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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, are equivalent to the movement of 5, 3, 4, and 2 charges respectively, giving
, , , and a b cI I I I
d b cI I Id
aI< < < .
2. (b). Under steady-state conditions, the current is the same in all parts of the wire. Thus, the drift velocity, given by d nq=v I , is inversely proportional to the cross-sectional area. A
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 increases, the slope (and hence V∆ 1 ) increases. Thus, the resistance decreases.
R
5. (b). Consider the expression for resistance: 2RA r
ρ ρπ
= =l l
. Doubling all linear
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.
6. (a). The resistance of the shorter wire is half that of the longer wire. The power dissipated,
( )2V R= ∆P , (and hence the rate of heating) will be greater for the shorter wire. Consideration of the expression 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.
2I R=P
69
70 CHAPTER 17
7. (b). a b c d e fI I I I I I= > = > = . Charges constituting the current leave the positive terminal
of the battery and then split to flow through the two bulbs; thus, aI
a cI I Ie= + . Because the potential difference is the same across the two bulbs and because the power delivered to a device is
V∆( )I V= ∆P
dI e
, the 60–W bulb with the higher power rating must carry the greater current, meaning that . 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
and
cI I> e
cI = fI I= . The two currents leaving the bulbs recombine to form the current back
into the battery, f d b+ =I I . I
8. (a) B, (b) B. Because the voltage across each resistor is the same, and the rate of energy delivered to a resistor is ( )2V R= ∆P , 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 ( )2R 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.
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. 55.21 10 m s−×
4. 143.75 10 electrons s×
6. 159 mA
8. 41.3 10 m s−×
10. 500 mA
12. (a) 1.82 m (b) 0.280 mm
14. 0.31 Ω
16. (Assumes a 400 W dryer used for about 10 min/d with electric energy costing about 10 cents/kWh.) ~ $1
18. (a) (b) 82.8 10 A× 71.8 10 A×
20. 31.4 10 C× °
22. 9.7 Ω
24. ( ) 131.1 10 °C −−×
26. (a) 0.65 mV (b) 1.1 mV
28. 7 24.90 10 m−×
30. 63.2°C
32. (a) $0.29 (b) $2.6
34. (a) 50 MW (b) 0.071 or 7.1%
36. (a) (b) 18 min 53.2 10 J×
38. (a) 184 W (b) 461°C
40. (a) 5.0 cents (b) 71%
42. (a) $1.61 (b) 0.583 cents (c) 41.6 cents
44. 3.36 h d
Current and Resistance 73
46. ( )2 8Any diameter and length related by 4.8 10 m ,d d −= ×l l 1.0 m=l 0.22 mmd =
such as and . Yes.
48. (a) (b) 4.80 s, lower potential energy (c) 0.040 0 s, converted to internal energy and light (d) $1.26, energy,
576 , 144 Ω Ω
8.94 10 J−×$1
50. (a) (b) 1.9 cents 58.6 10 J×
52. 89 Vµ
54. (a) 18 C (b) 3.6 A
56. (a) (b) 79.49 10 −× Ω 78.07 10 −× Ω
58. , 5.6 k (nichrome)Ω 4.4 k (carbon)Ω
60. No. The fuse should be rated at 3.87 A or less.
62. (a)
V∆ I R V I= ∆
1.5 V− 50.30 10 A−− × 55.0 10 × Ω
1.0 V− 50.20 10 A−− × 55.0 10 × Ω
0.50 V− 50.10 10 A−− × 55.0 10 × Ω
0.40 V+ 0.010 A + 40 Ω
0.50 V+ 0.020 A + 25 Ω
0.55 V+ 0.040 A + 14 Ω 0.70 V+ 0.072 A + 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.
64. (a) 9.3 m (b) 0.93 mm
66. (b) versus 1. for the more precise equation 1.420 Ω 418 Ω
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
( )
( ) ( )( )320
19
80.0 10 C s 10.0 min 60.0 s00 10 electrons
1.60 10 C
I tQn
e e
−
−
∆∆= =
×
×
min3.
= = ×
The negatively charged electrons move in the direction opposite to the conventional current flow.
17.2 The drift speed in a conductor of cross-sectional area A and carrying current I is dv 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,
( )( )( )
528 -3 19 6 2
2.50 C s5.21 10 m s
7.50 10 m 1.60 10 C 4.00 10 md
Iv
nqA−
− −= = = ×
× × ×
17.3 The current is Q V
It R
∆ ∆= =
∆. Thus, the change that passes is ( )V
Q tR
∆ ∆ =
∆ , giving
( ) ( )( )1.00 V0.100 A 20.0 s 2.00 C
10.0 Q t ∆ = ∆ = = Ω
17.4 ( )Q I t∆ = ∆ and the number of electrons is
( ) ( )( )6
14-19
60.0 10 C s 1.00 s3.75 10 electrons
1.60 10 CI tQ
e e
−×∆∆= = = = ×
×n
Current and Resistance 75
17.5 The period of the electron in its orbit is 2 rT vπ= , and the current represented by the orbiting electron is
( )( )( )6 19
311
2
2.19 10 m s 1.60 10 C1.05 10 C s 1.05 mA
2 5.29 10 m
e v eQI
t T rπ
π
−−
−
∆= = =
∆
× ×= = ×
×=
17.6 The mass of a single gold atom is
-22 -2523
197 g mol3.27 10 g 3.27 10 kg
6.02 10 atoms molatomA
Mm
N= = = × = ×
×
The number of atoms deposited, and hence the number of ions moving to the negative electrode, is
3
21-25
3.25 10 kg9.93 10
3.27 10 kgatom
mn
m
−×= = = ×
×
Thus, the current in the cell is
( )( )
( )( )
21 199.93 10 1.60 10 C0.159 A 159 mA
2.78 h 3 600 s 1 hQ ne
It t
−× ×∆= = = = =
∆ ∆
17.7 The drift speed of electrons in the line is ( )2 4d
I Iv
nqA n e dπ= = , or
( )
( )( ) ( )-4
228 3 19
4 1 000 A2.3 10 m s
8.5 10 m 1.60 10 C 0.020 mdv
π−= =
× ××
The time to travel the length of the 200-km line is then
3
-4 7
1 yr200 10 m27 yr
2.34 10 m s 3.156 10 sd
Lt
v× ∆ = = = × ×
76 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
A
A
density Nn
mass per atom M N Mρρ
= = = ,
or ( ) ( )( )23 3 6 3 3
28 36.02 10 mol 2.7 g cm 10 cm 1 m
6.0 10 m26.98 g mol
n × = = ×
The drift speed of the electrons in the wire is then
( )( )( )
428 3 19 6 2
5.0 C s1.3 10 m s
6.0 10 m 1.60 10 C 4.0 10 md
Iv
n e A−
− −= = = ×
× × ×
17.9 (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 d nqv I A= . Thus, the drift velocity is doubled
when the current is doubled.
17.10 120 V
0.500 A 500 mA240
VI
R∆
= = = =Ω
17.11 ( ) ( )6maxmax
80 10 AV I R −∆ = = ×
54.0 10 R = × Ω
R
Thus, if , ( )maxV∆ =
2 000 = Ω ( )
32 V
and if R , max 0.1V∆ = 6 V
17.12 The volume of the copper is
3
7 33 3
1.00 10 kg1.12 10 m
8.92 10 kg mm
Vdensity
−−×
= = = ××
L= ⋅ 7 31.12 10 m
Since, V A , this gives A L −⋅ = × (1)
Current and Resistance 77
(a) From L
RAρ
= , we find that
( )8
81.70 10 m3.40 10 m
0.500 A L L
Rρ −
− × Ω ⋅ = = = × Ω
(
L
Inserting this expression for A into equation (1) gives )8 2 7 33.40 10 m 1.12 10 mL− −× = × , which yields 1.82 mL =
(b) From equation (1), 2 71.12 10 m
4d
AL
π −×= =
( )
3
, or
( )
( )
7 3 7 3
4
4 1.12 10 m 4 1.12 10 m
1.82 m
2.80 10 m 0.280 mm
dLπ π
− −
−
× ×= =
= × =
17.13 From L
RAρ
= , we obtain 2
4d L
AR
π ρ= = , or
( )( )( )
8 24
4 5.6 10 m 2.0 10 m41.7 10 m 0.17 mm
0.050 L
dR
ρπ π
− −−
× Ω ⋅ ×= = = × =
Ω
17.14 ( )( )
( )
8
22 3
4 1.7 10 m 15 m0.31
4 1.024 10 m
L LR
A dρ ρ
π π
−
−
× Ω ⋅= = = = Ω
×
17.15 (a) 12 V
30 0.40 A
VI
∆= = = ΩR
(b) From, L
RAρ
= ,
( ) ( )22
430 0.40 10 m
4.7 10 m3.2 m
R AL
πρ
−
−
Ω ×⋅ = = = × Ω ⋅
78 CHAPTER 17
17.16 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 hr0.400 kW 10 365 d 24 kWh
d 60 minE t
= ∆ = = P
( )
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
$
24 kWh 0.10 $2.4 or ~$1kWh
cost E rate = × = =
17.17 The resistance is 9.11 V
0.253 36.0 A
VR
I∆
= = = Ω , so the resistivity of the metal is
( ) ( ) ( )
( )
22 38
4 0.253 2.00 10 m1.59 10 m
4 50.0 m
R dR AL L
π πρ
−−
⋅ Ω ×⋅= = = = × Ω ⋅
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:
( )( )1
10 cm 1 120 cm 40 cm 80 cm 0.80 m
LA
= = =
,
( )( )2
20 cm 1 110 cm 40 cm 20 cm 0.20 m
LA
= = =
,
and ( )( )3
40 cm 1 110 cm 20 cm 5.0 cm 0.050 m
LA
= = =
(a) ( )
( )( ) 8max 8
min min
6.0 V 0.80 m2.8 10 A
1.7 10 mV V
R L Aρ −
∆ ∆= = ×
× Ω ⋅I = =
(b) ( )
( )( ) 7min 8
max max
6.0 V 0.050 m1.8 10 A
1.7 10 mV V
R L Aρ −
∆ ∆= = ×
× Ω ⋅I = =
Current and Resistance 79
17.19 The volume of material, ( )20 0L r Lπ= = 0V A , in the wire is constant. Thus, as the wire is
stretched to decrease its radius, the length increases such that ( ) ( )20f fr L r Lπ π= 2
0 giving
( )2
20 00 0
0
4.00.25
r rL
r r
=
2
0 16ff
L L L
= = =
0L
The new resistance is then
( )
( ) ( )20 022 2
00
1616 4 256
4f f
ff f
L L L LR R
A r rrρ ρ ρ ρ
π ππ
= = = = = =
0 256 1.00 256= Ω Ω
17.20 Solving ( )0 1 T Tα= + − 0 R R for the final temperature gives
( ) ( )
300 1-3
0
140 19 20 C 1.4 10 C
4.5 10 C 19
R RT T
Rα −
− Ω − Ω= + = ° + = × °
× ° Ω
17.21 From Ohm’s law, i i f fV I R I R∆ = = , so the current in Antarctica is
( )( )
( )( ) ( )
( ) ( )
0 0
0 0
13
13
C 58.0 C 20.0 C1.00 1.98 A
C 88.0 C 20.0 C
iif i
f f
TRI I
R T
−−
−
+ − + −
° ° − ° = = ° − ° − °
1
1
1 3.90 10 A
1 3.90 10
i
R TI
R T
α
α
−
= =
+ ×
+ ×
17.22 The expression for the temperature variation of resistance, ( )0 01R R T Tα = + − with
, gives the temperature coefficient of resistivity of this material as
0 20 CT = °
( ) ( )( )4 1 °C0
0 0
10.55 10.00 7.9 10
10.00 90 C 20°CR R
R T Tα − −− Ω − Ω
= = =− Ω ° −
20 C= − °
( ) ( )
×
Then, at T , the resistance will be
( )( )4 110 °C 40 9.C− − × − ° 20 1R R= + 20 C 20 C 10.00 1 7.9 7 α − ° − ° = Ω + = Ω
80 CHAPTER 17
17.23 At 80°C,
( ) ( ) ( )( )13
0 0
5.0 V1 200 1 0.5 10 C 80 C 20 C
V VI
R R T Tα −−
∆ ∆= = =
+ − Ω + − × ° ° − °
or 22.6 10 A 26 mAI −= × =
17.24 If R T , then 41.0 at 20 C and 41.4 at 29.0 CR T= Ω = ° = Ω = ° ( )0 01 T Tα = + − R R gives
the temperature coefficient of resistivity of the material making up this wire as
( ) ( )( )
( ) 1 °C −30
0 0
41.4 41.0 1.1 10
41.0 29.0 C 20°CR R
R T Tα −− Ω − Ω
= = =− Ω ° −
×
17.25 ( )( )8
20 6 2
1.7 10 m 10.0 m5.67 10
3.00 10 mL
RAρ
−−
−
× Ω ⋅= = = ×
×Ω
(a) At ( )030.0 C, 1R Tα= ° = + −
( ) ( )
0T T R gives a resistance of
( )( )13 230.0 C 20.0 C 5.89 10 −− − ° ° − ° = × 0.056 7 1 3.9 10 CR = Ω + × Ω
(b) At ( )0 010.0 C, 1R Tα = ° = + −
( ) ( )
T R yields
T
( )( )13 210.0 C 20.0 C 5.45 10 −− − ° ° − ° = × 0.056 7 1 3.9 10 CR = Ω + × Ω
17.26 (a) At 20°C, the resistance of this copper wire is ( ) ( )20R L A L rρ ρ π= =
3.0 AI
and the
potential difference required to produce a current of = is
( )( ) ( )( )
80 0 22 2
m3.0 A 1.7 10 m
0 m
LV IR I
rρ
π−
−
∆ = = = × Ω ⋅
1.00
0.50 1π ×
or 4
0 6.5 10 V 0.65 mVV −∆ = × =
Current and Resistance 81
(b) At T , the potential difference required to maintain the same current in the copper wire is
200 C= °
( ) ( )0 0 0 01 1R IR T T V T Tα α = + − = ∆ + −
( )
V I∆ =
or ( )( )3 -165 mV 1 3.9 10 °C 200 C 20 C−+ × ° − °0.V∆ = 1.1 mV =
17.27 (a) The resistance at 20.0°C is
( )( )
( )
8
0 23
1.7 10 m 34.5 m3.0
0.25 10 m
LR
Aρ
π
−
−
× Ω ⋅= = =
×Ω
and the current will be 0
9.0 V3.0 A
3.0 V
IR∆
= = =Ω
(b) At 30.0°C,
Thus, the current is
( )
( ) ( )( )( )
0 0
13
1
3.0 1 3.9 10 C 30.0 C 20.0 C 3.1
R R T Tα
−−
= + −
= Ω + × ° ° − ° = Ω
9.0 V2.9 A
3.1 V
IR
∆= = =
Ω
17.28 The resistance of the heating element when at its operating temperature is
( ) ( )2 2120 V
13.7 1 050 W
VR
∆= = =
P
( )
Ω
From (1 T T )00
0 01L
R R T TA
ρα = + − = α + − , the cross-sectional area is
( )
( )( )( )( )( )
00
8
7 2
1
150 10 m 4.00 m
13.7
4.90 10 m
LA T T
R
A
1310 C 320 C 20.0 C1 0.40
ρ α
−
−
= + −
× Ω ⋅ −− × °= + ° − ° Ω
= ×
82 CHAPTER 17
17.29 (a) From R L Aρ= , the initial resistance of the mercury is
( )( )
( )
7
23
9.4 10 m 1.000 0 m1.2
1.00 10 m 4i
ii
LR
Aρ
π
−
−
× Ω ⋅= = = Ω
×
(b) Since the volume of mercury is constant, f f iL A LiV A= ⋅ = ⋅ gives the final cross-
sectional area as ( )f i i fA A L L= ⋅ . Thus, the final resistance is given by 2
f ff
f i i
L LR
A A L
ρ ρ= =
⋅. The fractional change in the resistance is then
( ) 2
1f
i
L
L
= −
2
1 1f i f f i i
i i i i
R R R L A L
R R L A
ρρ
− ⋅∆ = = − = −
2
4100.041 8.0 10
100.00− ∆ = − = ×
or a 0.080% increase
17.30 The resistance at 20.0°C is
( ) ( ) ( )
0 1-30
200.0 217
1 1+ 3.92 10 C 0 C 20.0 C
RR
T Tα −
Ω= = =
+ − × ° ° − °
(
Ω
Solving )0 01R R T Tα = + − for T gives the temperature of the melting potassium as
( ) ( )
00 13
0
253.8 217 20.0 C 63.2 C
3.92 10 C 217
R RT T
Rα −−
− Ω − Ω= + = ° + = °
× ° Ω
17.31 600 W
5.00 A120 V
IV
= = =∆P
and 120 V
24.0 5.00 A
VR
I∆
= = = Ω
17.32 (a) The energy used by a 100-W bulb in 24 h is ( )( ) ( )( )100 W 24 h 0.100 kW 24 h 2.4 kWhE t= ⋅∆ = = =P
( )
and the cost of this energy, at a rate of $0.12 per kilowatt-hour is ( )2.4 kWh $0.12 kWh $0.29cost E rate= ⋅ = =
Current and Resistance 83
(b) The energy used by the oven in 5.0 h is
( ) ( )( ) ( )3
1 kW20.0 C s 220 J C 5.0 h 22 kWh
10 J sE t I V t
= ⋅∆ = ∆ ⋅ ∆ = =
P
( )
and the cost of this energy, at a rate of $0.12 per kilowatt-hour is ( )22 kWh $0.12 kWh $2.6cost E rate= ⋅ = =
17.33 The maximum power that can be dissipated in the circuit is ( ) ( )( ) 3
max max 120 V 15 A 1.8 10 WV I= ∆ = = ×P Thus, one can operate at most 18 bulbs rated at 100 W per bulb.
17.34 (a) The power loss in the line is
( ) ( )( )22 71 000 A 0.31 km 160 km 5.0 10 W 50 MWloss I R = = Ω = × = P
(b) The total power transmitted is ( ) ( )( )3 8700 10 V 1 000 A 7.0 10 W 700 MWinput V I= ∆ = × = × =P
Thus, the fraction of the total transmitted power represented by the line losses is
50 MW
0.071700 MW
loss
input
fraction loss = = =P
P or 7.1%
17.35 The energy required to bring the water to the boiling point is ( ) ( )( )( ) 50.500 kg 4 186 J kg C 100 C 23.0 C 1.61 10 JE mc T= ∆ = ⋅° ° − ° = ×
( ) ( )
The power input by the heating element is ( )120 V 2.00 A 240 W 240 J sinput V I= ∆ = = =P
Therefore, the time required is
51.61 10 J 1 min
672 s 11.2 min240 J s 60 sinput
Et
× = = = = P
84 CHAPTER 17
17.36 (a) ( )( ) ( )( ) 590 W 1 h 90 J s 3 600 s 3.2 10 J= ⋅ = = = ×PE t
(b) The power consumption of the color set is ( ) ( )( )120 V 2.50 A 300 WV I= ∆ = =P
Therefore, the time required to consume the energy found in (a) is
5
33.2 10 J 1 min1.1 10 s 18 min
300 J s 60 sE
t× = = = × =
P
17.37 The energy input required is ( ) ( )( )( ) 51.50 kg 4 186 J kg C 50.0 C 10.0 C 2.51 10 JE mc T= ∆ = ⋅° ° − ° = ×
10.0 min 600 st
and, if this is to be added in ∆ = = , the power input needed is
52.51 10 J
419 W600 s
Et
×= = =
∆P
( )
The power input to the heater may be expressed as 2V R= ∆P
( )
, so the needed resistance is
( )2 2120 V
34.4 419 W
VR
∆= = = Ω
P
17.38 (a) At the operating temperature, ( ) ( )( )120 V 1.53 A 184 WV I= ∆ = =P
(b) From ( )0 1 T Tα= + − 0 R R , the temperature T is given by 00
0
R RRα
−= +T T . The
resistances are given by Ohm’s law as
( ) 120 V
1.53 AVI
∆= =R , and
( )00
0
120 V1.80 A
VR
I
∆= =
Therefore, the operating temperature is
( ) ( )
( )( )( )13
120 1.53 120 1.80461 C
10 C 120 1.80−−
−= °
× °20.0 C
0.400T = ° +
Current and Resistance 85
17.39 The resistance per unit length of the cable is
( )
25
22
2.00 W m2.22 10 m
300 A
I LRL L I
−= = = = × ΩP P
From R L Aρ= , the resistance per unit length is also given by R L Aρ= . Hence, the
cross-sectional area is 2r AR Lρπ = = , and the required radius is
( ) ( )
8
-5
1.7 10 m0.016 m 1.6 cm
2.22 10 mR Lρ
π
−× Ω ⋅= = = =
× Ωr
π
17.40 (a) The power input to the motor is ( ) ( )( )120 V 1.75 A 210 W 0.210 kWinput V I= ∆ = = =P
(
At a rate of $0.06/kWh, the cost of operating this motor for 4.0 h is
) ( )
( )( )( )
0.210 kW 4.0 h 6.0 cents kWh 5.0 cents
inputcost Energy used rate t rate= ⋅ = ⋅ ⋅
= =
P
(b) The efficiency is
( )( )0.20 hp 0.746 kW hp
0.710.210 kW
output
input
Eff = = =P
P or 71%
86 CHAPTER 17
17.41 The total power converted by the clocks is ( )( )6 82.50 W 270 10 6.75 10 W= × = ×P
(
and the energy used in one hour is )( )8 110 J× 26.75 10 W 3 600 s 2.43E t= ⋅ = × =P
The energy input required from the coal is
122.43 10 J
9.72 100.250coal
EE
efficiency×
= = = 12 J×
The required mass of coal is thus
12
6
9.72 10 J2.95
33.0 10 J kgcoalE
mheat of combustion
×= =
×
( )
510 kg= ×
or 53
1 metric ton2.95 10 kg 295 me
10 kgm
= × =
tric tons
17.42 (a) ( )( )( ) 440.0 W 14.0 d 24.0 h d 1.34 10 Wh 13.4 kWh= ⋅ = = × =P
( ) ( )
E t
( )13.4 kWh $0.120 kWh $1.61cost E rate= ⋅ = =
(b) ( )( )( ) 20.970 kW 3.00 min 1 h 60 min 4.85 10 kWh−= ⋅ = = ×P
( )
( )
E t
( )-24.85 10 kWh $0.120 kWh $0.005 83 0.583 cents
cost E rate= ⋅
= × = =
(c) ( )( )( )5.20 kW 40.0 min 1 h 60 min 3.47 kWh= ⋅ = =P
( ) ( )
E t
( )3.47 kWh $0.120 kWh $0.416 41.6 centscost E rate= ⋅ = = =
17.43 ( )( ) ( )( )0.180 kW 21 h and $0.070 0 kWhE t cost E rate t= ⋅ = = ⋅ = ⋅P P
( )( )
so ( )0.180 kW 21 h $0.070 0 kWh $0.26 26 centscost = = =
Current and Resistance 87
17.44 The energy used was 3$2002.50 10 kWh
$0.080 kWhcost
Erate
= = = ×
The total time the furnace operated was 32.50 10 kWh
104 h24.0 kW
E ×= = =P
t , and since
January has 31 days, the average time per day was
104 h
3.36 h d31.0 d
average daily operation = =
17.45 The energy saved is
and the monetary savings is
( ) ( )( ) 340 W 11 W 100 h 2.9 10 Wh 2.9 kWhhigh lowE t∆ = − ⋅ = − = × =P P
( )( )2.9 kWh $0.080 kWh $0.23 23 centssavings E rate= ∆ ⋅ = = =
17.46 The power required to warm the water to 100°C in 4.00 min is
( ) ( )( )( )
( )( )20.250 kg 4 186 J kg °C 100 C 20°C
3.5 10 W4.00 min 60 s 1 min
mc TQt t
⋅ ° −∆∆= = = = ×
∆ ∆P
( )
The required resistance (at 100°C) of the heating element is then
( )2 2
2
120 V41
3.5 10 WV
R∆
= = = Ω×P
( )
so the resistance at 20°C would be
( )( )0 -3 1
0
41 40
1 1+ 0.4 10 °C 100°C 20°CR
RT Tα −
Ω= = = Ω
+ − × −
We find the needed dimensions of a Nichrome wire for this heating element from
( )2 4R A dρ ρ π= =l l l0 , where is the length of the wire and d is its diameter. This
gives 2
440
dρ
π Ω =
l
( )
or ( ) ( )
82 8150 10 m
4.8 10 m10 10
dρ
π π
−− × Ω ⋅
= = = × Ω Ω l l l
88 CHAPTER 17
Thus,
( )2 8any combination of length and diameter satisfying the relation 4.8 10 md −= × l
1.0 m=l
will
be suitable. A typical combination might be
and 8 2 44.8 10 m 2.2 10 m 0.22 mmd − −= × = × =
Yes 3 cm
( ) ( ) ( ) ( )
, such heating elements could easily be made from less than 0.5 of Nichrome.
The volume of material required for the typical wire given above is
6 3
22 4 7 33
10 cm2.2 10 m 1.0 m 1.5 10 m 0.15 cm
1 mV A dπ π − −
= = = × = × =
l l 3
17.47 The energy that must be added to the water is
( ) ( ) ( ) 6
J 1 kWh200 kg 4 186 80 C 15 C 15 kWh
kg C 3.60 10 JE mc T
= ∆ = ° − ° = ⋅ ° ×
( )
and the cost is ( )15 kWh $0.080 kWh $1.2t E rate= ⋅ = =cos
17.48 (a) From ( )2V R= ∆
( )
P , the resistance of each bulb is
( )2 2120 V
576 25.0 Wdim
dim
VR
∆= = =
P
( )
Ω and
( )2 2120 V
144 100 Wbright
bright
VR
∆= = =P
Ω
(b) The current in the dim bulb is
25.0 W
0.208 A120 V
dimIV
= = =∆P
so the time for 1.00 C to pass through the bulb is
1.00 C
4.80 s0.208 A
Qt
I∆
∆ = = =
When the charge emerges from the bulb, it has lower potential energy
Current and Resistance 89
(c) The time for the dim bulb to dissipate 1.00 J of energy is
1.00 J
0.040 0 s25.0 Wdim
Et
∆∆ = = =
P
Electrical potential energy is transformed into internal energy and light
(d) In 30.0 days, the energy used by the dim bulb is ( )( )( ) 425.0 W 30.0 d 24.0 h d 1.80 10 Wh 18.0 kWhdimE t= ⋅ = = × =P
( )
and the cost is ( )18.0 kWh $0.070 0 kWh $1.26t E rate= ⋅ = =cos
The electric company sells energy , and the unit cost is
86
$0.070 0 1 kWh $1.94 10 Joule
kWh 3.60 10 Junit cost − = = × ×
17.49 From ( )2V R= ∆
( )
P , the total resistance needed is
( )2 220 V
8.3 48 W
VR
∆= = =
PΩ
Thus, from R L Aρ= , the length of wire required is
( )( )6 2
38
4.0 10 m1.1 10 m 1.1 km
3.0 10 mL
−
−
Ω ×= × =
× Ω ⋅
8.3R Aρ⋅
= =
17.50 (a) The power required by the iron is and the energy transformed in 20 minutes is
( ) ( )( ) 2120 V 6.0 A 7.2 10 WV I= ∆ = = ×√
( )2 5J 60 s7.2 10 20 min 8.6 10 J
s 1 minE t
= ⋅ = × = × √
90 CHAPTER 17
(b) The cost of operating the iron for 20 minutes is
( ) ( )56
1 kWh8.6 10 J $0.080 kWh $0.019 1.9 cents
3.60 10 J
cost E rate= ⋅
= × = = ×
17.51 Ohm’s law gives the resistance as ( )R V I= ∆ . From R L Aρ= , the resistivity is given by
( )R A Lρ = ⋅ . The results of these calculations for each of the three wires are
summarized in the table below.
( ) mL ( ) R Ω ( )mρ Ω ⋅
0.540 10.4 61.41 10−× 1.028 21.1 61.50 10−× 1.543 31.8 61.50 10−×
The average value found for the resistivity is
6av 1.47 10 m
3iρ
ρ −Σ= = × Ω ⋅
15ρ
which differs from the value of 80 10 m 1.50 10 m− 6−= × Ω ⋅ = × Ω ⋅ given in Table 17.1
by 2.0% .
17.52 The resistance of the 4.0 cm length of wire between the feet is
( )( )
( )
86
2
1.7 10 m 0.040 m1.79 10
0.011 mL
RAρ
π
−−
× Ω ⋅= = = ×
( )(
Ω ,
so the potential difference is )6 550 A 1.79 10 8.9 10 V 89 VV IR µ− −∆ = = × Ω = × =
Current and Resistance 91
17.53 (a) The total power you now use while cooking breakfast is ( )1 200 500 W 1.70 kW= + =P
( ) ( )
The cost to use this power for 0.500 h each day for 30.0 days is
( ) ( )h00 30.0 days $0.120 kWh $3.06
daycost
=
1.70 kW 0.5t rate= × ∆ × = P
(b) If you upgraded, the new power requirement would be: ( )2 400 500 W=2 900 W= +P
and the required current would be 2 900 W
26.4 A 20 A110 V
IV
= = = >∆P
No , your present circuit breaker cannot handle the upgrade.
17.54 (a) The charge passing through the conductor in the interval is represented by the area under the
graph given in Figure P17.54. This area consists of two rectangles and two triangles. Thus,
0 5.0 st≤ ≤ vs I t
( )( )( ) ( )
( )( ) ( )( )
1 2 1 2
5.0 s 0 2.0 A 0 4.0 s 3.0 s 6.0 A
1 1 3.0 s 2.0 s 6.0 A 2.0 A 5.0 s 6.0 A 2.0 A
2 2
18 C
rectangle rectangle triangle triangleQ A A A A
Q
∆ = + + +
= − − + −
+ − − + −
∆ =
2.0 A
4.0 s
−
−
(b) The constant current that would pass the same charge in 5.0 s is
18 C
3.6 A5.0 s
QI
t∆
= = =∆
92 CHAPTER 17
17.55 (a) From ( )V I= ∆√ , the current is 38.00 10 W
667 A12.0 V
IV
×= = =
∆P
(b) The time before the stored energy is depleted is
7
33
2.00 10 J2.50 10 s
8.00 10 J sstorageE
t×
= = = ××P
Thus, the distance traveled is ( )( )3 420.0 m s 2.50 10 s 5.00 10 m 50.0 kmd v t= ⋅ = × = × =
17.56 The volume of aluminum available is
3
5 33 3
115 10 kg4.26 10 m
2.70 10 kg mmass
Vdensity
−−×
= = = ××
(a) For a cylinder whose height equals the diameter, the volume is
2 3
4 4d d
V dπ π
= =
and the diameter is ( ) 1 3
5 31 3 4 4.26 10 m40.037 85 m
Vd
π π
− × = = =
The resistance between ends is then
( )
( )( )
87
2
4 2.82 10 m49.49 10
0.037 85 m4L d
RA dd
ρρ ρπ ππ
−−
× Ω ⋅= = = = = × Ω
(b) For a cube, V , so the length of an edge is
3L=
( ) ( )1 356 10 m 0.034 9 mL V −× =1 3 4.2= =
The resistance between opposite faces is
8
72.82 10 m8.07 10
0.034 9 mR
Lρ −
−× Ω ⋅= = = = = × Ω2
L LA Lρ ρ
Current and Resistance 93
17.57 The current in the wire is 15.0 V
150 A0.100
VR
I∆
= = =Ω
Then, from dv I nq=
( )
A , the density of free electrons is
( )( ) ( )22 4 19
150 A
3.17 10 m s 1.60 10 C 5.00 10 md
In
v e rπ π− −= =
× × × 3−
or 28 33.77 10 mn = ×
17.58 At temperature T, the resistance of the carbon wire is ( )0 01c c cR R T Tα = + − , and that
of the nichrome wire is ( )0 1n n nR R T Tα 0 = + −
( )
. When the wires are connected end to
end, the total resistance is ( )( )0T0 0 0 0c n c c n nR R Rα α+ + +
k
c nR R R R T= + = − If this is to have a constant value of 10.0 Ω as the temperature changes, it is necessary that (1) and
0 0 10.0 kc nR R+ =
0 0 0c c n nR R
Ω
α α+
( )10.0 k 0.4Ω −
=
0 010.0 kc nR R= Ω −
( )0 00.50 10 Cn nR − × ° +
(2) From equation (1), , and substituting into equation (2) gives
Solving this equation gives
( )1 13 30 10 C 0R− −− − × ° =
( )0 5.6 k nicnR = Ω hro
( )
me wire
Then, 0 10.0 k 5 irecR = Ω .6 k 4.4 k carbon w− Ω = Ω
17.59 (a) From ( )2V
R∆
=√ , the resistance is ( ) ( )2 2120 V
144 100 W
V∆R = = =
√Ω
(b) Solving L
RAρ
= for the length gives
( )( )2 2
-8 6 2
144 0.010 mm 1 m26 m
5.6 10 m 10 mmR A
Lρ
Ω ⋅= = = × Ω ⋅
94 CHAPTER 17
(c) The filament is tightly coiled to fit the required length into a small space
(d) From ( )0 01 T Tα = + − L L , where ( ) 164.5 10 Cα −−= × ° , the length at is
0 20 CT = °
( ) ( )( )( )00
25 m1 1 4.5 10 C
LL
T Tα −= = =
+ − + × °16
26 m
C 2 600 C 20−° ° −
17.60 Each speaker has a resistance of 4.00 R = Ω and can handle 60.0 W of power. From , the maximum safe current is
2I R=P
max
60.0 W4.00 R
= =Ω
√3.87 A=I
Thus, the system is not adequately protected by a 4.00 A fuse.
17.61 The cross-sectional area of the conducting material is ( )2 2outer innerA r rπ= −
Thus,
( )( )( ) ( )
5 2
2 22 2
3.5 10 m 4.0 10 m
1.2 10 m 0.50 10 m
LR
Aρ
π
−
− −
× Ω ⋅ ×= = Ω
× − ×
73.7 10 37 M= × Ω =
17.62 (a)
V∆ I R V I= ∆
1.5 V− 50.30 10 A−− × 55.0 10 × Ω
1.0 V− 50.20 10 A−− × 55.0 10 × Ω
0.50 V− 50.10 10 A−− × 55.0 10 × Ω
0.40 V+ 0.010 A + 40 Ω
0.50 V+ 0.020 A + 25 Ω
0.55 V+ 0.040 A + 14 Ω 0.70 V+ 0.072 A + 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 95
17.63 The power the beam delivers to the target is ( ) ( )( )6 34.0 10 V 25 10 A 1.0 10 WV I −= ∆ = × × = ×P 5
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 ( ) (m t c T )= ∆ ∆ ∆P as
( ) ( )( )
51.0 10 J s0.48 kg s
4 186 J kg C 50 Cmt c T
×∆= = =
∆ ∆ ⋅° °√
17.64 The volume of the material is
3
6 33 6 3
50.0 g 1 m6.36 10 m
7.86 g cm 10 cmmass
Vdensity
− = = = ×
L= ⋅
Since V A , the cross-sectional area of the wire is A V L=
(a) From 2L L L
RA V L Vρ ρ ρ
== = , the length of the wire is given by
( )( )6 3
-8
1.5 6.36 10 m9.3 m
11 10 mR V
Lρ
−Ω ×⋅= = =
× Ω ⋅
(b) The cross-sectional area of the wire is 2
4d V
AL
π= = . Thus, the diameter is
( )
( )
6 34
4 6.36 10 m49.3 10 m 0.93 mm
9.3 mV
dLπ π
−−
×= = = × =
17.65 (a) The cross-sectional area of the copper in the hollow tube is ( ) ( ) ( )( )3 40.080 m 2.0 10 m 1.6 10 mA circumference thickness − −= ⋅ = × = × 2
Thus, the resistance of this tube is
( )( )8
54 2
1.7 10 m 0.24 m2.6 10
1.6 10 mL
RAρ
−−
−
× Ω ⋅= = = × Ω
×
96 CHAPTER 17
(b) The mass may be written as ( ) ( )m density Volume density A L= ⋅ = ⋅ ⋅
From R L Aρ= , the cross-sectional area is A L Rρ= , so the expression for the mass becomes
( ) ( ) ( )( )223
1.7 1 500 m8 920 kg m 76 kg
Lm density
Rρ ⋅
= ⋅ = =Ω
810 m
4.5
−× Ω⋅
17.66 (a) At temperature T, the resistance is L
RAρ
= , where ( )0 01 T Tρ ρ α = + − ,
( )0 01L L T Tα′ = + − , and ( )0 0A A T T ( )2
01 1 2A T Tα α′ ′ 0 ≈ + − = + −
Thus,
( ) ( )
( )( )
( )( )0 01 1 T T R
T T
α α ⋅ + − = = + −
0 0
0 0
1 1
1 2
T T
T T
α
α α
′ ′ + − ⋅ + ′ ′ + −
00 0
0 1 2
T T T TLR
A
αρ + − −
(b) ( )( )
( )
8
0 00 2-3
0
1.70 10 m 2.00 m1.082
0.100 10
LA
ρ
π
−× Ω ⋅=
×
(
R = =
Then
Ω
)0 01R R T Tα = + −
( )
gives
( )( )310 C 80.0 C 1.420 R − × ° ° = 1.082 1 3.90= Ω +
( )
Ω
The more complex formula gives
( )( )
( )( )
6
6
1.420 1 17 10 C 80.0 C1.418
1 2 17 10 C 80.0 CR
−
−
Ω ⋅ + × ° ° = = + × ° °
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 97
17.67 Note that all potential differences in this solution have a value of 120 VV∆ = . 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
( )2
11
where 535 Wi
VR
∆= =P
P
2t i cR R R= +
Equation (1)
If each of the two conductors in the extension cord has resistance R , the total resistance in the path of the current (outside of the power source) is Equation (2) so the current which will exist is
c
tI V R= ∆ and the power that is delivered to the cleaner is
( ) ( )2 2 2 4
21 1
it t
V VV VR
R R R
∆ ∆∆ ∆= = = =
P P2
delivered iI RPt
Equation (3)
The resistance of a copper conductor of length l and diameter d is
( )
CuCu Cu 22
44cR
A ddρρ ρππ
= = =ll l
, maxcR
Thus, if is the maximum allowed value of , the minimum acceptable diameter of the conductor is
cR
Cumin
, max
4
c
dRρ
π=
l Equation (4)
(a) If R , then from Equations (2) and (1),
0.900 c = Ω
( ) ( ) ( )2 2
1
120 V.900 1.80 1.80
535 WV∆
Ω = + Ω = + ΩP
( )( )
2 0t iR R= +
and, from Equation (3), the power delivered to the cleaner is
( )
4
22
120 V470 W
V1.80 535 W
535 W
= =
+ Ω
120deliveredP
98 CHAPTER 17
(b) If the minimum acceptable power delivered to the cleaner is , then Equations (2) and (3) give the maximum allowable total resistance as
minP
( ) ( )4 2
, max , maxmin 1 min 1
2t i c
V VR R R
∆ ∆= + = =
P P P P
so
( ) ( ) ( ) ( )2 2 2
, max1 1min 1 min 1 min 1
1 1 12 2 2c i
V V VR R
∆ ∆ ∆= − = −
P PP P P P P P
525 W=( )
2V ∆=
1 −
When , then minP( )( )
2
, max
120 V 1 10.128
35525 W 535cR
2 5 W W= − =
Ω
and, from Equation (4), ( )( )
( )
8
min
4 1.7 10 m 15.0 m mm
0.128 d
π
−× Ω ⋅= =
Ω
532 W=
( )
1.60
When , then minP
( )( )
2120 V, max
532cR
1 10.037 9
2 535 W W 535 W
= − Ω
( )( )( )
=
and, 810 m 15.0 m
2.93 mm0.037 9 π
−× Ω ⋅= =
Ωmin
4d
1.7
