92
17.3
Chapter 17
The period of the electron in its orbit is T = 2π r v, and the current represented by the orbiting
electron is
I=
∆Q e v e
= =
∆t T 2π r
( 2 .19 × 10 m s) (1.60 × 10
2 π ( 5.29 × 10 m )
6
=
17.4
−19
−11
C
) = 1.05 × 10
−3
C s = 1.05 mA
If N is the number of protons, each with charge e, that hit the target in time ∆t, the average
current in the beam is I = ∆Q ∆t = Ne ∆t , giving
)
−6
I ( ∆t ) (125 × 10 C s ( 23.0 s )
N=
=
= 1.80 × 1016 protons
e
1.60 × 10 −19 C proton
17.5
17.6
(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.
The mass of a single gold atom is
matom =
M
197 g mol
=
= 3.27 × 10 −22 g = 3.27 × 10 −25 kg
N A 6.02 × 10 23 atoms mol
The number of atoms deposited, and hence the number of ions moving to the negative
electrode, is
3.25 × 10 −3 kg
m
=
= 9.93 × 10 21
matom 3.27 × 10 −25 kg
n=
Thus, the current in the cell is
I=
17.7
)
)
21
−19
∆Q ne ( 9.93 × 10 (1.60 × 10 C
=
=
= 0.159 A = 159 mA
∆t
∆t
( 2.78 h ) ( 3 600 s 1 h )
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 =
56157_17_ch17_p086-109.indd 92
L
200 × 10 3 m 
1 yr

=

 = 27 yr
−4
vd 2 .34 × 10 m s 3.156 × 10 7 s 
3/18/08 6:02:30 PM
Current and Resistance
17.8
93
Assuming that, on average, each aluminum atom contributes three electrons, the density of
charge carriers is three times the number of atoms per cubic meter. This is

density
3ρ
3N A ρ

,
n = 3
=
=
 mass per atom  M N A
M
or
n=
)
)
)
3 ( 6.02 × 10 23 mol ( 2 .7 g cm 3 (10 6 cm 3 1 m 3 
= 1.8 × 10 29 m 3
26.98 g mol
The drift speed of the electrons in the wire is then
vd =
17.9
(a)
I
5.0 C s
=
= 4.3 × 10 − 5 m s
29
3
n e A (1.8 × 10 m (1.60 × 10 −19 C ( 4.0 × 10 −6 m 2
)
)
)
Using the periodic table on the inside back cover of the textbook, we find
)
M Fe = 55.85 g mol = ( 55.85 g mol ) (1 kg 10 3 g = 55.85 × 10 −3 kg mol
(b)
From Table 9.3, the density of iron is ρFe = 7.86 × 10 3 kg m 3, so the molar density is
( molar density )Fe =
(c)
ρFe
7.86 × 10 3 kg m 3
=
= 1.41 × 10 5 mol m 3
M Fe 55.85 × 10 −3 kg mol
The density of iron atoms is
density of atoms = N A ( molar density )
atoms  
atoms

5 mol 
=  6.02 × 10 23
= 8.49 × 10 28
  1.41 × 10
3 


mol
m3
m
(d)
With two conduction electrons per iron atom, the density of charge carriers is
n = ( charge carriers atom ) ( density of atoms )
 electrons  
28 atoms 
29
3
= 2
  8.49 × 10
 = 1.70 × 10 electrons m

atom  
m3 
(e)
With a current of I = 30.0 A and cross-sectional area A = 5.00 × 10 −6 m 2 , the drift speed
of the conduction electrons in this wire is
vd =
17.10
56157_17_ch17_p086-109.indd 93
I
30.0 C s
=
= 2.21 × 10 −4 m s
29
−3
nqA (1.70 × 10 m (1.60 × 10 −19 C ( 5.00 × 10 −6 m 3
From Ohm’s law,
)
R=
)
)
∆V 1.20 × 10 2 V
=
= 8.89 Ω
13.5 A
I
3/18/08 6:02:31 PM
94
17.11
Chapter 17
)
( ∆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=
m
1.00 × 10 −3 kg
=
= 1.12 × 10 −7 m 3
density 8.92 × 10 3 kg m 3
Since V = A ⋅ L , this gives A ⋅ L = 1.12 × 10 −7 m 3.
(a)
From R =
[1]
ρL
, we find that
A
 1.70 × 10 −8 Ω ⋅ m 
 ρ
−8
A= L=
 L = ( 3.40 × 10 m L
 R
0.500 Ω

)
Inserting this expression for A into Equation [1] gives
( 3.40 × 10
(b)
−8
)
m L2 = 1.12 × 10 −7 m 3 , which yields
From Equation [1], A =
d=
L = 1.82 m
π d 2 1.12 × 10 −7 m 3
=
, or
4
L
4 (1.12 × 10 −7 m 3
πL
)=
4 (1.12 × 10 −7 m 3
π (1.82 m )
)
= 2 .80 × 10 −4 m = 0.280 mm
17.13
(a)
From Ohm’s law,
(b)
Using R = ρ
R=
∆V 1.20 × 10 2 V
=
= 13.0 Ω
9.25 A
I
L
and data from Table 17.1, the required length is found to be
A
)
2
(13.0 Ω ) π ( 0.791 × 10 −3 m
RA R (π r
L=
=
=
150 × 10 −8 Ω ⋅ m
ρ
ρ
17.14
17.15
R=
)
2
= 17.0 m
4 (1.7 × 10 −8 Ω ⋅ m ) (15 m )
ρL
ρL
=
=
= 0.31 Ω
2
A πd2 4
π (1.024 × 10 −3 m )
∆V
12 V
=
= 30 Ω
I
0.40 A
(a)
R=
(b)
From R =
ρL
,
A
)
2
−2


R ⋅ A ( 30 Ω ) π ( 0.40 × 10 m 
ρ=
=
= 4.7 × 10 − 4 Ω ⋅ m
L
3.2 m
56157_17_ch17_p086-109.indd 94
3/18/08 6:02:31 PM
Current and Resistance
17.16
Using R =
LCu
LAL
ρL
and data from Table 17.1, we have ρCu 2 = ρAl 2 , which reduces to
π rCu
π rAl
A
2
rAL
ρ
= AL
2
rCu ρCu
17.17
95
The resistance is R =
ρAL
2.82 × 10 −8 Ω ⋅ m
=
= 1.29
ρCu
1.70 × 10 −8 Ω ⋅ m
rAL
=
rCu
and yields
∆V 9.11 V
=
= 0.253 Ω, so the resistivity of the metal is
I
36.0 A
)
2
( 0.253 Ω ) π ( 2 .00 × 10 −3 m
R ⋅ A R ⋅ (π d 4
ρ=
=
=
4 ( 50.0 m )
L
L
)
2
= 1.59 × 10 −8 Ω ⋅ m
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
17.19
and
40 cm
1
1
 L
=
=
  =
A 3 (10 cm ) ( 20 cm ) 5.0 cm 0.050 m
(a)
I max =
( 6.0 V) ( 0.80 m )
∆V
∆V
= 2 .8 × 10 8 A
=
=
Rmin ρ ( L A )min 1.7 × 10 −8 Ω ⋅ m
(b)
I min =
( 6.0 V) ( 0.050 m )
∆V
∆V
=
=
= 1.8 × 10 7 A
1.7 × 10 −8 Ω ⋅ m
Rmax ρ ( L A )max
)
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
( )
)
2
r 
 r 
2
L f =  0  L0 =  0  L0 = ( 4.0 ) L0 = 16 L0
 0.25 r0 
 rf 
The new resistance is then
Rf = ρ
17.20
56157_17_ch17_p086-109.indd 95
Lf
Af
=ρ
Lf
πr
2
f
=ρ
16 L0
π ( r0 4 )
2
L 
2 
= 16 ( 4 )  ρ 02  = 256 R0 = 256 (1.00 Ω ) = 256 Ω
 π r0 
)
)
∆V = IR = ( 500 × 10 −3 A (1.0 × 10 6 Ω = 5 × 10 5 V
(a)
From Ohm’s law,
(b)
Rubber-soled shoes and rubber gloves can increase the resistance to current and help reduce
the likelihood of a serious shock.
3/18/08 6:02:32 PM
96
17.21
Chapter 17
If a conductor of length L has a uniform electric field E maintained within it, the potential difference between the ends of the conductor is ∆V = EL. But, from Ohm’s law, the relation between
the potential difference across a conductor and the current through it is ∆V = IR, where R = ρ L A.
Combining these relations, we obtain
∆V = EL = IR = I ( ρ L A )
17.22
E = ρ ( I A) = ρJ
or
Using R = R0 1 + α (T − T0 )  with R0 = 6.00 Ω at T0 = 20.0°C and α silver = 3.8 × 10 −3 ( °C )
(from Table 17.1 in the textbook), the resistance at T = 34.0°C is
−1
−1
R = ( 6.00 Ω ) 1 + 3.8 × 10 −3 ( °C ) ( 34.0°C − 20.0°C )  = 6.32 Ω
17.23
From Ohm’s law, ∆V = I i Ri = I f R f , so the current in Antarctica is
 R 1 + α (T − T )  
R 
0
i
0 

I f = Ii  i  = Ii  
 R0 1 + α T f − T0  
 Rf 


(
)
 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 ) 


17.24
(a)
Given: Aluminum wire with α = 3.90 × 10 −3 ( °C ) (see Table 17.1 in textbook), and
R0 = 30.0 Ω at T0 = 20.0°C. If R = 46.2 Ω at temperature T, solving
R = R0 1 + α (T − T0 )  gives the final temperature as
−1
T = T0 +
(b)
17.25
R0 ) − 1
( 46.2 Ω 30.0 Ω ) − 1 = 158°C
= 20.0°C +
−1
α
3.90 × 10 −3 ( °C )
The expansion of the cross-sectional area contributes slightly more than the expansion of
the length of the wire, so the answer would be slightly reduced.
For tungsten, the temperature coefficient of resistivity is α = 4.5 × 10 −3 ( °C ) . Thus, if
R0 = 15 Ω at T0 = 20°C, and R = 160 Ω at the operating temperature of the filament, solving
R = R0 1 + α (T − T0 )  for the operating temperature gives
−1
T = T0 +
17.26
(R
(R
R0 ) − 1
(160 Ω 15 Ω ) − 1 = 2.2 × 10 3 °C
= 20°C +
−1
α
4.5 × 10 −3 ( °C )
For aluminum, the resistivity at room temperature is ρ0 = 2.82 × 10 −8 Ω ⋅ m and the temperature
−1
coefficient of resistivity is α Al = 3.9 × 10 −3 ( °C ) . Thus, if at some temperature, the aluminum
has a resistivity of
ρ = 3 ( ρ0 )Cu = 3 (1.7 × 10 −8 Ω ⋅ m ) = 5.1 × 10 −8 Ω ⋅ m
solving ρ = ρ0 1 + α Al (T − T0 )  for that temperature gives
T = T0 +
56157_17_ch17_p086-109.indd 96
(ρ
ρ0 ) − 1
= 20°C +
α Al
 5.1 × 10 −8 Ω ⋅ m 
 2.82 × 10 −8 Ω ⋅ m  − 1
3.9 × 10 −3 ( °C )
−1
= 2.3 × 10 2 °C
3/18/08 6:02:33 PM
Current and Resistance
17.27
At 80°C,
I=
or
17.28
∆V
∆V
5.0 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
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=
17.29
97
(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 )
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 =
(b)
)
∆V 9.0 V
=
= 3.0 A
3.0 Ω
R0
At 30.0°C,
R = R0 1 + α (T − T0 ) 
(
= ( 3.0 Ω ) 1 + 3.9 × 10 −3 ( °C )

Thus, the current is I =
17.30
−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 )2 (120 V)2
=
P
1 050 W
= 13.7 Ω
ρL
From R = R0 1 + α (T − T0 )  = 0 1 + α (T − T0 )  , the cross-sectional area is
A
A=
=
ρ0 L
1 + α (T − T0 ) 
R 
(150 × 10
−8
)
Ω ⋅ m ( 4.00 m )
13.7 Ω
(
)
1 + 0.40 × 10 −3 ( °C )−1 ( 320°C − 20.0°C ) 


A = 4.90 × 10 −7 m 2
56157_17_ch17_p086-109.indd 97
3/18/08 6:02:37 PM
98
17.31
Chapter 17
(a)
From R = ρ L A, the initial resistance of the mercury is
−7
ρ Li ( 9.4 × 10 Ω ⋅ m ) (1.000 0 m )
Ri =
=
= 1.2 Ω
2
Ai
π (1.00 × 10 −3 m ) 4
(b)
Since the volume of mercury is constant, V = A f ⋅ L f = Ai ⋅ Li gives the final crosssectional area as A f = Ai ⋅ Li L f . Thus, the final resistance is given by
ρ Lf
ρ L2f
Rf =
=
. The fractional change in the resistance is then
Af
Ai ⋅ Li
(
R f − Ri
∆=
Ri
=
Rf
Ri
)
ρ L2f ( Ai ⋅ Li )
 Lf 
−1=   −1
ρ Li Ai
 Li 
2
−1=
2
 100.04 
− 1 = 8.0 × 10 −4 or a 0.080% increase
∆=
 100.00 
17.32
The resistance at 20.0°C is
R0 =
200.0 Ω
R
=
= 217 Ω
−
3
1 + α (T − T0 ) 1 +  3.92 × 10 ( °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 +
17.33
(a)
The power consumed by the device is P = I ( ∆V ), so the current must be
I=
17.34
R − R0
253.8 Ω − 217 Ω
= 63.3°C
= 20.0°C +
α R0
 3.92 × 10 −3 ( °C )−1  ( 217 Ω )


P
∆V
=
1.00 × 10 3 W
= 8.33 A
1.20 × 10 2 V
R=
∆V 1.20 × 10 2 V
=
= 14.4 Ω
I
8.33 A
(b)
From Ohm’s law, the resistance is
(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
(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
 100 3 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
56157_17_ch17_p086-109.indd 98
3/18/08 6:02:38 PM
Current and Resistance
17.35
The power required is
17.36
(a)
P
= I ( ∆V ) = ( 0.350 A ) ( 6.0 V) = 2.1 W
The power loss in the line is
Ploss = I 2 R = (1 000 A )2 ( 0.31
(b)
99
Ω km ) (160 km )  = 5.0 × 10 7 W = 50 MW
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 =
17.37
Ploss
Pinput
=
50 MW
= 0.071 or 7.1%
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
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
Js
Therefore, the time required is
t=
17.38
E
Pinput
=
1.61 × 10 5 J
 1 min 
= 672 s 
= 11.2 min
 60 s 
240 J s
(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=
17.39
E
P
=
3.2 × 10 5 J
 1 min 
= 18 min
= 1.1 × 10 3 s 
 60 s 
300 J 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
600 s
∆t
The power input to the heater may be expressed as
R=
56157_17_ch17_p086-109.indd 99
( ∆V )2
P
=
(120 V)2
419 W
P = ( ∆V )2
R, so the needed resistance is
= 34.4 Ω
3/18/08 6:02:38 PM
100
17.40
Chapter 17
(a)
At the operating temperature,
P = ( ∆V ) I = (120 V) (1.53 A ) =
(b)
184 W
R − R0
From R = R0 1 + α (T − T0 )  , the temperature T is given by T = T0 +
.
α R0
The resistances are given by Ohm’s law as
R=
( ∆V )
I
=
( ∆V )0 120 V
120 V
=
, and R0 =
1.53 A
I0
1.80 A
Therefore, the operating temperature is
T = 20.0°C +
17.41
(120 1.53) − (120 1.80 ) =
( 0.400 × 10−3 (°C)−1 ) (120 1.80 )
461°C
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
L
L
I
( 300 A )2
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 =
r=
17.42
(a)
ρ
, and the required radius is
R L
ρ
1.7 × 10 −8 Ω ⋅ m
=
= 0.016 m = 1.6 cm
π (R L)
π ( 2 .22 × 10 −5 Ω m )
The rating of the 12-V battery is I ⋅ ∆t = 55 A ⋅ h. Thus, the stored energy is
Energy stored = P ⋅ ∆t = ( ∆V ) I ⋅ ∆t = (12 V) ( 55 A ⋅ h ) = 660 W ⋅ h = 0.66 kWh
(b)
cost = ( 0.66 kWh ) ( $0.12 kWh ) = $0.079 = 7.9 cents
17.43
P = ( ∆V ) I = ( 75 × 10 −3 V) ( 0.20 × 10 −3 A ) = 1.5 × 10 −5
17.44
(a)
W = 15 × 10 −6 W = 15 µ W
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
56157_17_ch17_p086-109.indd 100
3/18/08 6:02:39 PM
Current and Resistance
17.45
101
The energy saved is
∆E =
(P
high
)
− 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
C − 20°C )
∆Q mc ( ∆T ) ( 0.250 kg ) ( 4 186 J kg ⋅ °C ) (100°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 )2
P
=
(120 V)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 = ρ0 L A = ρ0 L (π d 2 4 = 4 ρ0 L π d 2, where L is the length of the wire and d is
its diameter. This gives
)
)
 4 (150 × 10 −8 Ω ⋅ m 
 4ρ 
−8
d2 =  0  L = 
 L = ( 4.8 × 10 m L
π
π
R
40
Ω
(
)
 0


)
)
Thus, any combination of length and diameter satissfying the relation d 2 = ( 4.8 × 10 −8 m L
will be suitable. A typical combination might be
L = 3.0 m
and
d=
( 4.8 × 10
−8
)
m ( 3.0 m ) = 3.8 × 10 − 4 m = 0.38 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
π ( 3.8 × 10 − 4 m
 πd2 
=
V = AL = 
L
4
 4 
17.47
)
2
( 3.0 m ) = ( 3.4 × 10 −7
 10 6 cm 3 
= 0.34 cm 3
m3 
 1 m 3 
)
The energy that must be added to the water is

1 kWh 
J 
E = mc ( ∆T ) = ( 200 kg )  4 186
( 80°C − 15°C ) 
 = 15 kWh


3.60 × 10 6 J 
kg ⋅ °C 
and the cost is
56157_17_ch17_p086-109.indd 101
cost = E ⋅ rate = (15 kWh ) ( $0.080 kWh ) = $1.2
3/18/08 6:02:40 PM
102
17.48
Chapter 17
(a)
For tungsten, Table 17.1 from the textbook gives the resistivity at T0 = 20.0°C = 293 K
as ρ0 = 5.6 × 10 −8 Ω ⋅ m and the temperature coefficient of resistivity as
−1
α = 4.5 × 10 −3 ( °C ) = 4.5 × 10 −3 K −1. Thus, for a tungsten wire having a radius of
1.00 mm and a length of 15.0 cm, the resistance at T0 = 293 K is
R0 = ρ0
(b)
−2
)
)
2
= 2.7 × 10 −3 Ω
From Stefan’s law, the radiated power is P = σ AeT 4 , where A is the area of the radiating
surface. Note that since we are computing the radiated power, not the net energy gained
or lost as a result of radiation, the ambient temperature is not needed here. In the case of a
wire, this is the cylindrical surface area A = 2π rL . The temperature of the wire when it is
radiating a power of P = 75.0 W must be
14
 P 
T =
 σ Ae 
or
(c)
)
)
(15.0 × 10 m
L
L
= ρ0
= ( 5.6 × 10 −8 Ω ⋅ m
2
A
(π r
π (1.00 × 10 −3 m
14


75.0 W
=

−8
2
4
−3
 ( 5.669 × 10 W m ⋅ K 2π (1.00 × 10 m ( 0.150 m ) ( 0.320 ) 
)
)
T = 1.45 × 10 3 K
Assuming a linear temperature variation of resistance, the resistance of the wire at this
temperature is
)
)
)
R = R0 1 + α (T − T0 )  = ( 2.7 × 10 −3 Ω 1 + ( 4.5 × 10 −3 K −1 (1.45 × 10 3 K − 293 K 
giving
(d)
R = 1.7 × 10 −2 Ω
The voltage drop across the wire when it is radiating 75.0 W and has the resistance found in
2
part (c) above is given by P = ( ∆V ) R as
∆V = R ⋅ P =
(e)
(1.7 × 10
−2
)
Ω ( 75.0 W ) = 1.1 V
Tungsten bulbs release very little of the en
nergy consumed in the form of visible lightt ,
making them inefficient sources of light.
17.49
The battery is rated to deliver the equivalent of 60.0 amperes of current (i.e., 60.0 C/s) for 1 hour.
This is
Q = I ⋅ ∆t = ( 60.0 A ) (1 h ) = ( 60.0 C s ) ( 3 600 s ) = 2.16 × 10 5 C
56157_17_ch17_p086-109.indd 102
3/18/08 6:02:41 PM
Current and Resistance
17.50
103
The energy available in the battery is
Energy stored = P ⋅ t = ( ∆V ) I ⋅ t = ( ∆V ) ( I ⋅ t ) = (12 V) ( 900 A ⋅ h ) = 1.1 × 10 3 W ⋅ h
The two headlights together consume a total power of
required to completely discharge the battery is
∆t =
17.51
P
=
( ∆V2 )2
( ∆V1 )2
1.1 × 10 3 W ⋅ h
= 15 h
72 W
R
R
or
2
2


 6.0 V 
= 6.7 W
= (15 W ) 
 9.0 V 
 ∆V1 
P2 = P1  ∆V2 
The temperature variation of resistance is described by R = R0 1 + α (T − T0 )  , where R0 is the
−1
resistance at T0 = 20°C. Thus, if an aluminum wire α = 3.9 × 10 −3 ( °C )  has R = 2 R0, we
have 2 = 1 + α (T − 20°C ), and its temperature must be
T = 20°C +
17.53
=
Assuming a constant resistance, the power consumed by the device is proportional to the square
2
of the applied voltage, P = ( ∆V ) R . Thus,
P2
P1
17.52
Energy stored
P = 2 ( 36 W) = 72 W, so the time
1
1
2
= 20°C +
−1 = 2.8 × 10 °C
−3
α
3.9 × 10 ( °C )
From P = ( ∆V ) R , the total resistance needed is
2
R=
( ∆V )2 ( 20 V)2
=
P
48 W
= 8.3 Ω
Thus, from R = ρ L A, the length of wire required is
L=
17.54
)
−6
2
R ⋅ A ( 8.3 Ω ) ( 4.0 × 10 m
=
= 1.1 × 10 3 m = 1.1 km
3.0 × 10 −8 Ω ⋅ m
ρ
The resistance of the 4.0 cm length of wire between the feet is
R=
−8
ρ L (1.7 × 10 Ω ⋅ m ) ( 0.040 m )
=
= 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 µV
56157_17_ch17_p086-109.indd 103
3/18/08 6:02:41 PM
104
17.55
Chapter 17
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
10.4
1.41 × 10 −6
1.028
21.1
1.50 × 10 −6
1.543
31.8
1.50 × 10 −6
The average value found for the resistivity is
ρav =
Σρi
= 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% .
17.56
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
Solving this equation gives
R0 n = 5.6 kΩ
Then, R0 c = 10.0 kΩ − 5.6 kΩ = 4.4 kΩ
56157_17_ch17_p086-109.indd 104
( nichrome wire )
( carbon wire )
3/18/08 6:02:42 PM
Current and Resistance
17.57
(a)
105
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
I=
and the required current would be
P
∆V
=
2 900 W
= 26.4 A > 20 A
110 V
No , your present circuit breaker cannot handle the upgrade.
17.58
(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 P17.58. This area consists of two
rectangles and two triangles. Thus,
∆Q = Arectangle 1 + Arectangle 2 + Atriangle 1 + Atrriangle 2
= ( 5.0 s − 0 ) ( 2 .0 A − 0 ) + ( 4.0 s − 3.0 s ) ( 6.0 A − 2 .0 A )
+
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=
17.59
∆Q 18 C
=
= 3.6 A
∆t 5.0 s
P
8.00 × 10 3 W
= 667 A .
12 .0 V
(a)
From P = ( ∆V ) I, the current is I =
(b)
The time before the stored energy is depleted is
t=
Estorage
P
=
∆V
=
2 .00 × 10 7 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
56157_17_ch17_p086-109.indd 105
3/18/08 6:02:43 PM
106
17.60
Chapter 17
The volume of aluminum available is
V=
(a)
mass
115 × 10 −3 kg
=
= 4.26 × 10 −5 m 3
density 2 .70 × 10 3 kg m 3
For a cylinder whose height equals the diameter, the volume is
 πd2 
πd3
V =
d=

4
 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=
(b)
−8
4 ρ 4 ( 2 .82 × 10 Ω ⋅ m )
ρL
ρd
=
=
=
= 9.49 × 10 −7 Ω
A (π d 2 4 ) π d
π ( 0.037 85 m )
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=
17.61
ρ L ρ L ρ 2 .82 × 10 −8 Ω ⋅ m
= 2 = =
= 8.07 × 10 −7 Ω
A
L
L
0.034 9 m
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
17.62
150 A
I
=
−4
vd e ( π r 2
( 3.17 × 10 m s (1.60 × 10−19 C π ( 5.00 × 10−3 m
)
)
)
)
2
n = 3.77 × 10 28 m 3
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 =
P
R
=
60.0 W
= 3.87 A
4.00 Ω
Thus, the system is not adequately protected by a 4.00 A fuse.
56157_17_ch17_p086-109.indd 106
3/18/08 6:02:43 PM
Current and Resistance
17.63
)
2
2
− rinner
The cross-sectional area of the conducting material is A = π ( router
. Thus,
R=
17.64
107
( 3.5 × 105 Ω ⋅ m ) ( 4.0 × 10−2 m ) = 3.7 × 10 7 Ω = 37 MΩ
ρL
=
A π (1.2 × 10 −2 m )2 − ( 0.50 × 10 −2 m )2 


The volume of the material is
V=
mass
50.0 g  1 m 3 
= 6.36 × 10 − 6 m 3
=
density 7.86 g cm 3  10 6 cm 3 
Since V = A ⋅ L , the cross-sectional area of the wire is A = V L.
(a)
From R =
L=
(b)
ρ L ρ L ρ L2
, the length of the wire is given by
=
=
A V L
V
R ⋅V
=
ρ
m3
11 × 10 −8 Ω ⋅ m
The cross-sectional area of the wire is A =
4V
d=
=
πL
17.65
(1.5 Ω ) ( 6.36 × 10 −6
4 ( 6.36 × 10 −6 m 3
π ( 9.3 m )
)=
9.3 m
πd2 V
= . Thus, the diameter is
4
L
) = 9.3 × 10
−4
m = 0.93 mm
The power the beam delivers to the target is
P = ( ∆V ) I = ( 4.0 × 10 6
)
)
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
∆m
P =
1.0 × 10 5 J s
=
= 0.48 kg s
∆t c ( ∆T ) ( 4 186 J kg ⋅ °C ) ( 50°C )
17.66
(a)
ρL
, where ρ = ρ0 1 + α (T − T0 )  ,
A
2
L = L0 1 + α ′ (T − T0 )  , and A = A0 1 + α ′ (T − T0 )  ≈ A0 1 + 2 α ′ (T − T0 ) 
At temperature T, the resistance is R =
Thus,
R0 1 + α (T − T0 )  ⋅ 1 + α ′ (T − T0 ) 
 ρ L  1 + α (T − T0 )  ⋅ 1 + α ′ (T − T0 ) 
=
R= 0 0 
A
1 + 2 α ′ (T − T0 ) 
1 + 2 α ′ (T − T0 ) 
 0 
continued on next page
56157_17_ch17_p086-109.indd 107
3/18/08 6:02:44 PM
108
Chapter 17
(b)
R0 =
−8
ρ0 L0 (1.70 × 10 Ω ⋅ m ) ( 2 .00 m )
=
= 1.082 Ω
2
A0
π ( 0.100 × 10 −3 )
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
1 + 2 (17 × 10
−6
)
)
°C ( 80.0°C ) 
°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.
17.67
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
P1
where P1 = 535 W
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 + 2 Rc
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 )2 ( ∆V )4
= 2
Rt P1
P1
Equation [3]
The resistance of a copper conductor of length L and diameter d is
Rc = ρCu
4ρ L
L
L
= ρCu
= Cu2
2
πd
A
π
4
d
(
)
Thus, if Rc, max is the maximum allowed value of Rc, the minimum acceptable diameter of the
conductor is
dmin =
4 ρCu L
π Rc, max
Equation [4]
continued on next page
56157_17_ch17_p086-109.indd 108
3/18/08 6:02:45 PM
Current and Resistance
(a)
109
If Rc = 0.900 Ω, then from Equations [2] and [1],
Rt = Ri + 2 ( 0.900 Ω ) =
( ∆V )2
P1
(120 V)2
+ 1.80 Ω =
535 W
+ 1.80 Ω
and, from Equation [3], the power delivered to the cleaner is
Pdelivered =
(b)
(120 V)4
2

 (120 V)2
+ 1.80 Ω  ( 535 W )


 535 W
= 470 W
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 + 2 Rc, max =
( ∆V )4
PminP1
( ∆V )2
PminP1
=
so
Rc, max =
2
 1  ( ∆V )2
( ∆V )2  ( ∆V )2 
1  ( ∆V )
−
−
R
=
=



i
P1 
2 
2  PminP1
 2  PminnP1
When Pmin = 525 W, then Rc, max =
and, from Equation [4], dmin =
When
and
56157_17_ch17_p086-109.indd 109
Pmin = 532 W, then
dmin =
(120 V)2 


2
1
( 525 W) ( 535 W)
)
4 (1.7 × 10 −8 Ω ⋅ m (15.0 m )
Rc, max =
π ( 0.128 Ω )
(120 V)2 
2
)


4 (1.7 × 10 −8 Ω ⋅ m (15.0 m )
π ( 0.037 9 Ω )
−
1
PminP1
−
1
P1 
1 
 = 0.128 Ω
535 W 
= 1.60 mm
1
( 532 W) ( 535 W)
−
1 
 = 0.037 9 Ω
535 W 
= 2.93 mm
3/18/08 6:02:46 PM