Chapter 17
Current and Resistance
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× 1020 electrons
1.60 × 10−19 C
The negatively charged electrons move in the direction opposite
to the conventional current flow.
−3
17.3
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 
69
70 CHAPTER 17
17.8
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
density
ρ
N ρ
=
= A ,
mass per atom M NA
M
( 6.02 × 10
n=
23
mol ) ( 2 .7 g cm3 ) ( 106 cm3 1 m3 ) 
26.98 g mol
= 6.0 × 1028 m3
The drift speed of the electrons in the wire is then
vd =
17.9
I
5.0 C s
=
= 1.3 × 10 − 4 m s
28
3
n e A ( 6.0 × 10 m ) ( 1.60× 10−19 C) ( 4.0× 10−6 m2 )
(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.
17.11
( ∆V ) max = I max R = ( 80 × 10−6 A ) R
Thus, if R = 4.0 × 105 Ω , ( ∆V ) max = 32 V
and if R = 2 000 Ω , ( ∆V ) max = 0.16 V
17.13
From R =
d=
ρL
π d2 ρ L
, we obtain A =
, or
=
A
4
R
4ρ L
=
πR
(
)(
)
4 5.6 × 10 −8 Ω ⋅ m 2 .0 × 10−2 m
π ( 0.050 Ω )
= 1.7 × 10 −4 m = 0.17 mm
Current and Resistance
17.16
71
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
E=

( ∆t ) = ( 0.400 kW)  10

min  1 hr 


 ( 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 

17.19
or
~$1
The volume of material, V = AL0 = ( π r02 ) L0 , in the wire is constant. Thus, as the wire is
(
)
2
2
stretched to decrease its radius, the length increases such that π rf L f = ( π r0 ) L0 giving
2
2
r 
 r 
2
L f =  0  L0 =  0  L0 = ( 4.0 ) L0 = 16 L0
 rf 
r
0.25

0 
 
The new resistance is then
Rf = ρ
17.20
Lf
Af
Lf
πr
2
f
=ρ
16 L0
π ( r0 4 )
2
L 
2
= 16 ( 4 )  ρ 0 2  = 256R0 = 256( 1.00 Ω ) = 256 Ω
 π r0 
Solving R = R0 1 + α ( T − T0 )  for the final temperature gives
T = T0 +
17.23
=ρ
R − R0
140 Ω − 19 Ω
= 20 °C +
= 1.4 × 103 ° C
4.5 × 10 -3 ( ° C) −1  ( 19 Ω )
α R0


At 80°C,
I=
or
5.0 V
∆V
∆V
=
=
R
R0 1 + α ( T − T0 )  ( 200 Ω )  1 + ( − 0.5× 10−3 ° C−1 ) ( 80° C− 20° C) 


I = 2.6 × 10−2 A = 26 mA
72 CHAPTER 17
17.31
I=
∆V
and
17.33
=
600 W
= 5.00 A
120 V
R=
∆V 120 V
=
= 24.0 Ω
I
5.00 A
The maximum power that can be dissipated in the circuit is
max
= ( ∆V ) I max = ( 120 V ) ( 15 A ) = 1.8 × 103 W
Thus, one can operate at most 18 bulbs rated at 100 W per bulb.
17.39
The resistance per unit length of the cable is
I2
L 2 .00 W m
= 2 =
= 2 .22 × 10−5 Ω m
2
L
I
( 300 A)
R
=
L
From R = ρ L A , the resistance per unit length is also given by R L = ρ A . Hence, the
ρ
2
cross-sectional area is π r = A =
, and the required radius is
RL
r=
17.45
ρ
1.7 × 10−8 Ω ⋅ m
=
= 0.016 m = 1.6 cm
π ( R L)
π ( 2.22 × 10-5 Ω m )
The energy saved is
∆E =
(
high
−
low
) ⋅ 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
Current and Resistance
17.52
The resistance of the 4.0 cm length of wire between the feet is
−8
ρ L ( 1.7 × 10 Ω ⋅ m ) ( 0.040 m)
R=
=
= 1.79 × 10−6 Ω ,
2
A
π ( 0.011 m)
so the potential difference is
∆V = IR = ( 50 A ) ( 1.79 × 10 −6 Ω ) = 8.9 × 10 −5 V = 89 µ V
17.60
Each speaker has a resistance of R = 4.00 Ω and can handle 60.0 W of power. From
= I 2 R , the maximum safe current is
Imax =
R
=
60.0 W
= 3.87 A
4.00 Ω
Thus, the system is not adequately protected by a 4.00 A fuse.
73