Chapter 17
Current and Resistance
Quick Quizzes
1.
(d). Negative charges moving in one direction are equivalent to positive charges moving
in the opposite direction. Thus, I a , I b , I c , and I d are equivalent to the movement of 5, 3, 4,
and 2 charges respectively, giving I d < I b < I c < I a .
2.
(b). Under steady-state conditions, the current is the same in all parts of the wire. Thus,
the drift velocity, given by vd = I nqA , is inversely proportional to the cross-sectional area.
3.
(c), (d). Neither circuit (a) nor circuit (b) applies a difference in potential across the bulb.
Circuit (a) has both lead wires connected to the same battery terminal. Circuit (b) has a
low resistance path (a “short”) between the two battery terminals as well as between the
bulb terminals.
4.
(b). The slope of the line tangent to the curve at a point is the reciprocal of the resistance at
that point. Note that as ∆V increases, the slope (and hence 1 R ) increases. Thus, the
resistance decreases.
5.
6.
l
l
= ρ 2 . Doubling all linear
A
πr
dimensions increases the numerator of this expression by a factor of 2, but increases the
denominator by a factor of 4. Thus, the net result is that the resistance will be reduced to
one-half of its original value.
(b). Consider the expression for resistance: R = ρ
(a). The resistance of the shorter wire is half that of the longer wire. The power dissipated,
2
P = ( ∆V ) R , (and hence the rate of heating) will be greater for the shorter wire.
Consideration of the expression P = I 2 R might initially lead one to think that the reverse
would be true. However, one must realize that the currents will not be the same in the two
wires.
69
70
CHAPTER 17
7.
(b). I a = I b > I c = I d > I e = I f . Charges constituting the current I a leave the positive terminal
of the battery and then split to flow through the two bulbs; thus, I a = I c + I e . Because the
potential difference ∆V is the same across the two bulbs and because the power delivered
to a device is P = I ( ∆V ) , the 60–W bulb with the higher power rating must carry the
greater current, meaning that I c > I e . Because charge does not accumulate in the bulbs, all
the charge flowing into a bulb from the left has to flow out on the right; consequently
I c = I d and I e = I f . The two currents leaving the bulbs recombine to form the current back
into the battery, I f + I d = I b .
8.
(a) B, (b) B. Because the voltage across each resistor is the same, and the rate of energy
2
delivered to a resistor is P = ( ∆V ) R , the resistor with the lower resistance (that is, B)
dissipates more power. From Ohm’s law, I = ∆V R . Since the potential difference is the
same for the two resistors, B (having the smaller resistance) will carry the greater current.
Current and Resistance
71
Answers to Even Numbered Conceptual Questions
2.
In the electrostatic case in which charges are stationary, the internal electric field must be
zero. A nonzero field would produce a current (by interacting with the free electrons in
the conductor), which would violate the condition of static equilibrium. In this chapter we
deal with conductors that carry current, a nonelectrostatic situation. The current arises
because of a potential difference applied between the ends of the conductor, which
produces an internal electric field.
4.
The number of cars would correspond to charge Q. The rate of flow of cars past a point
would correspond to current.
6.
The 25 W bulb has the higher resistance. Because R = ( ∆V ) P , and both operate from
120 V, the bulb dissipating the least power has the higher resistance. The 100 W bulb
carries more current, because the current is proportional to the power rating of the bulb.
8.
An electrical shock occurs when your body serves as a conductor between two points
having a difference in potential. The concept behind the admonition is to avoid
simultaneously touching points that are at different potentials.
2
10.
The knob is connected to a variable resistor. As you increase the magnitude of the
resistance in the circuit, the current is reduced, and the bulb dims.
12.
Superconducting devices are expensive to operate primarily because they must be kept at
very low temperatures. As the onset temperature for superconductivity is increased
toward room temperature, it becomes easier to accomplish this reduction in temperature.
In fact, if room temperature superconductors could be achieved, this requirement would
disappear altogether.
14.
The amplitude of atomic vibrations increases with temperature, thereby scattering
electrons more efficiently.
72
CHAPTER 17
Answers to Even Numbered Problems
2.
5.21 × 10 −5 m s
4.
3.75 × 1014 electrons s
6.
159 mA
8.
1.3 × 10 − 4 m s
10.
500 mA
12.
(a)
14.
0.31 Ω
16.
~ $1 (Assumes a 400 W dryer used for about 10 min/d with electric energy costing about
10 cents/kWh.)
18.
(a)
20.
1.4 × 10 3 °C
22.
9.7 Ω
24.
1.1 × 10 −3 ( °C )
26.
(a)
28.
4.90 × 10 −7 m 2
30.
63.2°C
32.
(a)
34.
1.82 m
(b)
2.8 × 108 A
0.280 mm
(b)
1.8 × 107 A
(b)
1.1 mV
$0.29
(b)
$2.6
(a)
50 MW
(b)
0.071 or 7.1%
36.
(a)
3.2 × 10 5 J
(b)
18 min
38.
(a)
184 W
(b)
461°C
40.
(a)
5.0 cents
(b)
71%
42.
(a)
$1.61
(b)
0.583 cents
44.
3.36 h d
−1
0.65 mV
(c)
41.6 cents
Current and Resistance
46.
73
Any diameter d and length l related by d 2 = ( 4.8 × 10 −8 m )l,
such as l = 1.0 m and d = 0.22 mm . Yes.
48.
(a) 576 Ω , 144 Ω
(b) 4.80 s, lower potential energy
(c) 0.040 0 s, converted to internal energy and light
(d) $1.26, energy, $1.94 × 10 −8 J
50.
(a)
52.
89 µ V
54.
(a)
56.
(a)
58.
5.6 kΩ (nichrome) , 4.4 kΩ (carbon)
60.
No. The fuse should be rated at 3.87 A or less.
62.
(a)
8.6 × 10 5 J
(b)
1.9 cents
18 C
(b)
3.6 A
9.49 × 10 −7 Ω
(b)
8.07 × 10 −7 Ω
∆V
R = ∆V I
I
− 1.5 V
− 0.30 × 10
A
5.0 × 10 5 Ω
− 1.0 V
− 0.20 × 10 −5 A
5.0 × 10 5 Ω
− 0.50 V
− 0.10 × 10 −5 A
5.0 × 10 5 Ω
+ 0.40 V
+ 0.010 A
40 Ω
+ 0.50 V
+ 0.020 A
25 Ω
+ 0.55 V
+ 0.040 A
+ 0.70 V
+ 0.072 A
14 Ω
9.7 Ω
+ 0.75 V
+ 0.10 A
7.5 Ω
−5
(b) The resistance of the diode is very large when the applied potential difference has one
polarity and is rather small when the potential difference has the opposite polarity.
64.
(a)
66.
(b)
9.3 m
(b)
0.93 mm
1.420 Ω versus 1.418 Ω for the more precise equation
74
CHAPTER 17
Problem Solutions
17.1
The charge that moves past the cross section is ∆Q = I ( ∆t ) , and the number of electrons
is
n=
∆Q I ( ∆t )
=
e
e
( 80.0 × 10
=
C s ) ( 10.0 min ) ( 60.0 s min ) 
= 3.00 × 10 20 electrons
1.60 × 10 −19 C
The negatively charged electrons move in the direction opposite
to the conventional current flow.
ur
−3
v
17.2
2.50 C s
I
=
= 5.21 × 10 −5 m s
28
-3
nqA ( 7.50 × 10 m )( 1.60 × 10 −19 C )( 4.00 × 10 −6 m 2 )
The current is I =
∆Q ∆V
 ∆V 
=
. Thus, the change that passes is ∆Q = 
 ( ∆t ) , giving
∆t
R
 R 
 1.00 V 
∆Q = 
 ( ∆t ) = ( 0.100 A )( 20.0 s ) = 2.00 C
 10.0 Ω 
17.4
–
The drift speed in a conductor of cross-sectional area A and carrying current I is
vd = I nqA , where n is the number of free charge carriers per unit volume and q is the
charge of each of those charge carriers. In the given conductor,
vd =
17.3
ur
I
∆Q = I ( ∆t ) and the number of electrons is
n=
−6
∆Q I ( ∆t ) ( 60.0 × 10 C s ) ( 1.00 s )
=
=
= 3.75 × 1014 electrons
e
e
1.60 × 10 -19 C
Current and Resistance
17.5
The period of the electron in its orbit is T = 2π r v , and the current represented by the
orbiting electron is
I=
=
17.6
∆Q e v e
= =
∆t T 2π r
( 2.19 × 10
6
m s )( 1.60 × 10 −19 C )
2 π ( 5.29 × 10 −11 m )
= 1.05 × 10 −3 C s = 1.05 mA
The mass of a single gold atom is
matom =
197 g mol
M
=
= 3.27 × 10 -22 g = 3.27 × 10 -25 kg
23
N A 6.02 × 10 atoms mol
The number of atoms deposited, and hence the number of ions moving to the negative
electrode, is
n=
3.25 × 10 −3 kg
m
=
= 9.93 × 10 21
matom 3.27 × 10 -25 kg
Thus, the current in the cell is
21
−19
∆Q ne ( 9.93 × 10 )( 1.60 × 10 C )
I=
=
=
= 0.159 A = 159 mA
∆t ∆t
( 2.78 h ) ( 3 600 s 1 h )
17.7
The drift speed of electrons in the line is vd =
vd =
( 8.5 × 10
4 ( 1 000 A )
28
m
3
)(1.60 × 10
−19
I
I
=
, or
nqA n e (π d 2 4 )
C ) π ( 0.020 m )
2
= 2.3 × 10 -4 m s
The time to travel the length of the 200-km line is then
∆t =
1 yr
L
200 × 10 3 m 

=

 = 27 yr
-4
7
vd 2.34 × 10 m s  3.156 × 10 s 
75
76
17.8
CHAPTER 17
Assuming that, on average, each aluminum atom contributes one electron, the density of
charge carriers is the same as the number of atoms per cubic meter. This is
n=
or
n=
density
N ρ
ρ
=
= A ,
mass per atom M N A
M
( 6.02 × 10
23
mol ) ( 2.7 g cm 3 )( 10 6 cm 3 1 m 3 ) 
26.98 g mol
= 6.0 × 10 28 m 3
The drift speed of the electrons in the wire is then
vd =
17.9
5.0 C s
I
=
= 1.3 × 10 − 4 m s
−19
−6
28
3
2
n e A ( 6.0 × 10 m )( 1.60 × 10 C )( 4.0 × 10 m )
(a) The carrier density is determined by the physical characteristics of the wire, not the
current in the wire. Hence, n is unaffected .
(b) The drift velocity of the electrons is vd = I nqA . Thus, the drift velocity is doubled
when the current is doubled.
∆V 120 V
=
= 0.500 A = 500 mA
R
240 Ω
17.10
I=
17.11
( ∆V )max = I max R = ( 80 × 10 −6 A ) R
Thus, if R = 4.0 × 10 5 Ω , ( ∆V )max = 32 V
and if R = 2 000 Ω , ( ∆V )max = 0.16 V
17.12
The volume of the copper is
V=
1.00 × 10 −3 kg
m
=
= 1.12 × 10 −7 m 3
3
3
density 8.92 × 10 kg m
Since, V = A ⋅ L , this gives A ⋅ L = 1.12 × 10 −7 m 3
(1)
Current and Resistance
ρL
(a) From R =
, we find that
A
 1.70 × 10 −8 Ω ⋅ m 
ρ
−8
A =  L = 
 L = ( 3.40 × 10 m ) L
Ω
0.500
R


Inserting this expression for A into equation (1) gives
( 3.40 × 10
−8
m ) L2 = 1.12 × 10 −7 m 3 , which yields
(b) From equation (1), A =
π d2
4
4 ( 1.12 × 10 −7 m 3 )
d=
πL
=
=
L = 1.82 m
1.12 × 10 −7 m 3
, or
L
4 ( 1.12 × 10 −7 m 3 )
π ( 1.82 m )
= 2.80 × 10 −4 m = 0.280 mm
17.13
From R =
d=
17.14
R=
17.15
(a)
ρL
A
, we obtain A =
4ρ L
=
πR
ρL
A
=
R=
π d2
4
=
ρL
R
, or
4 ( 5.6 × 10 −8 Ω ⋅ m )( 2.0 × 10 −2 m )
ρL
π d2 4
π ( 0.050 Ω )
=
4 ( 1.7 × 10 −8 Ω ⋅ m ) ( 15 m )
π ( 1.024 × 10 −3 m )
2
= 1.7 × 10−4 m = 0.17 mm
= 0.31 Ω
∆V
12 V
=
= 30 Ω
I
0.40 A
(b) From, R =
ρL
A
,
2
−2


R ⋅ A ( 30 Ω ) π ( 0.40 × 10 m ) 
=
= 4.7 × 10 − 4 Ω ⋅ m
ρ=
L
3.2 m
77
78
17.16
CHAPTER 17
We assume that your hair dryer will use about 400 W of power for 10 minutes each day
of the year. The estimate of the total energy used each year is
min  1 hr 


E = P( ∆t ) = ( 0.400 kW )  10

 ( 365 d )  = 24 kWh
d
60
min




If your cost for electrical energy is approximately ten cents per kilowatt-hour, the cost of
using the hair dryer for a year is on the order of
$ 

cost = E × rate = ( 24 kWh )  0.10
 = $2.4
kWh 

17.17
The resistance is R =
or
~$1
∆V 9.11 V
=
= 0.253 Ω , so the resistivity of the metal is
I
36.0 A
−3
2
R ⋅ A R ⋅ (π d 4 ) ( 0.253 Ω ) π ( 2.00 × 10 m )
=
=
= 1.59 × 10 −8 Ω ⋅ m
ρ=
L
L
4 ( 50.0 m )
2
Thus, the metal is seen to be silver .
17.18
With different orientations of the block, three different values of the ratio L A are
possible. These are:
10 cm
1
1
L
,
=
=
  =
 A 1 ( 20 cm )( 40 cm ) 80 cm 0.80 m
20 cm
1
1
L
,
=
=
  =
 A  2 ( 10 cm )( 40 cm ) 20 cm 0.20 m
and
(a)
40 cm
1
1
L
=
=
  =
 A  3 ( 10 cm )( 20 cm ) 5.0 cm 0.050 m
I max =
(b) I min =
∆V
∆V
( 6.0 V )( 0.80 m )
=
=
= 2.8 × 10 8 A
−8
Rmin ρ ( L A )min
1.7 × 10 Ω ⋅ m
∆V
∆V
( 6.0 V )( 0.050 m )
=
=
= 1.8 × 107 A
Rmax ρ ( L A )max
1.7 × 10 −8 Ω ⋅ m
Current and Resistance
17.19
79
The volume of material, V = AL0 = (π r02 ) L0 , in the wire is constant. Thus, as the wire is
( )
stretched to decrease its radius, the length increases such that π rf2 L f = (π r02 ) L0 giving
2
r 
 r 
2
L f =  0  L0 =  0  L0 = ( 4.0 ) L0 = 16 L0
 rf 
0.25
r
0 

 
The new resistance is then
Rf = ρ
17.20
Lf
Af
=ρ
Lf
πr
2
f
=ρ
16 L0
π ( r0 4 )
2
L
2
= 16 ( 4 )  ρ 02
 π r0

 = 256 R0 = 256 ( 1.00 Ω ) = 256 Ω

Solving R = R0 1 + α ( T − T0 )  for the final temperature gives
T = T0 +
17.21
2
R − R0
140 Ω − 19 Ω
= 20°C +
= 1.4 × 10 3 °C
α R0
 4.5 × 10 -3 ( °C )−1  ( 19 Ω )


From Ohm’s law, ∆V = I i Ri = I f R f , so the current in Antarctica is
R
I f = Ii  i
 Rf

 R 1 + α ( T − T )  

0 
i
0  
 = Ii 

 R 1 + α T − T  

0 
f
 0

(
)
 1 +  3.90 × 10 −3 ( °C )−1  ( 58.0°C − 20.0°C ) 


 = 1.98 A
= ( 1.00 A ) 
 1 +  3.90 × 10 −3 ( °C )−1  ( −88.0°C − 20.0°C ) 




17.22
The expression for the temperature variation of resistance, R = R0 1 + α ( T − T0 )  with
T0 = 20°C , gives the temperature coefficient of resistivity of this material as
α=
R − R0
10.55 Ω − 10.00 Ω
=
= 7.9 × 10 −4 °C −1
R0 ( T − T0 ) ( 10.00 Ω )( 90°C − 20°C )
Then, at T = −20°C , the resistance will be
R = R20 1 + α ( −20°C − 20°C )  = ( 10.00 Ω ) 1 + ( 7.9 × 10 −4 °C −1 ) ( −40°C )  = 9.7 Ω
80
17.23
CHAPTER 17
At 80°C,
I=
I = 2.6 × 10 −2 A = 26 mA
or
17.24
∆V
∆V
5.0 V
=
=
R
R0 1 + α ( T − T0 )  ( 200 Ω ) 1 + ( − 0.5 × 10 −3 °C −1 ) ( 80°C − 20°C ) 


If R = 41.0 Ω at T = 20°C and R = 41.4 Ω at T = 29.0°C , then R = R0 1 + α ( T − T0 )  gives
the temperature coefficient of resistivity of the material making up this wire as
α=
17.25
R0 =
ρL
A
R − R0
41.4 Ω − 41.0 Ω
−1
=
= 1.1 × 10 −3 ( °C )
R0 ( T − T0 ) ( 41.0 Ω )( 29.0°C − 20°C )
(1.7 × 10
=
−8
Ω ⋅ m ) ( 10.0 m )
3.00 × 10 −6 m 2
= 5.67 × 10 −2 Ω
(a) At T = 30.0°C, R = R0 1 + α ( T − T0 )  gives a resistance of
(
R = ( 0.056 7 Ω ) 1 + 3.9 × 10 −3 ( °C )

−1
) ( 30.0°C − 20.0°C ) =
5.89 × 10 −2 Ω
(b) At T = 10.0°C, R = R0 1 + α ( T − T0 )  yields
(
R = ( 0.056 7 Ω ) 1 + 3.9 × 10 −3 ( °C )

17.26
−1
) (10.0°C − 20.0°C ) =
5.45 × 10 −2 Ω
(a) At 20°C, the resistance of this copper wire is R0 = ρ ( L A ) = ρ ( L π r 2 ) and the
potential difference required to produce a current of I = 3.0 A is
 L
∆V0 = IR0 = I  ρ 2
 πr
or
( 1.00 m )

−8
 = ( 3.0 A ) ( 1.7 × 10 Ω ⋅ m )
2

π ( 0.50 × 10 −2 m )
∆V0 = 6.5 × 10 −4 V = 0.65 mV
Current and Resistance
81
(b) At T = 200°C , the potential difference required to maintain the same current in the
copper wire is
∆V = IR = IR0 1 + α ( T − T0 )  = ∆V0 1 + α ( T − T0 ) 
∆V = ( 0.65 mV ) 1 + ( 3.9 × 10 −3 °C -1 ) ( 200°C − 20°C )  = 1.1 mV
or
17.27
(a) The resistance at 20.0°C is
R0 = ρ
−8
L ( 1.7 × 10 Ω ⋅ m ) ( 34.5 m )
=
= 3.0 Ω
2
A
π ( 0.25 × 10 −3 m )
and the current will be I =
∆V 9.0 V
=
= 3.0 A
R0 3.0 Ω
(b) At 30.0°C,
R = R0 1 + α ( T − T0 ) 
(
= ( 3.0 Ω ) 1 + 3.9 × 10 −3 ( °C )

Thus, the current is I =
17.28
−1
) ( 30.0°C − 20.0°C ) = 3.1 Ω
∆V 9.0 V
=
= 2.9 A
R
3.1 Ω
The resistance of the heating element when at its operating temperature is
R=
( ∆V )
P
2
=
( 120 V )
2
1 050 W
From R = R0 1 + α ( T − T0 )  =
A=
= 13.7 Ω
ρ0 L
1 + α ( T − T0 )  , the cross-sectional area is
A 
ρ0 L
1 + α ( T − T0 ) 
R 
(150 × 10
=
−8
Ω ⋅ m ) ( 4.00 m )
13.7 Ω
A = 4.90 × 10 −7 m 2
(
)
1 + 0.40 × 10 −3 ( °C )−1 ( 320°C − 20.0°C ) 


82
17.29
CHAPTER 17
(a) From R = ρ L A , the initial resistance of the mercury is
( 9.4 × 10 Ω ⋅ m ) (1.000 0 m ) =
π ( 1.00 × 10 m ) 4
−7
ρ Li
Ri =
=
Ai
2
−3
1.2 Ω
(b) Since the volume of mercury is constant, V = A f ⋅ L f = Ai ⋅ Li gives the final cross-
(
)
sectional area as A f = Ai ⋅ Li L f . Thus, the final resistance is given by
Rf =
ρ Lf
Af
∆=
=
ρ L2f
Ai ⋅ Li
R f − Ri
Ri
=
. The fractional change in the resistance is then
Rf
Ri
−1=
ρ L2f
( Ai ⋅ Li )
ρ Li Ai
2
 Lf 
−1=   −1
 Li 
2
 100.04 
−4
∆=
or a 0.080% increase
 − 1 = 8.0 × 10
 100.00 
17.30
The resistance at 20.0°C is
R0 =
200.0 Ω
R
=
= 217 Ω
1 + α ( T − T0 ) 1+  3.92 × 10 -3 ( °C )−1  ( 0°C − 20.0°C )


Solving R = R0 1 + α ( T − T0 )  for T gives the temperature of the melting potassium as
T = T0 +
17.31
I=
P 600 W
=
= 5.00 A
∆V 120 V
and
17.32
R − R0
253.8 Ω − 217 Ω
= 20.0°C +
= 63.2°C
α R0
 3.92 × 10 −3 ( °C )−1  ( 217 Ω )


R=
∆V 120 V
=
= 24.0 Ω
5.00 A
I
(a) The energy used by a 100-W bulb in 24 h is
E = P⋅∆t = ( 100 W )( 24 h ) = ( 0.100 kW )( 24 h ) = 2.4 kWh
and the cost of this energy, at a rate of $0.12 per kilowatt-hour is
cost = E ⋅ rate = ( 2.4 kWh ) ( $0.12 kWh ) = $0.29
Current and Resistance
(b) The energy used by the oven in 5.0 h is

 1 kW  
E = P⋅∆t =  I ( ∆V )  ⋅ ∆t = ( 20.0 C s )( 220 J C ) 
  ( 5.0 h ) = 22 kWh
3

 10 J s  
and the cost of this energy, at a rate of $0.12 per kilowatt-hour is
cost = E ⋅ rate = ( 22 kWh ) ( $0.12 kWh ) = $2.6
17.33
The maximum power that can be dissipated in the circuit is
Pmax = ( ∆V ) I max = ( 120 V )( 15 A ) = 1.8 × 10 3 W
Thus, one can operate at most 18 bulbs rated at 100 W per bulb.
17.34
(a) The power loss in the line is
Ploss = I 2 R = ( 1 000 A ) ( 0.31 Ω km ) ( 160 km )  = 5.0 × 107 W = 50 MW
2
(b) The total power transmitted is
Pinput = ( ∆V ) I = ( 700 × 10 3 V ) ( 1 000 A ) = 7.0 × 10 8 W = 700 MW
Thus, the fraction of the total transmitted power represented by the line losses is
fraction loss =
17.35
Ploss
50 MW
=
= 0.071 or 7.1%
Pinput 700 MW
The energy required to bring the water to the boiling point is
E = mc ( ∆T ) = ( 0.500 kg )( 4 186 J kg ⋅ °C ) ( 100°C − 23.0°C ) = 1.61 × 10 5 J
The power input by the heating element is
Pinput = ( ∆V ) I = ( 120 V ) ( 2.00 A ) = 240 W = 240 J s
Therefore, the time required is
t=
E
1.61 × 10 5 J
 1 min 
=
= 672 s 
 = 11.2 min
Pinput
240 J s
 60 s 
83
84
17.36
CHAPTER 17
(a)
E = P⋅t = ( 90 W )( 1 h ) = ( 90 J s )( 3 600 s ) = 3.2 × 10 5 J
(b) The power consumption of the color set is
P = ( ∆V ) I = ( 120 V ) ( 2.50 A ) = 300 W
Therefore, the time required to consume the energy found in (a) is
t=
17.37
E 3.2 × 10 5 J
=
= 1.1 × 10 3
P
300 J s
 1 min 
s
 = 18 min
 60 s 
The energy input required is
E = mc ( ∆T ) = ( 1.50 kg )( 4 186 J kg ⋅ °C ) ( 50.0°C − 10.0°C ) = 2.51 × 10 5 J
and, if this is to be added in ∆t = 10.0 min = 600 s , the power input needed is
P=
E 2.51 × 10 5 J
=
= 419 W
∆t
600 s
The power input to the heater may be expressed as P = ( ∆V ) R , so the needed
resistance is
2
R=
17.38
( ∆V )
P
2
=
( 120 V )
419 W
2
= 34.4 Ω
(a) At the operating temperature,
P = ( ∆V ) I = ( 120 V )( 1.53 A ) = 184 W
(b) From R = R0 1 + α ( T − T0 )  , the temperature T is given by T = T0 +
resistances are given by Ohm’s law as
R=
( ∆V )
I
=
( ∆V )0 120 V
120 V
=
, and R0 =
1.53 A
1.80 A
I0
Therefore, the operating temperature is
T = 20.0°C +
(120 1.53 ) − ( 120 1.80 ) =
( 0.400 × 10−3 (°C )−1 ) (120 1.80 )
461°C
R − R0
. The
α R0
Current and Resistance
17.39
The resistance per unit length of the cable is
R P I 2 P L 2.00 W m
=
= 2 =
= 2.22 × 10 −5 Ω m
2
L
L
I
( 300 A )
From R = ρ L A , the resistance per unit length is also given by R L = ρ A . Hence, the
cross-sectional area is π r 2 = A =
ρ
r=
17.40
π ( R L)
=
ρ
RL
, and the required radius is
1.7 × 10 −8 Ω ⋅ m
= 0.016 m = 1.6 cm
π ( 2.22 × 10 -5 Ω m )
(a) The power input to the motor is
Pinput = ( ∆V ) I = ( 120 V )( 1.75 A ) = 210 W = 0.210 kW
At a rate of $0.06/kWh, the cost of operating this motor for 4.0 h is
(
)
cost = ( Energy used ) ⋅ rate = Pinput ⋅t ⋅ rate
= ( 0.210 kW )( 4.0 h ) ( 6.0 cents kWh ) = 5.0 cents
(b) The efficiency is
Eff =
Poutput
Pinput
=
( 0.20 hp )( 0.746
kW hp )
0.210 kW
= 0.71 or 71%
85
86
17.41
CHAPTER 17
The total power converted by the clocks is
P= ( 2.50 W ) ( 270 × 10 6 ) = 6.75 × 108 W
and the energy used in one hour is
E = P⋅t = ( 6.75 × 108 W ) ( 3 600 s ) = 2.43 × 1012 J
The energy input required from the coal is
Ecoal =
2.43 × 1012 J
E
=
= 9.72 × 1012 J
0.250
efficiency
The required mass of coal is thus
m=
or
17.42
(a)
Ecoal
9.72 × 1012 J
=
= 2.95 × 10 5 kg
heat of combustion 33.0 × 10 6 J kg
 1 metric ton 
m = ( 2.95 × 10 5 kg ) 
 = 295 metric tons
3
 10 kg 
E = P⋅t = ( 40.0 W )( 14.0 d ) ( 24.0 h d ) = 1.34 × 10 4 Wh = 13.4 kWh
cost = E ⋅ ( rate ) = ( 13.4 kWh ) ( $0.120 kWh ) = $1.61
(b) E = P⋅t = ( 0.970 kW )( 3.00 min ) ( 1 h 60 min ) = 4.85 × 10 −2 kWh
cost = E ⋅ ( rate )
= ( 4.85 × 10 -2 kWh ) ( $0.120 kWh ) = $0.005 83 = 0.583 cents
(c)
E = P⋅t = ( 5.20 kW )( 40.0 min ) ( 1 h 60 min ) = 3.47 kWh
cost = E ⋅ ( rate ) = ( 3.47 kWh ) ( $0.120 kWh ) = $0.416 = 41.6 cents
17.43
E = P⋅t = ( 0.180 kW )( 21 h )
so
and
cost = E ⋅ rate = ( P⋅t ) ( $0.070 0 kWh )
cost = ( 0.180 kW )( 21 h )  ( $0.070 0 kWh ) = $0.26 = 26 cents
Current and Resistance
17.44
The energy used was E =
cost
$200
=
= 2.50 × 10 3 kWh
rate $0.080 kWh
E 2.50 × 10 3 kWh
= 104 h , and since
The total time the furnace operated was t = =
24.0 kW
P
January has 31 days, the average time per day was
average daily operation =
17.45
104 h
= 3.36 h d
31.0 d
The energy saved is
(
)
∆E = Phigh − Plow ⋅ t = ( 40 W − 11 W )( 100 h ) = 2.9 × 10 3 Wh = 2.9 kWh
and the monetary savings is
savings = ∆E ⋅ rate = ( 2.9 kWh )( $0.080 kWh ) = $0.23 = 23 cents
17.46
The power required to warm the water to 100°C in 4.00 min is
P=
∆Q mc ( ∆T ) ( 0.250 kg )( 4 186 J kg ⋅ °C ) ( 100°C − 20°C )
=
=
= 3.5 × 10 2 W
∆t
∆t
( 4.00 min ) ( 60 s 1 min )
The required resistance (at 100°C) of the heating element is then
R=
( ∆V )
P
( 120 V )
2
=
2
3.5 × 10 2 W
= 41 Ω
so the resistance at 20°C would be
R0 =
R
41 Ω
=
= 40 Ω
-3
1 + α ( T − T0 ) 1+ ( 0.4 × 10 °C −1 ) ( 100°C − 20°C )
We find the needed dimensions of a Nichrome wire for this heating element from
R0 = ρ l A = ρ l (π d 2 4 ) , where l is the length of the wire and d is its diameter. This
 4ρ  l
gives 40 Ω = 
 2
 π d
or


 150 × 10 −8 Ω ⋅ m 
ρ
−8
l
d2 = 
=


 l = ( 4.8 × 10 m )l
10
10
π
π
Ω
Ω
(
)
(
)




87
88
CHAPTER 17
Thus,
any combination of length and diameter satisfying the relation d 2 = ( 4.8 × 10 −8 m )l will
be suitable. A typical combination might be
l = 1.0 m
and
d = 4.8 × 10 −8 m 2 = 2.2 × 10 −4 m = 0.22 mm
Yes , such heating elements could easily be made from less than 0.5 cm 3 of Nichrome.
The volume of material required for the typical wire given above is
2
 10 6 cm 3
V = Al = (π d 2 )l = π ( 2.2 × 10 −4 m ) ( 1.0 m ) = ( 1.5 × 10 −7 m 3 ) 
3
 1m
17.47

3
 = 0.15 cm

The energy that must be added to the water is

 1 kWh
J 
E = mc ( ∆T ) = ( 200 kg )  4 186
 ( 80°C − 15°C ) 
6
kg
C
⋅
°
 3.60 × 10



 = 15 kWh
J
and the cost is cost = E ⋅ rate = ( 15 kWh ) ( $0.080 kWh ) = $1.2
17.48
(a) From P = ( ∆V ) R , the resistance of each bulb is
2
Rdim =
( ∆V )
Rbright =
2
=
Pdim
( ∆V )
( 120 V )
2
Pbright
=
2
25.0 W
( 120 V )
100 W
= 576 Ω and
2
= 144 Ω
(b) The current in the dim bulb is
I=
Pdim 25.0 W
=
= 0.208 A
∆V
120 V
so the time for 1.00 C to pass through the bulb is
∆t =
∆Q 1.00 C
=
= 4.80 s
I
0.208 A
When the charge emerges from the bulb, it has lower potential energy
Current and Resistance
(c) The time for the dim bulb to dissipate 1.00 J of energy is
∆t =
∆E
1.00 J
=
= 0.040 0 s
Pdim 25.0 W
Electrical potential energy is transformed into internal energy and light
(d) In 30.0 days, the energy used by the dim bulb is
E = Pdim ⋅t = ( 25.0 W )( 30.0 d ) ( 24.0 h d ) = 1.80 × 10 4 Wh = 18.0 kWh
and the cost is cost = E ⋅ rate = ( 18.0 kWh ) ( $0.070 0 kWh ) = $1.26
The electric company sells energy , and the unit cost is
 $0.070 0   1 kWh
unit cost = 

6
 kWh   3.60 × 10
17.49

−8
 = $1.94 × 10 Joule
J
From P = ( ∆V ) R , the total resistance needed is
2
R=
( ∆V )
2
P
=
( 20 V )
48 W
2
= 8.3 Ω
Thus, from R = ρ L A , the length of wire required is
L=
17.50
R⋅A
ρ
=
( 8.3 Ω ) ( 4.0 × 10 −6 m 2 )
3.0 × 10 −8 Ω ⋅ m
= 1.1 × 10 3 m = 1.1 km
(a) The power required by the iron is
√ = ( ∆V ) I = ( 120 V )( 6.0 A ) = 7.2 × 10 2 W
and the energy transformed in 20 minutes is
J 

 60 s  
5
E = √ ⋅ t =  7.2 × 10 2  ( 20 min ) 
  = 8.6 × 10 J
s 

 1 min  
89
90
CHAPTER 17
(b) The cost of operating the iron for 20 minutes is
cost = E ⋅ rate

 1 kWh
= ( 8.6 × 10 5 J ) 
6
 3.60 × 10

17.51

  ( $0.080 kWh ) = $0.019 = 1.9 cents
J 
Ohm’s law gives the resistance as R = ( ∆V ) I . From R = ρ L A , the resistivity is given by
ρ = R ⋅ ( A L ) . The results of these calculations for each of the three wires are
summarized in the table below.
L (m)
R (Ω)
ρ (Ω ⋅ m)
0.540
1.028
1.543
10.4
21.1
31.8
1.41 × 10 −6
1.50 × 10 −6
1.50 × 10 −6
The average value found for the resistivity is
ρav =
Σρ i
= 1.47 × 10 −6 Ω ⋅ m
3
which differs from the value of ρ = 150 × 10 −8 Ω ⋅ m = 1.50 × 10 −6 Ω ⋅ m given in Table 17.1
by 2.0% .
17.52
The resistance of the 4.0 cm length of wire between the feet is
R=
ρL
A
=
(1.7 × 10
−8
Ω ⋅ m ) ( 0.040 m )
π ( 0.011 m )
2
= 1.79 × 10 −6 Ω ,
so the potential difference is
∆V = IR = ( 50 A ) ( 1.79 × 10 −6 Ω ) = 8.9 × 10 −5 V = 89 µ V
Current and Resistance
17.53
(a) The total power you now use while cooking breakfast is
P = ( 1 200 + 500 ) W = 1.70 kW
The cost to use this power for 0.500 h each day for 30.0 days is



h 
cost =  P × ( ∆t )  × rate = ( 1.70 kW )  0.500
 ( 30.0 days )  ( $0.120 kWh ) = $3.06
day 



(b) If you upgraded, the new power requirement would be:
P = ( 2 400 + 500 ) W=2 900 W
and the required current would be
I=
2 900 W
P
=
= 26.4 A > 20 A
∆V
110 V
No , your present circuit breaker cannot handle the upgrade.
(a) The charge passing through the conductor in the interval
0 ≤ t ≤ 5.0 s is represented by the area under the I vs t
graph given in Figure P17.54. This area consists of two
rectangles and two triangles. Thus,
∆Q = Arectangle 1 + Arectangle 2 + Atriangle 1 + Atriangle 2
6
Current (A)
17.54
= ( 5.0 s − 0 ) ( 2.0 A − 0 ) + ( 4.0 s − 3.0 s ) ( 6.0 A − 2.0 A )
+
1 2 3 4 5
Time(s)
1
1
( 3.0 s − 2.0 s )( 6.0 A − 2.0 A ) + ( 5.0 s − 4.0 s ) ( 6.0 A − 2.0 A )
2
2
∆Q = 18 C
(b) The constant current that would pass the same charge in 5.0 s is
I=
2
∆Q 18 C
=
= 3.6 A
∆t 5.0 s
91
92
17.55
CHAPTER 17
(a) From √ = ( ∆V ) I , the current is I =
P 8.00 × 10 3 W
=
= 667 A
∆V
12.0 V
(b) The time before the stored energy is depleted is
t=
Estorage
P
=
2.00 × 107 J
= 2.50 × 10 3 s
8.00 × 10 3 J s
Thus, the distance traveled is
d = v ⋅ t = ( 20.0 m s ) ( 2.50 × 10 3 s ) = 5.00 × 10 4 m = 50.0 km
17.56
The volume of aluminum available is
V=
115 × 10 −3 kg
mass
=
= 4.26 × 10 −5 m 3
density 2.70 × 10 3 kg m 3
(a) For a cylinder whose height equals the diameter, the volume is
 π d2
V =
 4

π d3
=
d

4

 4V 
and the diameter is d = 

 π 
13
 4 ( 4.26 × 10 −5 m 3 ) 

=
π


13
= 0.037 85 m
The resistance between ends is then
R=
ρL
=
A
ρd
(π d 4 )
2
=
−8
4 ρ 4 ( 2.82 × 10 Ω ⋅ m )
=
= 9.49 × 10 −7 Ω
πd
π ( 0.037 85 m )
(b) For a cube, V = L3 , so the length of an edge is
L = (V )
13
= ( 4.26 × 10 −5 m )
13
= 0.034 9 m
The resistance between opposite faces is
R=
ρL
A
=
ρL
L2
=
ρ
L
=
2.82 × 10 −8 Ω ⋅ m
= 8.07 × 10 −7 Ω
0.034 9 m
93
Current and Resistance
17.57
The current in the wire is I =
∆V 15.0 V
=
= 150 A
R
0.100 Ω
Then, from vd = I nqA , the density of free electrons is
n=
or
17.58
I
vd e ( π r
2
=
) ( 3.17 × 10
150 A
−4
m s )( 1.60 × 10 −19 C ) π ( 5.00 × 10 −3 m )
2
n = 3.77 × 10 28 m 3
At temperature T, the resistance of the carbon wire is Rc = R0 c 1 + α c ( T − T0 )  , and that
of the nichrome wire is Rn = R0 n 1 + α n ( T − T0 )  . When the wires are connected end to
end, the total resistance is
R = Rc + Rn = ( R0 c + R0 n ) + ( R0 cα c + R0 nα n )( T − T0 )
If this is to have a constant value of 10.0 kΩ as the temperature changes, it is necessary
that
and
R0 c + R0 n = 10.0 kΩ
(1)
R0 cα c + R0 nα n = 0
(2)
From equation (1), R0 c = 10.0 kΩ − R0 n , and substituting into equation (2) gives
(10.0 kΩ − R0 n )  −0.50 × 10 −3 ( °C )−1  + R0 n 0.40 × 10 −3 ( °C )−1  = 0
R0 n = 5.6 kΩ
Solving this equation gives
Then, R0 c = 10.0 kΩ − 5.6 kΩ = 4.4 kΩ
17.59
(a) From √ =
( ∆V )
(b) Solving R =
L=
R⋅A
ρ
R
ρL
A
=
( nichrome wire )
( carbon wire )
2
, the resistance is R =
( ∆V )
√
2
=
( 120 V )
100 W
for the length gives
( 144 Ω ) ( 0.010 mm 2 )  1 m 2
5.6 × 10 Ω ⋅ m
-8

= 26 m
 6
2 
 10 mm 
2
= 144 Ω
94
CHAPTER 17
(c) The filament is tightly coiled to fit the required length into a small space
(d) From L = L0 1 + α ( T − T0 )  , where α = 4.5 × 10 −6 ( °C ) , the length at T0 = 20°C is
−1
L0 =
17.60
L
26 m
=
= 25 m
1 + α ( T − T0 ) 1 + 4.5 × 10 −6 ( °C )−1 ( 2 600°C − 20°C )
(
)
Each speaker has a resistance of R = 4.00 Ω and can handle 60.0 W of power. From
P = I 2 R , the maximum safe current is
I max =
√
60.0 W
=
= 3.87 A
R
4.00 Ω
Thus, the system is not adequately protected by a 4.00 A fuse.
17.61
2
2
The cross-sectional area of the conducting material is A = π ( router
− rinner
)
Thus,
R=
17.62
ρL
A
=
( 3.5 × 10
π ( 1.2 × 10

5
−2
Ω ⋅ m )( 4.0 × 10 −2 m )
m ) − ( 0.50 × 10
2
−2
m ) 

2
= 3.7 × 107 Ω = 37 MΩ
(a)
∆V
I
R = ∆V I
− 1.5 V
− 0.30 × 10 −5 A
5.0 × 10 5 Ω
− 1.0 V
− 0.20 × 10 −5 A
5.0 × 10 5 Ω
− 0.50 V
− 0.10 × 10 −5 A
5.0 × 10 5 Ω
+ 0.40 V
+ 0.010 A
40 Ω
+ 0.50 V
+ 0.020 A
25 Ω
+ 0.55 V
+ 0.040 A
+ 0.70 V
+ 0.072 A
14 Ω
9.7 Ω
+ 0.75 V
+ 0.10 A
7.5 Ω
(b) The resistance of the diode is very large when the applied potential difference has
one polarity and is rather small when the potential difference has the opposite
polarity.
Current and Resistance
17.63
95
The power the beam delivers to the target is
P = ( ∆V ) I = ( 4.0 × 106 V )( 25 × 10 −3 A ) = 1.0 × 10 5 W
The mass of cooling water that must flow through the tube each second if the rise in the
water temperature is not to exceed 50°C is found from P = ( ∆m ∆t ) c ( ∆T ) as
1.0 × 10 5 J s
∆m
√
=
=
= 0.48 kg s
∆t c ( ∆T ) ( 4 186 J kg ⋅ °C ) ( 50°C )
17.64
The volume of the material is
V=
50.0 g  1 m 3
mass
=

density 7.86 g cm 3  10 6 cm 3

3
−6
 = 6.36 × 10 m

Since V = A ⋅ L , the cross-sectional area of the wire is A = V L
(a) From R =
L=
ρL
A
R ⋅V
ρ
=
=
ρL
V L
ρ L2
=
V
, the length of the wire is given by
( 1.5 Ω ) ( 6.36 × 10 −6 m 3 )
11 × 10 -8 Ω ⋅ m
= 9.3 m
(b) The cross-sectional area of the wire is A =
d=
17.65
4V
=
πL
4 ( 6.36 × 10 −6 m 3 )
π ( 9.3 m )
π d2
4
=
V
. Thus, the diameter is
L
= 9.3 × 10 −4 m = 0.93 mm
(a) The cross-sectional area of the copper in the hollow tube is
A = ( circumference ) ⋅ ( thickness ) = ( 0.080 m ) ( 2.0 × 10 −3 m ) = 1.6 × 10 −4 m 2
Thus, the resistance of this tube is
R=
ρL
A
=
(1.7 × 10
−8
Ω ⋅ m ) ( 0.24 m )
1.6 × 10 −4 m 2
= 2.6 × 10 −5 Ω
96
CHAPTER 17
(b) The mass may be written as m = ( density ) ⋅ Volume = ( density ) ⋅ A ⋅ L
From R = ρ L A , the cross-sectional area is A = ρ L R , so the expression for the mass
becomes
m = ( density ) ⋅
17.66
ρ L2
R
= ( 8 920 kg m
3
)
(1.7 × 10
⋅
(a) At temperature T, the resistance is R =
ρL
A
−8
Ω ⋅ m ) ( 1 500 m )
4.5 Ω
2
= 76 kg
, where ρ = ρ 0 1 + α ( T − T0 )  ,
L = L0 1 + α ′ ( T − T0 )  , and A = A0 1 + α ′ ( T − T0 )  ≈ A0 1 + 2α ′ ( T − T0 ) 
2
Thus,
ρ L
R= 0 0
 A0
(b) R0 =
ρ0 L0
A0
R0 1 + α ( T − T0 )  ⋅ 1 + α ′ ( T − T0 ) 
 1 + α ( T − T0 )  ⋅ 1 + α ′ ( T − T0 ) 
=

1 + 2α ′ ( T − T0 ) 
1 + 2α ′ ( T − T0 ) 

(1.70 × 10 Ω ⋅ m ) ( 2.00 m ) = 1.082 Ω
π ( 0.100 × 10 )
−8
=
-3 2
Then R = R0 1 + α ( T − T0 )  gives
R = ( 1.082 Ω ) 1 + ( 3.90 × 10 −3 °C ) ( 80.0°C )  = 1.420 Ω
The more complex formula gives
R=
( 1.420 Ω ) ⋅ 1 + ( 17 × 10 −6 °C ) ( 80.0°C )
1 + 2 ( 17 × 10 −6 °C ) ( 80.0°C ) 


= 1.418 Ω
Note: Some rules for handing significant figures have been deliberately violated in
this solution in order to illustrate the very small difference in the results obtained
with these two expressions.
Current and Resistance
17.67
97
Note that all potential differences in this solution have a value of ∆V = 120 V .
First, we shall do a symbolic solution for many parts of the problem, and then enter the
specified numeric values for the cases of interest.
From the marked specifications on the cleaner, its internal resistance (assumed constant)
is
Ri =
( ∆V )
2
where P1 = 535 W
P1
Equation (1)
If each of the two conductors in the extension cord has resistance Rc , the total resistance
in the path of the current (outside of the power source) is
Rt = Ri + 2Rc
Equation (2)
so the current which will exist is I = ∆V Rt and the power that is delivered to the cleaner
is
2
Pdelivered
 ∆V 
 ∆V 
= I Ri = 
 Ri = 

 Rt 
 Rt 
2
2
( ∆V )
P1
2
=
( ∆V )
4
Equation (3)
Rt2 P1
The resistance of a copper conductor of length l and diameter d is
4ρ l
l
l
Rc = ρ Cu = ρ Cu
= Cu2
2
A
(π d 4 ) π d
Thus, if Rc , max is the maximum allowed value of Rc , the minimum acceptable diameter
of the conductor is
dmin =
4 ρ Cu l
π Rc , max
Equation (4)
(a) If Rc = 0.900 Ω , then from Equations (2) and (1),
Rt = Ri + 2 ( 0.900 Ω ) =
( ∆V )
2
P1
+ 1.80 Ω =
( 120 V )
2
535 W
+ 1.80 Ω
and, from Equation (3), the power delivered to the cleaner is
Pdelivered =
( 120 V )
4
2
 ( 120 V )2

+ 1.80 Ω  ( 535 W )

 535 W

= 470 W
98
CHAPTER 17
(b) If the minimum acceptable power delivered to the cleaner is Pmin , then Equations
(2) and (3) give the maximum allowable total resistance as
Rt , max = Ri + 2Rc , max =
( ∆V )
4
Pmin P1
=
( ∆V )
2
Pmin P1
so
Rc , max =
2
2
2
 1  ( ∆V ) 2
1  ( ∆V )
( ∆V )  ( ∆V )  1
1
− Ri  = 
−
− 

=

2  Pmin P1
2  Pmin P1 P1 
P1 
 2  Pmin P1
When Pmin = 525 W , then Rc , max =
and, from Equation (4), dmin =
2
( 120 V ) 
2

1
( 525 W )( 535 W )
4 ( 1.7 × 10 −8 Ω ⋅ m ) ( 15.0 m )
π ( 0.128 Ω )
When Pmin = 532 W , then
Rc , max =
and,
2
( 120 V ) 
2
dmin =

1
( 532 W )( 535 W )
−
1 
 = 0.037 9 Ω
535 W 

4 ( 1.7 × 10 −8 Ω ⋅ m ) ( 15.0 m )
π ( 0.037 9 Ω )
= 2.93 mm
−
1 
 = 0.128 Ω
535 W 

= 1.60 mm