Physics 9 Fall 2009
Homework 9 - Solutions
1. Chapter 36 - Exercise 15.
What are VR and VC if the emf frequency in the
figure is 10 kHz?
————————————————————————————————————
Solution
The voltage are VR = IR, and VC = IXC , where the capitative reactance is XC =
1/ωC = 1/2πf C. The peak current is I = √ E20 2 . So, the peak current is
R +XC
10
E0
=q
I=p
= 0.04 A.
2
2
R + XC
1502 + (2π (104 ) 80 × 10−9 )−2
So, we can now find
VR = IR
= 0.04 × 150
= 6.0 V
0.04
VC = IXC = 2π(104 )(80×10
=
8.0 V
−9 )
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2. Chapter 36 - Exercise 27.
A series RLC circuit consists of a 50 Ω resistor, a 3.3 mH inductor, and a 480 nF
capacitor. It is connected to an oscillator with a peak voltage of 5.0 V. Determine the
impedance, the peak current, and the phase angle at frequencies (a) 3000 Hz, (b) 4000
Hz, and (c) 5000 Hz.
————————————————————————————————————
Solution
The impedance is
Z=
q
s
R2
2
+ (XC − XL ) =
R2
+ 2πf L −
1
2πf C
2
,
−1 XL −XC
=
while hthe peak current
is
I
=
E
/Z,
and
the
phase
angle
is
φ
=
tan
0
R
i
C
tan−1 2πf L−1/2πf
. So, all we have to do is to plug in each of values for the different
R
frequencies. Doing so, we get the following table
3000 Hz
Z
70 Ω
I 0.072 A
φ
−44◦
4000 Hz
50 Ω
0.100 A
0◦
2
5000 Hz
62 Ω
0.080 A
37◦
3. Chapter 36 - Exercise 31.
The motor of an electric drill draws a 3.5 A current at the power-line voltage of 120 V
rms. What is the motor’s power if the current lags the voltage by 20◦ ?
————————————————————————————————————
Solution
The average power supplied by the motor is Psource = Irms Vrms cos φ, where Irms is the
current drawn by the motor (3.5 A, in this case). So, we just plug everything in.
Psource =
Irms Vrms cos φ
= (3.5) (120) cos 20◦
=
400 W.
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4. Chapter 36 - Problem 37.
(a) Evaluate VC in the figure at emf frequencies
1, 3, 10, 30, and 100 kHz.
(b) Graph VC versus frequency. Draw a smooth
curve through your five points.
————————————————————————————————————
Solution
(a) The voltage across the capacitor is VC = IXC , where I = √
1
ωC
=
1
.
2πf C
E0
,
2
R2 +XC
So,
E0
VC = IXC =
2πf C
q
R2 +
1
(2πf C)2
E0
=q
.
2
1 + (2πf RC)
So, plugging in the values gives
f (kHz) VC (V)
1
9.95
3
9.57
10
7.05
30
3.15
100
0.990
The graph is seen to the right. Notice that
the voltage approaches zero as the frequency in(b)
creases. The circuit acts like a low-pass filter,
blocking out higher frequencies.
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and XC =
5. Chapter 36 - Problem 38.
(a) Evaluate VR in the figure at emf frequencies
100, 300, 1000, 3000, and 10,000 Hz.
(b) Graph VR versus frequency. Draw a smooth
curve through your five points.
————————————————————————————————————
Solution
(a) The voltage across the resistor is VR = IR, where I = √
1
.
2πf C
So,
E0
.
VR = IR = q
R2 + (2πf1C)2
So, plugging in the values gives
f (Hz) VR (V)
100
1.00
300
2.89
1000
7.09
3000
9.49
10,000
9.95
The graph is seen to the right. Notice that the
voltage approaches the full 10 V as the frequency
(b) increases, while for low frequencies the voltage
drops. The circuit acts like a high-pass filter,
blocking out lower frequencies.
5
E0
,
2
R2 +XC
and XC =
1
ωC
=
6. Chapter 36 - Problem 42.
(a) What is the peak current supplied by the
emf in the figure?
(b) What is the peak voltage across the 3.0 µF
capacitor?
————————————————————————————————————
Solution
(a) The circuit can be redrawn in terms of a single equivalent capacitor. The two
parallel capacitors combine together to give an equivalent capacitance of 6 µF,
which then combines with the 3 µF capacitor for a net capacitance of 2 µF.
E0
Then, the current is I = XVC = 1/2πf
= 2πf CE0 . Thus, the peak current is
C
−9
I = 2πf CE0 = 2π (200) (2 × 10 ) (10) = 25 mA.
(b) All the current passes though the 3 µF capacitor, and so the peak voltage is
Ceqv
E0 =
VC = IXC = 2πfI C . Notice that since I = 2πf Ceqv , then VC = 2πfI C = 2πf
2πf C
Ceqv
E0 .
C
So, we find
VC =
Ceqv
2
E0 = × 10 = .67 × 10 = 6.7 V.
C
3
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7. Chapter 36 - Problem 43.
You have a resistor and a capacitor of unknown values. First, you charge the capacitor
and discharge it through the resistor. By monitoring the capacitor voltage on an
oscilloscope, you see that the voltage decays to half its initial value in 2.5 ms. You
then use the resistor and capacitor to make a low-pass filter. What is the crossover
frequency fc ?
————————————————————————————————————
Solution
1
ωc
= 2πRC
. The voltage across a capacitor falls off
The crossover frequency is fc = 2π
−t/RC
exponentially fast as V = V0 e
. If the voltage falls to half its initial value in a
time t0 , then
V0
1
V (t0 ) =
= V0 e−t0 /RC ⇒ = e−t0 /RC .
2
2
1
,
Taking natural logs of both sides and solving gives RC = lnt02 . Then, since fc = 2πRC
we have
ln 2
fc =
.
2πt0
With numbers,
0.69
ln 2
=
≈ 44 Hz.
fc =
2πt0
2π (2.5 × 10−3 )
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8. Chapter 36 - Problem 51.
For the circuit in the figure,
(a) What is the resonance frequency, in both
rad/s and Hz?
(b) Find VR and VC at resonance.
(c) How can VC be larger than E0 ? Explain.
————————————————————————————————————
Solution
(a) The resonance frequency is ωR =
ωc =
fc =
√1
LC
1
√
2π LC
=
=
√1 ,
LC
and so fR =
1
10−3 ×10−6
√ 1
2π 10−3 ×10−6
√
ωc
2π
=
1
√
.
2π LC
Thus,
= 3.2 × 104 rad/s.
=
5 × 103 Hz.
(b) At resonance, XC = XL , and so Z = R, giving I = E0 /R, or 1 amp. This gives
VR = IR = 1 × 10 = 10 V. Furthermore, VC = IXC = I/ωR C. We found the
resonance frequency in part (a), and so substituting in gives VC = 3.2×1014 ×10−6 =
32 V.
(c) We know that, for the instantaneous values, vR + vC + vL = E, but this isn’t true
for the peak values. At resonance, vL and vC cancel each other out, but can each
be big, if they are compensated by an equally big vR . This lets the peak values
be larger than the peak current, due to the cancellation.
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9. Chapter 36 - Problem 57.
Show that the impedance of a series RLC circuit can be written
q
2
Z = R2 + ω 2 L2 (1 − ω02 /ω 2 ) .
————————————————————————————————————
Solution
q
1
The impedance Z = R2 + (XL − XC )2 , where XL ≡ ωL, and XC = ωC
. Now,
2
2
1
(XL − XC )2 = ωL − ωC
= ω 2 L2 1 − ω21LC . But, LC = ω12 , where ω0 is the
0
ω2
natural resonant frequency. Thus, (XL − XC )2 = ω 2 L2 1 − ω02 . So, we find that
s
Z=
R2
+
9
ω 2 L2
ω02
1− 2 .
ω
10. Chapter 36 - Problem 68.
(a) Show that the average power loss in a series RLC circuit is
Pavg =
2
R
ω 2 Erms
2.
ω 2 R2 + L2 (ω 2 − ω02 )
(b) Prove that the energy dissipation is a maximum at ω = ω0 .
————————————————————————————————————
Solution
(a) The average power dissipated is Pavg = Irms Erms cos φ, where Irms = Erms /Z, and
2
R/Z 2 . Now,
cos φ = R/Z. Thus, Pavg = Erms
Z 2 = R2 + (XL − XC )2
1 2
= R2 + ωL − ωC
2
1 2
= R2 + Lω2 ω 2 − LC
2
2
= R2 + Lω2 (ω 2 − ω02 ) ,
since 1/LC = ω02 . So,
Pavg =
2
R
Erms
R2 +
L2
ω2
2
(ω 2 − ω02 )
=
2
Rω 2
Erms
2
R2 ω 2 + L2 (ω 2 − ω02 )
(b) The power dissipated depends on the frequency, ω. It is a maximum when
d
P = 0. So, taking the derivative gives
dω avg
2
2
2
2Erms
Rω R2 ω 2 + L2 (ω 2 − ω02 ) − Erms
Rω 2 (2ωR2 + 2L2 (ω 2 − ω02 ) · 2ω)
= 0.
2
2 2
2
2
2
2
R ω + L (ω − ω0 )
2
This reduces to (ω 2 − ω02 ) − 2ω 2 (ω 2 − ω02 ) = 0, which is solved by ω = ω0 . So,
the power is maximized when ω = ω0 , i.e., when the system is at resonance!
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