf of 10 V is connected
to a 20-μF capacitor. Find (a) the peak current and (b) the rms current.
Picture the Problem We can use Ipeak = εpeak/XC and XC = 1/ωC to express Ipeak as
a function of εpeak, f, and C. Once we’ve evaluate Ipeak, we can use
Irms = Ipeak/ 2 to find 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
Undriven Circuits Containing Capacitors, Resistors and
Inductors
29 •
(a) What is the period of oscillation of an LC circuit consisting of an
ideal 2.0-mH inductor and a 20-μF capacitor? (b) A circuit that oscillates consists
solely of an 80-μF capacitor and a variable ideal inductor. What inductance is
needed in order to tune this circuit to oscillate at 60 Hz?
Picture the Problem We can use T = 2π/ω and ω = 1
f) to L and C.
LC to relate T (and hence
Deleted:
X L = 2π (6.00 kHz )(1
Alternating-Current Circuits
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=
177
ω
1
LC
(1)
(2.0 mH )(20 μF) =
1.3 ms
1
T2
=
4π 2C 4π 2 f 2C
(
4π 60 s
2
1
) (80 μF)
−1 2
= 88 mH
33 ••• An inductor and a capacitor are connected, as shown in Figure 29-30.
Initially, the switch is open, the left plate of the capacitor has charge Q0. The
switch is then closed. (a) Plot both Q versus t and I versus t on the same graph,
and explain how it can be seen from these two plots that the current leads the
charge by 90º. (b) The expressions for the charge and for the current are given by
Equations 29-38 and 29-39, respectively. Use trigonometry and algebra to show
that the current leads the charge by 90º.
Picture the Problem Let Q represent the instantaneous charge on the capacitor
and apply Kirchhoff’s loop rule to obtain the differential equation for the circuit.
We can then solve this equation to obtain an expression for the charge on the
capacitor as a function of time and, by differentiating this expression with respect
to time, an expression for the current as a function of time. We’ll use a
spreadsheet program to plot the graphs.
Apply Kirchhoff’s loop rule to a
clockwise loop just after the switch
is closed:
Because I = dQ dt :
The solution to this equation is:
Q
dI
+L =0
C
dt
L
d 2Q Q
d 2Q 1
+
=
0
or
+
Q=0
dt 2 LC
dt 2 C
Q(t ) = Q0 cos(ωt − δ )
where ω =
1
LC
178
Chapter 29
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°.
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⎠
⎝
Use this identity to rewrite equation
(1):
π⎞
⎛
I = −ωQ0 sin ωt = ωQ0 cos⎜ ωt + ⎟
2⎠
⎝
Thus, the current leads the charge by
90°.
(1)
Driven RL Circuits
A coil that has a resistance of 80.0 Ω has an impedance of 200 Ω when
35 ••
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=
179
Z 2 − R2
2πf
(200 Ω )2 − (80.0 Ω )2
2π (1.00 kHz )
= 29.2 mH
39 ••
A coil that has a resistance R and an inductance L has a power factor
equal to 0.866 when driven at a frequency of 60 Hz. What is the coil’s power
factor if it is driven at 240 Hz?
Picture the Problem We can use the definition of the power factor to find the
relationship between XL and R when the coil is driven at a frequency of 60 Hz and
then use the definition of XL to relate the inductive reactance at 240 Hz to the
inductive reactance at 60 Hz. We can then use the definition of the power factor to
determine its value at 240 Hz.
Using the definition of the power
factor, relate R and XL:
cos δ =
Square both sides of the equation
to obtain:
cos 2 δ =
R
=
Z
R
R + X L2
2
(1)
R2
R + X L2
2
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
− 1⎟⎟ = 13 R 2
X L2 (60 Hz ) = 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
180
Chapter 29
Dividing the second of these
equations by the first and simplifying
yields:
X L2 ( f' ) 4πf' 2 L2
f' 2
=
=
X L2 ( f ) 4πf 2 L2
f2
or
2
⎛ f' ⎞
X L2 ( f' ) = ⎜⎜ ⎟⎟ X L2 ( f )
⎝ f ⎠
2
Substitute numerical values to
obtain:
⎛ 240 s −1 ⎞ 2
⎟ X L (60 Hz )
X L2 (240 Hz ) = ⎜⎜
−1 ⎟
⎝ 60 s ⎠
⎛ 1 ⎞ 16
= 16⎜ R 2 ⎟ = R 2
⎝3 ⎠ 3
Substitute in equation (1) to obtain:
(cos δ )240 Hz
=
R
R2 +
16 2
R
3
=
3
19
= 0.397
Figure 29-33 shows a load resistor that has a resistance of RL = 20.0 Ω
41 ••
connected to a high-pass filter consisting of an inductor that has inductance
L = 3.20-mH and a resistor that has resistance R = 4.00-Ω. The output of the ideal
ac generator is given by ε = (100 V) cos(2πft). Find the rms currents in all three
branches of the circuit if the driving frequency is (a) 500 Hz and (b) 2000 Hz.
Find the fraction of the total average power supplied by the ac generator that is
delivered to the load resistor if the frequency is (c) 500 Hz and (d) 2000 Hz.
Picture the Problem 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 + V2 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.
Alternating-Current Circuits
181
r
I RL
The phasor diagram for the currents
in the circuit is:
r
I
δ
r
IL
Adding the voltage phasors to the
diagram gives:
r
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)
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
182
Chapter 29
Solving for Z yields:
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 :
Z = R 2 + Z 22 + 2 RZ 2 cos δ
I peak =
(4)
ε peak
(5)
Z
1
I peak
2
I rms =
(6)
⎛
⎞
20.0 Ω
⎟⎟
⎝ 2π (500 Hz )(3.20 mH ) ⎠
δ = tan −1 ⎜⎜
⎛ 20.0 Ω ⎞
⎟⎟ = 63.31°
= tan −1 ⎜⎜
⎝ 10.053 Ω ⎠
Solving equation (2) for Z2 yields:
Substitute numerical values and
evaluate Z2:
1
Z2 =
−2
L
R + X L−2
1
Z2 =
(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° = 11.36 Ω
Substitute numerical values in
equation (5) and evaluate I peak :
I peak =
100 V
= 8.806 A
11.36 Ω
Substitute for I peak in equation
I rms =
1
(8.806 A ) = 6.23 A
2
(6) and evaluate I rms :
The maximum and rms values of
V2 are given by:
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 =
Alternating-Current Circuits
The rms values of I RL ,rms and
I RL ,rms =
I L ,rms are:
V2,rms
RL
=
183
55.929 V
= 2.80 A
20.0 Ω
and
I L ,rms =
V2,rms 55.929 V
=
= 5.56 A
XL
10.053 Ω
X L = 40.2 Ω , δ = 26.4° , 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:
⎛
P
= ⎜1 + R
PRL + PR ⎜⎝ PRL
PRL
−1
2
⎞
⎛
⎞
⎟ = ⎜1 + I rms R ⎟
2
⎜ I R ,rms RL ⎟
⎟
L
⎠
⎝
⎠
−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 ) ⎠
(d) Substitute numerical values for f = 2000 Hz to obtain:
−1
PRL
PRL + PR
f = 2000 Hz
⎛ (3.28 A )2 (4.00 Ω ) ⎞
⎟ = 0.800 = 80.0%
= ⎜⎜1 +
2
⎟
⎝ (2.94 A ) (20.0 Ω ) ⎠
Filters and Rectifiers
A slowly varying voltage signal V(t) is applied to the input of the high47 ••
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.
Comment [EPM1]: DAVID: The
solution is lacking in two areas. First, the
statement that the voltage drop across the
resistor is small compared to the voltage
drop across the capacitor needs to be
justified. Second, the restriction of the
voltage signal to one frequency needs to
be dropped.
Chapter 29
184
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.
Because there is no significant
change in the voltage signal during
one time constant:
dVC
dV
= 0 ⇒ RC C = 0
dt
dt
Substituting for RC
dVC
yields:
dt
Consequently, the potential
difference across the resistor is given
by:
Vpeak cos ωt − VC − RC
Comment [EPM2]: DAVID: The
solution is lacking in two areas. First, the
statement that the voltage drop across the
resistor is small compared to the voltage
drop across the capacitor needs to be
justified. Second, the restriction of the
voltage signal to one frequency needs to
be dropped.
Vin peak cos ωt − VC = 0
and
VC = Vin peak cos ωt
V R = RC
[
dVC
d
= RC Vin peak cos ωt
dt
dt
]
Show that the average power dissipated in the resistor of the high-pass
V in2 peak
.
filter of Problem 44 is given by Pave =
−2
2R ⎡1+ (ω RC ) ⎤
⎣
⎦
49
••
Picture the Problem We can express the instantaneous power dissipated in the
resistor and then use the fact that the average value of the square of the cosine
function over one cycle is ½ to establish the given result.
The instantaneous power P(t)
dissipated in the resistor is:
P (t ) =
2
Vout
R
Alternating-Current Circuits
The output voltage Vout is:
From Problem 44:
Substitute in the expression for P(t)
to obtain:
Vout = VH cos(ωt − δ )
Vin peak
VH =
1 + (ωRC )
−2
VH2
cos 2 (ωt − δ )
R
Vin2 peak
=
cos 2 (ωt − δ )
−2
R 1 + (ωRC )
P (t ) =
[
Because the average value of the
square of the cosine function over
one cycle is ½:
185
Pave =
]
[
Vin2 peak
2 R 1 + (ωRC )
−2
]
51 ••
The circuit shown in Figure 29-36 is an example of a low-pass filter.
(Assume that the output is connected to a load that draws only an insignificant
amount of current.) (a) If the input voltage is given by Vin = Vin peak cos ωt, show
that the output voltage is Vout = VL cos(ωt – δ) where V L = V in peak
1+ (ω RC ) .
2
(b) Discuss the trend of the output voltage in the 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
r
Vapp = 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.
(a) Express Vapp :
r
VR
r
Vapp
ωt
φ δ
r
VC
Vapp = Vin peak cos ωt
where Vin peak = I peak Z
and Z 2 = R 2 + X C2
Vout = VC is given by:
Vout = VC , peak cos φ
= I peak X C cos φ
(1)
186
Chapter 29
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 )
⎛ 1 ⎞ ωC
2
R +⎜
⎟
⎝ ωC ⎠
Vin peak
2
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.
55 ••• The circuit shown in Figure 29-37 is a trap filter. (Assume that the
output is connected to a load that draws only an insignificant amount of current.)
(a) Show that the trap filter acts to reject signals in a band of frequencies centered
at 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
Alternating-Current Circuits
187
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.
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
= I peak Z trap cos(ωt − δ )
=
Using equation (2) to substitute for Z
yields:
Noting that Vout = 0 provided
Z trap = 0, set Z trap = 0 to obtain:
Substituting for XL and XC yields:
(2)
Vout =
Vpeak
Z
Z trap cos(ωt − δ )
Vpeak
2
R 2 + Z trap
Z trap cos(ωt − δ )
Z trap = X L − X C = 0
ωL −
1
= 0 ⇒ω =
ωC
1
LC
188
Chapter 29
(b) Let the bandwidth Δω be:
Δω = ω − ω trap
Let the frequency bandwidth be
defined by the frequency at which
Z trap = R . Then:
ωL −
Because ωtrap =
1
:
LC
For ω ≈ ωtrap:
(3)
1
2
= R ⇒ ω LC − 1= ωRC
ωC
⎛ ω
⎜
⎜ω
⎝ trap
2
⎞
⎟ − 1 = ωRC
⎟
⎠
2
⎛ ω 2 − ω trap
⎜
⎜ ω trap
⎝
⎞
⎟ ≈ ω trap RC
⎟
⎠
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 − ωtrap
≈ 2ωtrap (ω − ωtrap )
Substitute in equation (3) to obtain:
Δω = ω − ωtrap =
2
RCωtrap
2
=
R
2L
Driven RLC Circuits
2
63 ••
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 capacitor, and
2
(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:
(b) and (c) If R = 0, then:
Pav =
2
2
Rε rms
Rε rms
=
=
Z2
R2
2
ε rms
Pav =
2
2
(0)ε rms
Rε rms
=
= 0
Z2
( X L − X C )2
R
Alternating-Current Circuits
189
Remarks: Recall that there is no energy dissipation in an ideal inductor or
capacitor.
65 ••
Find (a) the Q factor and (b) the resonance width (in hertz) for the
circuit in Problem 64. (c) What is the power factor when ω = 8000 rad/s?
Picture the Problem The Q factor of the circuit is given by 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=
(b) Substitute numerical values in
equation (2) and evaluate Δf:
Δf =
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
7.07 ×103 rad/s
= 80 Hz
2π (14.1)
(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 2 + ⎜ ωL −
⎟
ωC ⎠
⎝
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 μF) ⎟⎠
⎝
(
)
(
)
2
= 0.27
190
Chapter 29
69 ••
In the circuit shown in Figure 29-42 the ideal generator produces an
rms voltage of 115 V when operated at 60 Hz. What is the rms voltage between
points (a) A and B, (b) B and C, (c) C and D, (d) A and C, and (e) B and D?
Picture the Problem We can find the rms current in the circuit and then use it to
find the potential differences across each of the circuit elements. We can use
phasor diagrams and our knowledge of the phase shifts between the voltages
across the three circuit elements to find the voltage differences across their
combinations.
(a) Express the potential difference
between points A and B in terms of
I rms and XL:
V AB = 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
=
R + (X L − X C )
2
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 ) + (51.648 Ω − 106.10 Ω )2
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:
(c) Express the potential difference
between points C and D in terms of
I rms and XC:
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
VCD = I rms X C = (1.5556 A )(106.10 Ω )
= 165.05 V = 0.17 kV
Alternating-Current Circuits
(d) The voltage across the inductor
leads the voltage across the resistor
as shown in the phasor diagram to
the right:
191
r
V AC
r
V AB
r
V BC
Use the Pythagorean theorem to find
VAC:
2
2
+ V BC
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
+ V BC
V BD = VCD
=
(165.05 V )2 + (77.780 V )2
= 182.46 V = 0.18 kV
The Transformer
A rms voltage of 24 V is required for a device whose impedance is 12
79 •
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
(1)
192
Chapter 29
Solve for and evaluate the ratio
N2/N1:
N 2 V2, rms 24 V
1
=
=
=
N1 V1, rms 120 V
5
(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
Express the current in the primary
winding in terms of the voltage
across it and its impedance:
I 2, rms =
V2, rms
I1, rms =
N 2 V2, rms
N1 Z 2
Substitute for I2, rms to obtain:
Substitute numerical values and
evaluate I1, rms:
Z2
⎛ 5 ⎞⎛ 120 V ⎞
⎟⎟ = 50 A
I1 = ⎜ ⎟⎜⎜
⎝ 1 ⎠⎝ 12 Ω ⎠
General Problems
85 ••
Figure 29-45 shows the voltage versus time for a square-wave voltage
source. If V0 = 12 V, (a) what is the rms voltage of this source? (b) If this
alternating waveform is rectified by eliminating the negative voltages, so that only
the positive voltages remain, what is the new rms voltage?
Picture the Problem The average of any quantity over a time interval ΔT is the
integral of the quantity over the interval divided by ΔT. We can use this definition
to find both the average of the voltage squared, 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
Comment [DN3]: Needs checking …
the 4e gets 10 A
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 by integrating V2
from t = 0 to t = 12 ΔT and dividing
(V )
2
av
1
ΔT
2
V2
= 0
ΔT
∫ dt =
0
by ΔT:
Substitute to obtain:
Vrms =
1
2
V02 =
V0
2
193
V02 12 ΔT 1 2
[t ] 0 = 2 V0
ΔT
=
12 V
2
= 8.5 V
89 ••• A circuit consists of an ac generator, a capacitor and an ideal
inductor⎯all connected in series. The emf of the generator is given by
ε peak cos ωt . (a) Show that the charge on the capacitor obeys the equation
L
d 2Q Q
+ = ε peak cos ωt . (b) Show by direct substitution that this equation is
dt 2 C
satisfied by Q = Qpeak cos ωt where Qpeak = −
(
ε peak
L ω 2 − ω02
can be written as I = I peak cos(ωt − δ ) , where I peak =
) . (c) Show that the current
ωε peak
Lω −ω
2
2
0
=
ε peak
XL − XC
and
δ = –90º for ω < ω0 and δ = 90º for ω > ω0, where ω0 is the resonance frequency.
Picture the Problem In Part (a) we can apply Kirchhoff’s loop rule to obtain the
2nd order differential equation relating the charge on the capacitor to the time. In
Part (b) we’ll assume a solution of the form 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
C
dt
=0
194
Chapter 29
Substitute for ε and rearrange the
differential equation to obtain:
L
dI Q
+ = ε max cos ωt
dt C
Because I = dQ dt :
L
d 2Q Q
+ = ε max cos ωt
dt 2 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
dt 2
Substitute for
dQ
d 2Q
and
in the
dt
dt 2
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:
Solving for Qpeak yields:
Factor L from the denominator
and substitute for 1/LC to obtain:
Qpeak
cos ωt
C
= ε peak cos ωt
− ω 2 LQpeak +
Qpeak =
Qpeak =
Qpeak
C
= ε peak
ε peak
− ω 2L +
1
C
ε peak
1 ⎞
⎛
L⎜ − ω 2 +
⎟
LC
⎝
⎠
= −
(
ε peak
L ω 2 − ω02
)
Alternating-Current Circuits
(c) From (a) and (b) we have:
I=
=
dQ
= −ωQpeak sin ωt
dt
ωε peak
sin ωt = I peak sin ωt
L(ω 2 − ω02 )
= 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 > ω0, XL > XC and the current lags the voltage by 90° (δ = 90°).
If ω < ω0, XL < XC and the current leads the voltage by 90°(δ = −90°).
195
196
Chapter 29