27 Current and Resistance CHAPTER OUTLINE 27.1 27.2 27.3 27.4 27.5 27.6
Electric Current Resistance A Model for Electrical Conduction Resistance and Temperature Temperature Superconductors Electrical Power
ANSWERS TO QUESTIONS Q27.1
Voltage is a measure of potential potentia l difference, not of current. “Surge” implies a flow—and only charge, in coulombs, can flow through a system. It would also be correct to say that the victim carried a certain current, in amperes.
Q27.2
Geometry and resistivity. resistivity. In turn, the resistivity of the material depends on the temperature.
*Q27.3
A A = 3 r L / A B. Then A A / A B = 1/3, (i) We require r L / answer (f ). (ii) πr A2 / πr B2 = 1/3 gives r A / r r B = 1/ 3, answer (e).
*Q27.4 Originally, R =
ρ A
. Finally, Finally, R f =
ρ ( / 3) ρ R = = . 3 A 9A 9
Answer (b). Q27.5
The conductor conductor does not follow follow Ohm’s Ohm’s law, law, and must have have a resistivity that is current-dependent, or more likely temperature-dependent.
Q27.6
The amplitude amplitude of atomic atomic vibrations increases with temperature. temperature. Atoms can then then scatter scatter electrons more efficiently.
Q27.7
(i) The current current density increases, so the the drift drift speed must must increase. Answer (a). (ii) Answer (a).
Q27.8
The resistance of copper increases with temperature, while the resistance of silicon decreases with increasing temperature. The conduction electrons are scattered more by vibrating atoms when copper heats up. Silicon’s charge carrier density increases as temperature increases and more atomic electrons are promoted to become conduction electrons.
*Q27.9
In a normal normal metal, suppose that we could proceed proceed to a limit of zero resistance resistance by lengthening lengthening the average time between collisions. The classical model of conduction then suggests that a constant applied voltage would cause constant acceleration of the free electrons. The drift speed and the current would increase steadily in time. It is not the situation envisioned in the question, but we can actually switch to zero resistance by substituting a superconducting wire for the normal metal. In this case, the drift velocity of electrons is established by vibrations of atoms in the crystal lattice; the maximum current is limited; and it becomes impossible to establish a potential difference across the superconductor.
Q27.10
Because there are so so many many electrons electrons in a conductor (approximately 10 28 electrons/m 3) the average velocity of charges is very slow. When you connect a wire to a potential difference, you establish an electric field everywhere in the wire nearly instantaneously, instantaneously, to make electrons start drifting everywhere all at once. 101
102
Chapter 27
*Q27.11 Action (a) makes the current three times larger.
(b) causes no change in current. (c) corresponds to a current 3 times larger. larger. (d) R is 1/4 as large, so current is 4 times larger. larger. (e) R is 2 times larger, larger, so current is half as large. (f) R increases by a small percentage as current has a small decrease. (g) Current decreases by a large factor. factor. The ranking is then d > a > c > b > f > e > g.
*Q27.12 R A P A
*Q27.13 R A P A
=
ρ L A π (d A / 2 )2
=
ρ 2 L B π ( 2 d B / 2)2
=
1 ρ L B 2 π ( d B / 2) 2
=
R B
= I A ∆V = (∆V ) 2 / RA = 2( ∆V )2 / RB = 2P B =
ρ A L A
=
2 ρ B L A
2 Answer (e).
= 2 R B
= I A ∆V = (∆V ) 2 / RA = ( ∆V )2 / 2 RB = P B / 2
Answer (f ).
*Q27.14 (i) Bulb (a) must have higher resistance so that it will carry less current and have lower power.
(ii) Bulb (b) carries more current. coulombs. The ampere–hour rating rating is the quantity *Q27.15 One ampere–hour is (1 C/s)(3 600 s) = 3 600 coulombs. of charge that the battery can lift though its nominal potential difference. Answer (d). Q27.16
Choose the voltage of the the power supply supply you will use to drive drive the heater. heater. Next calculate the required resistance R as
∆V 2 P
. Knowing the resistivity ρ of the material, choose a combination
of wire length and cross-sectional area to make
= R . ρ A
You will have to pay for less
material if you make both and A smaller, but if you go too far the wire will have too little surface area to radiate away the energy; then the resistor will melt.
SOLUTIONS TO PROBLEMS Electric Current
Section 27.1
P27.1
I =
∆Q ∆t
N =
∆Q = I ∆t = ( 30.0 × 10 −6 A) ( 40.0 s) = 1.20 × 10 −3 C
1. 20 20 × 10 −3 C = = 7.50 × 1015 electrons −19 e 1. 60 60 × 10 C elec electr tron on
Q
Current and Resistance
P27.2
= 107.9 g
The molar mass of silver
mole and the volume V is
V = ( area ) ( thickness ) = ( 700 × 10
The mass of silver deposited is mAg
103
−4
m 2 ) ( 0.133 × 10 −3 m ) = 9.31 × 10 −6 m 3
= ρ V = (10.5 × 10 3 kg
m 3 ) ( 9.31 × 10 −6 m 3 ) = 9.78 × 10 −2 kg.
And the number of silver atoms deposited is
6. 02 × 10 23 atoms 1 000 g = 5. 45 × 10 23 atoms 107.9 g 1 kg
−2
N = ( 9. 78 × 10 kg )
∆V
I =
R
=
∆Q
∆t =
I
12.0 V 1.80 Ω
=
Ne
Cs
( 5.45 × 10 23 ) (1. 60 × 10 −19 C) = 1.31 × 10 4 s =
=
I
= 6.67 A = 6.67
6.67 C s
3.64 h
t
P27.3
Q ( t ) =
∫ Idt = I τ (1 − e
− t τ
0
)
0
−1
) =
(a)
Q (τ ) = I 0 τ (1 − e
(b)
Q (10τ ) = I 0 τ (1 − e
(c)
Q (∞) = I 0τ (1 − e
P27.4
The period of revolution for the sphere is T =
q = 4t
3
) = ( 0.999 95 ) I τ
0
−∞
revolving charge is I = P27.5
−10
( 0.632 ) I0 τ
) =
q
I0 τ
qω
=
T
2π
2π , and the average current represented by this ω
.
+ 5t + 6 2
A = (2.00 cm
2
1.00 m ) 100 cm = 2.00 × 10 −4 m 2
(a)
I (1.00 s ) =
(b)
J =
P27.6
I =
I A
=
dt t =1.00 s
= (12t 2 + 5 ) t =1.00 s =
17.0 A 2. 00 × 10 −4 m 2
=
q=
dt
∫ dq = ∫ Idt = ∫ (100 A) sin
120π t dt s
0
I
(a)
J =
(b)
From J
(c)
From I =
A
=
17.0 A
85.0 kA m 2
1 240 s
dq
q =
P27.7
dq
− 100 C π + 100 C cos − cos 0 = = 2 120π 120π 8. 00 × 10 −6 A
π (1. 00 × 10
−3
m)
= nevd , we have
2
=
n=
0.265 C
2.55 A m 2 J evd
=
2.55 A m 2 = 5.31 × 1010 m −3 . −19 8 (1.60 × 10 C) ( 3.00 × 10 m s )
∆ Q ∆Q N A e ( 6.02 × 10 23 ) (1.60 × 10 −19 C) = = = , we have ∆t = ∆t I I 8.00 × 10 −6 A
(This is about 382 years!)
1. 20 × 1010 s .
104
Chapter 27
*P27.8
(a) (b)
J =
(a)
5.00 A
=
2
−3
=
99.5 kA m 2
π ( 4.00 × 10 m ) Current is the same and current density is smaller. Then I = 5.00 A , 1 1 4 2 4 2 J 2 = J 1 = 9.95 × 10 A/m = 2.49 × 10 A/m 4 4 A2
P27.9
I A
= 4 A1
π r22
or
= 4 π r12
so
= 2 r 1 =
r 2
The speed of each deuteron is given by 1
kg ) v 2 and
The time between deuterons passing a stationary point is t in 10.0 × 10 −6 C s = 1.60 × 10 −19 C t or
1 2 mv 2 7 m s v = 1. 38 × 10 K
( 2.00 × 106 ) (1.60 × 10−19 J ) = 2 ( 2 × 1.67 × 10 −27
0.800 cm
=
I =
q
t −14 s t = 1. 60 × 10
So the distance between them is vt = (1.38 × 10 7 m s ) (1.60 × 10 −14 s ) = 2.21 × 10 −7 m . (b)
One nucleus will put its nearest neighbor at potential V =
keq
=
r
N ⋅ m 2 C2 ) (1.60 × 10 −19 C )
(8.99 × 109
2.21 × 10
−7
m
= 6.49 × 10 −3
V
This is very small compared to the 2 MV accelerating potential, so repulsion within the beam is a small effect. P27.10
We use I = nqAvd n is the number of charge carriers per unit volume, and is identical to the number of atoms per unit volume. We assume a contribution of 1 free electron per atom in the relationship above. For aluminum, which has a molar mass of 27, we know that Avogadro’s number of atoms, N A, has a mass of 27.0 g. Thus, the mass per atom is 27.0 g N A
n =
Thus,
=
27.0 g 6. 02 × 10 23
density of aluminum mass per atom
n = 6.02 × 10 d
=
d
=
Therefore,
v
or,
v
Section 27.2
= 4.49 × 10 −23 g
I nqA
=
22
2.70 g cm 3 4. 49 × 10 −23 g atom
=
atoms cm 3
atom
= 6.02 × 1028 atoms
m3
5.00 A
( 6.02 × 10
28
m
−3
) (1.60 × 10
−19
−6
C) ( 4. 00 × 10 m
0.130 mm s
Resistance
P27.11
∆V = IR
and
R =
ρ A
∆V =
2
1.00 m A = ( 0.600 mm ) = 6.00 × 10 −7 m 2 1 000 mm 2
:
I ρ A
:
I =
∆VA ρ
=
( 0.900 V) ( 6.00 × 10 −7 m 2 ) ( 5.60 × 10 −8 Ω ⋅ m ) (1.50 m )
I = 6.43 A
2
)
= 1.30 × 10 −4
m s
Current and Resistance
P27.12
I =
P27.13
(a)
∆V R
120 V 240 Ω
=
M
we obtain:
A =
V =
M
ρ d
,
Thus,
500 mA
= ρdV = ρ d A where
Given
Thus,
(b)
= 0.500 A =
M
ρ d
=
Taking r r ≡ resistivity,
MR
ρr ρ d
1.00 × 10 −3 ) ( 0.500 ) ( = (1.70 × 10 −8 ) ( 8.92 × 10 3 )
or
π r 2 =
r =
M
π ρ d
r d ≡ mass density,
=
(a)
(b)
(c)
J
I =
= σ E
Section 27.3
*P27.16 (a)
=
A
=
ρ r M ρ d
=
ρr ρ d 2 M
1.82 m
ρ d 1. 00 × 10 −3
π ( 8.92 × 10 3 ) (1.82 )
r = 1.40 × 10
diameter
ρ
=
A
−1 13 4 ρ 4 (10 Ω ⋅ m ) (10 m ) ~ = ~1018 2 −3 π d 2 π (10 m )
−8 −3 4 ρ 4 (1.7 × 10 Ω ⋅ m ) (10 m ) R = ~ ~10 −7 2 2 − 2 π d π ( 2 × 10 m )
I ~
P27.15
ρr
=
−4
m
280 µ m
Suppose the rubber is 10 cm long and 1 mm in diameter. R =
R =
M
The diameter is twice this distance: P27.14
105
Ω
Ω
∆V
10 2 V ~ 18 ~10 −16 A 10 Ω R
10 2 V ~10 9 A −7 10 Ω so
σ =
J E
=
6.00 × 10 −13 A m 2 100 V m
=
6.00 × 10 −15 ( Ω ⋅ m )
−1
A Model for Electrical Conduction The density of charge carriers n is set by the material and is unaffected .
(b)
The current density is proportional to current according to J =
(c)
For larger current density in J
(d)
The time between collisions τ =
= nevd the drift speed vd mσ nq 2
I A
so it doubles .
doubles .
is unchanged as long as σ does not change due to a
temperature change in the conductor.
106
Chapter 27
P27.17
ρ =
m 2
nq τ
We take the density of conduction electrons from an Example in the chapter text.
τ =
so
v
d
9.11 × 10 −31 ) ( = = 2.47 × 10 −14 s −8 −19 2 28 (1.70 × 10 ) ( 8.46 × 10 ) (1.60 × 10 )
m
ρ nq 2
=
qE m
τ −4
1.60 × 10 −19 ) E ( 2.47 × 10 −14 ) ( =
gives
7.84 × 10
Therefore,
E = 0.180 V m
9. 11 × 10 −31
Resistance and Temperature
Section 27.4
P27.18
R = R0 1 + α ( ∆T ) gives
140
Solving,
∆T = 1.42 × 10 3°C = T − 20.0°C 3 T = 1. 44 × 10 °C
And the final temperature is
Ω = (19.0 Ω ) 1 + ( 4.50 × 10 −3
− T 0 ) = ( 2.82 × 10 −8 Ω ⋅ m ) 1 + 3.90 × 10 −3 ( 30.0°) =
P27.19
(a)
ρ = ρ0 1 + α (T
(b)
J =
(c)
π d 2 I = JA = J = (6.35 × 10 6 A 4
(d)
n =
v
E
ρ
=
(e)
26.98 g ( 2.70 × 106 g
any ∆T
10
m 3 )
π 1.00 × 10−4 m 2 ( ) = m2 ) 4
= 6.02 × 10 28 electrons m2 )
( 6.02 × 10 28 electrons m 3 ) (1.60 × 10−19 C)
Ω=
Ω=
3.15 × 10 −8
=
49.9 mA
m3
659 µ m s
V m ) ( 2.00 m ) = 0.400 V
3.5 × 10 −5 Ω ⋅ m 1 π (1.5 × 10 −3 m )2
+
1.5 × 10 −6 Ω ⋅ m 2 π (1.5 × 10 −3 m )2
and for
−6 3.5 × 10 −5 Ω ⋅ m 1 ∆ T −3 ∆T 1.5 × 10 Ω ⋅ m 2 + 1 − 0.5 × 10 1 + 0.4 × 10 −3 −3 − 2 3 2 °C π (1.5 × 10 m ) °C π (1.5 × 10 m )
simplifying gives 10 = 4.951 5 1 + 0.212 21 2 and
0 = – 2.475 7 × 10–3 1 + 8.488 3 × 10–5 2
These conditions are just sufficient to determine 1 and 2. The design goal can be met. We have 2 = 29.167 1
and
1
Ω⋅m
6.35 × 10 6 A m 2
( 6.35 × 10 6 A
∆V = E = ( 0.200
*P27.20 We require 10
=
6.02 × 10 23 electrons
ne
0.200 V m 3.15 × 10 −8 Ω ⋅ m
= J = d
°C) ∆T
so
10 = 4.951 5 1 + 0.212 21 (29.167 1)
= 10/11.141 = 0.898 m = 1
2
= 26.2 m
Current and Resistance
P27.21
R = R0 [1 + α T ] R − R0
= R0α ∆T
R − R0
= α ∆T = ( 5.00 × 10 −3 ) 25.0 =
R0 P27.22
For aluminum,
ρ
R =
A
Section 27.5
α E = 3.90 × 10 −3°C−1
(Table 27.2)
α = 24.0 × 10 −6°C−1
(Table 19.1)
ρ0 (1 + α E ∆T ) (1 + α ∆T )
=
0.125
A (1 + α ∆T )
2
= R0
1.39 (1 + α E ∆T ) = (1.234 Ω ) = 1.71 Ω (1 + α ∆T ) 1.002 4
Superconductors
Problem 50 in Chapter 43 can be assigned with this section.
Electrical Power
Section 27.6 P
P27.23
I =
and R =
P27.24
P
∆V
=
600 W 120 V
∆V I
=
=
5.00 A
120 V 5.00 A
=
24.0
Ω
= I ∆V = 500 × 10 −6 A (15 × 10 3 V ) =
7.50 W
*P27.25 The energy that must be added to the water is
Q = mc∆T = (109 kg ) ( 4 186 J kg°C ) ( 29.0°C ) = 1.32 × 10 J 7
Thus, the power supplied by the heater is P
=
W
∆ t
=
1. 32 × 10 7 J = 8 820 W ∆ t 25 × 60 s Q
=
and the resistance is R =
*P27.26 (a)
efficiency = I =
( ∆V )2 ( 220 V )2 P
=
8 820 W
=
mechanical power output
5.49
Ω
.
= 0.900 = 2.50 hp(746 W/1 hp) (120 V) I
total power input
1 860 J/s = 2 070 J/s = 17.3 A 0.9(120 V) 120 J/C
(b)
energy input = P input ∆t = (2 070 J/s) 3 (3 600 s) = 2. 24 × 10 J
(c)
cost = 2.24 × 107 J
7
S/ 0.16 k 1 kWh 103
J h = $ 0.995 W s 3 600 s
107
108
P27.27
Chapter 27
P P 0
2
( ∆V )2 R ∆V
2
140 = = = = 1.361 2 ( ∆V0 ) R ∆V 0 120
P − P 0 P (100%) = − 1 (100%) = (1.3661 − 1)100% = ∆% = P 0 P 0 P27.28
36.1%
The battery takes in energy by electric transmission
∆ t = ( ∆V ) I ( ∆t ) = 2.3
P
J C (13.5 × 10 −3 C s ) 4.2 h
3600 s = 469 J 1 h
It puts out energy by electric transmission
( ∆V ) I ( ∆t ) = 1.6 J C (18 × 10 −3 C s ) 2.4 h
3600 s 1 h = 249 J
= useful output = 249 J =
(a)
efficiency
(b)
The only place for the missing energy to go is into internal energy: 469 J = 249 J + ∆ E int
∆ E int = (c)
total input
0.530
469 J
221 J
We imagine toasting the battery over a fire with 221 J of heat input: Q = mc∆T
∆T = P27.29
P
Q mc
= I ( ∆V ) = ρ
=
kg°C 975 J
221 J 0.015 kg
( ∆V )2 R
R =
(b)
R = R0 [1 + α ∆T ] = 24.2 P
P27.30
R =
ρ A
=
so
( ∆V )2 (110 )2 R
=
(110 V) R = = 24.2 Ω (500 W)
= 500 W
(a)
15.1°C 2
A
=
35.6
=
=
Ω ) π ( 2.50 × 10 −4 1. 50 × 10 −6 Ω ⋅ m
RA ( 24.2
ρ
=
m)
2
=
3.17 m
Ω 1 + (0.400 × 10−3 ) (1180) = 35.6 Ω 340 W
(1.50 × 10 −6 Ω ⋅ m ) 25.0 m = = 298 Ω 2 π ( 0.200 × 10 −3 m )
∆V = IR = (0.500 A)( 298 Ω) = 149 V ∆V
(a)
E =
(b)
P
(c)
R = R0 1 + α (T − T 0 ) = 298
=
149 V 25.0 m
=
5.97 V m
= ( ∆V ) I = (149 V ) ( 0.500 A ) =
74.6 W
Ω 1 + (0.400 × 10−3 °C) 320°C = 337 Ω
∆V ( 149 V ) = = 0.443 A R ( 337 Ω ) P = ( ∆V ) I = (149 V ) ( 0.443 A ) =
I =
66.1 W
Current and Resistance
P27.31
(a)
(b) *P27.32 (a)
109
⋅ ∆U = q ( ∆V ) = It ( ∆V ) = (55.0 A ⋅ h) (12.0 V) 1 C 1 J 1 W s 1 A ⋅ s 1 V ⋅ C 1 J = 660 W ⋅ h = 0.660 kWh Cost
= 0.660 kWh $0.060 0 = 1 kWh
3.96¢
The resistance of 1 m of 12-gauge copper wire is R =
ρ
=
A
ρ π ( d 2 )
2
4 ρ = 2 π d
=
4 (1.7 × 10 −8
Ω ⋅ m )1 m
π ( 0.205 3 × 10
−2
m)
The rate of internal energy production is P = I ∆V
2
= 5.14 × 10 −3 Ω
= I 2 R = ( 20 A )2 5.14 × 10 −3 Ω =
2.05 W .
2
(b)
P Al P Al P Cu
= I =
2
R=
I 4 ρ Al
π d 2
ρ Al ρ Cu
P Al
=
2.82 × 10 −8 Ω ⋅ m 2.05 W = 3.41 W 1.7 × 10 −8 Ω ⋅ m
Aluminum of the same diameter will get hotter than copper. It would not be as safe. If it is surrounded by thermal insulation, it could get much hotter than a copper wire. P27.33
The energy taken in by electric transmission for the fluorescent lamp is
∆t = 11
P
3 600 s = 3.96 × 106 J 1 h $0.08 k W ⋅ s h = $0.088 J kWh 1000 J 3 6 00 s
J s (100 h )
cost = 3.96 × 10 6
For the incandescent bulb, 3 600 s ∆t = 40 W (100 h ) = 1. 44 × 10 7
P
1h
J
$0.08 = $0.32 3.6 × 106 J
cost = 1. 44 × 10 7 J
saving = $0.32 − $0.088 = $0.232 P27.34
The total clock power is
( 270 × 106 clocks ) 2.50
From e =
W out Qin
J s 3 600 s = 2.43 × 1012 J h clock 1 h
, the power input to the generating plants must be: Qin
∆ t
=
Wout e
∆ t
=
2.43 × 1012 J h 0.250
= 9. 72 × 1012
J h
and the rate of coal consumption is Rate = ( 9.72 × 1012 J h )
1.00 kg coal = 2.95 × 105 33.0 × 106 J
kg coal h = 295 metric ton h
110
P27.35
Chapter 27
P
= I ( ∆V ) = (1.70 A ) (110 V ) = 187 W = ( 0.187 kW ) ( 24.0 h ) = 4.49 kWh.
Energy used in a 24-hour day
$0.060 0 = $0.269 = kWh
Therefore daily cost = 4.49 kWh
P27.36
P
26.9¢ .
= I ∆V = ( 2.00 A ) (120 V ) = 240 W
∆ E int = ( 0.500 kg ) ( 4 186 J ∆E int 1.61 × 10 5 J ∆t = = = P
240 W
kg ⋅ °C ) ( 77.0°C ) = 161 kJ 672 s
P27.37
At operating temperature,
(a)
P
(b)
Use the change in resistance to find the final operating temperature of the toaster.
= I ∆V = (1.53 A ) (120 V ) =
184 W
R = R0 (1 + α ∆T )
120 1.53
∆T = 441°C
T = 20.0°C + 441°C = 461°C
=
120 1 + 0.400 × 10 −3 1.80
(
) ∆T
P27.38
You pay the electric company for energy transferred in the amount E = P ∆ t .
(a)
∆ t = 40 W ( 2 weeks)
7 d 86 400 s 1 J = 48.4 MJ 1 week 1 d 1 W ⋅ s
P
7 d 24 h k = 13.4 kWh 1 week 1 d 1 000
∆ t = 40 W ( 2 weeks)
P
7 d 24 h k 0.12 $ = $1.61 kWh 1 week 1 d 1000
∆ t = 40 W ( 2 weeks)
P
1 h k 0.12 $ = $0.005 82 kWh 60 min 1000
∆ t = 970 W ( 3 min )
= 0.582¢
(b)
P
(c)
P
P27.39
Consider a 400-W blow dryer used for ten minutes daily for a year. The energy transferred to the dryer is
1 h k 0.12 $ = $0.416 kWh 60 min 1000
∆ t = 5 200 W ( 40 min )
∆ t = ( 400
P
J s ) ( 600 s d ) ( 365 d ) ≈ 9 × 10 7 J
1 kWh ≈ 20 kWh 3.6 × 106 J
We suppose that electrically transmitted energy costs on the order of ten cents per kilowatt-hour. Then the cost of using the dryer for a year is on the order of
Cost ≈ ( 20 kWh ) ( $0.10 kWh ) = $2 ~$1
Current and Resistance
111
Additional Problems *P27.40 (a)
I =
∆ V
R =
(b)
(c)
so
R
( ∆V )
2
=
( 120 V )
2
=
576 Ω
25.0 W 25.0 W Q P I = = = 0.208 A = ∆V 120 V ∆t 1.00 C ∆t = = 4.80 s 0.208 A P
= I ∆V =
P
R =
and
=
( ∆V )
( ∆V )2
2
P
=
R
( 120 V )2 100 W
=
144
Ω
1.00 C
∆t
The charge itself is the same. It comes out at a location that is at lower potential. 1.00 J ∆U 1.00 J ∆t = = 0.040 0 s = P = 25.0 W = 25.0 W ∆t ∆t The energy itself is the same. It enters the bulb by electrical transmission and leaves by heat and electromagnetic radiation.
(d)
∆U = P ∆t = ( 25.0
J s ) (86 400 s d ) ( 30.0 d ) = 64.8 × 10 6 J
The electric company sells energy .
$0.070 0 k W ⋅ s h = kWh 1000 J 3 600 s $0.070 0 kWh Cost per joule = = $1. 94 × 10 −8 J 6 kWh 3.60 × 10 J Cost = 64. 8 × 10 6 J
$1.26
*P27.41 The heater should put out constant power P
=
Q
∆t
=
− Ti ) ( 0.250 kg) ( 4 186 J) (100°C − 20°C) 1 min = 60 s = 349 ∆t kg ⋅ °C ( 4 min )
mc (T f
Js
Then its resistance should be described by P
= ( ∆V ) I =
2
( ∆V ) ( ∆V )
R =
R
( ∆V ) P
2
=
(120 J C ) 349 J s
= 41.3 Ω
Its resistivity at 100°C is given by
ρ = ρ0 1 + α ( T
− T 0 ) = (1.50 × 10 −6 Ω ⋅ m ) 1 + 0.4 × 10 −3 (80 ) = 1.55 × 10 −6 Ω ⋅ m
Then for a wire of circular cross section R = ρ
A
= ρ
π r
2
= ρ
(
41.3 Ω = 1.55 × 10 −6 2
d
= 2.09 × 10 +7
4
π d 2
Ω ⋅ m)
m
4
π d 2
or
2
d
= ( 4.77 × 10 −8
m
)
One possible choice is = 0.900 m and d = 2.07 × 10 −4 m. If and d are made too small, the surface area will be inadequate to transfer heat into the water fast enough to prevent overheating of the filament. To make the volume less than 0.5 cm 3 , we want and d less π d 2 −6 −8 3 2 than those described by = 0.5 × 10 m . Substituting d = 4. 77 × 10 m gives 4 π 2 −8 ( 4.77 × 10 m ) = 0.5 × 10 −6 m3 , = 3.65 m and d = 4.18 × 10 −4 m. Thus our answer is: 4 Any diameter d and length related by d 2 = 4. 77 × 10 −8 m would have the right resistance. One possibility is length 0.900 m and diameter 0.207 mm, but such a small wire might overheat rapidly if it were not surrounded by water. The volume can be less than 0.5 cm 3 .
(
(
)
)
112
P27.42
Chapter 27
1 1 Q2 Q∆V i = . 2 2 C When the switch is closed, charge Q distributes itself over the plates of C and 3C in parallel, presenting equivalent capacitance 4 C . Then the final potential difference is
=
The original stored energy is U i (a)
∆V f =
Q
4C
for both.
The smaller capacitor then carries charge C ∆V f
(b)
carries charge 3C
Q
4C
2
Q 1 capacitor possesses energy 3C 2 4C
4
. The larger capacitor
2
( ∆V ) f
2
=
1 Q C 2 4C
Q2
=
32C
. The larger
3Q 2 . 32C
=
3Q 2 Q 2 = . The loss of potential energy is the energy 32C 32C 8C 2 2 3Q 2 Q = Q + ∆ E int ∆ E int = appearing as internal energy in the resistor: 8C 2C 8C
(d)
P27.43
Q
C = 4C
3Q . 4
=
1 The smaller capacitor stores final energy C 2
(c)
Q
=
Q
The total final energy is
2
+
ρ
(a)
∫
Separating variables,
ρ 0
ρ = α (T − T 0 ) ρ 0
ln
1 d ρ ρ dT
α =
We begin with the differential equation
d ρ
ρ
ρ = ρ 0 e
and
T
= ∫ α dT T 0
α ( T − T 0 )
.
From the series expansion e x ≈ 1 + x , ( x << 1) , we have
(b)
ρ ≈ ρ0 1 + α ( T − T 0 ) . P27.44
We find the drift velocity from
v
v
d
=
=
I
=
2
nqπ r
x t
t =
*P27.45 From ρ =
RA
= nqvd A = nqvd π r 2
1000 A 8.46 × 10
x v
I
=
=
28
(1.60 × 10
200 × 10 3 m 2.35 × 10 −4 m s
( ∆V ) A
ρ = 1. 47 × 10 −6
m
−3
I
Ω⋅m
−19
C ) π (10
= 8.50 × 108 s =
we compute
−2
m)
= 2.35 × 10 −4 m
2
s
27.0 yr
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
(m)
. With its uncertainty range from 1.41 to 1. 50, this average value agrees
with the tabulated value of 1.50 × 10 −6
Ω ⋅ m in Table 27.2.
Current and Resistance
P27.46
2 wires R =
→ = 100 m
0.108 Ω (100 m ) = 0.036 0 300 m
Ω
(a)
( ∆V )home = ( ∆V )line − IR = 120 − (110 ) ( 0.036 0 ) =
(b)
P
(c)
P wires
= I ( ∆V ) = (110 A ) (116 V ) =
(b)
dV ˆ i= dx
E=−
R =
(c)
I =
(d)
J =
(e)
ρ J
ρ A
( 0 − 4.00 ) V = ( 0. 500 − 0 ) m
R
8.00 ˆi V m
8
=
4.00 V 0.637 Ω
=
−4
2
π (1.00 × 10 m ) are both in the x direction.
2
= 2.00 × 108 A
= ( 4.00 × 10 −8 Ω ⋅ m ) ( 2.00 × 108 A
*P27.48 (a)
E=−
(b)
R =
(c)
I =
(d)
J =
(e)
ρ J =
0.637
Ω
6.28 A
6.28 A
=
A
436 W
4.00 × 10 − Ω ⋅ m ) ( 0.500 m ) ( = = − π (1. 00 × 10 m ) 4
∆V I
−
116 V
12.8 kW
= I 2 R = (110 A )2 ( 0.036 0 Ω ) =
*P27.47 (a)
113
m2
m2
=
200 MA m 2
The field and the current
) = 8.00 V m = E
dV (x ) ˆ V ˆ i= i dx L
ρ A
∆V R I A
=
π d 2 2
=
V
V π d
4 ρ L V
=
L
4 ρ L
ρ L
=
E
The field and the current are both in the x direction.
114
Chapter 27
P27.49
(a)
P
= I ∆V
so I =
(b)
∆t =
P
∆V
∆U P
=
=
8. 00 × 10 3 W 12.0 V
2. 00 × 10 7 J 8.00 × 103 W
=
667 A
= 2.50 × 10 3
s
and ∆ x = v∆ t = ( 20.0 m s ) ( 2.50 × 10 3 s ) = 50.0 km
*P27.50 (a)
We begin with
R =
which reduces to
R =
For copper:
ρ 0
(b)
ρ A
ρ 0 1 + α (T
− T0 ) 0 1 + α ′ (T − T0 ) , A0 1 + 2α ′ (T − T 0 )
=
R0 1 + α (T
− T0 ) 1 + α ′ (T − T 0 ) 1 + 2α ′ (T − T 0 )
= 1. 70 × 10 −8 Ω ⋅ m, α = 3.90 × 10 −3°C−1 , and
α ′ = 17.0 × 10 −6°C−1 R0
=
ρ 0 0 A0
(1.70 × 10 −8 ) ( 2.00 ) = 2 = 1.08 Ω π ( 0.100 × 10 −3 )
The simple formula for R gives: R = (1.08 Ω) 1 + (3.90 × 10 °C −3
−1
) (100°C − 20.0°C) =
1.420
Ω
while the more complicated formula gives: R =
=
( 1.08 Ω ) 1 + ( 3.90 × 10 −3°C −1 ) ( 80.0°C ) 1 + (17.0 × 10 −6°C −1 ) ( 80.0°C )
1 + 2 (17.0 × 10 −6°C−1 ) ( 80.0°C) 1.418 Ω
The results agree to three digits. The variation of resistance with temperature is typically a much larger effect than thermal expansion in size.
Current and Resistance
P27.51
Let a be the temperature coefficient at 20.0°C, and α ′ be the temperature coefficient at 0°C.
Then ρ = ρ 0 1 + α (T − 20.0°C), and ρ = ρ ′ 1 + α ′ ( T − 0°C ) must both give the correct
115
resistivity at any temperature T. That is, we must have:
ρ0 [1 + α ( T
− 20.0°C) ] = ρ ′ [1 + α ′ ( T − 0°C) ]
Setting T = 0 in equation (1) yields:
ρ ′ = ρ0 1 − α ( 20.0°C) ,
and setting T = 20.0°C in equation (1) gives:
ρ0
= ρ ′ [1 + α ′ ( 20.0°C) ]
Put ρ ′ from the first of these results into the second to obtain:
ρ0 Therefore
1 + α ′ ( 20.0 °C) =
which simplifies to
= ρ0 [1 − α ( 20.0°C) ][1 + α ′ ( 20 .0°C) ]
1 1 − α ( 20.0°C)
α ′ =
α [1 − α ( 20.0°C) ]
From this, the temperature coefficient, based on a reference temperature of 0°C, may be computed for any material. For example, using this, Table 27.2 becomes at 0°C : Material
Temp Coefficients at 0°C
Silver
4.1 × 10 −3 °C
Copper
4.2 × 10 −3 °C
Gold
3.6 × 10−3 °C
Aluminum
4.2 × 10 −3 °C
Tungsten
4.9 × 10 −3 °C
Iron
5.6 × 10−3 °C
Platinum
4.25 × 10 −3 °C
Lead
4.2 × 10 −3 °C
Nichrome
0.4 × 10−3 °C
Carbon Germanium Silicon
−0.5 × 10 −3 ° C −24 × 10−3 °C −30 × 10−3 °C
(1)
116
Chapter 27
P27.52
(a)
A thin cylindrical shell of radius r , thickness dr , and length L contributes resistance dR =
ρd A
ρ dr ( 2π r ) L
=
ρ dr = 2π L r
The resistance of the whole annulus is the series summation of the contributions of the thin shells: R =
(b)
ρ 2π L
r b
dr
∫ r
r ρ ln b 2π L r a
=
r a
In this equation
∆V = ρ
2π L
I
ρ =
we solve for *P27.53
r b r a
ln
2π L ∆V I ln ( rb r a )
The original resistance is Ri = r L / Ai. i The new length is L = Li + d L = Li(1 + d ). Constancy of volume implies AL = A L i i The new resistance is R =
ρL A
=
so A =
Ai Li L
ρ Li (1 + δ ) = Ri (1 + δ ) 2 Ai / (1 + δ )
=
Ai Li Li (1 + δ )
Ai
=
(1 + δ )
= Ri (1 + 2δ + δ 2 ).
The result is exact if the assumptions are precisely true. Our derivation contains no approximation steps where delta is assumed to be small. P27.54
Each speaker receives 60.0 W of power. Using I =
P
R
=
60.0 W 4.00 Ω
(a)
∆V = −E ⋅ or ∆V = − IR = − E ⋅ I =
(b)
P27.56
dq dt
=
E ⋅ R
= I 2 R, we then have
= 3.87 A
The system is not adequately protected P27.55
P
=
A
ρ
dV
since the
fuse should be set to melt at 3.87 A, or lesss .
= −E ⋅ dx
E ⋅ =
A dV dV E = −σ A = σ A A ρ dx dx
Current flows in the direction of decreasing voltage. Energy flows by heat in the direction of decreasing temperature.
From the geometry of the longitudinal section of the resistor shown in the figure, we see that
( b − r )
=
y
r = ( a − b)
From this, the radius at a distance y from the base is For a disk-shaped element of volume dR =
Using the integral formula
∫ (audu+ b)
2
=−
h
ρ dy : π r 2 1 a ( au + b)
(b − a)
R =
,
h y h
+b
FIG. P27.56
ρ dy 2 π 0 ( a − b ) ( y h ) + b
R =
∫
ρ h π ab
Current and Resistance
P27.57
R =
R =
117
∫ ρ Adx = ∫ ρ wydx where y = y + y L− y x 2
1
1
y2 − y1 L ∫ 0 y1 + [( y2 − y1 ) L] x = w ( y2 − y1 ) ln y1 + L x 0
L
ρ w
ρ L
dx
FIG. P27.57
y2 w ( y2 − y1 ) y1 ρ L
R =
ln
*P27.58 A spherical layer within the shell, with radius r and thickness dr , has resistance
dR =
ρ dr 4 π r 2
The whole resistance is the absolute value of the quantity R =
b
r b
a
r a
ρ dr 4π r 2
∫ dR = ∫
r b
ρ r −1 = 4π −1 r
=−
a
ρ 1 − 4π r a
1 ρ 1 1 + = − rb 4π ra r b
*P27.59 Coat the surfaces of entry and exit with material of much higher conductivity than the bulk mate-
rial of the object. The electric potential will be essentially uniform over each of these electrodes. Current will be distributed over the whole area where each electrode is in contact with the resistive object. P27.60
(a)
The resistance of the dielectric block is R = The capacitance of the capacitor is C = Then RC =
κ ∈0 σ C
d κ ∈0 A
σ A
σ
d
= ρκ ∈0 =
∈ = κ 0
ρ A
κ∈0 A d
=
d
σ A
.
.
is a characteristic of the material only.
Ω ⋅ m (3.78) 8.85 × 10 −12 C2 = 14 × 10−9 F N ⋅ m2 75 × 1016
(b)
R =
P27.61
(a)
Think of the device as two capacitors in parallel. The one on the left has κ 1 A1
= + x . 2
κ 1 ∈0 A1 d
(b)
C
Ω = 1,
The equivalent capacitance is
κ 2 ∈0 A2
+
1.79 × 1015
d
=
∈0 κ ∈0 ∈0 + x + − x = ( + 2 x + κ − 2κ x ) d 2 d 2 2d
The charge on the capacitor is Q = C ∆V Q=
∈0 ∆V 2d
( + 2 x + κ − 2κ x )
The current is I =
dQ dt
=
dQ dx dx dt
=
∈0 ∆V 2d
( 0 + 2 + 0 − 2κ ) v = −
∈0 ∆V v d
(κ − 1)
The negative value indicates that the current drains charge from the capacitor. Positive current is clockwise
∈0 ∆V v d
(κ − 1) .
118
Chapter 27
e∆V = I 0 exp − 1 k BT
P27.62
I
with I 0
= 1.00 × 10 −9
and R =
∆V I
A, e = 1.60 × 10 −19 C, and k B = 1.38 × 10 −23 J K
The following includes a partial table of calculated values and a graph for each of the specified temperatures.
(i)
For T = 280 K:
∆V (V)
I ( A)
R ( Ω)
0. 400
0. 015 6 25. 6
0.440
0.081 8
5.3 38
0.480
0.429
1.12
0.520
2.25
0.232
0.560 11.8
0.047 6
0.600
0.009 7
61.6
FIG. P27.62(i)
(ii)
For T = 300 K:
∆V (V)
I ( A)
0. 400
0. 005
77. 3
0. 440
0. 024
18. 1
0.480
0.114
4.22
0. 520
0. 534
0. 973
0.560
2.51
0.223
0.600 11.8
R ( Ω)
0.051
FIG. P27.62(ii)
(iii) For T = 320 K:
∆V ( V)
I ( A)
R ( Ω)
0.400
0.002 0
0.440
0.008 4
52.5
0.480
0.035 7
13.4
0.520
0.152
3.42
0.560
0.648
0.864
0.600
2.76
0.217
203
FIG. P27.62(iii)
Current and Resistance
P27.63
The volume of the gram of gold is given by ρ = V =
m
ρ
=
10 −3 kg 19.3 × 10 3 kg m 3
A = 2.16 × 10 R =
P27.64
ρ A
=
−11
= 5.18 × 10 −8
m V
m 3 = A ( 2.40 × 10 3 m )
m2
2. 44 × 10 −8
Ω ⋅ m ( 2.4 × 103 2.16 × 10 −11 m 2
The resistance of one wire is
m)
=
2.71 × 10 6
Ω
0.500 Ω (100 mi) = 50.0 Ω. mi
The whole wire is at nominal 700 kV away from ground potential, but the potential difference between its two ends is IR = (1 000 A ) ( 50.0
Ω ) = 50.0 kV
Then it radiates as heat power
P27.65
R = R0 1 + α (T
In this case, I =
− T 0 ) I 0
10
,
P
= ( ∆V ) I = ( 50.0 × 10 3 V ) (1 000 A ) =
50.0 MW .
1R 1 I = T 0 + − 1 = T 0 + 0 − 1 α R0 α I 1 9 T = T 0 + ( 9 ) = 20° + =
so
T
so
α
0.004 50 °C
2 020°C
ANSWERS TO EVEN PROBLEMS P27.2
3.64 h
P27.4
qw /2p
P27.6
0.265 C
P27.8
(a) 99.5 kA/m2 0.800 cm
P27.10
0.130 mm/s
P27.12
500 mA
P27.14
(a) ~1018 Ω
P27.16
(a) no change
P27.18
1.44 × 103 °C
P27.20
She can meet the design goal by choosing 1 = 0.898 m and 2 = 26.2 m.
P27.22
1.71 Ω
P27.24
7.50 W
(b) Current is the same, current density is smaller. 5.00 A, 24.9 kA/m2,
(b) ~10−7 Ω (b) doubles
(c) ~100 aA, ~1 GA (c) doubles
(d) no change
119
120
Chapter 27
P27.26
(a) 17.3 A
(b) 22.4 MJ
P27.28
(a) 0.530
P27.30
(a) 5.97 V/m
P27.32
(a) 2.05 W (b) 3.41 W. It would not be as safe. If surrounded by thermal insulation, it would get much hotter than a copper wire.
P27.34
295 metric ton h
P27.36
672 s
P27.38
(a) $1.61
P27.40
(a) 576 Ω and 144 Ω (b) 4.80 s. The charge itself is the same. It is at a location that is lower in potential. (c) 0.040 0 s. The energy itself is the same. It enters the bulb by electric transmission and leaves by heat and electromagnetic radiation. (d) $1.26, energy, 1.94 × 10−8 $/J
P27.42
(a) Q /4C (b) Q /4 and 3Q /4
P27.44
8.50 × 108 s = 27.0 yr
P27.46
(a) 116 V
P27.48
L in the x direction (a) E = V / (e) See the solution.
P27.50
(a) See the solution.
P27.52
(a) R =
P27.54
No. The fuses should pass no more than 3.87 A.
P27.56
See the solution.
P27.58
See the solution.
P27.60
(b) 1.79 PΩ
P27.62
See the solution.
P27.64
50.0 MW
(b) 221 J
(c) $0.995
(c) 15.1°C
(b) 74.6 W
(b) $0.005 82
(b) 12.8 kW
ρ r ln b 2π L r a
(c) 66.1 W
(c) $0.416
(c) Q2 /32C and 3Q2 /32C (d) 3Q2 /8C
(c) 436 W (b) R = 4 r L/ pd / r L 2 (c) I = V p d 2 /4 r L (d) J = V
Ω nearly agrees with 1.420 Ω. 2π L ∆V ρ =
(b) 1.418 (b)
I ln ( rb r a )
