Chapter 33 Solutions
33.1
∆v(t) = ∆Vmax sin(ω t) = 2 ∆Vrms sin(ω t) = 200 2 sin[2 π (100t)] = (283 V) sin (628 t)
33.2
∆Vrms =
(a)
P=
(120 V)2
(∆Vrms )2
→R=
= 193 Ω
R
75.0 W
(b)
R=
(120 V)2
= 144 Ω
100 W
Each meter reads the rms value.
33.3
∆Vrms =
I rms =
33.4
170 V
= 120 V
2
(a)
100 V
= 70.7 V
2
∆Vrms 70.7 V
=
= 2.95 A
24.0 Ω
R
∆vR = ∆Vmax sin ω t
∆vR = 0.250 ( ∆Vmax ) ,
so
sin ω t = 0.250, or ω t = sin −1 (0.250)
The smallest angle for which this is true is ω t = 0.253 rad. Thus, if t = 0.010 0 s ,
ω=
(b)
The second time when ∆vR = 0.250 ( ∆Vmax ) , ω t = sin −1 (0.250) again. For this occurrence,
ω t = π − 0.253 rad = 2 .89 rad (to understand why this is true, recall the identity sin( π − θ ) = sin θ
from trigonometry). Thus,
t=
33.5
0.253 rad
= 25.3 rad/s
0.010 0 s
2.89 rad
= 0.114 s
25.3 rad s
iR = I max sin ω t
becomes
(0.00700)ω = sin–1(0.600) = 0.644
Thus,
and
0.600 = sin(ω 0.00700)
ω = 91.9 rad/s = 2π f
so
f = 14.6 Hz
© 2000 by Harcourt, Inc. All rights reserved.
Chapter 33 Solutions 271
P = I rms ( ∆Vrms ) and ∆Vrms = 120 V for each bulb (parallel circuit), so:
33.6
I1 = I 2 =
I3 =
∆Vrms 120 V
P1
150 W
=
= 1.25 A , and R1 =
=
= 96.0 Ω = R2
1.25 A
∆Vrms 120 V
I1
∆Vrms
P3
100 W
120 V
=
= 0.833 A , and R3 =
=
= 144 Ω
0.833 A
∆Vrms 120 V
I3
∆Vmax = 15.0 V
33.7
I max =
and Rtotal = 8.20 Ω + 10.4 Ω = 18.6 Ω
∆Vmax 15.0 V
=
= 0.806 A = 2 I rms
18.6 Ω
Rtotal
2
 0.806 A 
2
P speaker = I rms
Rspeaker = 
 (10.4 Ω) = 3.38 W

2 
For Imax = 80.0 mA,
33.8
Irms =
80.0 mA
2
= 56.6 mA
Vrms
50.0 V
(XL)min =
Irms = 0.0566 A = 884 Ω
XL
884 Ω
XL = 2π f L → L = 2π f ≥ 2π (20.0) ≥ 7.03 H
33.9
(a)
XL =
L=
(b)
33.10
XL
13.3
=
= 0.0424 H = 42.4 mH
2 π (50.0)
ω
XL =
ω=
∆Vmax 100
=
= 13.3 Ω
7.50
I max
∆Vmax 100
=
= 40.0 Ω
2.50
I max
XL
40.0
=
= 942 rad/s
L
42.4 × 10 − 3
At 50.0 Hz,
I max =
 X

L 60.0 Hz
50.0
=
XL = 2 π ( 50.0 Hz)L = 2 π ( 50.0 Hz)
( 54.0 Ω) = 45.0 Ω
 2 π (60.0 Hz)  60.0


∆Vmax
=
XL
2 ( ∆Vrms )
=
XL
2 (100 V )
= 3.14 A
45.0 Ω
© 2000 by Harcourt, Inc. All rights reserved.
272 Chapter 33 Solutions
iL (t ) =
33.11
[
(80.0 V ) sin (65.0 π )(0.0155) − π 2
∆Vmax
sin (ω t − π 2) =
ωL
(65.0 π rad s) 70.0 × 10−3 H
(
)
]
iL (t ) = ( 5.60 A ) sin(1.59 rad) = 5.60 A
ω = 2 π f = 2 π (60.0 / s) = 377 rad / s
33.12
XL = ω L = (377 / s)(0.0200 V ⋅ s / A) = 7.54 Ω
I rms =
∆Vrms 120 V
=
= 15.9 A
7.54 Ω
XL
I max = 2 I rms = 2 (15.9 A) = 22.5 A
 2 π (60.0) 1 s 
i(t) = I max sin ω t = (22.5 A)sin 
⋅
 = (22.5 A) sin 120° = 19.5 A

s
180 
V ⋅ s

(19.5 A)2 = 3.80 J
U = 21 Li 2 = 21 0.0200

A 
L=
33.13
N ΦB
where Φ B is the flux through each turn.
I
N Φ B, max =
33.14
(a)
XC =
(
2 ∆VL, rms
1
:
2π f C
2π f
XC ∝
f > 41.3 Hz
1
, so X(44) = 21 X(22): XC < 87.5 Ω
C
I max = 2 I rms =
33.15
max
=
120 V ⋅ s  T ⋅ C ⋅ m   N ⋅ m   J 
= 0.450 T · m2


2 π (60.0)  N ⋅ s   J   V ⋅ C 
1
< 175 Ω
2 π f (22.0 × 10 −6 )
1
<f
2 π (22.0 × 10 −6 )(175)
(b)
)=
N Φ B, max = LI B,
2 ( ∆Vrms )
= 2 ( ∆Vrms ) 2 π f C
XC
(a)
I max = 2 (120 V)2 π (60.0 / s)(2.20 × 10 − 6 C / V) = 141 mA
(b)
I max = 2 (240 V)2 π (50.0 / s)(2.20 × 10 − 6 F) = 235 mA
(
XL ∆VL, max
ω
XL
)
Chapter 33 Solutions 273
[
]
33.16
Qmax = C ( ∆Vmax ) = C
2 ( ∆Vrms ) =
2 C ( ∆Vrms )
33.17
I max = ( ∆Vmax )ω C = (48.0 V)(2 π )(90.0 s −1 )(3.70 × 10 − 6 F) = 100 mA
33.18
XC =
1
1
=
= 2.65 Ω
ω C 2 π (60.0 / s)(1.00 × 10 − 3 C / V)
vC (t) = ∆Vmax sin ω t , to be zero at t = 0
iC =
33.19
(a)
∆Vmax
sin(ω t + φ ) =
XC


2 (120 V)
60 s -1
sin 2 π
+ 90.0° = (64.0 A)sin(120° + 90.0°) = – 32.0 A
-1
2.65 Ω
 180 s

XL = ω L = 2π (50.0)(400 × 10- 3) = 126 Ω
XC =
1
1
=
= 719 Ω
ω C 2 π (50.0)(4.43 × 10 −6 )
Z = R 2 + (XL − XC )2 = 500 2 + (126 − 719)2 = 776 Ω
∆Vmax = Imax Z = (250 × 10- 3)(776) = 194 V
(b)
φ = tan −1 

ωL =
33.20
X L − XC 
 126 − 719 
= tan −1
= – 49.9°
 500 

R
1
→ω =
ωC
Thus, the Current leads the voltage.
1
1
=
= 1.75 × 10 4 rad / s
−6
LC
(57.0 × 10 )(57.0 × 10 −6 )
ω
f = 2π = 2.79 kHz
33.21
(a)
XL = ω L = 2π (50.0 s-1)(250 × 10-3 H) = 78.5 Ω
(b)
XC =
(c)
Z = R 2 + (XL − XC )2 = 1.52 kΩ
(d)
I max =
(e)
φ = tan −1 

[
]
1
= 2 π (50.0 s −1 )(2.00 × 10 −6 F)
ωC
−1
= 1.59 kΩ
∆Vmax
210 V
=
= 138 mA
Z
1.52 × 10 3 Ω
X L − XC 
−1
 = tan (−10.1) = – 84.3°
R
© 2000 by Harcourt, Inc. All rights reserved.
274 Chapter 33 Solutions
33.22
(a)
Z = R 2 + ( XL − XC ) = 68.0 2 + (16.0 − 101) = 109 Ω
2
XL = ω L = (100)(0.160) = 16.0 Ω
XC =
1
1
=
= 101 Ω
ω C (100) 99.0 × 10 −6
(
)
(b)
I max =
∆Vmax 40.0 V
=
= 0.367 A
109 Ω
Z
(c)
tan φ =
XL − XC 16.0 − 101
=
= −1.25:
68.0
R
φ = − 0.896 rad = − 51.3°
Imax = 0.367 A
ω = 100 rad/s
φ = – 0.896 rad = – 51.3°
XL = 2 π f L = 2 π (60.0)(0.460) = 173 Ω
33.23
XC =
(a)
1
1
=
= 126 Ω
2 π f C 2 π (60.0) 21.0 × 10 − 6
(
tan φ =
)
XL − XC 173 Ω − 126 Ω
=
= 0.314
150 Ω
R
φ = 0.304 rad = 17.4°
(b)
33.24
Since XL > XC , φ is positive; so voltage leads the current .
XC =
1
1
=
= 1.33 × 108 Ω
2 π f C 2 π (60.0 Hz)(20.0 × 10 −12 F)
Z = (50.0 × 10 3 Ω)2 + (1.33 × 108 Ω)2 ≈ 1.33 × 108 Ω
Irms =
∆Vrms
5000 V
=
= 3.77 × 10–5 A
Z
1.33 × 108 Ω
( ∆Vrms )body = Irms Rbody = (3.77 × 10− 5 A)(50.0 × 10 3 Ω) =
1.88 V
Chapter 33 Solutions 275
XC =
33.25
1
1
=
= 49.0 Ω
ω C 2 π (50.0)(65.0 × 10 −6 )
XL = ω L = 2 π (50.0)(185 × 10 −3 ) = 58.1 Ω
Z = R 2 + (XL − XC )2 = (40.0)2 + (58.1 − 49.0)2 = 41.0 Ω
I max =
∆Vmax 150
=
= 3.66 A
41.0
Z
(a)
∆VR = Imax R = (3.66)(40) = 146 V
(b)
∆VL = Imax XL = (3.66)(58.1) = 212.5 = 212 V
(c)
∆VC = Imax XC = (3.66)(49.0) = 179.1 V = 179 V
(d) ∆VL – ∆VC = 212.5 – 179.1 = 33.4 V
R = 300 Ω
33.26
XL = 200 Ω
 500 −1 
XL = ω L = 2 π
s (0.200 H) = 200 Ω
 π

XC =
(
)
1
  500 −1 

11.0 × 10 −6 F 
= 2 π
s


ωC 
π

Z = R 2 + ( XL − XC ) = 319 Ω
2
33.27
(a)
and
−1
XL - XC = 109 Ω
= 90.9 Ω
φ = tan −1 

XC = 90.9 Ω
X L − XC 
= 20.0°

R
XL = 2 π (100 Hz)( 20.5 H) = 1.29 × 10 4 Ω
Z=
∆Vrms
200 V
=
= 50.0 Ω
4.00 A
I rms
(XL − XC )2 = Z2 − R 2 = (50.0 Ω)2 − (35.0 Ω)2
XL − XC = 1.29 × 10 4 Ω −
(b)
1
= ± 35.7 Ω
2 π (100 Hz)C
(
{
C = 123 nF or 124 nF
)
∆VL,rms = I rms XL = ( 4.00 A ) 1.29 × 10 4 Ω = 51.5 kV
Notice that this is a very large voltage!
© 2000 by Harcourt, Inc. All rights reserved.
Z
φ
R = 300 Ω
276 Chapter 33 Solutions
XL = ω L = [(1000 / s)(0.0500 H)] = 50.0 Ω
33.28
[
]
XC = 1/ ω C = (1000 / s)(50.0 × 10 − 6 F)
−1
= 20.0 Ω
Z = R 2 + (XL − XC )2
Z = (40.0)2 + (50.0 − 20.0)2 = 50.0 Ω
(a)
I rms = ( ∆Vrms ) / Z = 100 V / 50.0 Ω
I rms = 2.00 A
X − XC 
φ = Arctan  L


R
φ = Arctan
33.29
30.0 Ω
= 36.9°
40.0 Ω
(b)
P = ( ∆Vrms ) I rms cos φ = 100 V(2.00 A) cos 36.9° = 160 W
(c)
2
P R = I rms
R = (2.00 A)2 40.0 Ω = 160 W
ω = 1000 rad/s,
R = 400 Ω,
∆Vmax = 100 V,
 1 
ω L = 500 Ω , 
 = 200 Ω
 ω C
C = 5.00 × 10– 6 F,
L = 0.500 H
2

1 
Z = R2 +  ω L −
= 400 2 + 300 2 = 500 Ω
ω C 

I max =
∆Vmax 100
=
= 0.200 A
500
Z
The average power dissipated in the circuit is
P=
(0.200 A)2
(400 Ω) = 8.00 W
2
 I2 
2
P = I rms
R =  max  R
 2 
Chapter 33 Solutions 277
Goal Solution
An ac voltage of the form ∆v = (100 V ) sin(1000 t ) is applied to a series RLC circuit.
C = 5.00 µ F, and L = 0.500 H, what is the average power delivered to the circuit?
G: Comparing
∆v = (100 V ) sin(1000 t ) with
∆Vmax = 100 V
and
If R = 400 Ω,
∆v = ∆Vmax sin ω t, we see that
ω = 1000 s -1
Only the resistor takes electric energy out of the circuit, but the capacitor and inductor will impede the
current flow and therefore reduce the voltage across the resistor. Because of this impedance, the
average power dissipated by the resistor must be less than the maximum power from the source:
P max =
( ∆Vmax )2 = (100 V)2
2( 400 Ω)
2R
= 12.5 W
2
R, where I rms = ∆Vrms / Z.
O: The actual power dissipated by the resistor can be found from P = I rms
A : ∆Vrms =
100
= 70.7 V
2
In order to calculate the impedance, we first need the capacitive and inductive reactances:
XC =
1
1
=
= 200 Ω
ω C (1000 s -1 )(5.00 × 10 −6 F)
Then,
I rms =
and
(
)
XL = ω L = 1000 s -1 (0.500 H) = 500 Ω
Z = R 2 + (XL − XC )2 = (400 Ω)2 + (500 Ω − 200 Ω)2 = 500 Ω
∆Vrms 70.7 V
=
= 0.141 A
500 Ω
Z
and
2
P = I rms
R = (0.141 A ) ( 400 Ω) = 8.00 W
2
L : The power dissipated by the resistor is less than 12.5 W, so our answer appears to be reasonable. As
with other RLC circuits, the power will be maximized at the resonance frequency where XL = XC so
that Z = R . Then the average power dissipated will simply be the 12.5 W we calculated first.
33.30
Z = R 2 + ( X L − XC )
2
or
( X L − XC ) =
(XL − XC ) = (75.0 Ω)2 − ( 45.0 Ω)2
φ = tan −1 

I rms =
Z2 − R2
= 60.0 Ω
X L − XC 
 60.0 Ω 
= tan −1
= 53.1°
 45.0 Ω 

R
∆Vrms 210 V
=
= 2.80 A
75.0 Ω
Z
P = ( ∆Vrms ) I rms cos φ = ( 210 V )( 2.80 A ) cos( 53.1˚ ) = 353 W
© 2000 by Harcourt, Inc. All rights reserved.
278 Chapter 33 Solutions
33.31
(a)
P = I rms (∆Vrms )cos φ = (9.00)(180) cos(– 37.0°) = 1.29 × 10 3 W
2
P = I rms
R
(b)
tan φ =
X L − XC
R
so
1.29 × 10 3 = (9.00)2 R
becomes
tan( − 37.0°) =
R = 16.0 Ω
so
XL – XC = – 12.0 Ω
XL = ω L = 2 π (60.0 / s)(0.0250 H) = 9.42 Ω
*33.32
Z = R 2 + (XL − XC )2 =
(20.0)2 + (9.42)2
Ω = 22.1 Ω
∆Vrms 120 V
=
= 5.43 A
Z
22.1 Ω
(a)
I rms =
(b)
φ = tan −1 (9.42 / 20.0) = 25.2°
(c)
We require φ = 0. Thus, XL = XC:
9.42 Ω =
and
C = 281 µF
(d)
Pb = Pd
or
( ∆Vrms )d =
33.33
X L − XC
:
16
and
power factor = cos φ = 0.905
so
( ∆Vrms )b ( Irms )b cos φb =
R( ∆Vrms )b ( I rms )b cos φ b =
1
2 π (60.0 s −1 )C
( ∆Vrms )d 2
R
(20.0 Ω)(120 V )( 5.43 A)(0.905) =
109 V
Consider a two-wire transmission line:
R1
I rms
P
100 × 106 W
=
=
= 2.00 × 10 3 A
∆Vrms 50.0 × 10 3 V
∆Vrms
2
2
loss = (0.010 0)P = I rms
R line = I rms
(2R1 )
Thus, R1
(0.010 0)P = (0.010 0)(100 × 10
=
2
2 I max
But
R1 =
ρl
or
A
Therefore
d=
4ρ l
=
π R1
(
R1
6
2 2.00 × 10 A
A=
(
3
RL
)
W
2
) = 0.125 Ω
π d 2 ρl
=
R1
4
)(
4 1.70 × 10 − 8 Ω ⋅ m 100 × 10 3 m
π (0.125 Ω)
) = 0.132 m =
132 mm
Chapter 33 Solutions 279
33.34
Consider a two-wire transmission line:
P
=
∆Vrms
I rms
and
power loss =
R1
2
I rms
R line
2
 P 
P
Thus, 
 ( 2R1 ) =
100
 ∆Vrms 
R1 =
ρ d ( ∆Vrms )
=
A
200 P
R1 =
or
2
or
A=
∆Vrms
( ∆Vrms )2
R1
200 P
800ρ P d
2r =
π ( ∆Vrms )
2
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 
 1
 2R + 2R 
−1
=R
RL
π ( 2r )2 200ρ P d
=
4
( ∆Vrms )2
and the diameter is
33.35
P
=
100
∆Vrms
( ∆V rms )2
and the power is
R1
R
RL
R1
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
1
 1
Req = R + 
+ 
 3R R 
−1
=
7R
and
4
P=
( ∆V rms )2
The overall time average power is:
33.36
At resonance,
1
= 2π f L
2π f C
Req
[(∆V
=
rms
4( ∆V rms )
7R
)2
2
][
R + 4( ∆V rms ) 7R
2
2
and
1
( 2 π f )2 L
=C
The range of values for C is 46.5 pF to 419 pF
33.37
ω 0 = 2 π (99.7 × 106 ) = 6.26 × 108 rad / s =
C=
1
LC
1
1
=
= 1.82 pF
ω 0 2 L (6.26 × 108 )2 (1.40 × 10 − 6 )
© 2000 by Harcourt, Inc. All rights reserved.
]=
11( ∆V rms )
14 R
2
280 Chapter 33 Solutions
L = 20.0 mH, C = 1.00 × 10–7, R = 20.0 Ω, ∆Vmax = 100 V
33.38
33.39
(a)
The resonant frequency for a series –RLC circuit is
(b)
At resonance,
I max =
(c)
From Equation 33.36,
Q=
(d)
∆VL, max = XL I max = ω 0 L I max = 2.24 kV
f=
1
2π
1
= 3.56 kHz
LC
∆Vmax
= 5.00 A
R
ω 0L
= 22.4
R
The resonance frequency is ω 0 = 1
LC . Thus, if ω = 2ω 0 ,
L
 2 
XL = ω L = 
L = 2
 LC 
C
and
XC =
Z = R 2 + ( XL − XC ) = R 2 + 2.25( L C )
so
I rms =
2
1
LC 1 L
=
=
2 C
ωC
2C
∆Vrms
=
Z
∆Vrms
R + 2.25( L C )
2
and the energy dissipated in one period is Q = P ∆t:
Q=
( ∆Vrms )2 R  2π  = ( ∆Vrms )2 RC π
(
R 2 + 2.25( L C )  ω  R 2C + 2.25 L
4 π ( ∆Vrms ) RC LC
2
)
LC =
4R 2C + 9.00 L
With the values specified for this circuit, this gives:
2
Q=
33.40
(
) (10.0 × 10 H)
F ) + 9.00(10.0 × 10 H)
4 π ( 50.0 V ) (10.0 Ω) 100 × 10 −6 F
(
4(10.0 Ω) 100 × 10 −6
2
The resonance frequency is ω 0 = 1
32
−3
12
−3
LC .
L
 2 
XL = ω L = 
L = 2
 LC 
C
Then Z = R 2 + ( XL − XC ) = R 2 + 2.25( L C )
2
= 242 mJ
Thus, if ω = 2ω 0 ,
1
LC 1 L
=
=
2 C
ωC
2C
and
XC =
so
I rms =
∆Vrms
=
Z
and the energy dissipated in one period is
Q = P ∆t =
( ∆Vrms )2 R  2π  = ( ∆Vrms )2 RC π
(
R 2 + 2.25( L C )  ω  R 2C + 2.25 L
)
LC =
4 π ( ∆Vrms ) RC LC
2
4R 2C + 9.00 L
∆Vrms
R + 2.25( L C )
2
Chapter 33 Solutions 281
ω0 =
For the circuit of problem 22,
*33.41
For the circuit of problem 23,
1
=
LC
(160 × 10
1
−3
)(
H 99.0 × 10 −6 F
(
(a)
1 120 V = 9.23 V
∆V 2, rms = 13
(
)
(b)
∆V1, rms I1, rms = ∆V 2, rms I 2, rms
)
−3
ω 0 L ( 251 rad s) 160 × 10 H
=
= 0.591
R
68.0 Ω
Q=
ω 0L
L
1 L
1
460 × 10 −3 H
=
=
=
= 0.987
R
R LC R C 150 Ω 21.0 × 10 −6 F
(120 V)( 0.350 A) = (9.23 V)I 2, rms
I 2, rms =
(c)
42.0 W
= 4.55 A for a transformer with no energy loss
9.23 V
P = 42.0 W from (b)
( ∆Vout )max = N2 ( ∆Vin )max = 
N
33.43
1
( ∆Vout )rms =
33.44
2000 
(170 V) = 971 V
350 
(971 V)
= 687 V
2
(a)
(∆V2, rms ) = NN21 (∆V1, rms )
(b)
I1, rms ∆V1, rms = I 2, rms ∆V 2, rms
(c)
0.950 I1, rms ∆V1, rms = I 2, rms ∆V 2, rms
(
)
(
N2 =
(
)
(
= 251 rad s
Q=
The circuit of problem 23 has a sharper resonance.
33.42
)
)
)
(2200)(80)
= 1600 windings
110
I1, rms =
(1.50)(2200)
= 30.0 A
110
I1, rms =
(1.20)(2200)
= 25.3 A
110(0.950)
© 2000 by Harcourt, Inc. All rights reserved.
282 Chapter 33 Solutions
The rms voltage across the transformer primary is
33.45
(
N1
∆V 2, rms
N2
so the source voltage is
∆V s, rms = I1, rms Rs +
The secondary current is
(∆V2, rms ) ,
Then ∆V s, rms =
and
33.46
Rs =
(
)
N1RL
N1RL
(
N 2 ∆V 2, rms
(
N2
∆V1, rms
N1
)
+
)
(
I 2, rms ∆V 2, rms = 0.900 I1, rms ∆V1, rms
(
)
∆V 2, rms
)
)
)  = 5(50.0 Ω)  80.0 V − 5(25.0 V) =


2(25.0 V) 
2

87.5 Ω
)
 120 V 
(120 V )
 24.0 Ω 
I 2, rms = 54.0 mA
10.0 × 10 3 V
= 185 kΩ
0.054 A
(c)
Z2 =
(a)
R = (4.50 × 10 − 4 Ω / m)(6.44 × 10 5 m) = 290 Ω
I 2, rms
=
(
N 2 ∆V 2, rms
= I1, rms
N1
RL
N 2 ∆V 2, rms 10.0 × 10 3 V
=
=
= 83.3
120 V
N1 ∆V1, rms
(b)
)
)
N2

N1 ∆V 2, rms
 ∆V s, rms −

N2

∆V 2, rms =
(
(
N1 ∆V 2, rms
(a)
(
(
N1
∆V 2, rms
N2
so the primary current is
RL
N 2 ∆V 2, rms Rs
I 2, rms 10.0 × 10 3 V = 0.900
33.47
)
and
I rms =
5.00 × 106 W
P
=
= 10.0 A
∆Vrms
5.00 × 10 5 V
2
P loss = I rms R = (10.0 A)2 (290 Ω) = 29.0 kW
(b)
(c)
P loss 2.90 × 10 4
=
= 5.80 × 10 − 3
P
5.00 × 106
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 3 V)2
= 17.5 kW, far below the required 5 000 kW
2 ⋅ 2(290 Ω)
Chapter 33 Solutions 283
33.48
(a)
(b)
For the filter circuit,
∆Vout
=
∆Vin
At f = 600 Hz ,
XC =
and
∆Vout
=
∆Vin
R + XC2
1
1
=
= 3.32 × 10 4 Ω
2 π f C 2 π (600 Hz) 8.00 × 10 −9 F
(
)
3.32 × 10 4 Ω
(90.0 Ω)2 + (3.32 × 10 4
Ω
)
2
≈ 1.00
1
1
=
= 33.2 Ω
3
2 π f C 2 π 600 × 10 Hz 8.00 × 10 −9 F
At f = 600 kHz,
XC =
and
∆Vout
=
∆Vin
For this RC high-pass filter,
33.49
XC
2
(
)(
33.2 Ω
(90.0 Ω)2 + (33.2 Ω)2
∆Vout
=
∆Vin
)
= 0.346
R
R + XC2
2
(a)
(a)
When
then
∆Vout
= 0.500,
∆Vin
0.500 Ω
(0.500 Ω)2 + XC2
= 0.500 or XC = 0.866 Ω
(b)
If this occurs at f = 300 Hz, the capacitance is
C=
(b)
1
1
=
= 6.13 × 10 − 4 F = 613 µ F
2 π f XC 2 π ( 300 Hz)(0.866 Ω)
With this capacitance and a frequency of 600 Hz,
XC =
(
1
2 π (600 Hz) 6.13 × 10 − 4 F
∆Vout
=
∆Vin
R
R 2 + XC2
=
)
(c)
= 0.433 Ω
0.500 Ω
(0.500 Ω)2 + (0.433 Ω)2
= 0.756
© 2000 by Harcourt, Inc. All rights reserved.
Figures for Goal
Solution
284 Chapter 33 Solutions
Goal Solution
The RC high-pass filter shown in Figure 33.22 has a resistance R = 0.500 Ω. (a) What capacitance gives an
output signal that has one-half the amplitude of a 300-Hz input signal? (b) What is the gain ( ∆V out / ∆Vin )
for a 600-Hz signal?
G: It is difficult to estimate the capacitance required without actually calculating it, but we might expect a
typical value in the µ F to pF range. The nature of a high-pass filter is to yield a larger gain at higher
frequencies, so if this circuit is designed to have a gain of 0.5 at 300 Hz, then it should have a higher
gain at 600 Hz. We might guess it is near 1.0 based on Figure (b) above.
O: The output voltage of this circuit is taken across the resistor, but the input sees the impedance of the
resistor and the capacitor. Therefore, the gain will be the ratio of the resistance to the impedance.
∆Vout
=
∆Vin
A:
(a)
(b)
R
R + (1 ω C )
2
2
When
∆V out / ∆Vin = 0.500
solving for C gives
C=
At 600 Hz, we have
ω = ( 2 π rad) 600 s -1
so
∆Vout
=
∆Vin
1
2
 ∆Vin 
ωR 
 −1
 ∆Vout 
(
=
1
(2 π )(300 Hz)(0.500 Ω) (2.00)2 − 1
= 613 µ F
)
0.500 Ω

(0.500 Ω)2 + 
 (1200 π

1
rad / s) (613 µ F ) 
2
= 0.756
L : The capacitance value seems reasonable, but the gain is considerably less than we expected. Based o n
our calculation, we can modify the graph in Figure (b) to more transparently represent the
characteristics of this high-pass filter, now shown in Figure (c). If this were an audio filter, it would
reduce low frequency “humming” sounds while allowing high pitch sounds to pass through. A low
pass filter would be needed to reduce high frequency “static” noise.
33.50
∆V1 = I
(r + R)2 + XL2 ,
and
∆V 2 = I R 2 + XL2
r = 20.0 Ω
Thus, when ∆V1 = 2 ∆V 2
(r + R)2 + XL2 = 4( R 2 + XL2 )
or
(25.0 Ω)
which gives
XL = 2 π f (0.250 H) =
2
+ XL2
L = 250 mH
∆V2
∆V1
= 4( 5.00 Ω) +
2
4XL2
R =5.00 Ω
625 − 100
Ω and f = 8.42 Hz
3
Chapter 33 Solutions 285
∆Vout
=
∆Vin
*33.51
(a)
R
R + ( X L − XC )
2
(8.00 Ω)2
1
=
4
At 200 Hz:
2

(8.00 Ω)2 + 400 π L −

(8.00 Ω)
At 4000 Hz:
1 

400 π C 
2
2
2


1
2
+ 8000 π L −
 = 4(8.00 Ω)
8000
π
C


400π L −
For the high frequency half-voltage point,
8000π L −
Solving Equations (1) and (2) simultaneously gives
C = 54.6 µ F
(b)
When XL = XC ,
∆Vout  ∆Vout 
=
= 1.00

∆Vin  ∆Vin  max
(c)
XL = XC requires
f0 =
1
2 π LC
∆Vout R 1
= =
Z 2
∆Vin
(d) At 200 Hz,
=
1
2π
(5.80 × 10
and
XC > X L ,
−4
)(
H 5.46 × 10 − 5 F
XL - XC
so the phasor diagram is as shown:
= 894 Hz
R
φ
or
Z
φ
At f 0 , XL = XC
so
∆Vout and ∆Vin have a phase difference of 0°
or
and
∆Vout
∆Vin
∆Vout leads ∆Vin by 60.0°
∆Vout R 1
= =
Z 2
∆Vin
[2]
L = 580 µ H
and
so
Thus, φ = cos −1
X L − XC > 0
 1
= 60.0°
 2
XL - XC
∆Vin
Z
or
φ
R
φ
∆Vout
∆Vout lags ∆Vin by 60.0°
(∆Vout, rms ) = ( 21 ∆Vin, rms )
P=
2
At 200 Hz and at 4 kHz,
At f 0 , P =
(f)
)
[1]
1
= + 13.9 Ω
8000π C
R
1
φ = − cos −1   = − cos −1  
 Z
 2
At 4000 Hz,
(e)
1
= − 13.9 Ω
400π C
At the low frequency, XL − XC < 0 . This reduces to
R
2
=
R
(
1 1 ∆V
in, max
2 2
(∆Vout,rms )2 = (∆Vin,rms )2 = 21 (∆Vin,max )2 = (10.0 V)2 =
We take: Q =
R
ω 0L
R
R
=
R
(
2(8.00 Ω)
)
−4
2 π f 0 L 2 π (894 Hz) 5.80 × 10 H
=
= 0.408
8.00 Ω
R
© 2000 by Harcourt, Inc. All rights reserved.
R
)
2
=
6.25 W
(10.0 V )2 =
8(8.00 Ω)
1.56 W
286 Chapter 33 Solutions
33.52
( ∆Vout )1
=
( ∆Vin )1
Now
33.53
∆Vout
=
∆Vin
For a high-pass filter,
R
 1 
R2 + 

 ωC
( ∆Vout )2
( ∆Vin )2
and
2
( ∆Vin )2 = ( ∆Vout )1
R
 1 
R2 + 

 ωC
R
=
( ∆Vout )2
( ∆Vin )1
so
2
 1 
R2 + 

 ωC
=
2
R2
 1 
R2 + 

 ωC
2
=
1
 1 
1+ 

 ω RC 
Rewrite the circuit in terms of impedance as shown in Fig. (b).
Find:
∆Vout =
From Figure (c),
∆V ab =
ZR
∆V ab
ZR + ZC
ZR
∆Vin
[1]
[
ZC
]
a
]
(ZR + ZC ) ZR + ZC||(ZR + ZC )
∆Vin
∆Vab
∆Vout
∆Vin
Figure (b)
 1

1
ZR 
+

Z
Z
+
Z
R
C
 C
=
−1

 1
 
1
(ZR + ZC )ZR +  Z + Z + Z  
 C


R
C
ZR
∆Vin
∆Vout
=
∆Vin
∆Vout
=
∆Vin
 1

− R 2ω C j
3R − 
ωC

− ĵ
ωC
R
(3R)
2
 1

+
− R 2ω C
ω
C


∆Vab
b
R
R
1
where we used = −j.
j
 1 
2
3R − 
 j+ R ω C j
 ω C
=
ZR
Figure (c)
−j
where j= −1
ωC
R
a
ZC
ZC
∆Vout
ZR ZC
ZR
=
=
∆Vin
ZC ( ZC + ZR ) + ZR ( ZR + 2ZC ) 3ZR + ZC + ( ZR )2 ZC
Now, ZR = R and ZC =
∆Vout
ZR
b
−1
or
∆Vout
b
Figure (a)
ZR ZC || ( ZR + ZC )
[
ZC
ZR
ZC
ZC || ( ZR + ZC )
∆Vin
ZR + ZC || ( ZR + ZC )
So Eq. [1] becomes ∆Vout =
a
2
2
=
(
Z
1.00 × 10 3
3.00 × 10 3
)
2
+ (1592 − 628)
= 0.317
2
Chapter 33 Solutions 287
33.54
The equation for
y = mx + b ) is:
∆v(t ) =
[
( ∆v)2
[
( ∆v)
2
2( ∆Vmax ) t
− ∆Vmax
T
ω0 =
2
( ∆Vmax )
1 T
∆v
t
dt =
(
)
[
]
∫
0
T
T
]
=
]
( ∆Vmax )2  T  [2t T − 1] 3
=
ave
ave
∆Vrms =
33.55
∆v(t ) during the first period (using
 2
T
[
( ∆v)2
]
ave
=
3
( ∆Vmax )2
3
2
∫0

 T t − 1 dt
t=T
=
( ∆Vmax )2 (+1)3 − (−1)3 = ( ∆Vmax )2
[
]
t=0
=
2
T2
6
3
∆Vmax
3
1
1
=
= 2000 s −1
−6
LC
(0.0500 H)(5.00 × 10 F)
so the operating frequency of the circuit is ω =
Using Equation 33.35,
P=
P=
( ∆Vrms )2 Rω 2
(
R 2ω 2 + L2 ω 2 − ω 02
(400)2 (8.00)(1000)2
[
ω0
= 1000 s −1
2
(8.00)2 (1000)2 + (0.0500)2 (1.00 − 4.00) × 106
]
2
)
(Q ≈ 12.5)
2
= 56.7 W
Figure for Goal
Solution
Goal Solution
A series RLC circuit consists of an 8.00-Ω resistor, a 5.00-µ F capacitor, and a 50.0-mH inductor. A variable
frequency source applies an emf of 400 V (rms) across the combination. Determine the power delivered
to the circuit when the frequency is equal to one half the resonance frequency.
G: Maximum power is delivered at the resonance frequency, and the power delivered at other
frequencies depends on the quality factor, Q. For the relatively small resistance in this circuit, we
could expect a high Q = ω 0 L R . So at half the resonant frequency, the power should be a small
2
fraction of the maximum power, P av, max = ∆Vrms
R = ( 400 V ) 8 Ω = 20 kW.
2
O: We must first calculate the resonance frequency in order to find half this frequency. Then the power
delivered by the source must equal the power taken out by the resistor. This power can be found
2
R where I rms = ∆Vrms / Z.
from P av = I rms
© 2000 by Harcourt, Inc. All rights reserved.
288 Chapter 33 Solutions
A : The resonance frequency is
f0 =
1
=
2 π LC 2 π
1
(0.0500 H)( 5.00 × 10 −6
F
)
= 318 Hz
The operating frequency is f = f 0 / 2 = 159 Hz . We can calculate the impedance at this frequency:
XL = 2 π f L = 2 π (159 Hz)(0.0500 H) = 50.0 Ω
and
XC =
1
1
=
= 200 Ω
2 π f C 2 π (159 Hz) 5.00 × 10 -6 F
(
)
Z = R 2 + (XL − XC )2 = 8.00 2 + (50.0 − 200)2 Ω = 150 Ω
I rms =
So,
∆Vrms 400 V
=
= 2.66 A
150 Ω
Z
The power delivered by the source is the power dissipated by the resistor:
P av = I rms 2 R = (2.66 A)2 (8.00 Ω) = 56.7 W
L : This power is only about 0.3% of the 20 kW peak power delivered at the resonance frequency. The
significant reduction in power for frequencies away from resonance is a consequence of the relatively
high Q -factor of about 12.5 for this circuit. A high Q is beneficial if, for example, you want to listen
to your favorite radio station that broadcasts at 101.5 MHz, and you do not want to receive the signal
from another local station that broadcasts at 101.9 MHz.
R=
∆V
12.0 V
I = 0.630 A = 19.0 Ω
The impedance of the circuit is Z =
∆Vrms
24.0 V
=
= 42.1 Ω
0.570 A
I rms
The resistance of the circuit is
33.56
Z 2 = R 2 + ω 2 L2
L=
33.57
1
1
Z2 − R2 =
(42.1)2 − (19.0)2 = 99.6 mH
377
ω
(a)
When ω L is very large, the bottom branch carries negligible current. Also, 1/ω C will be
negligible compared to 200 Ω and 45.0 V/200 Ω = 225 mA flows in the power supply and the
top branch.
(b)
Now 1/ω C → ∞ and ω L → 0 so the generator and bottom branch carry 450 mA
Chapter 33 Solutions 289
33.58
(a)
With both switches closed, the current goes only through
generator and resistor.
i(t) =
∆Vmax
cos ω t
R
1 ( ∆Vmax )
2
R
(b)
P=
(c)
i(t) =
2
∆Vmax
cos [ω t + Arctan(ω L / R)]
R 2 + ω 2 L2
1 

ω L−
 0
ω 0C 
0 = φ = Arctan 



R


(d) For
We require ω 0 L =
1
, so
ω0 C
(e)
At this resonance frequency,
(f)
U = 21 C ( ∆VC ) = 21 C I 2 XC 2
1
ω 02 L
Z= R
2
2
U max = 21 CI max
XC 2 = 21 C
(g)
C=
U max =
1 LI 2
2 max
=
(h) Now ω = 2ω 0 =
1L
2
( ∆Vmax )2
R2
1
=
ω 0 2C 2
( ∆Vmax )2 L
2R 2
( ∆Vmax )2
R2
2
LC
1 


L 1 L
ωL −
2 C − 2 C 

 3 L
ωC 
So φ = Arctan 
 = Arctan 
 = Arctan 

 2R C 




R
R




(i)
Now ω L =
1 1
2 ωC
ω=
1
2 LC
=
ω0
2
© 2000 by Harcourt, Inc. All rights reserved.
290 Chapter 33 Solutions
33.59
(a)
As shown in part (b),
circuit (a) is a high-pass filter
and
circuit (b) is a low-pass filter .
∆Vin
(b)
For circuit (a),
For circuit (b),
33.60
(a)
(b)
I R, rms =
∆Vout
=
∆Vin
RL2 + ( XL − XC )
2
=
As ω → 0,
∆Vout
≈ ω RLC ≈ 0
∆Vin
As ω → ∞ ,
∆Vout
≈1
∆Vin
∆Vout
=
∆Vin
+ ( X L − XC )
RL2 + (ω L)
RL2 + (ω L − 1 ω C )
2
Circuit (a)
(high-pass filter)
XC
RL2
2
=
As ω → 0,
∆Vout
≈1
∆Vin
As ω → ∞ ,
∆Vout
1
≈ 2
≈0
∆Vin
ω LC
∆V out
∆Vin
1 ωC
RL2
+ (ω L − 1 ω C )
2
Circuit (b)
(low-pass filter)
∆Vrms 100 V
=
= 1.25 A
80.0 Ω
R
The total current will lag
phasor diagram at the right.
I L, rms =
IR
∆V
φ
the applied voltage as seen in the
IL
I
∆Vrms
100 V
=
= 1.33 A
XL
2 π 60.0 s −1 (0.200 H)
(
Thus, the phase angle is:
*33.61
RL2 + XL2
∆Vout
2
)
 I L, rms 
−1  1.33 A 
φ = tan −1 
 = tan  1.25 A  = 46.7°
I
 R, rms 
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 10¢ per kWh). Suppose the transmission line is at 20 kV. Then
I rms =
(20 000)(500 W)
P
=
∆Vrms
20 000 V
~10 3 A
If the transmission line had been at 200 kV, the current would be only ~10 2 A .
Chapter 33 Solutions 291
L = 2.00 H, C = 10.0 × 10– 6 F, R = 10.0 Ω, ∆v(t) = (100 sin ω t)
33.62
(a)
The resonant frequency ω0 produces the maximum current and thus the maximum power
dissipation in the resistor.
ω0 =
1
1
=
= 224 rad/s
LC
(2.00)(10.0 × 10 − 6 )
( ∆Vmax )2
(b)
P=
(c)
I rms =
2R
=
∆Vrms
=
Z
2
I rms
R=
∆Vrms

1 
R2 +  ω L −
ω C 

( )
1 2
I rms
2
(100)2
= 500 W
2(10.0)
max
2
R
∆Vrms
R
and
( Irms )max =
or
( ∆Vrms )2 R = 1 ( ∆Vrms )2 R
Z2
R2
2
2

1 
R2 +  ω L −
= 2R 2
ω C 

This occurs where Z 2 = 2R 2:
ω 4 L2C 2 − 2Lω 2C − R 2ω 2C 2 + 1 = 0
L2C 2ω 4 − (2LC + R 2C 2 )ω 2 + 1 = 0
or
[(2.00) (10.0 × 10 ) ]ω − [2(2.00)(10.0 × 10
−6 2
2
4
−6
]
) + (10.0)2 (10.0 × 10 − 6 )2 ω 2 + 1 = 0
Solving this quadratic equation, we find that
ω 2 = 51 130,
ω 1 = 48 894 = 221 rad/s
ω 2 = 51 130 = 226 rad/s
and
48 894
R = 200 Ω, L = 663 mH, C = 26.5 µ F, ω = 377 s −1 , ∆Vmax = 50.0 V
33.63
ω L = 250 Ω,
(a)
I max =
 1 
 ω C  = 100 Ω,


Z = R 2 + ( XL − XC ) = 250 Ω
2
∆Vmax 50.0 V
=
= 0.200 A
250 Ω
Z
φ = tan −1 

X L − XC 
= 36.8°

R
(∆V leads I)
(b)
∆V R, max = I max R = 40.0 V at φ = 0°
(c)
∆VC, max =
(d)
∆VL, max = I maxω L = 50.0 V at φ = + 90.0°
I max
= 20.0 V at φ = – 90.0°
ωC
(I leads ∆V)
(∆V leads I)
© 2000 by Harcourt, Inc. All rights reserved.
292 Chapter 33 Solutions
2
P = I rms
R=
*33.64
2
 ∆Vrms 
R,
 Z 
(120 V )2
Z
2
( 40.0 Ω) :
(120)2 ( 40.0)
250 =
( 40.0)
1=
250 W =
so
2

1
+ 2 π f (0.185) −

2 π f 65.0 × 10 − 6

(
)




2
2304 f 2
1600 f 2 + 1.3511 f 4 − 5692.3 f 2 + 5 995 300
f2=
6396.3 ±
(6396.3)2 − 4(1.3511)( 5 995 300)
2(1.3511)
Z = R 2 + (ω L − 1 ω C )
2
576 000 f 2
and
250 =
so
1.3511 f 4 − 6396.3 f 2 + 5 995 300 = 0
(
1600 f 2 + 1.1624 f 2 − 2448.5
)
2
= 3446.5 or 1287.4
f = 58.7 Hz or 35.9 Hz
33.65
(a)
From Equation 33.39,
N1 ∆V1
=
N 2 ∆V 2
Let output impedance
Z1 =
so that
N1 Z1I1
=
N 2 Z2 I 2
∆V1
I1
and the input impedance Z2 =
But from Eq. 33.40,
N1
=
N2
So, combining with the previous result we have
(b)
N1
=
N2
IR =
33.66
Z1
=
Z2
∆Vrms
;
R
8000
= 31.6
8.00
IL =
∆Vrms
;
ωL
IC =
(a)
I rms = I R2 + (IC − I L )2 = ∆Vrms
(b)
tan φ =
∆Vrms
(ω C)−1
1 
 1  
+ ωC −
 R 2  
ω L 
 1

IC − I L
1 
1
= ∆Vrms 
−


X
X
IR
/
R
∆V


L
rms
 C
 1
1 
tan φ = R 
−

X
X
L
 C
2
Z1
Z2
∆V 2
I2
I1 ∆V 2 N 2
=
=
I 2 ∆V1 N1
Chapter 33 Solutions 293
33.67
(a)
I rms = ∆Vrms
1 
1 
+ ωC −
2
ω L 
R

2
∆Vrms → ( ∆Vrms )max when ω C =
f=
f=
(b)
1
ωL
1
2 π LC
1
2 π 200 × 10
−3
H)(0.150 × 10 − 6 F)
= 919 Hz
IR =
∆Vrms 120 V
=
= 1.50 A
80.0 Ω
R
IL =
∆Vrms
120 V
=
= 1.60 A
ωL
(374 s −1 )(0.200 H)
IC = ∆Vrms (ω C) = (120 V)(374 s −1 )(0.150 × 10 − 6 F) = 6.73 mA
(c)
I rms = I R2 + (IC − I L )2 = (1.50)2 + (0.00673 − 1.60)2 = 2.19 A
(d)
I − I 
0.00673 − 1.60 
φ = tan −1  C L  = tan −1 
 = – 46.7°
1.50

 IR 
The current is lagging the voltage .
33.68
(a)
tan φ =
∆VL I (ω L) ω L
=
=
∆V R
IR
R
Thus, R =
(b)
(
L
∆Vin
)
200 s (0.500 H)
ωL
= 173 Ω
=
tan( 30.0°)
tan φ
−1
∆Vout ∆V R
=
= cos φ
∆Vin
∆Vin
R
∆V = IZ
∆VL = IXL
∆Vout = ( ∆Vin ) cos φ = (10.0 V ) cos 30.0° = 8.66 V
© 2000 by Harcourt, Inc. All rights reserved.
φ
∆VR = IR
∆V out
294 Chapter 33 Solutions
33.69
(a)
XL = XC = 1884 Ω
L=
f = 2000 Hz
when
XL
1884 Ω
=
= 0.150 H
2 π f 4000 π rad s
XL = 2 π f (0.150 H) XC =
C=
and
1
(2π f )( 4.22 × 10
−8
F
)
(b)
33.70
f (Hz)
X L (Ω)
XC (Ω)
Z (Ω)
300
600
800
1000
1500
2000
3000
4000
6000
10000
283
565
754
942
1410
1880
2830
3770
5650
9420
12600
6280
4710
3770
2510
1880
1260
942
628
377
12300
5720
3960
2830
1100
40
1570
2830
5020
9040
ω0 =
1
= 1.00 × 106 rad s
LC
For each angular frequency, we
find
Z = R 2 + (ω L − 1/ ω C )
then
I = (1.00 V ) / Z
and
P = I 2 (1.00 Ω)
2
The full width at half maximum is:
∆f =
∆ω (1.0005 − 0.9995)ω 0
=
2π
2π
∆f =
1.00 × 10 3 s −1
= 159 Hz
2π
while
1.00 Ω
R
=
= 159 Hz
2 π L 2 π 1.00 × 10 −3 H
(
)
Z=
1
1
=
(2π f )XC ( 4000π rad s)(1884 Ω)
= 42.2 nF
( 40.0 Ω)2 + (XL − XC )2
Impedence, Ω
ω ω0
ω L (Ω )
0.9990
0.9991
0.9993
0.9995
0.9997
0.9999
1.0000
1.0001
1.0003
1.0005
1.0007
1.0009
1.0010
999.0
999.1
999.3
999.5
999.7
999.9
1000
1000.1
1000.3
1000.5
1000.7
1000.9
1001
1
ωC
(Ω )
1001.0
1000.9
1000.7
1000.5
1000.3
1000.1
1000.0
999.9
999.7
999.5
999.3
999.1
999.0
Z (Ω )
2.24
2.06
1.72
1.41
1.17
1.02
1.00
1.02
1.17
1.41
1.72
2.06
2.24
P = I 2 R (W)
0.19984
0.23569
0.33768
0.49987
0.73524
0.96153
1.00000
0.96154
0.73535
0.50012
0.33799
0.23601
0.20016
Chapter 33 Solutions 295
∆Vout
=
∆Vin
33.71
(a)
R
R 2 + (1 ω C )
2
=
R
C
R 2 + (1 2 π f C )
2
∆Vin
∆Vout 1
1
=
when
=R 3
2
∆Vin
ωC
Hence, f =
∆Vout
R
1
ω
=
= 1.84 kHz
2 π 2 π RC 3
Log Gain versus Log Frequency
(b)
0
-1
Log∆V
out / ∆ V in
-2
-3
-4
0
1
2
3
Log f
© 2000 by Harcourt, Inc. All rights reserved.
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5
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