48
••
A 300-turn solenoid has a radius equal to 2.00 cm; a length equal to 25.0 cm, and a
1000-turn solenoid has a radius equal to 5.00 cm and is also
25.0-cm long. The two solenoids are coaxial, with one nested completely inside the other. What is their mutual inductance?
Picture the Problem We can find the mutual inductance of the two coaxial solenoids using M
2 , 1

 m2
 
0 n
2
.
I
1 n
1

 r
1
2
Substitute numerical values and evaluate M
2,1
:
M
2 , 1


4
 
10

7
N/A
2
 

300
0.250
m




1000
0.250
m



0 .
250 m 0 .
0200 m

2 
1 .
89 mH
49
••
[SSM] An insulated wire that has a resistance of 18.0

/m and a length of 9.00 m will be used to construct a resistor. First, the wire is bent in half and then the doubled wire is wound on a cylindrical form ( Figure 28-50) to create a 25.0-cm-long helix that has a diameter equal to 2.00 cm. Find both the resistance and the inductance of this wire-wound resistor.
Picture the Problem Note that the current in the two parts of the wire is in opposite directions. Consequently, the total flux in the coil is zero. We can find the resistance of the wire-wound resistor from the length of wire used and the resistance per unit length.
Because the total flux in the coil is zero:
0 L

Express the total resistance of the wire:
R

18 .
0
Ω m
L
Substitute numerical values and evaluate R :
R



18 .
0
Ω m

9 .
00 m


162 Ω
50
••
You are given a length  of wire that has radius a and are told to wind it into an inductor in the shape of a helix that has a circular cross section of radius r . The windings are to be as close together as possible without overlapping. Show that the self-inductance of this inductor is L

1
4

0 r  a .
Picture the Problem The wire of length  and radius a is shown in the diagram, as is the inductor constructed with this wire and whose inductance L is to be found. We can use the equation for the self-inductance of a cylindrical inductor to derive an expression for L.

The self-inductance of an inductor with length d , cross-sectional area A , and number of turns per unit length n is:
The number of turns N is given by:
L
 
0 n
2
Ad (1)
Assuming that a << r, the length of the wire  is related to N and r :
Solving for d yields:
N
 d
2 a
 n

 
N

2
 r


N d

1
2 a d
2 a

2
 r

 a r d d
 a 
 r
Substitute for d, A, and n in equation
(1) to obtain: L
 
0
2
1 a
2   

 a  r



1
4

0 r  a
51
•
Using the result of Problem 50, calculate the self-inductance of an inductor wound from 10 cm of wire that has a diameter of 1.0 mm into a coil that has a radius of 0.25 cm.
Picture the Problem We can substitute numerical values in the expression derived in Problem 50to find the self-inductance of the inductor.
From Problem 50 we have:
L


0 rd
4 a
Substitute numerical values and evaluate L :

L


4
 
10

7 2
N/A
4

0 .
50
 
0 .
25 mm
 cm

10 cm


0 .
16

H
61
••
[SSM] In the circuit shown in Figure 28-54, let

0
= 12.0 V,
R = 3.00

, and L = 0.600 H. The switch, which was initially open, is closed at time t = 0. At time t = 0.500 s, find ( a ) the rate at which the battery supplies energy, ( b ) the rate of Joule heating in the resistor, and ( c ) the rate at which energy is being stored in the inductor.
Picture the Problem We can find the current using I

I

1
 e
 t


, where f
I f
=

0
/ R, and

= L / R, and its rate of change by differentiating this expression with respect to time.
Express the dependence of the current on I f
and

:
Evaluating I f
and

yields:
I

I f

1
 e
 t


I f


0
R

1 2 .
0 V
3.00
Ω

4.00
A and
 
L
R

0 .
600 H
3.00
Ω

0.200
s
Substitute for I f
and

to obtain:
Express dI / dt :
I
  
4 .
00 A
 
1
 e
 t 0 .
200 s
 dI dt



4 .
00 A
 
 e
 t 0 .
200 s


5 .
00 s

1


20 .
0 A/s
 e
 t 0 .
200 s
P

I

0
( a ) The rate at which the battery supplies energy is given by:
Substituting for I and

0
yields:
The rate at which the battery supplies energy at t = 0.500 s is:
P
  


4 .
00
48 .
0
A
W




1
1

 e e
 t
 t
0 .
200 s
0 .
200 s

 
12 .
0 V

P

0 .
500 s

48 .
0 W
 
1
 e

0 .
500 0 .
200 s

44 .
1 W
( b ) The rate of Joule heating is:
Substitute for I and R and simplify to obtain:
P
J

I
2
R
P
J


 
4 .
00 A


4 8.0
W



1
1

 e e
 t
 t
0 .
200 s
0 .
200 s


2
2
 
3 .
00 Ω


The rate of Joule heating at t = 0.500 s is:
( c ) Use the expression for the magnetic energy stored in an inductor to express the rate at which energy is being stored:
P
J

0 .
500 s
 
4 8.0
W
 
1
 e

0 .
500 s 0 .
200 s
2


40 .
4 W dU
L dt
 d dt

1
2
LI 2


LI dI dt

Substitute for L, I, and dI / dt to obtain: dU
L dt



0 .
600 H

4 .
00 A
 
1
 e
 t

48 .
0 W
 
1
 e
 t
0 .
200 s
0 .
200 s
 e
 t 0 .
200 s
 
20 .
0 A/s
 e
 t 0 .
200 s
Evaluate this expression for t = 0.500 s: dU
L dt

 t

0 .
500 s


48 .
0 W
 
1
 e

0 .
500 s 0 .
200 s
 e

0 .
500 s 0 .
200 s 
3 .
62 W
Remarks: Note that, to a good approximation, dU
L
/dt = P

P
J
.
3
•
[SSM] If the frequency in the circuit shown in Figure 29-27 is doubled, the inductance of the inductor will ( a ) double, ( b ) not change, ( c ) halve, ( d ) quadruple.
Determine the Concept The inductance of an inductor is determined by the details of its construction and is independent of the frequency of the circuit. The inductive reactance, on the other hand, is frequency dependent. ( b ) is correct.
5
•
If the frequency in the circuit in Figure 29-28 is doubled, the capacitive reactance of the circuit will ( a ) double, ( b ) not change, ( c ) halve,
( d ) quadruple.
Determine the Concept The capacitive reactance of an capacitor varies with the frequency according to X
C

1

C .
Hence, doubling

will halve X
C
.
( c ) is correct.
34
••
A circuit consists of a resistor, an ideal 1.4-H inductor and an ideal
60-Hz generator, all connected in series. The rms voltage across the resistor is
30 V and the rms voltage across the inductor is 40 V. ( a ) What is the resistance of the resistor?
( b ) What is the peak emf of the generator?
Picture the Problem We can express the ratio of V
R
to V
L
and solve this expression for the resistance R of the circuit. In ( b ) we can use the fact that, in an
LR circuit, V
L
leads V
R
by 90

to find the ac input voltage.
( a ) Express the potential differences across R and L in terms of the common current through these components:
V
L

IX
L and
V
R

IR

I

L

Divide the second of these equations by the first to obtain:
Substitute numerical values and evaluate R :
V
R
V
L

IR
I

L

R

L

R



V
R
V
L



L
R



30 V
40 V


2


60 s

1
 
1 .
4 H


V peak

2 V rms

2 V
R
2 
V
L
2
0 .
40 kΩ
( b ) Because V
R
leads V
L
by 90

in an LR circuit:
Substitute numerical values and evaluate V peak
:
V peak

2

30 V
 
40 V

2 
71 V
35
••
[SSM] A coil that has a resistance of 80.0

has an impedance of 200

when driven at a frequency of 1.00 kHz. What is the inductance of the coil?
Picture the Problem We can solve the expression for the impedance in an LR circuit for the inductive reactance and then use the definition of X
L
to find L .
Express the impedance of the coil in terms of its resistance and inductive reactance:
Solve for X
L
to obtain:
Z

X
L

R
2 
X
2
L
Z
2 
R
2
Express X
L
in terms of L : X
L

2
 fL
Equate these two expressions to obtain:
Substitute numerical values and evaluate L :
2
 fL
L



Z
2 
R
2 
L

200
2

Ω

1
 
.
00
80 .
0 kHz

Ω

2
Z
2 
2
 f
R
2

29 .
2 mH
41
••
[SSM] Figure 29-33 shows a load resistor that has a resistance of
R
L
= 20.0

connected to a high-pass filter consisting of an inductor that has inductance L =
3.20-mH and a resistor that has resistance R = 4.00-

. The output of the ideal ac generator is

given by

= (100 V) cos(2
 ft ). Find the rms currents in all three branches of the circuit if the driving frequency is ( a ) 500 Hz and
( b ) 2000 Hz. Find the fraction of the total average power supplied by the ac generator that is delivered to the load resistor if the frequency is ( c ) 500 Hz and ( d ) 2000 Hz.
Picture the Problem


V
1

V
2
, where V
1
is the voltage drop across R and V
2
is the voltage drop across the parallel combination of L and R
L
. relation for the phasors. For the parallel combination I


I

R
L


I

L

V
1


V
2
. Also, is the
V
1
is in phase with I and V
2
is in phase with I
R
L
. First draw the phasor diagram for the currents in the parallel combination, then add the phasors for the voltages to the diagram.
The phasor diagram for the currents in the circuit is:
Adding the voltage phasors to the diagram gives:
The maximum current in the inductor, I
2, peak
, is given by: tan
 is given by:
I
2 , peak where

V
Z
2

2
2 , peak
Z
2

(1)
R
L

2 
X
L

2
(2) tan
 
I
L , peak
I
R , peak

R
X
L
L

V
2 , peak
V
2 , peak
X

R
L

L

2
R
L
 fL
L
R
L

Solve for
 to obtain:
  tan

1


R
L
2
 fL

 (3)

Apply the law of cosines to the triangle formed by the voltage phasors to obtain:

2 peak
 2
V
1 , peak

V
2
2
, peak

2 V
1 , peak
V
2 , peak cos
 or
I
2 peak
Z
2 
I
2 peak
R
2 
I
2 peak
Z
2
2 
2 I peak
RI peak
Z
2 cos

Dividing out the current squared yields:
Solving for Z yields:
Z
2
Z


R
2 
Z
2
2 
2 RZ
2 cos

R
2 
Z
2
2 
2 RZ
2 cos

(4)
The maximum current I peak in the circuit is given by:
I rms
is related to I peak
according to:
( a ) Substitute numerical values in equation (3) and evaluate

:
I peak

 peak
Z
(5)
I rms

1
2
I peak
(6)
 
 tan

1


2
 
500
20 .
0
Hz

Ω
3 .
20 mH


 tan

1


20 .
0
Ω
10 .
053
Ω



63 .
31

Solving equation (2) for Z
2
yields:
Substitute numerical values and evaluate Z
2
:
Z
2

1
R
L

2 
X
L

2
Z
2

1

20 .
0
Ω
 
2 

10 .
053
Ω
 
2

8 .
982 Ω
Substitute numerical values and evaluate Z :
Z


4 .
00
Ω
 
8 .
982
Ω

2 
2

4 .
00
Ω

8 .
982
Ω
 cos 63 .
31
 
11 .
36
Ω
Substitute numerical values in equation (5) and evaluate I peak
:
I peak

100 V
11.36


8 .
806 A

Substitute for I peak
in equation
(6) and evaluate I : rms
The maximum and rms values of
V
2
are given by:
The rms values of I
R
L
, rms
and
I
L , rms are:
( b ) Proceed as in ( a ) with f = 2000 Hz to obtain:
I rms

1
2

8 .
806 A
 
6 .
23 A
V
2, peak


I Z peak

8 .
806
2
A

8 .
982 Ω


79 .
095 V and
V
2 , rms

1
2
V
2 , peak

1
2

79 .
095 V


55 .
929 V
I
R
L
, rms

V
2 , rms
R
L and
I
L , rms

V
2 , rms
X
L

55 .
929
20.0
Ω
V

55 .
929 V
1 0 .
053
Ω


2 .
80 A
5 .
53 A
X
L
Z

40 .
2

21 .
6


,
,

I peak

26 .
4

,

Z
4 .
64 A
2

17
, and
.
9

,
I rms

V
2 , max
3 .
28 A ,

83 .
0 V , V
2 , rms

58 .
7 V ,
I
R
L
, rms

2 .
94 A , and I
L , rms

1 .
46 A
( c ) The power delivered by the ac source equals the sum of the power dissipated in the two resistors. The fraction of the total power delivered by the source that is dissipated in load resistor is given by:
P
R
L
P
R
L

P
R


1

P
P
R
R
L

1


1

I
2
I
2 rms
R
L
, rms
R
R
L

1
Substitute numerical values for f = 500 Hz to obtain:
P
R
L
P
R
L

P
R f

500 Hz



1



6 .
23
2 .
80
A
A
 
4 .
00
20 .
0

 
 


1

0 .
502

50 .
2 %

( d ) Substitute numerical values for f = 2000 Hz to obtain:
P
R
L
P
R
L

P
R f

2000 Hz



1



3 .
28
2 .
94
A
A
 
 
4 .
00
20 .
0



 


1

0 .
800

80 .
0 %