Chapter 17
Current and Resistance
Problem Solutions
17.1
The charge that m oves past the cross section is
is
Q I t
n
e
e
80.0 10
3
C s
t , and the num ber of electrons
Q I
10.0 min 60.0 s min
1.60 10
19
3.00 1020 electrons
C
The negatively charged electrons m ove in the d irection opposite
to the conventional current flow .
17.2
(a) From Exam ple 17.2 in the textbook, the d ensity of charge carriers (electrons) in a
copper w ire is n 8.46 1028 electrons m3 . With A
r 2 and q e , the d rift speed of
electrons in this w ire is
vd
(b)
I
nq A
I
ne
3.70 C s
r
2
28
8.46 10 m
-3
1.60 10
19
C
1.25 10
3
m
2
5.57 10
5
ms
The drift speed is smaller because more electrons are being conducted. To create the
sam e current, therefore, the d rift speed need not be as great.
109
110
17.3
CH APTER 17
The period of the electron in its orbit is T
orbiting electron is
Q
t
I
e
ve
T
2 r
2.19 106 m s 1.60 10
2
17.4
C
1.05 10
m
3
Cs
1.05 mA
If N is the num ber of protons, each w ith charge e, that hit the target in tim e
average cu rrent in the beam is I
Q t Ne t , giving
I
N
17.5
5.29 10
11
19
2 r v , and the current represented by the
125 10
t
e
6
1.60 10
C s 23.0 s
19
t , the
1.80 1016 protons
C proton
(a) The carrier d ensity is d eterm ined by the physical characteristics of the w ire, not the
current in the w ire. H ence, n is unaffected .
(b) The d rift velocity of the electrons is vd
I nqA . Thus, the d rift velocity is doubled
w hen the current is d oubled .
17.6
The m ass of a single gold atom is
M
NA
matom
197 g mol
6.02 1023 atoms mol
3.27 10-22 g 3.27 10-25 kg
The num ber of atom s d eposited , and hen ce the num ber of ions m oving to the negative
electrod e, is
n
m
matom
3.25 10 3 kg
3.27 10-25 kg
9.93 1021
Thus, the current in the cell is
I
Q
t
ne
t
9.93 1021 1.60 10
19
2.78 h 3 600 s 1 h
C
0.159 A
159 mA
Current and Resistance
17.7
I
nqA
The d rift speed of electrons in the line is vd
I
4 1000 A
vd
8.5 10
28
3
m
1.60 10
19
C
0.020 m
d2 4
ne
111
, or
2.3 10-4 m s
2
The tim e to travel the length of the 200-km line is then
L
vd
t
17.8
200 103 m
1 yr
-4
2.34 10 m s 3.156 107 s
27 yr
Assum ing that, on average, each alum inum atom contributes three electrons, the d ensity
of charge carriers is three tim es the num ber of atom s per cubic m eter. This is
n 3
density
mass per atom
3 6.02 1023 mol
or
n
3N A
M
3
M NA
,
2.7 g cm3 106 cm3 1 m3
1.8 1029 m3
26.98 g mol
The d rift speed of the electrons in the w ire is then
vd
17.9
I
ne A
5.0 C s
1.8 10
29
m
3
1.60 10
19
C 4.0 10
6
m
2
4.3 10
5
ms
(a) Using the period ic table on the insid e back cover of the textbook, w e find
M Fe
55.85 g mol
55.85 g mol 1 kg 103 g
(b) From Table 9.3, the d ensity of iron is
molar density
Fe
Fe
M Fe
Fe
55.85 10
3
kg mol
7.86 103 kg m3 , so the m olar d ensity is
7.86 103 kg m 3
55.85 10 3 kg mol
1.41 105 mol m 3
(c) The d ensity of iron atom s is
density of atoms
N A molar density
6.02 1023
atoms
mol
1.41 105
mol
m3
8.49 1028
atoms
m3
112
CH APTER 17
(d ) With tw o cond uction electrons per iron atom , the d ensity of charge carriers is
n
charge carriers atom density of atoms
2
electrons
atom
atoms
m3
8.49 1028
1.70 1029 electrons m3
(e) With a current of I 30.0 A and cross-sectional area A 5.00 10
speed of the cond uction electrons in this w ire is
vd
17.10
17.11
17.12
I
nqA
30.0 C s
1.70 10
From Ohm ’ s law ,
V
I max R
max
29
Thus, if R 4.0 105
,
and if R 2 000
V
,
6
1.60 10
19
C 5.00 10
1.20 102 V
13.5 A
V
I
R
80 10
m
3
6
m3
max
max
8.89
32 V
0.16 V
The volum e of the copper is
V
m
density
1.00 10 3 kg
8.92 103 kg m3
1.12 10
7
m3
A L , this gives A L 1.12 10 7 m3
Since, V
A
[1]
L
, w e find that
A
(a) From R
R
L
1.70 10 8
0.500
m
L
3.40 10
8
m L
Inserting this expression for A into Equation [1] gives
3.40 10
8
m L2 1.12 10
7
m3 , w hich yield s
L
m2 , the d rift
2.21 10
A R
V
6
1.82 m
4
ms
Current and Resistance
d2
4
(b) From Equation [1], A
d
R
(a)
2
d 4
V
I
1.20 102 V
9.25 A
V
I
R
0.791 10 3 m
13.0
rAl2
2
rCu
8
8
13.0
3
m
2
17.0 m
m
m 15 m
1.024 10
12 V
0.40 A
0.31
2
30
L
,
A
R A
L
Using R
m3
0.280 mm
4 1.7 10
(b) From , R
17.16
m
r2
R
L
R
7
, or
1.82 m
150 10
L
A
m3
L
and d ata from Table 17.1, the required length is found to be
A
RA
L
17.15
4
7
4 1.12 10
(a) From Ohm ’ s law ,
(b) Using R
17.14
m3
L
2.80 10
17.13
7
4 1.12 10
1.12 10
L
113
30
0.40 10
2
m
2
4.7 10
3.2 m
L
and d ata from Table 17.1, w e have
A
Al
Cu
and yield s
rAl
rCu
Al
Cu
4
Cu
m
LCu
2
rCu
2.82 10 8
1.70 10 8
Al
LAl
, w hich red uces to
rAl2
m
m
1.29
114
17.17
CH APTER 17
V
I
The resistance is R
d2 4
R
R A
L
9.11 V
36.0 A
, so the resistivity of the m etal is
0.253
0.253
2.00 10
L
3
m
2
1.59 10
4 50.0 m
8
m
Thus, the m etal is seen to be silver .
17.18
With d ifferent orientations of the block, three d ifferent values of the ratio L A are
possible. These are:
L
A
L
A
L
A
and
(a)
I max
(b) I min
17.19
1
10 cm
20 cm 40 cm
1
80 cm
1
,
0.80 m
2
20 cm
10 cm 40 cm
1
20 cm
1
,
0.20 m
3
40 cm
10 cm 20 cm
1
5.0 cm
1
0.050 m
V
6.0 V 0.80 m
V
L A
Rmin
V
1.7 10
min
2.8 108 A
m
6.0 V 0.050 m
V
L A
Rmax
8
1.7 10
max
The volum e of m aterial, V
AL0
8
m
1.8 107 A
r02 L0 , in the w ire is constant. Thus, as the w ire is
stretched to d ecrease its rad ius, the length increases such that
2
Lf
r0
rf
L0
r0
0.25r0
rf2 L f
r02 L0 giving
2
L0
2
4.0 L0 16L0
The new resistance is then
Rf
Lf
Af
Lf
2
f
r
16L0
r0 4
2
16 4
2
L0
r02
256R0
256 1.00
256
Current and Resistance
17.20
(a) From Ohm ’ s law ,
V
IR
500 10
3
A 1.0 106
115
5 105 V
(b) Rubber-soled shoes and rubber gloves can increase the resistance to current and
help red uce the likelihood of a serious shock.
17.21
If a cond uctor of length L has a uniform electric field E m aintained w ithin it, the
potential d ifference betw een the end s of the cond uctor is V EL . But, from Ohm ’ s
law , the relation betw een the potential d ifference across a cond uctor and the current
through it is V IR where R
L A . Com bining these relations, w e obtain
V
17.22
EL IR I
Using R R0 1
or
L A
T T0
w ith R0
6.00
E
I A
at T0
20.0 °C and
(from Table 17.1 in the textbook), the resistance at T
R
17.23
6.00
1 3.8 10
From Ohm ’ s law ,
If
Ii
V
Ri
Rf
Ii Ri
Ii
3
°C
R0 1
Ti T0
R0 1
Tf
T0
3
C
1
3.90 10
1
3.90 10
(a) Given: Alum inum w ire w ith
R0
30.0
R R0 1
T
T0
at T0
R R0
3.8 10
3
1
6.32
3
C
1
58.0 C 20.0 C
1.98 A
1
88.0 C 20.0 C
3.90 10
3
1
°C
(See Table 17.1 in textbook), and
at tem perature T, solving
gives the final tem perature as
1
°C
34.0 C is
34.0°C 20.0 C
20.0 C . If R 46.2
T T0
silver
I f R f , so the current in Antarctica is
1.00 A
17.24
1
J
20.0°C
46.2
30.0
3.90 10
3
1
°C
1
158°C
(b) The expansion of the cross-sectional area contributes slightly m ore than the
expansion of the length of the w ire, so the answ er w ould be slightly red uced .
116
17.25
CH APTER 17
For tungsten, the tem perature coefficient of resistivity is
R0 15
at T0
20°C , and R 160
solving R R0 1
T
17.26
1
20°C
160
15
1
3
1
tem perature coefficient of resistivity is
°C
3
3.9 10
tem perature, the alum inum has a resistivity of
solving
0
T
3 1.7 10
1
Al
0
T0
T T0
1
8
m
Al
5.1 10
8
2.82 10
0
1
°C
8
m and the
. Thus, if at som e
m
20°C
5.1 10 8
2.82 10 8
3.9 10
3
m
m
°C
1
2.3 102 °C
1
At 80°C,
I
or
17.28
. Thus, if
for that tem perature gives
Al
17.27
1
2.2 103 °C
For alum inum , the resistivity at room tem perature is
0 Cu
°C
at the operating tem perature of the filam ent,
4.5 10
3
3
for the operating tem perature gives
T T0
R R0
T0
4.5 10
I
If R 41.0
V
R
R0 1
2.6 10
at T
2
A
V
T T0
5.0 V
200
1
0.5 10
3
C
1
80 C 20 C
26 mA
20 C and R 41.4
at T
29.0 C, then R R0 1
T T0
the tem perature coefficient of resistivity of the m aterial m aking up this w ire as
R R0
R0 T T0
41.4
41.0
41.0
29.0 C 20°C
1.1 10
3
°C
1
gives
Current and Resistance
17.29
(a) The resistance at 20.0°C is
1.7 10
L
A
R0
8
m 34.5 m
0.25 10
3
m
V
R0
and the current w ill be I
3.0
2
9.0 V
3.0
3.0 A
(b) At 30.0°C,
R
R0 1
T T0
3.0
1
3.9 10
C
V
R
Thus, the current is I
17.30
3
30.0 C 20.0 C
9.0 V
3.1
3.1
2.9 A
The resistance of the heating elem ent w hen at its operating tem perature is
V
R
From R
2
2
120 V
P
1050 W
R0 1
L
1
R
13.7
0L
1
A
T T0
0
A
8
m 4.00 m
13.7
A
4.90 10
(a) From R
Ri
T T0
, the cross-sectional area is
T T0
150 10
17.31
1
7
1
0.40 10
3
C
1
m2
L A , the initial resistance of the m ercury is
Li
Ai
9.4 10
7
m 1.000 0 m
1.00 10
3
m
2
4
1.2
320 C 20.0 C
117
118
CH APTER 17
(b) Since the volum e of m ercury is constant, V
sectional area as Af
L2f
Lf
Rf
Af
Ai Li
Rf
Ri
17.32
L2f
1
Ai Li
1
Li Ai
1
Li
2
1
R
T T0
1
Solving R R0 1
T
8.0 10
4
or a 0.080% increase
200.0
1+ 3.92 10
T T0
R R0
R0
T0
-3
C
P
V
0 C 20.0 C
for T gives the tem perature of the m elting potassium as
253.8
20.0 C
3.92 10
1.00 103 W
1.20 102 V
217
1
3
217
1
C
(a) The pow er consum ed by the d evice is P
I
I
63.3 C
217
V , so the current m ust be
8.33 A
(b) From Ohm ’ s law , the resistance is
17.34
2
Lf
The resistance at 20.0°C is
R0
17.33
. The fractional change in the resistance is then
Ri
100.04
100.00
Ai Li gives the final cross-
Li L f . Thus, the final resistance is given by
Ai
Rf
Ri
Af L f
R
V
I
1.20 102 V
8.33 A
14.4
(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 kilow att-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
E P
t
I
V
t
20.0 C s 220 J C
1 kW
103 J s
5.0 h
and the cost of this energy, at a rate of $0.12 per kilow att-hour is
cost
17.35
E rate
The pow er required is
22 kWh $0.12 kWh
P I
V
$2.6
0.350 A 6.0 V
2.1 W
22 kWh
119
120
17.36
CH APTER 17
(a) The pow er loss in the line is
Ploss I 2 R
1000 A
2
0.31
km 160 km
5.0 107 W
50 MW
(b) The total pow er transm itted is
Pinput
V I
700 103 V 1000 A
7.0 108 W 700 MW
Thus, the fraction of the total transm itted pow er represented by the line losses is
fraction loss
17.37
Ploss
Pinput
50 MW
700 MW
0.071 or 7.1%
The energy required to bring the w ater to the boiling point is
E mc
T
0.500 kg 4186 J kg C 100 C 23.0 C
1.61 105 J
The pow er input by the heating elem ent is
Pinput
V I
120 V 2.00 A
240 W 240 J s
Therefore, the tim e required is
t
17.38
(a)
E
Pinput
E P t
1.61 105 J
240 J s
90 W 1 h
672 s
1 min
60 s
11.2 min
3.2 105 J
90 J s 3 600 s
(b) The pow er consum ption of the color set is
P
V I
120 V 2.50 A
300 W
Therefore, the tim e required to consum e the energy found in (a) is
t
E
P
3.2 105 J
300 J s
1.1 103 s
1 min
60 s
18 min
Current and Resistance
17.39
The energy input required is
E mc
T
and , if this is to be ad d ed in
P
2.51 105 J
600 s
E
t
t 10.0 min 600 s , the pow er input need ed is
419 W
The pow er input to the heater m ay be expressed as P
resistance is
R
17.40
V
2.51 105 J
1.50 kg 4186 J kg C 50.0 C 10.0 C
2
120 V
P
V
2
R , so the need ed
2
419 W
34.4
(a) At the operating tem perature,
P
V I
120 V 1.53 A
(b) From R R0 1
T T0
184 W
, the tem perature T is given by T
resistances are given by Ohm ’ s law as
R
V
I
V
120 V
, and R0
1.53 A
0
I0
120 V
1.80 A
Therefore, the operating tem perature is
T
20.0 C
120 1.53
0.400 10
3
120 1.80
C
1
120 1.80
461 C
T0
R R0
. The
R0
121
122
17.41
CH APTER 17
The resistance per unit length of the cable is
R
L
From R
P I 2 P L 2.00 W m
I2
L
r2
A
R L
1.7 10
r
R L
Energy stored P
8
2.22 10
t
m
-5
P
V I
17.44
(a)
E P t
cost
(b) E P t
cost
0.016 m
m
V I
0.66 kWh $0.12 kWh
17.43
t
E P t
cost
1.6 cm
12 V 55 A h
$0.079
75 10 3 V 0.20 10 3 A
660 W h
5
W 15 10
6
W
15 W
1.34 104 Wh 13.4 kWh
13.4 kWh $0.120 kWh
0.970 kW 3.00 min 1 h 60 min
$1.61
4.85 10 2 kWh
E rate
$0.005 83
5.20 kW 40.0 min 1 h 60 min
E rate
0.66 kWh
7.9 cents
1.5 10
40.0 W 14.0 d 24.0 h d
E rate
A . H ence, the
t 55 A h . Thus, the stored energy is
4.85 10-2 kWh $0.120 kWh
(c)
m
, and the required rad ius is
(a) The rating of the 12-V battery is I
(b) cost
5
L A , the resistance per unit length is also given by R L
cross-sectional area is
17.42
300 A
2.22 10
2
3.47 kWh $0.120 kWh
0.583 cents
3.47 kWh
$0.416
41.6 cents
Current and Resistance
17.45
123
The energy saved is
Phigh Plow t
E
40 W 11 W 100 h
2.9 103 Wh 2.9 kWh
and the m onetary savings is
savings
17.46
E rate
2.9 kWh $0.080 kWh
$0.23
23 cents
The pow er required to w arm the w ater to 100°C in 4.00 m in is
mc
Q
t
P
T
0.250 kg 4 186 J kg °C 100 C 20°C
t
3.5 102 W
4.00 min 60 s 1 min
The required resistance (at 100°C) of the heating elem ent is then
V
R
2
P
120 V
2
41
3.5 102 W
so the resistance at 20°C w ould be
R0
R
T T0
1
41
-3
1+ 0.4 10 °C
1
40
100°C 20°C
We find the need ed d im ensions of a N ichrom e w ire for this heating elem ent from
R0
d 2 4 4 0 L d 2 , w here L is the length of the w ire and d is its
0L A
0L
d iam eter. This gives
d
2
4
0
R0
L
4 150 10
40
8
m
L
4.8 10
8
m L
Thus, any combination of length and diameter satisfying the relation d 2
4.8 10
8
m L w ill
be suitable. A typical com bination m ight 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 elem ents could easily be m ad e from less than 0.5 cm3 of N ichrom e.
The volum e of m aterial required for the typical w ire given above is
124
CH APTER 17
V
17.47
AL
m
2
3.0 m
4
3.4 10
7
m3
106 cm3
1 m3
0.34 cm3
The energy that m ust be ad d ed to the w ater is
E
mc
T
200 kg
and the cost is cost
17.48
4
3.8 10
d2
L
4
4186
E rate
J
kg C
1 kWh
3.60 106 J
80 C 15 C
15 kWh $0.080 kWh
15 kWh
$1.2
(a) For tungsten, Table 17.1 from the textbook gives the resistivity at
m and the tem perature coefficient of
T0 20.0°C 293 K as 0 5.6 10 8
resistivity as
4.5 10
3
°C
1
4.5 10 3 K 1 . Thus, for a tungsten w ire having a
rad ius of 1.00 m m and a length of 15.0 cm long, the resistance at T0
R0
0
L
A
L
0
r
5.6 10
2
8
m
15.0 10
2
1.00 10
3
m
m
2
2.7 10
293 K is
3
(b) From Stefan’ s law , the rad iated pow er is P
AeT 4 , w here A is the area of the
rad iating surface. N ote that since w e are com puting the rad iated pow er, not the net
energy gained or lost as a result of rad iation, the am bient tem perature is not need ed
here. In the case of a w ire, this is the cylind rical surface area A 2 rL . The
tem perature of the w ire w hen it is rad iating a pow er of P 75.0 W m ust be
T
or T
14
14
P
75.0 W
5.669 10-8 W m 2 K 4 2
Ae
1.00 10
3
m 0.150 m 0.320
1.45 103 K
(c) Assum ing a linear tem perature variation of resistance, the resistance of the w ire at
this tem perature is
R
giving
R0 1
R
T T0
1.7 10
2.7 10
2
3
1
4.5 10
3
K
1
1.45 103 K 293 K
Current and Resistance
125
(d ) The voltage d rop across the w ire w hen it is rad iating 75.0 W and has the resistance
found in part (c) above is given by P
R P
V
(e)
1.7 10
2
V
75.0 W
2
R as
1.1 V
Tungsten bulbs release very little of the energy consumed in the form of visible light ,
m aking them inefficient sources of light.
17.49
The battery is rated to d eliver the equivalent of 60.0 am peres of current (i.e., 60.0 C/ s)
for 1 hour. This is
Q I
17.50
t
60.0 A 1 h
2.16 105 C
60.0 C s 3 600 s
The energy available in the battery is
Energy stored P t
V I t
V I t
12 V 90 A h
The tw o head lights together consum e a total pow er of P
required to com pletely d ischarge the battery is
t
17.51
Energy stored
P
1.1 103 W h
72 W
72 W , so the tim e
2 36 W
15 h
Assum ing a constant resistance, the pow er consum ed by the d evice is proportional to
the square of the applied voltage, P
P2
P1
17.52
1.1 103 W h
V2
V1
2
2
R
or
R
V
2
P2 P1
R . Thus,
V2
V1
2
15 W
6.0 V
9.0 V
The tem perature variation of resistance is d escribed by R R0 1
is the resistance at T0
20°C . Thus, if an alum inum w ire
R 2R0 , w e have 2 1
T
20 C
1
1
3.9 10
3
°C
1
2.8 10 2 °C
6.7 W
T T0
3.9 10
T 20°C , and its tem perature m ust be
20 C
2
3
°C
, w here R0
1
has
126
17.53
CH APTER 17
From P
2
V
2
V
R
17.54
2
20 V
P
8.3
48 W
Thus, from R
L
R , the total resistance need ed is
L A , the length of w ire required is
R A
8.3
6
4.0 10
3.0 10
8
m2
1.1 103 m
m
1.1 km
The resistance of the 4.0 cm length of w ire betw een the feet is
R
1.7 10
L
A
8
m 0.040 m
0.011 m
1.79 10
2
6
,
so the potential d ifference is
V
17.55
IR
50 A 1.79 10
6
8.9 10
Ohm ’ s law gives the resistance as R
5
V
89 V
V I . From R
L A , the resistivity is given
by
R A L . The results of these calculations for each of the three w ires are
sum m arized in the table below .
R
L m
0.540
1.028
1.543
10.4
21.1
31.8
m
1.41 10 6
1.50 10 6
1.50 10 6
The average value found for the resistivity is
i
av
3
1.47 10
6
w hich d iffers from the value of
by 2.0% .
m
150 10 8
m 1.50 10 6
m given in Table 17.1
Current and Resistance
17.56
At tem perature T, the resistance of the carbon w ire is Rc
the nichrom e w ire is Rn
R0n 1
T T0
n
R0c 1
T T0
c
127
, and that of
. When the w ires are connected end to end ,
the total resistance is
R Rc
Rn
R0c
R0n
R0c
R0n
c
If this is to have a constant value of 10.0 k
that
R0c
and
R0c
T T0
as the tem perature changes, it is necessary
[1]
R0n 10.0 k
c
R0n
n
[2]
0
From Equation [1], R0c 10.0 k
10.0 k
17.57
n
R0 n
0.50 10
R0n , and substituting into Equation [2] gives
3
Solving this Equation gives
R0n
Then, R0c 10.0 k
4.4 k
5.6 k
C
1
5.6 k
R0 n 0.40 10
3
C
1
0
nichrome wire
carbon wire
(a) The total pow er you now use w hile cooking breakfast is
P
1200 500 W 1.70 kW
The cost to use this pow er for 0.500 h each d ay for 30.0 d ays is
cost
P
t
rate
1.70 kW 0.500
h
day
30.0 days
$0.120 kWh
(b) If you upgrad ed , the new pow er requirem ent w ould be:
P
2 400 500 W=2 900 W
and the required current w ould be
I
P
V
2 900 W
110 V
26.4 A
No , your present circuit breaker cannot hand le the upgrad e.
20 A
$3.06
128
17.58
CH APTER 17
(a) The charge passing through the cond uctor in the interval
0 t 5.0 s is represented by the area und er the I vs t
graph given in Figure P17.58. This area consists of tw o
rectangles and tw o triangles. Thus,
Q
Arectangle 1
Arectangle 2
Atriangle 1
5.0 s 0 2.0 A 0
Atriangle 2
4.0 s 3.0 s 6.0 A 2.0 A
1
3.0 s 2.0 s 6.0 A 2.0 A
2
Q
1
5.0 s 4.0 s 6.0 A 2.0 A
2
18 C
(b) The constant current that w ould pass the sam e charge in 5.0 s is
Q
t
I
17.59
(a) From P
18 C
5.0 s
3.6 A
V I , the current is I
P
V
8.00 103 W
12.0 V
667 A
(b) The tim e before the stored energy is d epleted is
t
Estorage
P
2.00 107 J
8.00 103 J s
2.50 103 s
Thus, the d istance traveled is
d
v t
20.0 m s 2.50 103 s
5.00 104 m
50.0 km
Current and Resistance
17.60
The volum e of alum inum available is
V
115 10 3 kg
2.70 103 kg m 3
mass
density
4.26 10
5
m3
(a) For a cylind er w hose height equals the d iam eter, the volum e is
d2
d
4
V
d3
4
13
4V
and the d iam eter is d
4 4.26 10
5
m3
13
0.037 85 m
The resistance betw een end s is then
R
L
A
d
2
V
d
d 4
8
m
9.49 10
0.037 85 m
7
L3 , so the length of an ed ge is
(b) For a cube, V
L
4 2.82 10
4
13
4.26 10
5
m
13
0.034 9 m
The resistance betw een opposite faces is
R
17.61
L
A
L
2
L
L
2.82 10 8
m
0.034 9 m
The current in the w ire is I
Then, from vd
n
or
n
V
R
7
150 A
I nqA , the d ensity of free electrons is
I
vd e
15.0 V
0.100
8.07 10
150 A
r
2
3.17 10
3.77 1028 m3
4
m s 1.60 10
19
C
5.00 10
3
m
2
129
130
17.62
CH APTER 17
Each speaker has a resistance of R 4.00
P I 2 R , the m axim um safe current is
P
I max
R
60.0 W
4.00
and can hand le 60.0 W of pow er. From
3.87 A
Thus, the system is not adequately protected by a 4.00 A fuse.
17.63
The cross-sectional area of the cond ucting m aterial is A
2
router
2
rinner
Thus,
R
17.64
3.5 105
L
A
1.2 10
2
m 4.0 10
m
2
2
m
2
0.50 10
m
3.7 107
2
37 M
The volum e of the m aterial is
V
Since V
mass
density
50.0 g
1 m3
7.86 g cm3 106 cm3
6.36 10
6
m3
A L , the cross-sectional area of the w ire is A V L
(a) From R
L
L
A
R V
L
V L
1.5
L2
, the length of the w ire is given by
V
6.36 10
11 10-8
6
m3
m
9.3 m
(b) The cross-sectional area of the w ire is A
d
4V
L
4 6.36 10
6
9.3 m
m3
9.3 10
4
d2
4
V
. Thus, the d iam eter is
L
m
0.93 mm
Current and Resistance
17.65
131
The pow er the beam d elivers to the target is
P
4.0 106 V 25 10 3 A
V I
1.0 105 W
The m ass of cooling w ater that m ust flow through the tube each second if the rise in the
w ater tem perature is not to exceed 50°C is found from P
m t c T as
P
m
t
17.66
c
1.0 105 J s
4186 J kg C 50 C
T
(a) At tem perature T, the resistance is R
L L0 1
, and A
T T0
0.48 kg s
L
, w here
A
A0 1
T T0
0
2
1
A0 1 2
T T0
,
T T0
Thus,
L
A0
0 0
(b) R0
1
L
A0
0 0
R
T T0
1 2
1.70 10
8
1.082
m 2.00 m
T T0
1
T T0
R0 1
T T0
0.100 10-3
Then R R0 1
R
1
2
T T0
1
1 2
T T0
T T0
1.082
gives
3
3.90 10
C 80.0 C
1.420
The m ore com plex form ula gives
1.420
1
17 10
R
1 2 17 10
6
6
C 80.0 C
1.418
C 80.0 C
N ote: Som e rules for hand ing significant figures have been d eliberately viola ted in
this solution in ord er to illustrate the very sm all d ifference in the results obtained
w ith these tw o expressions.
132
17.67
CH APTER 17
N ote that all potential d ifferences in this solution have a value of V 120 V .
First, w e shall d o a sym bolic solution for m any parts of the problem , and then enter the
specified num eric values for the cases of interest.
From the m arked specifications on the cleaner, its internal resistance (assum ed constant)
is
2
V
Ri
where P1 535 W
P1
Equation [1]
If each of the tw o cond uctors in the extension cord has resistance Rc , the total resistance
in the path of the current (outsid e of the pow er source) is
Rt
Ri
Equation [2]
2Rc
so the current w hich w ill exist is I
is
Pdelivered I 2 Ri
V
Rt
2
V Rt and the pow er that is d elivered to the cleaner
V
Rt
Ri
2
V
P1
2
V
4
Equation [3]
Rt2P1
The resistance of a copper cond uctor of length L and d iam eter d is
4 Cu L
L
L
Rc
Cu
Cu
2
A
d2
d 4
Thus, if Rc, max is the m axim um allow ed value of Rc , the m inim um acceptable d iam eter
of the cond uctor is
4 Cu L
Rc, max
dmin
(a) If Rc
Rt
0.900
Ri
Equation [4]
, then from Equations [2] and [1],
V
2 0.900
2
1.80
P1
120 V
2
535 W
1.80
and , from Equation [3], the pow er d elivered to the cleaner is
120 V
Pdelivered
120 V
535 W
4
470 W
2
2
1.80
535 W
Current and Resistance
133
(b) If the m inim um acceptable pow er d elivered to the cleaner is Pmin , then Equations
[2] and [3] give the m axim um allow able total resistance as
Rt , max
Ri
2Rc, max
Rc, max
1
2
V
4
V
2
V
PminP1
PminP1
so
When Pmin
2
PminP1
Ri
and ,
dmin
V
4 1.7 10
2
8
525 W 535 W
8
2
2
m 15.0 m
0.037 9
2
1
m 15.0 m
0.128
120 V
V
P1
2
2
532 W , then Rc , max
4 1.7 10
2
PminP1
120 V
525 W , then Rc , max
and , from Equation [4], d min
When Pmin
V
1
2
1
1
PminP1 P1
1
535 W
0.128
1.60 mm
1
532 W 535 W
2.93 mm
2
1
535 W
0.037 9