Download 23.1 Induced Emf and Magnetic Flux

OpenStax College Physics
Instructor Solutions Manual
Chapter 23
CHAPTER 23: ELECTROMAGNETIC
INDUCTION, AC CIRCUITS, AND
ELECTRICAL TECHNOLOGIES
23.1 INDUCED EMF AND MAGNETIC FLUX
1.
What is the value of the magnetic flux at coil 2 in Figure 23.56 due to coil 1?
Solution Using the equation Φ  BA cos , we can calculate the flux through coil 2, since the
coils are perpendicular: Φ  BA cos  BA cos 90  0
2.
What is the value of the magnetic flux through the coil in Figure 23.56(b) due to the
wire?
Solution
Φ  BA cos  BA cos 90  0
23.2 FARADAY’S LAW OF INDUCTION: LENZ’S LAW
3.
Referring to Figure 23.57(a), what is the direction of the current induced in coil 2: (a)
If the current in coil 1 increases? (b) If the current in coil 1 decreases? (c) If the current
in coil 1 is constant? Explicitly show how you follow the steps in the Problem-Solving
Strategy for Lenz's Law.
Solution
(a) CCW
(b) CW
(c) no current in coil 2
4.
Referring to Figure 23.57(b), what is the direction of the current induced in the coil:
(a) If the current in the wire increases? (b) If the current in the wire decreases? (c) If
the current in the wire suddenly changes direction? Explicitly show how you follow the
steps in the Problem-Solving Strategy for Lenz’s Law.
Solution
(a) CCW
(b) CW
(c) CW
5.
Referring to Figure 23.58, what are the directions of the currents in coils 1, 2, and 3
(assume that the coils are lying in the plane of the circuit): (a) When the switch is first
closed? (b) When the switch has been closed for a long time? (c) Just after the switch
is opened?
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Solution
(a) (i) CCW (ii) CCW (iii) CW
(b) (i) no current (ii) no current (iii) no current
(c) (i) CW (ii) CW (iii) CCW
6.
Repeat the previous problem with the battery reversed.
Solution
(a) (i) CW (ii) CW (iii) CCW
(b) (i) no current (ii) no current (iii) no current
(c) (i) CCW (ii) CCW (iii) CW
7.
Verify that the units of Φ / t are volts. That is, show that 1 T  m 2 / s  1 V .
Solution
The units of
ΔΦ
will be:
Δt
ΔΦ  T  m 2  N A  m m 2   N  m  N  m  V so that 1 T  m 2 /s  1 V
 s  A s
Δt 
s
C


8.
Solution
9.
Solution
Suppose a 50-turn coil lies in the plane of the page in a uniform magnetic field that is
directed into the page. The coil originally has an area of 0.250 m 2 . It is stretched to
have no area in 0.100 s. What is the direction and magnitude of the induced emf if
the uniform magnetic field has a strength of 1.50 T?
EN


ΔΦ NBA (50)1.50 T  0.250 m 2


 187.5 V CW  188 V CW
Δt
t
0.100 s
(a) An MRI technician moves his hand from a region of very low magnetic field
strength into an MRI scanner’s 2.00 T field with his fingers pointing in the direction of
the field. Find the average emf induced in his wedding ring, given its diameter is 2.20
cm and assuming it takes 0.250 s to move it into the field. (b) Discuss whether this
current would significantly change the temperature of the ring.


ΔΦ N B A (1)2.00 T   1.10 10-2 m 2


 3.04 10-3 V  3.04 mV
Δt
t
0.250 s
2
V
(b) Since power goes as P 
, the smaller the resistance the larger the power and
R
the more heat transferred to the ring. So, as a lower limit on the ring, estimate
R  1.00 m . In this case, the power would be: P  9.24 mW . This power exists
for 1 of a second, so the heat transferred will be:
4
Q  Pt  9.24  10 3 W 0.250 s   2.31 mJ .
This is not a significant amount of heat.
(a) E  N
2
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10.
Solution
Instructor Solutions Manual
Chapter 23
Referring to the situation in the previous problem: (a) What current is induced in the
ring if its resistance is 0.0100  ? (b) What average power is dissipated? (c) What
magnetic field is induced at the center of the ring? (d) What is the direction of the
induced magnetic field relative to the MRI’s field?
V 3.04 10 3 V

 0.304 A
R
0.0100 Ω
(b) P  I 2 R  0.304 A 2 0.0100 Ω   0.924 mW
(a) I 
(c) B 
0 I

4π  10 T.m/A 0.304 A  1.74  10
21.10  10 m 
7
5
T
2r
(d) Looking in the direction of the magnetic field, the induced current is
counterclockwise.
11.
Solution
12.
Solution
13.
Solution
2
An emf is induced by rotating a 1000-turn, 20.0 cm diameter coil in the Earth’s
5.00  10 5 T magnetic field. What average emf is induced, given the plane of the coil
is originally perpendicular to the Earth’s field and is rotated to be parallel to the field
in 10.0 ms?
EN

 
ΔΦ NBA (1000) 5.00  10 -5 T   10.0  10 -2 m


Δt
t
10.0  10 -3 s



2
 0.157 V
A 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms,
originally having its plane perpendicular to a uniform magnetic field. (This is 60 rev/s.)
Find the magnetic field strength needed to induce an average emf of 10,000 V.
EN


ΔΦ NBA
Et (10,000 V) 4.17  10-3 s

B

 0.425 T
Δt
t
NA
500 0.250 m2
Approximately how does the emf induced in the loop in Figure 23.57(b) depend on the
distance of the center of the loop from the wire?
The magnetic field due to an infinite straight wire is
emf induced in the loop will be proportional to
wire.
μ0 I
, so the flux and hence the
2π r
1
, where r is the distance from the
r
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Chapter 23
14.
A lightning bolt produces a rapidly varying magnetic field. If the bolt strikes the earth
vertically and acts like a current in a long straight wire, it will induce a voltage in a
loop aligned like that in Figure 23.57(b). What voltage is induced in a 1.00 m diameter
loop 50.0 m from a 2.00 106 A lightning strike, if the current falls to zero in
25.0 μs ? (b) Discuss circumstances under which such a voltage would produce
noticeable consequences.
Solution
(a) We know E 0  
NΦ
, where the minus sign means that the emf creates the
t
current and magnetic field that opposes the change in flux, and Φ  BA  πr 2 B .
Since the only thing that varies in the magnetic flux is the magnetic field, we can
I
2
then say that Φ  r B . Now, since B  0 the change in magnetic field
2d
 I
occurs because of a change in the current, or B  0 . Finally, substituting
2d
NΦ
these into the equation E 0  
gives:
t
Nr 2  0 I
 0 Nr 2 I
E0  

2dt
2dt
4  10



T  m/A 10.500 m   2.00  10 6 A
 251 V
250.0 m  2.50  10 5 s
(b) An example is the alternator in your car. If you were driving during a lightning
storm and this large bolt of lightning hit at 50.0 m away, it is possible to fry the
alternator of your battery because of this large voltage surge. In addition, the hair
at the back of your neck would stand on end because it becomes statically
charged.

7
2


23.3 MOTIONAL EMF
15.
Use Faraday’s law, Lenz’s law, and RHR-1 to show that the magnetic force on the
current in the moving rod in Figure 23.11 is in the opposite direction of its velocity.
I
Solution
l
R
I
v
Because the flux through the loop (into the page) is increasing, a magnetic field out
of the page is induced. This requires the CCW current in the loop. The force on the
current in the rod, from the B field (into the page), is to the left.
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Chapter 23
16.
If a current flows in the Satellite Tether shown in Figure 23.12, use Faraday’s law,
Lenz’s law, and RHR-1 to show that there is a magnetic force on the tether in the
direction opposite to its velocity.
Solution
The flux through the loop (into the page) is increasing because the loop is getting
larger and enclosing more magnetic field. Thus, a magnetic field (out of the page) is
induced to oppose the change in flux from the original field. Using RHR-2, point your
fingers out of the page within the loop, then your thumb points in the
counterclockwise direction around the loop, so the induced magnetic field is
produced by the induction of a counterclockwise current in the circuit. Finally, using
RHR-1, putting your right thumb in the direction of the current and your fingers into
the page (in the direction of the magnetic field), your palm points to the left. This
means the magnetic force on the wire is to the left (in the direction opposite to its
velocity).
17.
(a) A jet airplane with a 75.0 m wingspan is flying at 280 m/s. What emf is induced
between wing tips if the vertical component of the Earth’s field is 3.00  10 5 T ? (b) Is
an emf of this magnitude likely to have any consequences? Explain.
Solution
18.
Solution
(a) E  Blv  3.00  10 5 T 75.0 m280 m/s   0.630 V
(b) No, this is a very small emf.
(a) A nonferrous screwdriver is being used in a 2.00 T magnetic field. What maximum
emf can be induced along its 12.0 cm length when it moves at 6.00 m/s? (b) Is it likely
that this emf will have any consequences or even be noticed?
(a) E  Blv  2.00 T0.120 m6.00 m/s   1.44 V
(b) No.
19.
At what speed must the sliding rod in Figure 23.11 move to produce an emf of 1.00 V
in a 1.50 T field, given the rod’s length is 30.0 cm?
Solution
E  Blv  v 
20.
The 12.0 cm long rod in Figure 23.11 moves at 4.00 m/s. What is the strength of the
magnetic field if a 95.0 V emf is induced?
Solution
E  Blv  B 
21.
Prove that when B ,  , and v are not mutually perpendicular, motional emf is given
by emf  Bv sin  . If v is perpendicular to B , then  is the angle between  and B .
If  is perpendicular to B , then  is the angle between v and B .
E
1.00 V

 2.22 m/s
Bl 1.50 T 0.300 m 
E
95.0 V

 197.9 T  198 T
lv 0.120 m 4.00 m/s 
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Solution
Instructor Solutions Manual
Chapter 23
Case i lB
B
B
B

v
Side view of Figure 23.11 with angled field.
Case ii vB

v
End view of Figure 23.11 with angled field.
ΔΦ BA cos90   
Δx
in either case E 

 B sin  l
 Blv sin 
Δt
t
Δt
22.
In the August 1992 space shuttle flight, only 250 m of the conducting tether
considered in Example 23.2 could be let out. A 40.0 V motional emf was generated in
the Earth’s 5.00 105 T field, while moving at 7.80 103 m/s . What was the angle
between the shuttle’s velocity and the Earth’s field, assuming the conductor was
perpendicular to the field?
Solution


40.0 V
E  Blv sin     sin 1 
  24.2
5
3
 5.00  10 T 250 m 7.80  10 m/s 
23.
Integrated Concepts Derive an expression for the current in a system like that in
Figure 23.11, under the following conditions. The resistance between the rails is R ,
the rails and the moving rod are identical in cross section A and have the same
resistivity  . The distance between the rails is l, and the rod moves at constant speed
v perpendicular to the uniform field B . At time zero, the moving rod is next to the
resistance R .
Solution
From R 

l
A
the resistance of the rails is

2 vt
.
A

l
. The total resistance, R tot , is then
A
 2vt  l   RA
E
BlvA

; E  Blv  I 

A
Rtot  2vt  l   RA
The resistance of the moving bar is
Rtot

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24.
Integrated Concepts The Tethered Satellite in Figure 23.12 has a mass of 525 kg and
is at the end of a 20.0 km long, 2.50 mm diameter cable with the tensile strength of
steel. (a) How much does the cable stretch if a 100 N force is exerted to pull the
satellite in? (Assume the satellite and shuttle are at the same altitude above the
Earth.) (b) What is the effective force constant of the cable? (c) How much energy is
stored in it when stretched by the 100 N force?
Solution
2.00 10 4 m 100 N 
 LA  LF


(a) F   

A 2.10 1011 N/m 2 π  1.25 10 3 m
 L 
F
100 N
(b) F  kL  k 

 51.54 N/m
L 1.94 m
1
2
(c) v  kL2  0.551.54 N/m 1.94 m   97.0 J
2
25.
Solution



 

2
 1.94 m
Integrated Concepts The Tethered Satellite discussed in this module is producing 5.00
kV, and a current of 10.0 A flows. (a) What magnetic drag force does this produce if
the system is moving at 7.80 km/s? (b) How much kinetic energy is removed from the
system in 1.00 h, neglecting any change in altitude or velocity during that time?
(c) What is the change in velocity if the mass of the system is 100,000 kg? (d) Discuss
the long term consequences (say, a week-long mission) on the space shuttle’s orbit,
noting what effect a decrease in velocity has and assessing the magnitude of the
effect.
(a) F  IlB  10.0 A2.00 10 4 m5.00 10 5 T   10.0 N
(b) W  Fx  Fvt  10.0 N 7.80  10 3 m/s 3600 s   2.81 108 J
(c) 1 mv' 2  1 mv 2  W ;
2
2
2W 

v '  v 2 
m 

12



2
2 2.81  10 8 J   7.79964  10 3 m/s
  7.80  10 3 m/s 

1.00  10 5 kg 

Δv  v-v'  0.36 m/s  0;
Note that five digits are needed to see the difference between v and v .
(d) For a week-long mission (168 hours), the change in velocity will be 60 m/s, or
approximately 1%. In general, a decrease in velocity would cause the orbit to start
spiraling inward because the velocity would no longer be sufficient to keep the
circular orbit. The long term consequences are that the shuttle would require a
little more fuel to maintain the desired speed, otherwise the orbit would spiral
slightly inward.


12
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23.4 EDDY CURRENTS AND MAGNETIC DAMPING
26.
Make a drawing similar to Figure 23.14, but with the pendulum moving in the
opposite direction. Then use Faraday’s law, Lenz’s law, and RHR-1 to show that
magnetic force opposes motion.
Solution
Pivot
v
B in
v
F
F
As the plate enters from the right, flux increases, so an eddy current is set up
(Faraday’s Law) in the counter-clockwise direction (Lenz’s Law). The force on the left
side of the current loop is to the right. When the plate leaves the field the eddy
current is now clockwise and again there is a force to the right on the current.
27.
A coil is moved through a magnetic field as shown in Figure 23.59. The field is uniform
inside the rectangle and zero outside. What is the direction of the induced current and
what is the direction of the magnetic force on the coil at each position shown?
Solution (a) The magnetic field is zero and not changing, so there is no current and therefore
no force on the coil.
(b) The magnetic field is increasing out of the page, so the induced magnetic field is
into the page, created by an induced clockwise current. This current creates a
force to the left.
(c) The magnetic field is not changing, so there is no current and therefore no force
on the coil.
(d) The magnetic field is decreasing out of the page, so the induced magnetic field is
out of the page, created by an induced counterclockwise current. This current
creates a force to the right.
(e) The magnetic field is zero and not changing, so there is no current and therefore
no force on the coil.
23.5 ELECTRIC GENERATORS
28.
Calculate the peak voltage of a generator that rotates its 200-turn, 0.100 m diameter
coil at 3600 rpm in a 0.800 T field.
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Solution
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Chapter 23
 2π rad  1 min 

  377 rad/s;
 1 rev  60 s 
  3600 rev/min 
E0  NAB  200π 0.050 m  0.800 T 377 rad/s   474 V
2
29.
At what angular velocity in rpm will the peak voltage of a generator be 480 V, if its
500-turn, 8.00 cm diameter coil rotates in a 0.250 T field?
Solution
E0  NAB
 
30.
Solution
E0
480 V

 763.9 rad/s  7.30  103 rpm
2
2
NR B 500π 0.0400 m  0.250 T 
What is the peak emf generated by rotating a 1000-turn, 20.0 cm diameter coil in the
Earth’s 5.00 105 T magnetic field, given the plane of the coil is originally
perpendicular to the Earth’s field and is rotated to be parallel to the field in 10.0 ms?
 π/2  rad

 157.08 rad/s
t 0.0100 s
2
E0  NAB  1000π 0.100 m  5.00  10 -5 T 157.08 rad/s   0.247 V



31.
What is the peak emf generated by a 0.250 m radius, 500-turn coil that is rotated
one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a
uniform magnetic field. (This is 60 rev/s.)
Solution
Using the information given in Exercise 23.12:
1
1
  rev  2 rad  and t  4.17  10 3 s,
4
4
2
N  500; A   r 2   0.250 m  ; and B  0.425 T,
 1 / 42  rad

 376.7 rad/s. Therefore,
we get:  
t
4.17  10 -3 s
2
E0  NAB  500 0.250 m 0.425 T376.7 rad/s   1.57  10 4 V  15.7 kV
32.
(a) A bicycle generator rotates at 1875 rad/s, producing an 18.0 V peak emf. It has a
1.00 by 3.00 cm rectangular coil in a 0.640 T field. How many turns are in the coil? (b)
Is this number of turns of wire practical for a 1.00 by 3.00 cm coil?
Solution
4
2
(a) A  0.0100 m 0.0300 m   3.00  10 m
E
18.0 V
E0  NAB  N  0 
 50.0
-4
2
ABω 3.00  10 m 0.640 T 1875 rad/s 
(b) Yes, 50 turns is practical for a coil of this size.
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33.
Solution
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Chapter 23
Integrated Concepts This problem refers to the bicycle generator considered in the
previous problem. It is driven by a 1.60 cm diameter wheel that rolls on the outside
rim of the bicycle tire. (a) What is the velocity of the bicycle if the generator’s angular
velocity is 1875 rad/s? (b) What is the maximum emf of the generator when the
bicycle moves at 10.0 m/s, noting that it was 18.0 V under the original conditions? (c)
If the sophisticated generator can vary its own magnetic field, what field strength will
it need at 5.00 m/s to produce a 9.00 V maximum emf?
(a) v  r  8.00  10 3 m1875 rad/s   15.0 m/s
E'
'
 v' 
 10.0 m/s 
(b) E0  NAB  0   E '0  E0    18.0 V 
  12.0 V
E0

v
 15.0 m/s 



rE0
NABv
8.00 10 3 m 9.00 V 
(c) E0  NAB 
B

 0.960 T
r
NAv 50 3.00 10 4 m 2 5.00 m/s 

34.
(a) A car generator turns at 400 rpm when the engine is idling. Its 300-turn, 5.00 by
8.00 cm rectangular coil rotates in an adjustable magnetic field so that it can produce
sufficient voltage even at low rpms. What is the field strength needed to produce a
24.0 V peak emf? (b) Discuss how this required field strength compares to those
available in permanent and electromagnets.
Solution
 1 min 
  41.89 rad/s;
(a)   400 rev/min 2 rad/rav 
 60 s 
E
24.0 V
E0  NAB  B 0 
 0.477 T
NAω 3000.0500 m 0.0800 m 41.89 rad/s 
(b) This field strength is small enough that it can be obtained using either a
permanent magnet or an electromagnet.
35.
Show that if a coil rotates at an angular velocity  , the period of its AC output is
2 /  .
Solution
The output voltage of the coil will be positive for half of a revolution and negative for
the other half. So one period corresponds to one revolution

1 2
  2f  f 
;T  
2
f

36.
A 75-turn, 10.0 cm diameter coil rotates at an angular velocity of 8.00 rad/s in a 1.25
T field, starting with the plane of the coil parallel to the field. (a) What is the peak
emf? (b) At what time is the peak emf first reached? (c) At what time is the emf first
at its most negative? (d) What is the period of the AC voltage output?
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 23
B
B
B
 t

B
E  E0 sin  and   90  t
(a) E0  NAB  75 0.0500 m2 1.25 T8.00 rad/s   5.89 V
(b) E  E0 sin   E0 sin 90  ωt   E0 cos t.
We know this is the correct expression for E since E must change sign when

  t  , E  E0 when t  0  peak emf is when t  0 (just after the
2
generator is turned on).
π
π
 0.393 s
(c) E   E0 when ωt  π  t  
ω 8.00 rad/s
2
2π
(d) T 

 0.785 s
 8.00 rad/s
37.
Solution
38.
(a) If the emf of a coil rotating in a magnetic field is zero at t  0 , and increases to its
first peak at t  0.100 ms , what is the angular velocity of the coil? (b) At what time
will its next maximum occur? (c) What is the period of the output? (d) When is the
output first one-fourth of its maximum? (e) When is it next one-fourth of its
maximum?
T
2
2π
 0.100 ms  T  0.400 ms , so that  

 1.57  10 4 rad/s
4
4
T
4.00  10 s
(b) 0.100 ms  T  0.500 ms
(c) T  0.400 ms
E
1
(d) E  0  E0 sin ωt  sin ωt   ωt  0.2527 rad
4
4
0.25272 rad
t 
 1.6110 5 s
4
1.57  10 rad/s
(e) t  T  1.6110 5 s  4.00 10 4 s  4.16 10 4 s
(a)
Unreasonable Results A 500-turn coil with a 0.250 m 2 area is spun in the Earth’s
5.00  10 5 T field, producing a 12.0 kV maximum emf. (a) At what angular velocity
must the coil be spun? (b) What is unreasonable about this result? (c) Which
assumption or premise is responsible?
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 23
(a) E0  NAB 
E
1.20  10 4 V
 0 
 1.92  10 6 rad/s  1.83  10 7 rpm
NAB 500 0.250 m 2 5.00  10 5 T
(b) This angular velocity is unreasonably high, higher than can be obtained for any
mechanical system.
(c) The assumption that a voltage as great as 12.0 kV could be obtained is
unreasonable.



23.6 BACK EMF
39.
Suppose a motor connected to a 120 V source draws 10.0 A when it first starts. (a)
What is its resistance? (b) What current does it draw at its normal operating speed
when it develops a 100 V back emf?
Solution
(a) R 
40.
A motor operating on 240 V electricity has a 180 V back emf at operating speed and
draws a 12.0 A current. (a) What is its resistance? (b) What current does it draw when
it is first started?
Solution
(a) R 
41.
What is the back emf of a 120 V motor that draws 8.00 A at its normal speed and 20.0
A when first starting?
Solution
V 120 V

 6.00 , so that
I
20 A
120  V   8.00 A 6.00    48.0 V
V   120 V  48.0 V  72.0 V
42.
The motor in a toy car operates on 6.00 V, developing a 4.50 V back emf at normal
speed. If it draws 3.00 A at normal speed, what current does it draw when starting?
V 120 V

 6.00 
I 20.0 A
V 120 V  100 V
(b) I  
 3.33 A
R
6.00 
V 240 V  180 V

 5.00 
I
12.0 A
V
240 V
(b) I  
 48.0 A
R 5.00 
R
OpenStax College Physics
Instructor Solutions Manual
Chapter 23
V 6.00 V  4.50 V

 0.500 , so that
I
3.00 A
V
6.00 V
I 
 12.0 A
R 0.500 
Solution
R
43.
Integrated Concepts The motor in a toy car is powered by four batteries in series,
which produce a total emf of 6.00 V. The motor draws 3.00 A and develops a 4.50 V
back emf at normal speed. Each battery has a 0.100  internal resistance. What is
the resistance of the motor?
Solution Since the resistors are in series, we know the total internal resistance of the batteries
is R  40.100  . Therefore,
E V
E V
6.00 V  4.50 V
I
, so that R  R 
 R 
 40.100 Ω   0.100 Ω
R  R
I
3.00 A
23.7 TRANSFORMERS
44.
A plug-in transformer, like that in Figure 23.29, supplies 9.00 V to a video game
system. (a) How many turns are in its secondary coil, if its input voltage is 120 V and
the primary coil has 400 turns? (b) What is its input current when its output is 1.30 A?
Solution
V 
 9.00 V 
(a) N s  N p  s   400 
  30.0
V
120
V


p
 
N 
 30 
2
(b) I p  I s  s   1.30 A 
  9.75  10 A
 400 
 Np 
45.
An American traveler in New Zealand carries a transformer to convert New Zealand’s
standard 240 V to 120 V so that she can use some small appliances on her trip. (a)
What is the ratio of turns in the primary and secondary coils of her transformer? (b)
What is the ratio of input to output current? (c) How could a New Zealander traveling
in the United States use this same transformer to power her 240 V appliances from
120 V?
Solution
(a)
Np
Ns
Ip

Vp
Vs
Np

240 V
 2.00
120 V

 0.500
Is Ns
(c) By hooking it up backwards, that is, by using input as output and vice versa.
(b)
OpenStax College Physics
Instructor Solutions Manual
Chapter 23
46.
A cassette recorder uses a plug-in transformer to convert 120 V to 12.0 V, with a
maximum current output of 200 mA. (a) What is the current input? (b) What is the
power input? (c) Is this amount of power reasonable for a small appliance?
Solution
I p N p Vp
(a) I  N  V , so that
s
s
s
 Vp 
 12.0 V 
2
I p  I s    0.200 A 
  2.00  10 A  20.0 mA
 120 V 
 Vs 
(b) Pin  I pVp  0.200 A 120 V  2.40 W
(c) Yes, this amount of power is quite reasonable for a small appliance.
47.
(a) What is the voltage output of a transformer used for rechargeable flashlight
batteries, if its primary has 500 turns, its secondary 4 turns, and the input voltage is
120 V? (b) What input current is required to produce a 4.00 A output? (c) What is the
power input?
Solution
N 
 4 
(a) Vs  Vp  s   120 V 
  0.96 V
 500 
 Np 
N 
 4 
2
(b) I p  I s  s   4.00 A 
  3.20  10 A  32.0 mA
 500 
 Np 
(c) P  IV  3.20  10 2 A120 V  3.84 W
48.
(a) The plug-in transformer for a laptop computer puts out 7.50 V and can supply a
maximum current of 2.00 A. What is the maximum input current if the input voltage is
240 V? Assume 100% efficiency. (b) If the actual efficiency is less than 100%, would
the input current need to be greater or smaller? Explain.
Solution
V 
N s Vs
 7.50 V 
  I p  I s  s   2.00 A 
  0.0625 A  0.063 A


I s N p Vp
 240 V 
 Vp 
(b) If the efficiency is less than 100%, the input current would need to be greater
because more power would need to be input into the transformer to produce the
required power output.
(a)
49.
Ip

A multipurpose transformer has a secondary coil with several points at which a
voltage can be extracted, giving outputs of 5.60, 12.0, and 480 V. (a) The input
voltage is 240 V to a primary coil of 280 turns. What are the numbers of turns in the
parts of the secondary used to produce the output voltages? (b) If the maximum input
current is 5.00 A, what are the maximum output currents (each used alone)?
OpenStax College Physics
Instructor Solutions Manual
Chapter 23
Solution
V 
 5.60 V 
(a) N s  N p  s   280
  6.53;
 240 V 
 Vp 
280 12 V   14.0; 280 480 V   560
 240 V 
 240 V 
 Vp 
 240 V 
(b) I s  I p    5.0 A 
  214 A
 5.60 V 
 Vs 
5.0 A  240 V   100 A; 5.0 A  240 V   2.50 A
 12 V 
 480 V 
50.
A large power plant generates electricity at 12.0 kV. Its old transformer once
converted the voltage to 335 kV. The secondary of this transformer is being replaced
so that its output can be 750 kV for more efficient cross-country transmission on
upgraded transmission lines. (a) What is the ratio of turns in the new secondary
compared with the old secondary? (b) What is the ratio of new current output to old
output (at 335 kV) for the same power? (c) If the upgraded transmission lines have
the same resistance, what is the ratio of new line power loss to old?
Solution
(a)
N s Vs N 's V 's
N' V '
750 kV
 ;

 s  s 
 2.2
N p Vp N p Vp
Ns
Vs 335 kV
(b)
I 's Vp V 's Vs 335 kV



 0.45
I s Vp Vs V 's 750 kV
(c) Ploss  I 2 R 
51.
P' I ' 2

 0.20
P I2
If the power output in the previous problem is 1000 MW and line resistance is 2.00 
, what were the old and new line losses?
Solution
(a)
P 1.000  109 W

 2985 A
V
3.35  105 V
2
 2985 A  2.00 Ω   1.78  10 7 W  17.8 MW
Old : P  IV  I 
 Ploss
New : Uses 7.50  105 V  I  2985 A, Ploss  3.56 MW
52.
Unreasonable Results The 335 kV AC electricity from a power transmission line is fed
into the primary coil of a transformer. The ratio of the number of turns in the
secondary to the number in the primary is N s / N p  1000 . (a) What voltage is
induced in the secondary? (b) What is unreasonable about this result? (c) Which
assumption or premise is responsible?
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 23
N 
(a) Vs  Vp  s   3.35 10 5 V 1000  335 MV
 Np 
(b) The result is way too high, well beyond the breakdown voltage of air over
reasonable distances.
(c) The input voltage is too high.


23.8 ELECTRICAL SAFETY: SYSTEMS AND DEVICES
54.
Integrated Concepts A short circuit to the grounded metal case of an appliance occurs
as shown in Figure 23.60. The person touching the case is wet and only has a 3.00 k
resistance to earth/ground. (a) What is the voltage on the case if 5.00 mA flows
through the person? (b) What is the current in the short circuit if the resistance of the
earth/ground wire is 0.200  ? (c) Will this trigger the 20.0 A circuit breaker supplying
the appliance?
Solution (a) V  IR  5.00  10 3 A 3000 Ω  15.0 V
V
15.0 V
 75.0 A
(b) I  
R 0.200 Ω
(c) Yes.
23.9 INDUCTANCE
55.
Two coils are placed close together in a physics lab to demonstrate Faraday’s law of
induction. A current of 5.00 A in one is switched off in 1.00 ms, inducing a 9.00 V emf
in the other. What is their mutual inductance?
Solution
Using the equation E2   M
I1
, where the minus sign is an expression of Lenz’s
t
law, we can calculate the mutual inductance between the two coils:
t
1.00  10 3 s
M  E2
 9.00 V 
 1.80 mH
I 1
5.00 A

56.
Solution
57.

If two coils placed next to one another have a mutual inductance of 5.00 mH, what
voltage is induced in one when the 2.00 A current in the other is switched off in 30.0
ms?
E2   M


I1
5.00 10 3 H 2.00 A 

 0.333 V
t
3.00 10 2 s
The 4.00 A current through a 7.50 mH inductor is switched off in 8.33 ms. What is the
emf induced opposing this?
OpenStax College Physics
Solution
58.
Solution
59.
Solution
EL
Instructor Solutions Manual

Chapter 23

I
7.50 10 3 H 4.00 A 

 3.60 V
t
8.33 10 3 s
A device is turned on and 3.00 A flows through it 0.100 ms later. What is the selfinductance of the device if an induced 150 V emf opposes this?
EL


I
t 150 V  1.00 10 4 s
LE

 5.00 mH
t
I
3.00 A
Starting with emf 2   M
E1   M
I1
, show that the units of inductance are (V  s)/A    s .
t
t E1   s  V /A
I 2
 M  
I 2 
t
from V  IR  V  A  Ω  M   s  Ω
60.
Solution
Camera flashes charge a capacitor to high voltage by switching the current through
an inductor on and off rapidly. In what time must the 0.100 A current through a 2.00
mH inductor be switched on or off to induce a 500 V emf?
EL


I
LI
2.00  10 3 H 0.100 A 
 t 

 4.00  10 -7 s
t
E
500 V
61.
A large research solenoid has a self-inductance of 25.0 H. (a) What induced emf
opposes shutting it off when 100 A of current through it is switched off in 80.0 ms? (b)
How much energy is stored in the inductor at full current? (c) At what rate in watts
must energy be dissipated to switch the current off in 80.0 ms? (d) In view of the
answer to the last part, is it surprising that shutting it down this quickly is difficult?
Solution
(a) Using the equation E  L
I
, we have
t
100 A   3.125 10 4 V  31.3 kV
I
EL
 25.0 H 
t
8.00 102 s
1
1
(b) Using Eind  LI 2   25.0 H 100 A 2  1.25  10 5 J
2
2
E
(c) Using the equation P 
,we have
t
E 1.25 105 J
P

 1.563 10 6 W  1.56 MW
-2
t 8.00 10 s
(d) No, it is not surprising since this power is very high.
OpenStax College Physics
62.
Solution
Instructor Solutions Manual
Chapter 23
(a) Calculate the self-inductance of a 50.0 cm long, 10.0 cm diameter solenoid having
1000 loops. (b) How much energy is stored in this inductor when 20.0 A of current
flows through it? (c) How fast can it be turned off if the induced emf cannot exceed
3.00 V?
(a) L 
0 N 2 A

 0 N 2 r 2

4π  10
7

T  m/A 1000 π0.0500 m 
 19.7 mH
0.500 m
2
2
l
l
1
2
(b) Eind  LI 2  0.5 1.974  10  2 H 20.0 A   3.95 J
2
I
LI
1.974  10 2 H 20.0 A 
(c) E  L
 t 

 0.132 s  t  0.132 s
t
E
3.00 V



63.
Solution

A precision laboratory resistor is made of a coil of wire 1.50 cm in diameter and 4.00
cm long, and it has 500 turns. (a) What is its self-inductance? (b) What average emf is
induced if the 12.0 A current through it is turned on in 5.00 ms (one-fourth of a cycle
for 50 Hz AC)? (c) What is its inductance if it is shortened to half its length and
counter-wound (two layers of 250 turns in opposite directions)?
(a) L 
0 N 2 A
; A   r 2   0.00750 m   1.767  10  4 m 2
2
l
2
4π  10 7 T  m/A 500  1.77  10  4 m 2
L
 1.39 mH
0.0400 m
I 1.388 10 -3 H 12.0 A 
(b) E  L

 3.33 V
t
5.00 10 -3 s
(c) L  0 . Two 250 turn coils in opposite directions will cancel each other.

64.
Solution



The heating coils in a hair dryer are 0.800 cm in diameter, have a combined length of
1.00 m, and a total of 400 turns. (a) What is their total self-inductance assuming they
act like a single solenoid? (b) How much energy is stored in them when 6.00 A flows?
(c) What average emf opposes shutting them off if this is done in 5.00 ms (one-fourth
of a cycle for 50 Hz AC)?
(a) L 
 0 N 2 A  0 N 2 r 2

l
l
4π 10

7


T  m/A 400 π 4.00 10 3 m
1.00 m

2


2
 1.0110 5 H
1 2
2
LI  0.5 1.01  10 5 H 6.0 A   1.82  10  4 J
2
I 1.01  10 5 H 6.0 A 
(c) E  L

 12.12 mV
t
5.00  10 3 s
(b) Eind 
OpenStax College Physics
65.
Solution
Instructor Solutions Manual
Chapter 23
When the 20.0 A current through an inductor is turned off in 1.50 ms, an 800 V emf is
induced, opposing the change. What is the value of the self-inductance?
EL


I
Et 800 V  1.50  10 3 s
L

 60.0 mH
t
I
20.0 A
66.
How fast can the 150 A current through a 0.250 H inductor be shut off if the induced
emf cannot exceed 75.0 V?
Solution
EL
67.
Integrated Concepts A very large, superconducting solenoid such as one used in MRI
scans, stores 1.00 MJ of energy in its magnetic field when 100 A flows. (a) Find its selfinductance. (b) If the coils “go normal,” they gain resistance and start to dissipate
thermal energy. What temperature increase is produced if all the stored energy goes
into heating the 1000 kg magnet, given its average specific heat is 200 J/kg  C ?
Solution
2 Eind 2 1.00  10 6 J
1 2
 200 H
(a) Eind  LI  L  2 
2
I
100 A 2
E
1.00  10 6 J
(b) E  mcT  T 

 5.00 C
mc 1000 kg 200 J/kg C
68.
Unreasonable Results A 25.0 H inductor has 100 A of current turned off in 1.00 ms.
(a) What voltage is induced to oppose this? (b) What is unreasonable about this
result? (c) Which assumption or premise is responsible?
Solution
(a) E  L
I
LI 0.250 H 150 A 
 t 

 0.500 s  t  0.500 s
t
E
75.0 V


100 A   2.50  10 6 V
I
 25.0 H 
t
1.00  10 3 s
(b) The voltage is so extremely high that arcing would occur and the current would
not be reduced so rapidly.
(c) It is not reasonable to shut off such a large current in such a large inductor in such
an extremely short time.
23.10 RL CIRCUITS
69.
If you want a characteristic RL time constant of 1.00 s, and you have a 500  resistor,
what value of self-inductance is needed?
Solution
Using the equation  
L
, we know L  R  1.00 s500 Ω  500 H.
R
OpenStax College Physics
Instructor Solutions Manual
Chapter 23
70.
Your RL circuit has a characteristic time constant of 20.0 ns, and a resistance of
5.00 M . (a) What is the inductance of the circuit? (b) What resistance would give
you a 1.00 ns time constant, perhaps needed for quick response in an oscilloscope?
Solution
(a)  
71.
A large superconducting magnet, used for magnetic resonance imaging, has a 50.0 H
inductance. If you want current through it to be adjustable with a 1.00 s characteristic
time constant, what is the minimum resistance of system?
Solution

72.
Verify that after a time of 10.0 ms, the current for the situation considered in Example
23.9 will be 0.183 A as stated.
Solution
I  I 0e
73.
Suppose you have a supply of inductors ranging from 1.00 nH to 10.0 H, and resistors
ranging from 100.0  to 1000.0 M . What is the range of characteristic RL time
constants you can produce by connecting a single resistor to a single inductor?
Solution



L
 L  R  5.00  10 6 Ω 2.00  10 8 s  0.100 H
R
L
0.100 H
(b) R  
 1.00  10 8 Ω  100 MΩ
9
 1.00  10 s
L
L 50.0 H
R 
 50.0 Ω
R
 1.00 s
 max 
t

3
3
 10.0 Ae 1.0010 s  2.5010 s   0.183 A
Lmax
L
10.0 H
1.00  10 -9 H

 100 s;  min  min 
 1.00  10 -15 s
6
Rmin 0.100 
Rmax 1.00  10 
74.
(a) What is the characteristic time constant of a 25.0 mH inductor that has a
resistance of 4.00  ? (b) If it is connected to a 12.0 V battery, what is the current
after 12.5 ms?
Solution
(a)  
L 2.50 10 -2 H

 6.25 10 -3 s
R
4.00 
V 12.0 V
(b) I 0  
 3.00 A, so that
R 4.00 Ω
I  I 0 1  e

75.
t



  3.00 A 1  e 1.2510 2 s  6.25103 s   2.59 A

What percentage of the final current I 0 flows through an inductor L in series with a
resistor R , three time constants after the circuit is completed?
OpenStax College Physics
Instructor Solutions Manual
Chapter 23
Solution
t
I 
 1  e    1  e 3  0.9502

I0 
The current is 95.0% of the final current after 3 time constants.
76.
The 5.00 A current through a 1.50 H inductor is dissipated by a 2.00  resistor in a
circuit like that in Figure 23.44 with the switch in position 2. (a) What is the initial
energy in the inductor? (b) How long will it take the current to decline to 5.00% of its
initial value? (c) Calculate the average power dissipated, and compare it with the
initial power dissipated by the resistor.
Solution
(a) Eind 
1 2
2
LI  0.51.50 H 5.00 A   18.75 J  18.8 J
2
t
L 1.50 H
I  I 0 e  ;  
 0.750 s
R 2.00 
t
 I 
I
t
 e   ln    
I0

 I0 
(b)
 I 
or t   ln    0.750 s  ln 0.0500  2.247 s  2.25 s
 I0 
(c) E  1 LI 2  0.5150 H 0.05005.00 A 2  4.69 J
f
f
2
E  E f 18.75 J  4.69 J
Pave  i

 6.25 W
t
2.25 s
P
2
Pi  I 02 R  5.00 A  2.00 Ω   50.0 W  i  8.00
Pave
77.
Solution
(a) Use the exact exponential treatment to find how much time is required to bring
the current through an 80.0 mH inductor in series with a 15.0  resistor to 99.0% of
its final value, starting from zero. (b) Compare your answer to the approximate
treatment using integral numbers of  . (c) Discuss how significant the difference is.
L 8.00  10 -2 H

 5.333  10 -3 s, and
R
15.0 
t
t
I 
I  I 0 1  e   
 1  e  



I0 

(a)
e
t

 I
 1  
 I0

 

  I
t   ln 1  
  I0

   5.333  10 -3 s ln 1  0.990   2.4561  10 -2 s  24.6 ms



OpenStax College Physics
Instructor Solutions Manual
Chapter 23
(b) Using the result from part (a),
t
 4.61, t  5  55.33  10 3 s   2.667  10  2 s  26.7 ms

26.7 ms
(c)
 1.085  9% difference
24.6 ms
So, using integer values is off by about 9% which is greater than the inherent
uncertainty in the given parameters.
78.
(a) Using the exact exponential treatment, find the time required for the current
through a 2.00 H inductor in series with a 0.500  resistor to be reduced to 0.100%
of its original value. (b) Compare your answer to the approximate treatment using
integral numbers of  . (c) Discuss how significant the difference is.
Solution
L 2.00 H
(a)   R  0.500   4.00 s;
t
 I 
I  I 0 e   t   ln    4.00 s  ln 1.00 10 -3   27.6 s
 I0 
t
(b) Using the result from part (a),  6.91, t  7  74.00 s   28.00 s  28.0 ms

28.00 s
(c)
 1.013  1% difference . So, using integer values is off by about 1% which
27.63 s
is larger than the inherent uncertainty in the given parameters.
23.11 REACTANCE, INDUCTIVE AND CAPACITIVE
79.
At what frequency will a 30.0 mH inductor have a reactance of 100  ?
Solution
X L  2fL  f 
80.
What value of inductance should be used if a 20.0 k reactance is needed at a
frequency of 500 Hz?
Solution
X L  2 fL  L 
81.
What capacitance should be used to produce a 2.00 M reactance at 60.0 Hz?
Solution
C
82.
At what frequency will an 80.0 mF capacitor have a reactance of 0.250  ?
XL
100 Ω

 531 Hz
2L 2π 3.00 10 2 H




XL
2.00 10 4 Ω

 6.37 H
2f
2π 500 Hz 
1
1

 1.326 10 9 F  1.33 nF
6
2fX C 2π 60.0 Hz  2.00 10 Ω


OpenStax College Physics
Instructor Solutions Manual
Chapter 23
Solution
XC 
83.
(a) Find the current through a 0.500 H inductor connected to a 60.0 Hz, 480 V AC
source. (b) What would the current be at 100 kHz?
Solution
1
1
1
 f 

 7.96 Hz
2fC
2CX C 2π 8.00  10 -2 F 0.250  


2
(a) X L  2fL  2π60.0 Hz 0.500 H   1.885  10 Ω, so that
V
480 V
I

 2.55 A
X L 1.885  10 2 Ω


5
5
(b) X L  2π 1.00  10 Hz 0.500 H   3.142  10 Ω, so that
V
480 V
I

 1.53 mA
X L 3.142  10 5 Ω
84.
(a) What current flows when a 60.0 Hz, 480 V AC source is connected to a 0.250 F
capacitor? (b) What would the current be at 25.0 kHz?
Solution
(a) X C 
1
1

 1.061  10 4 , so that
-7
2fC 2π 60.0 Hz  2.50  10 F
V
480 V
I

 45.2 mA
X C 1.061  10 4 Ω,
1
 2.546  101 , so that
(b) X C 
4
-7
2π 2.50  10 Hz 2.50  10 F
V
480 V
I

 18.8 A
X C 2.546  101 





85.
A 20.0 kHz, 16.0 V source connected to an inductor produces a 2.00 A current. What is
the inductance?
Solution
X L  2fL; I 
L
86.
Solution
V
V

X L 2 fL
V
16.0 V

 63.7 μH
2fL 2π 2.00  10 4 Hz 2.00 Hz 


A 20.0 Hz, 16.0 V source produces a 2.00 mA current when connected to a capacitor.
What is the capacitance?
XC 
1
V
I
2.00  10 -3 A
;I 
 2VfC  C 

 0.995 μF
2fC
XC
2Vf
2π 16.0 V 20.0 Hz 
OpenStax College Physics
Instructor Solutions Manual
Chapter 23
87.
(a) An inductor designed to filter high-frequency noise from power supplied to a
personal computer is placed in series with the computer. What minimum inductance
should it have to produce a 2.00 k reactance for 15.0 kHz noise? (b) What is its
reactance at 60.0 Hz?
Solution
(a) Using the equation X L  2fL ,
X
2.00 103 Ω  2.122 102 H  21.2 mH
X L  2fL, or L  L 
2f
2π 1.50 10 4 Hz 
(b) Again using X L  2fL ,
X L  2fL  2π60.0 Hz 2.122 10-2 H   8.00 Ω
88.
The capacitor in Figure 23.55(a) is designed to filter low-frequency signals, impeding
their transmission between circuits. (a) What capacitance is needed to produce a
100 k reactance at a frequency of 120 Hz? (b) What would its reactance be at 1.00
MHz? (c) Discuss the implications of your answers to (a) and (b).
Solution
(a) X C 
1

2fC
1
1
C

 1.326  10 -8 F  13.3 nF
5
2fX C


2π 120 Hz 1.00  10 
1
1
(b) X C 

 12.0 
6
2fC 2π 1.00 10 Hz 1.326 10 -8 F
(c) This implies that the capacitor impedes low frequencies significantly more than
high frequencies, and therefore does what it is designed to do.





89.
The capacitor in Figure 23.55(b) will filter high-frequency signals by shorting them to
earth/ground. (a) What capacitance is needed to produce a reactance of 10.0 m for
a 5.00 kHz signal? (b) What would its reactance be at 3.00 Hz? (c) Discuss the
implications of your answers to (a) and (b).
Solution
(a) X C 
1
1
1
C 

 3.18 mF
3
2fC
2fX C
2π 5.00 10 Hz 1.00 10 2 
1
1
(b) X C 

 16.7 
2fC 2π 3.00 Hz 3.18 10 -3 F
(c) The reactance from the capacitor at high frequencies is much smaller than that at
lower frequencies, so the capacitor does indeed allow the high frequencies to
short to ground much better than the low frequencies, therefore working as it is
designed to.



OpenStax College Physics
Instructor Solutions Manual
Chapter 23
90.
Unreasonable Results In a recording of voltages due to brain activity (an EEG), a 10.0
mV signal with a 0.500 Hz frequency is applied to a capacitor, producing a current of
100 mA. Resistance is negligible. (a) What is the capacitance? (b) What is
unreasonable about this result? (c) Which assumption or premise is responsible?
Solution
(a) X C 
1
V
I
0.100 A
;I 
 2fCV  C 

 3.18 F
2fC
XC
2fV 2π0.500 Hz  1.00  10 2 V
(b) That is an enormous capacitance. A capacitor that size would fill a large room.
(c) It is unreasonable to assume such a small voltage could achieve such a current at
the low frequency given. Capacitive resistance is very large at low frequencies
unless the capacitance is very large.


23.12 RLC SERIES AC CIRCUITS
92.
Solution
An RL circuit consists of a 40.0  resistor and a 3.00 mH inductor. (a) Find its
impedance Z at 60.0 Hz and 10.0 kHz. (b) Compare these values of Z with those
found in Example 23.12, in which there was also a capacitor.
(a)


X L  2fL  2π60.0 Hz  3.00  10 -3 H  1.131 Ω;
X C  0 since there is no capacitor.

Z  R2  X L  X C 
  40.0
2 12
2
 1.131` 
 40.02 Ω (4 sig fig. to show difference )



2 12

X ' L  2f ' L  2π 1.00  10 4 Hz 3.00  10 -3 H  188.5 Ω, so


Z '  40.0 Ω   188.5 Ω 
 193 Ω
(b) At 60 Hz, with a capacitor, Z  531  , over 13 times as high as without the
capacitor. The capacitor makes a large difference at low frequencies. At 10 kHz,
with a capacitor Z  190  , about the same as without the capacitor. The
capacitor has a smaller effect at high frequencies.
93.
2
2 12
An RC circuit consists of a 40.0  resistor and a 5.00 F capacitor. (a) Find its
impedance at 60.0 Hz and 10.0 kHz. (b) Compare these values of Z with those found
in Example 23.12, in which there was also an inductor.
OpenStax College Physics
Solution
(a)
Instructor Solutions Manual
Chapter 23
1
1

 530.52  ;
2fC 2π 60.0 Hz 5.00 10 -6 F
X L  0 since there is no inductor.
At 60 Hz : X C 

  40.0 Ω
Z  R 2  X L  X C 
At 10 kHz : X 'C 
2 12
 530.52`Ω 

2 12
2
 532 Ω
1
1

 3.183  ;
4
2f ' C 2π 1.00 10 Hz 5.00 10 -6 F


Z  40.0    3.183  
 40.1 Ω
(b) At 60 Hz Z is about the same with or without the inductor because the capacitor
dominates at low frequency. At 10.0 kHz Z is much less without the inductor
because the inductor dominates at high frequencies.
94.
Solution
2 12
2
An LC circuit consists of a 3.00 H inductor and a 5.00 F capacitor. (a) Find its
impedance at 60.0 Hz and 10.0 kHz. (b) Compare these values of Z with those found
in Example 23.12, in which there was also a resistor.
(a)


At 60 Hz : X L  2fL  2π60.0 Hz  3.00 10 -3 H  1.131 Ω;
XC 
1
1

 530.52  ;
2fC 2π 60.0 Hz  5.00 10 -6 F



  1.131   530.52` 
Z  R 2  X L  X C 
2 12

2 12


 529 Ω
At 10 kHz : X ' L  2f ' L  2π 1.00 10 4 Hz 3.00 10 -3 H  188.5 Ω;
X 'C 
1
1

 3.183  ;
4
2f ' C 2π 1.00 10 Hz 5.00 10 -6 F





Z  188.5   3.183  
 185 Ω
(b) These values are close to those obtained in Example 23.12 because at low
frequency the capacitor dominates and at high frequency the inductor dominates.
So in both cases the resistor makes little contribution to the total impedance.
2 12
95.
What is the resonant frequency of a 0.500 mH inductor connected to a 40.0 F
capacitor?
Solution
Using f 0 
96.
1
2 LC

2
5.00 10
1
-4

H 4.00 10 F
-5

 1.125 10 3 Hz  1.13 kHz
To receive AM radio, you want an RLC circuit that can be made to resonate at any
frequency between 500 and 1650 kHz. This is accomplished with a fixed 1.00 H
inductor connected to a variable capacitor. What range of capacitance is needed?
OpenStax College Physics
Solution
f0 
Cmin 
Cmax 
97.
Solution
Solution
1
2 LC
C 
1
;
4 Lf 02
2
1

-6

-6

4 1.00  10 H 1.650  10 Hz
1
2
6

4 1.00  10 H 5.00  10 Hz
2
5

2

2
 9.304  10 -9 F  9.30 nF
 1.013  10 -7 F  101.3 nF  9.30 nF  101 nF
1
1

-9
2 Lmin C min 2 1.00 10 H 1.00 10 -12 F
1

 0.159 Hz
12
2 10.0 H 0.100 F
f 0 max 


1
f0 
 f 02 

Solution

12
 5.03 GHz
What capacitance do you need to produce a resonant frequency of 1.00 GHz, when
using an 8.00 nH inductor?
1

4 2 LC
1
2 LC
1
C

2
4 Lf 02 4 2 8.00  10 -9 H 1.00  10 9 Hz
99.
Chapter 23
Suppose you have a supply of inductors ranging from 1.00 nH to 10.0 H, and
capacitors ranging from 1.00 pF to 0.100 F. What is the range of resonant frequencies
that can be achieved from combinations of a single inductor and a single capacitor?
f 0 min
98.
Instructor Solutions Manual

 3.17  10 -12 F  3.17 pF

2
What inductance do you need to produce a resonant frequency of 60.0 Hz, when
using a 2.00 F capacitor?
f0 
1
2 LC
L
1
1

 3.52 H
2
2
2
4 Cf 0
4 2.00  10 -6 F 60.0 Hz 

2

100.
The lowest frequency in the FM radio band is 88.0 MHz. (a) What inductance is
needed to produce this resonant frequency if it is connected to a 2.50 pF capacitor?
(b) The capacitor is variable, to allow the resonant frequency to be adjusted to as high
as 108 MHz. What must the capacitance be at this frequency?
Solution
(a) f 0 
(b) C 
1
2 LC
L
1
1

2
4 Cf 0 4 2 2.50 10 -12 F 8.80 10 7 Hz

2
1
1

2
4 Lf 0
4 2 1.308 10 -6 H 1.08 108 Hz
2




2
 1.66 pF

2
 1.31 μH
OpenStax College Physics
101.
Solution
Instructor Solutions Manual
Chapter 23
An RLC series circuit has a 2.50  resistor, a 100 H inductor, and an 80.0 F
capacitor. (a) Find the circuit’s impedance at 120 Hz. (b) Find the circuit’s impedance
at 5.00 kHz. (c) If the voltage source has Vrms  5.60 V , what is I rms at each
frequency? (d) What is the resonant frequency of the circuit? (e) What is I rms at
resonance?
(a) X L  2fL  2π120 Hz 1.00 10 -4 H   7.540 10 -2 Ω
1
1
XC 

 16.58 
2 f C 2π 120 Hz 8.00 10 -5 F
Z  R 2  X L  X C  
2
2.50 Ω 2  7.54 10 2 Ω  16.58 Ω 2
(b) X L  2π5.00 10 3 Hz 1.00 10 -4 H   3.142 Ω
1
XC 
 3.979 10 -1 
3
-4
2π 5.00  10 Hz 8.00 10 F
Z
2.50 Ω 2  3.142 Ω  3.979 10 1 Ω 2
 16.7 Ω
 3.71 Ω
Vrms
5.60 V

 0.336 A
Z
16.69 Ω
5.60 V
and for f  5.00 kHz , I rms 
 1.51 A
3.712 Ω
(c) For f  120 Hz , I rms 
(d) f 0 
1
2 LC


1

2 1.00 10 H 8.00 10 F
-7
-5

 5.63 10 4 Hz  56.3 kHz
(e) At resonance, X L  X R , so that Z  R thus: I rms 
102.
Vrms 5.60 V

 2.24 A
R
2.50 Ω
An RLC series circuit has a 1.00 k resistor, a 150 H inductor, and a 25.0 nF
capacitor. (a) Find the circuit’s impedance at 500 Hz. (b) Find the circuit’s impedance
at 7.50 kHz. (c) If the voltage source has Vrms  408 V , what is I rms at each
frequency? (d) What is the resonant frequency of the circuit? (e) What is I rms at
resonance?
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 23
-4
(a) X L  2fL  2π500 Hz 1.50  10 H   0.471 Ω
1
1
XC 

 1.273  10 4 
-8
2fC 2π 500 Hz 2.50  10 F

Z  R 2  X L  X C 
  1000 Ω 
2 12
2
 0.4714 Ω  12,730 Ω 

2 12
(b) X L  2π7500 Hz 1.50  10 H   7.069 Ω
1
XC 
 848.8 
2π 7500 Hz 2.50  10 -8 F
 12.8 kΩ
-4


Z  1000 Ω   7069.2 Ω  848.8 Ω 
 6.30 kΩ
408 V
V
408 V
(c) (i) I rms  rms 
 64.8 mA
 31.9 mA ; (ii) I rms 
6300 Ω
Z
12,800 Ω
1
1
(d) f 0 

 82.2 kHz
12
-4
2 LC 2 1.50 10 H 2.50 10 8 F
(e) I rms 
2
2 12
Vrms
408 V

 0.408 A
R
1000 Ω
103.
An RLC series circuit has a 2.50  resistor, a 100 H inductor, and an 80.0 F
capacitor. (a) Find the power factor at f  120 Hz . (b) What is the phase angle at
120 Hz? (c) What is the average power at 120 Hz? (d) Find the average power at the
circuit’s resonant frequency.
Solution
(a) From Exercise 23.101, we know Z : cos  
R 2.50 

 0.150
Z 16.69 
R
(b)   cos 1    cos 1 0.1498  81.4
Z
(c) Pave  I rms Vrms cos   3.355  10 1 A 5.60 V 0.1498  0.281 W
(d) Pave 
104.
2
Vrms
5.60 V 2  12.5 W

R
2.50 Ω
An RLC series circuit has a 1.00 k resistor, a 150 H inductor, and a 25.0 nF
capacitor. (a) Find the power factor at f  7.50 kHz . (b) What is the phase angle at
this frequency? (c) What is the average power at this frequency? (d) Find the average
power at the circuit’s resonant frequency.
OpenStax College Physics
Solution
Instructor Solutions Manual
(a) From Exercise 23.102, we know Z : cos  
Chapter 23
R 1000 

 0.159
Z 6300 
R
(b)   cos 1    cos 1 0.1587   80.9
Z
(c) Pave  I rms Vrms cos   6.48  10 2 A 408 V   26.4 W
2
2
Vrms

408 V 


 166.46 W  166 W
R
1000 Ω
(d) Pave
105.
An RLC series circuit has a 200  resistor and a 25.0 mH inductor. At 8000 Hz, the
phase angle is 45.0 . (a) What is the impedance? (b) Find the circuit’s capacitance. (c)
If Vrms  408 V is applied, what is the average power supplied?
Solution
(a) cos  
R
R
200 
Z 

 283 Ω
Z
cos  cos 45


-2
3
(b) X L  2fL  2π8000 Hz  2.50  10 H  1.26  10 Ω

Z  R2  X L  X C 

2 12

so that  X L  X C   Z 2  R 2
2
 X C  X L  Z 2  R2


12
X C  1.2566  10 3 Ω  282.84 Ω   200 Ω 
XC 
Solution
 1.0566  10 3 Ω
1
1
1
C 

 18.8 nF
2fC
2fX C
2π 8000 Hz  1.0566  10 3 

(c) Pave  I rms Vrms 
106.

2 12
2

2
Vrms
408 V 2  589 W

Z
282.84 Ω
Referring to Example 23.14, find the average power at 10.0 kHz.



X L  2fL  2π 1.00  10 4 Hz 3.00  10 -3 H  1.89  10 2 Ω
XC 
1
1

 3.18 
4
2fC 2π 1.00  10 Hz 5.00  10 -6 F



Z  R2  X L  X C 
  40.0 Ω
2 12
2

 188.5 Ω  3.183 Ω 

2 12
 189.6 Ω, so that
Vrms
120 V

 0.6330 A and
Z
189.6 Ω
40.0 Ω   16.0 W
R
 I rms Vrms cos   I rms Vrms    0.633 A 120 V 
189.6 Ω 
Z
I rms 
Pave
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