33 Alternating Current Circuits CHAPTER OUTLINE 33.1 AC Sources 33.2 Resistors in an AC Circuit 33.3 Inductors in an AC Circuit 33.4 Capacitors in an AC Circuit 33.5 The RLC Series Circuit 33.6 Power in an AC Circuit 33.7 Resonance in a Series RLC Circuit 33.8 The Transformer and Power Transmission 33.9 Rectifiers and Filters ANSWERS TO QUESTIONS *Q33.1 (i) Answer (d). ∆ ∆ V V avg = max 2 (ii) Answer (c). The average of the squared voltage is ∆ ∆ V V [ ] ( ) = ( ) 2 2 2 avg . max Then its square root is ∆ ∆ V V rms = max 2 *Q33.2 Answer (c). AC ammeters and voltmeters read rms values. With an oscilloscope you can read a maximum voltage, or test whether the average is zero. *Q33.3 (i) Answer (f). The voltage varies between +170 V and −170 V. (ii) Answer (d). (iii) 170V/ 2 = 120 V. Answer (c). Q33.4 The capacitive reactance is proportional to the inverse of the frequency. At higher and higher frequen- cies, the capacitive reactance approaches zero, making a capacitor behave like a wire. As the frequency goes to zero, the capacitive reactance approaches infinity—the resistance of an open circuit. Q33.5 The second letter in each word stands for the circuit element. For an inductor L, the emf ε leads the current I—thus ELI. For a capacitor C, the current leads the voltage across the device. In a circuit in which the capacitive reactance is larger than the inductive reactance, the current leads the source emf—thus ICE. Q33.6 The voltages are not added in a scalar form, but in a vector form, as shown in the phasor diagrams throughout the chapter. Kirchhoff’s loop rule is true at any instant, but the voltages across different circuit elements are not simultaneously at their maximum values. Do not forget that an inductor can induce an emf in itself and that the voltage across it is 90° ahead of the current in the circuit in phase. Q33.7 In an RLC series circuit, the phase angle depends on the source frequency. At very low frequency the capacitor dominates the impedance and the phase angle is near −90°. The phase angle is zero at the resonance frequency, where the inductive and capacitive reactances are equal. At very high frequencies f approaches +90°. *Q33.8 (i) Inductive reactance doubles when frequency doubles. Answer (f ). (ii) Capacitive reactance is cut in half when frequency doubles. Answer (b). (iii) The resistance remains unchanged. Answer (d). *Q33.9 At resonance the inductive reactance and capacitive reactance cancel out. Answer (c). 247
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33Alternating Current Circuits
CHAPTER OUTLINE
33.1 AC Sources33.2 Resistors in an AC Circuit33.3 Inductors in an AC Circuit33.4 Capacitors in an AC Circuit33.5 The RLC Series Circuit33.6 Power in an AC Circuit33.7 Resonance in a Series RLC
Circuit33.8 The Transformer and Power
Transmission33.9 Rectifi ers and Filters
ANSWERS TO QUESTIONS
*Q33.1 (i) Answer (d). ∆∆
VV
avg = max
2 (ii) Answer (c). The average of the squared voltage is
∆∆
VV[ ]( ) =
( )2
2
2avg.max Then its square root is
∆∆
VV
rms = max
2
*Q33.2 Answer (c). AC ammeters and voltmeters read rms values. With an oscilloscope you can read a maximum voltage, or test whether the average is zero.
*Q33.3 (i) Answer (f). The voltage varies between +170 V and −170 V. (ii) Answer (d). (iii) 170V/ 2 = 120 V. Answer (c).
Q33.4 The capacitive reactance is proportional to the inverse of the frequency. At higher and higher frequen-cies, the capacitive reactance approaches zero, making a capacitor behave like a wire. As the frequency goes to zero, the capacitive reactance approaches infi nity—the resistance of an open circuit.
Q33.5 The second letter in each word stands for the circuit element. For an inductor L, the emf ε leads the current I—thus ELI. For a capacitor C, the current leads the voltage across the device. In a circuit in which the capacitive reactance is larger than the inductive reactance, the current leads the source emf—thus ICE.
Q33.6 The voltages are not added in a scalar form, but in a vector form, as shown in the phasor diagrams throughout the chapter. Kirchhoff’s loop rule is true at any instant, but the voltages across different circuit elements are not simultaneously at their maximum values. Do not forget that an inductor can induce an emf in itself and that the voltage across it is 90° ahead of the current in the circuit in phase.
Q33.7 In an RLC series circuit, the phase angle depends on the source frequency. At very low frequency the capacitor dominates the impedance and the phase angle is near −90°. The phase angle is zero at the resonance frequency, where the inductive and capacitive reactances are equal. At very high frequencies f approaches +90°.
*Q33.8 (i) Inductive reactance doubles when frequency doubles. Answer (f). (ii) Capacitive reactance is cut in half when frequency doubles. Answer (b). (iii) The resistance remains unchanged. Answer (d).
*Q33.9 At resonance the inductive reactance and capacitive reactance cancel out. Answer (c).
*Q33.10 At resonance the inductive reactance and capacitive reactance add to zero. tan−1(XL−X
C)/R = 0.
Answer (c).
*Q33.11 (a) The circuit is in resonance. (b) 10 Ω/20 Ω = 0.5 (c) The resistance of the load could be increased to make a greater fraction of the emf’s power go to the load. Then the emf would put out a lot less power and less power would reach the load.
Q33.12 The person is doing work at a rate of P = Fv cosθ. One can consider the emf as the “force” that moves the charges through the circuit, and the current as the “speed” of the moving charges. The cosθ factor measures the effectiveness of the cause in producing the effect. Theta is an angle in real space for the vacuum cleaner and phi is the analogous angle of phase difference between the emf and the current in the circuit.
*Q33.13 The resonance is high-Q, so at 1 000 Hz both XL and X
C are equal and much larger than R. Now X
C
at 500 Hz is twice as large as at 1 kHz. And XL at 1.5 kHz is 1.5 times larger than at 1 kHz. Again,
XC at 1 500 Hz is two-thirds as large as at 1 kHz. And X
L at 500 Hz is half as large as at 1 kHz. The
resistance does not change with frequency. The ranking is then a > f > b = e > c > d > g = h = i.
Q33.14 In 1881, an assassin shot President James Garfi eld. The bullet was lost in his body. Alexander Graham Bell invented the metal detector in an effort to save the President’s life. The coil is preserved in the Smithsonian Institution. The detector was thrown off by metal springs in Garfi eld’s mattress, a new invention itself. Surgeons went hunting for the bullet in the wrong place. Garfi eld died.
Q33.15 No. A voltage is only induced in the secondary coil if the fl ux through the core changes in time.
Q33.16 The Q factor determines the selectivity of the radio receiver. For example, a receiver with a very low Q factor will respond to a wide range of frequencies and might pick up several adjacent radio stations at the same time. To discriminate between 102.5 MHz and 102.7 MHz requires a high-Q circuit. Typically, lowering the resistance in the circuit is the way to get a higher quality resonance.
*Q33.17 In its intended use, the transformer takes in energy by electric transmission at 12 kV and puts out nearly the same energy by electric transmission at 120 V. With the small generator putting energy into the secondary side of the transformer at 120 V, the primary side has 12 kV induced across it. It is deadly dangerous for the repairman.
Section 33.1 AC Sources
Section 33.2 Resistors in an AC Circuit
P33.1 ∆ ∆ ∆v t V t V t( ) = ( ) = ( ) =max sin sin sinω ω π2 200 2 2 1rms 000 283 628t t( )⎡⎣ ⎤⎦ = ( ) ( ) V sin
*P33.31 Consider a two-wire transmission line taking in power P
IVrms
rms
.= P∆
Then power loss = =I Rrms2
line
P100
.
Thus,P P
∆VR
rms
⎛⎝⎜
⎞⎠⎟
( ) =2
12100
or RV
1
2
200= ( )∆ rms
P
RA
V1
2
200= =
( )ρ ∆ rms
P or A
r
V= ( )
=( )
π ρ2
4
2002
2
P ∆ rms
and the diameter is 2800
2rV
=( )
ρπ
P ∆
(a) 2800 1 7 10 18 0008
rV
= × ( )( )
−( . Ω∆
m) 20 000 W m
π 22 39 5= ⋅. V m/∆V
(b) The diameter is inversely proportional to the potential difference.
(c) 2r = 39.5 V⋅m/1 500 V = 2.63 cm
(d) ∆V = 39.5 V⋅m/0.003 m = 13.2 kV
*P33.32 (a) XL = ωL = 2p (60/s) 0.1 H = 37.7 Ω
Z = (1002 + 37.72)1/2 = 107 Ω power factor = cosf = 100/107 = 0.936
(b) The power factor cannot in practice be made 1.00. If the inductor were removed or if the generator were replaced with a battery, so that either L = 0 or f = 0, the power factor would be 1, but we would not have a magnetic buzzer.
(c) We want resonance, with f = 0. We insert a capacitor in series with
XL = X
C so 37.7 Ω = 1 s/2pC 60 and C = 70.4 mF
P33.33 One-half the time, the left side of the generator is positive, the top diode conducts, and the bottom diode switches off. The power supply sees resistance
1
2
1
2
1
R RR+⎡
⎣⎢⎤⎦⎥
=−
and the power is∆V
Rrms( )2
.
The other half of the time the right side of the generator is positive, the upper diode is an open circuit, and the lower diode has zero resistance. The equivalent resistance is then
R RR R
Req = + +⎡
⎣⎢⎤⎦⎥
=−1
3
1 7
4
1
and P =( )
=( )∆ ∆V
R
V
Rrms
eq
rms
2 24
7
The overall time average power is: ∆ ∆ ∆V R V R Vrms rms rms
Ploss rms2 A kW= = ( ) ( ) =I R 10 0 290 29 02. .Ω
(b) PPloss = ×
×= × −2 90 10
5 00 105 80 10
4
63.
..
(c) It is impossible to transmit so much power at such low voltage. Maximum power transfer occurs when load resistance equals the line resistance of 290 Ω, and is
4 50 10
2 2 29017 5
3 2.
.×( )
⋅ ( ) =V
kW,Ω
far below the required 5 000 kW.
Section 33.9 Rectifi ers and Filters
P33.44 (a) Input power = 8 W
Useful output power = = ( ) =I V∆ 0 3 2 7. .A 9 V W
P33.50 The equation for ∆v( )t during the fi rst period (usingy mx b= + ) is:
∆∆
∆
∆ ∆
v
v v
tV t
TV
Tt
( ) = ( ) −
( )⎡⎣
⎤⎦ = ( )⎡
2
12
maxmax
avg⎣⎣ ⎤⎦ = ( ) −
⎡⎣⎢
⎤⎦⎥∫ ∫2
0
2 2
0
21dt
V
T Tt dt
T T∆ max
∆∆
v( )⎡⎣ ⎤⎦ =( ) ⎛
⎝⎞⎠
−[ ]=
=2
2 3
02
2 1
3avg
V
T
T t T
t
t T
max ==( )
+( ) − −( )⎡⎣ ⎤⎦ =( )
=
∆ ∆
∆ ∆
V V
V
max max
2
3 3
2
61 1
3
rms vv( )⎡⎣ ⎤⎦ =( )
=2
2
3 3avg
max∆ ∆V Vmax
*P33.51 (a) Z 2 = R2 + (XL – X
C)2 760 2 = 4002 + (700 − X
C)2 417 600 = (700 − X
C)2
There are two values for the square root. We can have 646.2 = 700 − XC or −646.2 = 700 − X
C.
XC can be 53.8 Ω or it can be 1.35 kΩ.
(b) If we were below resonance, with inductive reactance 700 Ω and capacitive reactance 1.35 kΩ, raising the frequency would increase the power. We must be above resonance,
with inductive reactance 700 Ω and capacitive reactance 53.8 Ω.
than the resonance frequency for the inductive reactance to be the greater.
(b) It is possible to determine the values for L and C, because we have three independent equations in the three unknowns L, C, and the unknown angular frequency w. The equations are
2 0002 = 1/LC 12 = wL and 8 = 1/wC
(c) We eliminate w = 12/L to have 8 wC = 1 = 8(12/L)C = 96C/L so L = 96C
Then 4 000 000 = 1/96 C 2 so C = 51.0 mF and L = 4.90 mH
*P33.55 The lowest-frequency standing-wave state is NAN. The distance between the clamps we
represent as d d= =NN .λ2
The speed of transverse waves on the string is v = = =fT
f dλµ
2 .
The magnetic force on the wire oscillates at 60 Hz, so the wire will oscillate in resonance at 60 Hz.
Td
0 01960 4
2 2
. kg ms= ( ) T d= ⋅( )274 2kg m s2
Any values of T and d related according to this expression will work, including
if m Nd T= =0 200 10 9. . . We did not need to use the value of the current and magnetic
fi eld. If we assume the subsection of wire in the fi eld is 2 cm wide, we can fi nd the rms value of the magnetic force:
F I B TB = = ( )( )( ) = sin . . sin .θ 9 0 02 0 015 3 90 2 7A m ° 55 mN
So a small force can produce an oscillation of noticeable amplitude if internal friction is small.
*P33.56 φ ω ω= −⎛⎝
⎞⎠
−tan/1 1L C
Rchanges from −90° for w = 0 to 0
at the resonance frequency to +90° as w goes to infi nity. The slope of the graph is df/dw :
d
d L C
R
RL
C
φω ω ω ω
=+ −⎛
⎝⎜
⎞⎠⎟
− −⎛⎝⎜
⎞⎠⎟
1
11
1 11
12 2
/( ) At resonance
we have w0L = 1/w
0C and LC = 1/w
02.
Substituting, the slope at the resonance point is
d
d RL
CLC
L
R
Qφω ωω0
1
1 0
1 1 2 22
0
=+
+⎛⎝
⎞⎠ = =
P33.57 (a) When ωL is very large, the bottom branch carries negligible current. Also, 1
ωCwill be
negligible compared to 200 Ω and45 0
225. V
200mA
Ω= fl ows in the power supply and the
top branch.
(b) Now 1
ωC→ ∞ and ωL → 0 so the generator and bottom branch carry 450 mA .
P33.60 Suppose each of the 20 000 people uses an average power of 500 W. (This means 12 kWh per day, or $36 per 30 days at 10c| per kWh.) Suppose the transmission line is at 20 kV. Then
IVrms
rms
W
VA= =
( )( )P∆
20 000 500
20 000103~
If the transmission line had been at 200 kV, the current would be only ~ .102 A
P33.61 R = 200 Ω, L = 663 mH, C = 26 5. F,µ ω = −377 1s , ∆Vmax .= 50 0 V
ω L = 250 Ω,1
100ωC
⎛
⎝⎜
⎞
⎠⎟ = Ω, Z R X XL C= + −( ) =2 2
250 Ω
(a) IV
Zmaxmax .
.= = =∆
Ω50 0
0 200V
250A
φ = −⎛⎝
⎞⎠ =−tan .1 36 8
X X
RL C ° (∆V leads I )
(b) ∆V I RR,max V= =max .40 0 at φ = 0°
(c) ∆VI
CC ,
max .max V= =ω
20 0 at φ = −90 0. ° (I leads ∆V )
(d) ∆V I LL , max . max V= =ω 50 0 at φ = +90 0. ° (∆V leads I )
P33.62 L = 2 00. H, C = × −10 0 10 6. F, R = 10 0. ,Ω ∆v t t( ) = ( )100sinω
(a) The resonant frequencyω0 produces the maximum current and thus the maximum power delivery to the resistor.
ω0 6
1 1
2 00 10 0 10224= =
( ) ×( )=
−LC . .rad s
(b) P =( )
= ( )( ) =
∆V
Rmax
.
2 2
2
100
2 10 0500 W
(c) IV
Z
V
R L Crms
rms rms= =+ − ( )( )
∆ ∆2 2
1ω ω
and IV
Rrmsrms( ) =
max
∆
I R I Rrms2
rms2= ( )1
2 max or
∆ ∆V
ZR
V
RRrms rms( )
=( )2
2
2
2
1
2
This occurs where Z R2 22= : R LC
R2
2
212+ −
⎛⎝⎜
⎞⎠⎟
=ωω
ω ω ω4 2 2 2 2 2 22 1 0L C L C R C− − + = or L C LC R C2 2 4 2 2 22 1 0ω ω− +( ) + =
*P33.66 An RLC series circuit, containing a 35.0-Ω resistor, a 205-mH inductor, a capacitor, and a power supply with rms voltage 200 V and frequency 100 Hz, carries rms current 4.00 A. Find the capacitance of the capacitor.
We solve for C 2500 = 352 + (129 − 1/628C)2 1275 = (129 − 1/628C)2
There are two possibilities: 35.7 = 129 − 1/628C and −35.7 = 129 − 1/628C 1/628C = 93.1 or 1/628C = 164.5
P33.12 (a) greater than 41 3. Hz (b) less than 87 5. Ω
P33.14 2C V∆ rms( )
P33.16 –32.0 A
P33.18 2 79. kHz
P33.20 (a) 109 Ω (b) 0 367. A (c) Imax .= 0 367 A, ω = 100 rad s, φ = −0 896. rad
P33.22 19.3 mA
P33.24 (a) 146 V (b) 212 V (c) 179 V (d) 33 4. V
P33.26 Cutting the plate separation in half doubles the capacitance and cuts in half the capacitive reactance to X
C/2. The new impedance must be half as large as the old impedance for the new
current to be doubled. For the new impedance we then have
(R2 + [R − XC/2]2)1/2 = 0.5(R2 + [R − X
C]2)1/2. Solving yields X
C = 3R.
P33.28 353 W
P33.30 (a) 5 43. A (b) 0 905. (c) 281 Fµ (d) 109 V
P33.32 (a) 0.936 (b) Not in practice. If the inductor were removed or if the generator were replaced with a battery, so that either L = 0 or f = 0, the power factor would be 1, but we would not have a magnetic buzzer. (c) 70.4 mF
P33.52 Only one value for R and only one value for C are possible. Two values for L are possible. R = 99.6 Ω, C = 24.9 mF, and L = 164 mH or 402 mH
P33.54 (a) Higher. At the resonance frequency XL = X
C. As the frequency increases, X
L goes up and
XC goes down. (b) It is. We have three independent equations in the three unknowns L, C, and
the certain f. (c) L = 4.90 mH and C = 51.0 mF
P33.56 See the solution.
P33.58 (a) i tV
Rt( ) =
∆ max cosω (b) P =( )∆V
Rmax
2
2 (c) i t
V
R Lt
L
R( ) =
++ ⎛
⎝⎞⎠
⎡⎣⎢
⎤⎦⎥
−∆ max cos tan2 2 2
1
ωω ω
(d) CL
= 1
02ω
(e) Z R= (f)∆V L
Rmax( )2
22 (g)
∆V L
Rmax( )2
22 (h) tan− ⎛
⎝⎜⎞⎠⎟
1 3
2R
L
C (i) 1
2LC
P33.60 ~103 A
P33.62 (a) 224 rad s (b) 500 W (c) 221 rad s and 226 rad s
P33.64 The frequency could be either 58.7 Hz or 35.9 Hz. We can be either above or below resonance.
P33.66 An RLC series circuit, containing a 35.0-Ω resistor, a 205-mH inductor, a capacitor, and a power supply with rms voltage 200 V and frequency 100 Hz, carries rms current 4.00 A. Find the capacitance of the capacitor. Answer: It could be either 17.1 mF or 9.67 mF.