serway1 (1)

27

Current and Resistance

CHAPTER OUTLINE ANSWERS TO QUESTIONS

27.1 Electric Current Q27.1 Voltage is a measure of potential difference, not of
27.2 Resistance current. “Surge” implies a flow—and only charge, in
27.3 A Model for Electrical Conduction coulombs, can flow through a system. It would also be
27.4 Resistance and Temperature correct to say that the victim carried a certain current, in
27.5 Superconductors amperes.
27.6 Electrical Power

Q27.2 Geometry and resistivity. In turn, the resistivity of the
material depends on the temperature.

*Q27.3 (i) We require rL /AA = 3rL /AB. Then AA/AB = 1/3,
*Q27.4 answer (f).

(ii) πrA2/πrB2 = 1/3 gives rA /rB = 1/ 3, answer (e).

Originally, R = ρ . Finally, Rf = ρ( /3) = ρ = R.
A 3A 9A 9

Answer (b).

Q27.5 The conductor does not follow Ohm’s law, and must have a resistivity that is current-dependent,
or more likely temperature-dependent.

Q27.6 The amplitude of atomic vibrations increases with temperature. Atoms can then scatter electrons
more efficiently.

Q27.7 (i) The current density increases, so the drift speed 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 metal, suppose that we could proceed to a limit of zero resistance by 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 many electrons in a conductor (approximately 1028 electrons/m3) 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, 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.
(d) R is 1/4 as large, so current is 4 times larger.
(e) R is 2 times 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.
The ranking is then d > a > c > b > f > e > g.

*Q27.12 RA = π ρLA = π ρ2LB = 1 ρLB = RB
(dA / 2)2 (2dB / 2)2 2 π (dB/2)2 2

PA = IA∆V = (∆V )2/RA = 2(∆V )2/RB = 2PB Answer (e).

*Q27.13 RA = ρA L = 2ρB L = 2RB
A A

PA = IA∆V = (∆V )2/RA = (∆V )2/ 2RB = PB / 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.

*Q27.15 One ampere–hour is (1 C/s)(3 600 s) = 3 600 coulombs. The ampere–hour rating is the quantity
of charge that the battery can lift though its nominal potential difference. Answer (d).

Q27.16 Choose the voltage of the power supply you will use to drive the heater. Next calculate the

required resistance R as ∆V 2 . Knowing the resistivity ρ of the material, choose a combination

P ⎛ R⎞
⎝⎜ ρ ⎟⎠
of wire length and cross-sectional area to make ⎛ ⎞ = . You will have to pay for less
⎝ A⎠

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

Section 27.1 Electric Current

P27.1 I = ∆Q ( )∆Q = I ∆t = 30.0 × 10−6 A (40.0 s) = 1.20 × 10−3 C
∆t

N = Q = 1.60 1.20 × 10−3 C = 7.50 × 1015 electrons
e × 10−19 C electron

Current and Resistance 103

P27.2 The molar mass of silver = 107.9 g mole and the volume V is

P27.3 ( )( )V = (area)(thickness) = 700 × 10−4 m2 0.133 × 10−3 m = 9.31 × 10−6 m3
P27.4 ( )( )The mass of silver deposited is mAg = ρV = 10.5 × 103 kg m3 9.31 × 10−6 m3 = 9.78 × 10−2 kg.
P27.5
P27.6 And the number of silver atoms deposited is
P27.7
( )N = ⎛ 6.02 × 1023 atoms⎞ ⎛ 1 000 g⎞
9.78 × 10−2 kg ⎜⎝ 107.9 g ⎟⎠ ⎝⎜ 1 kg ⎟⎠ = 5.45 × 1023 atoms

I = ∆V = 12.0 V = 6.67 A = 6.67 C s
R 1.80 Ω

( )( )∆t = ∆Q = Ne =
5.45 × 1023 1.60 × 10−19 C 3.64 h
= 1.31 × 104 s =
II 6.67 C s

t

( )∫Q (t ) = Idt = I0τ 1 − e−t τ
0

( )(a) Q (τ ) = I0τ 1 − e−1 = (0.632) I0τ

( )(b) Q (10τ ) = I0τ 1 − e−10 = (0.999 95) I0τ

( )(c) Q(∞) = I0τ 1− e−∞ = I0τ

The period of revolution for the sphere is T = 2π , and the average current represented by this
ω

revolving charge is I = q = qω .
T 2π

q = 4t 3 + 5t + 6

( )A = ⎛ 1.00 m ⎞ 2 × 10−4
2.00 cm2 ⎜⎝ 100 cm ⎠⎟ = 2.00 m2

(a) I (1.00 s) = dq = (12t2 + 5) = 17.0 A
dt t=1.00 s t =1.00 s

(b) J = I = 17.0 A m2 = 85.0 kA m2
A 2.00 × 10−4

I = dq 1 240 s ⎛ 120π t ⎞
dt 0 ⎝⎜ s ⎟⎠
∫ ∫ ∫q = = (100 A) sin
dq = Idt dt

q = −100 C ⎢⎣⎡cos ⎛ π ⎞ − cos 0 ⎦⎥⎤ = +100 C = 0.265 C
120π ⎝⎜ 2 ⎠⎟ 120π

J = I = 8.00 × 10−6 A = 2.55 A m2
A 1.00 × 10−3 m
( )(a) π 2

From J = nevd, we have n= J = 2.55 A m2 = 5.31 × 1010 m−3 .
evd 1.60 × 10−19 C 3.00 × 108
( )( )(b)
ms

( )( )(c)
From I = ∆Q , we have ∆t = ∆Q = NAe = 6.02 × 1023 1.60 × 10−19 C = 1.20 × 1010 s .
∆t II 8.00 × 10−6 A

(This is about 382 years!)

104 Chapter 27

J= I = 5.00 A = 99.5 kA m2
Aπ 4.00 × 10−3
( )*P27.8 (a) m 2

(b) Current is the same and current density is smaller. Then I = 5.00 A ,

J2 = 1 = 1 9.95 × 104 A/m2 = 2.49 × 104 A/m2
4 J1 4

A2 = 4A1 or πr22 = 4πr12 so r2 = 2r1 = 0.800 cm

P27.9 (a) The speed of each deuteron is given by K = 1 mv2
2
( )( ) ( )2.00 × 106 1.60 × 10−19 J = 1 2 × 1.67 × 10−27 kg v2 and
2 v = 1.38 × 107 m s
The time between deuterons passing a stationary point is t in I=q

10.0 × 10−6 C s = 1.60 × 10−19 C t or t
t = 1.60 × 10−14 s

( )( )So the distance between them is vt = 1.38 × 107 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
8.99 × 109 N ⋅ m2 C2 1.60 × 10−19 C = 6.49 × 10−3 V
2.21 × 10−7 m

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, NA, has a mass of 27.0 g. Thus, the mass per atom is

27.0 g = 27.0 g = 4.49 × 10−23 g atom
NA 6.02 × 1023

Thus, n = density of aluminum = 2.70 g cm 3
mass per atom 4.49 × 10−23 g atom

n = 6.02 × 1022 atoms cm3 = 6.02 × 1028 atoms m3

= I = 5.00 A = 1.30 × 10−4 m s
nqA 1.60 × 10−19 C
Therefore, ( )( )( )vd
6.02 × 1028 m−3 4.00 × 10−6 m2

or, vd = 0.130 mm s

Section 27.2 Resistance

P27.11 ∆V = IR
and R = ρ :
A = (0.600 mm )2 ⎛ 1.00 m ⎞ 2 = 6.00 × 10−7 m2
A ⎝⎜ 1 000 mm ⎠⎟

( ( ) )I
∆V = I ρ : = ∆VA = (0.900 V) 6.00 × 10−7 m2
A ρ 5.60 × 10−8 Ω ⋅ m (1.50 m)

I = 6.43 A

Current and Resistance 105

P27.12 I = ∆V = 120 V = 0.500 A = 500 mA
R 240 Ω

P27.13 (a) Given M = ρdV = ρd A where r ≡ mass density,
we obtain: d

Thus, A= M Taking r ≡ resistivity, R = ρr = ρr = ρr ρd 2
ρd r A M ρd M

( )= MR =
( )( )ρrρd
1.00 × 10−3 (0.500) = 1.82 m

1.70 × 10−8 8.92 × 103

(b) V = M , or π r2 = M
ρd ρd

M 1.00 × 10−3 r = 1.40 × 10−4 m
π ρd diameter = 280 µm
π 8.92 × 103 (1.82)
Thus, ( )r = =

The diameter is twice this distance:

P27.14 (a) Suppose the rubber is 10 cm long and 1 mm in diameter.

( ( )( ) )R =
ρ = 4ρ ~ 4 1013 Ω ⋅ m 10−1 m = ~1018 Ω
A π d2 π 10−3 m 2

( ( )() )(b)
R = 4ρ ~ 4 1.7 × 10−8 Ω ⋅ m 10−3 m ~10−7 Ω
π d2 π 2 × 10−2 m 2

(c) I = ∆V 102 V ~10−16 A
R ~ 1018 Ω

102 V ~109 A
I ~ 10−7 Ω

P27.15 J = σ E so σ = J = 6.00 × 10−13 A m2 = 6.00 × 10−15 (Ω ⋅ m)−1

E 100 V m

Section 27.3 A Model for Electrical Conduction

*P27.16 (a) The density of charge carriers n is set by the material and is unaffected .
(b)
(c) The current density is proportional to current according to J = I so it doubles .
(d) A

For larger current density in J = nevd the drift speed vd doubles .

The time between collisions τ = mσ is unchanged as long as σ does not change due to a
nq2

temperature change in the conductor.

106 Chapter 27

P27.17 ρ = m We take the density of conduction electrons from an Example in the chapter text.
nq2τ

( )(( )() )so
τ = m = 1.70 × 10−8 9.11 × 10−31 1.60 × 10−19 2 = 2.47 × 10−14 s
ρnq2 8.46 × 1028

vd = qE τ
m

( ) ( )7.84 × 10−4 =
gives 1.60 × 10−19 E 2.47 × 10−14
Therefore, 9.11 × 10−31

E = 0.180 V m

Section 27.4 Resistance and Temperature

P27.18 R = R0 ⎡⎣1 + α (∆T )⎦⎤ gives ( )140 Ω = (19.0 Ω) ⎣⎡1 + 4.50 × 10−3 °C ∆T ⎦⎤

Solving, ∆T = 1.42 × 103°C = T − 20.0°C
And the final temperature is T = 1.44 × 103°C

( )P27.19 (a) ρ = ρ0 ⎣⎡1 + α (T − T0 )⎤⎦ = 2.82 × 10−8 Ω ⋅ m ⎣⎡1 + 3.90 × 10−3 (30.0°)⎦⎤ = 3.15 × 10−8 Ω ⋅ m

(b) J = E = 0.200 V m m = 6.35 × 106 A m2
ρ 3.15 × 10−8 Ω ⋅

( ) ( )(c) ⎢⎡ π 2⎤
⎛π d2 ⎞ ⎢⎣ 1.00 × 10−4 m ⎥
I = JA = J ⎜ ⎟ = 6.35 × 106 A m2 4 ⎦⎥ = 49.9 mA
⎝4⎠

(d) n = 6.02 × 1023 electrons = 6.02 × 1028 electrons m3

( )⎡⎣26.98 g 2.70 ×106 g m3 ⎦⎤

( ( )( ) )vd
= J = 6.35 × 106 A m2 = 659 µm s
ne 6.02 × 1028 electrons m3 1.60 × 10−19 C

(e) ∆V = E = (0.200 V m)(2.00 m) = 0.400 V

*P27.20 We require 10 Ω = 3.5 × 10−5 Ω ⋅ m 1 + 1.5 × 10−6 Ω ⋅ m 2 and for
π (1.5 × 10−3 m)2 π (1.5 × 10−3 m)2

any ∆T 10 Ω = 3.5 × 10−5 Ω ⋅ m 1 ⎜⎝⎛1 − 0.5 × 10−3 ∆T ⎞ + 1.5 × 10−6 Ω ⋅ m 2 ⎝⎜⎛1 + 0.4 × 10−3 ∆T ⎞
π (1.5 × 10−3 m)2 °C ⎟⎠ π (1.5 × 10−3 m)2 °C ⎟⎠

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 so 10 = 4.951 5 1 + 0.212 21 (29.167 1)

and 1 = 10/11.141 = 0.898 m = 1 2 = 26.2 m

Current and Resistance 107

P27.21 R = R0 [1 + αT ]
P27.22
R − R0 = R0α∆T

( )R − R0 = α∆T = 5.00 × 10−3 25.0 = 0.125
R0

For aluminum, αE = 3.90 × 10−3°C−1 (Table 27.2)

α = 24.0 × 10−6°C−1 (Table 19.1)

R= ρ = ρ0 (1 + αE ∆T ) (1 + α∆T ) = R0 (1 + αE ∆T ) = (1.234 Ω ) ⎝⎜⎛ 1.39 ⎞ = 1.71 Ω
A A (1 + α∆T )2 (1 + α∆T ) 1.002 ⎠⎟
4

Section 27.5 Superconductors
Problem 50 in Chapter 43 can be assigned with this section.

Section 27.6 Electrical Power

P27.23 I = P = 600 W = 5.00 A

∆V 120 V

P27.24 and R = ∆V = 120 V = 24.0 Ω
I 5.00 A

( )P = I ∆V = 500 × 10−6 A 15 × 103 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 × 107 J

Thus, the power supplied by the heater is

P = W = Q = 1.32 × 107 J = 8 820 W

∆ t ∆ t 25 × 60 s

and the resistance is R= (∆V )2 = (220 V)2 = 5.49 Ω .
P
8 820 W

*P27.26 (a) efficiency = mechanical power output = 0.900 = 2.50 hp(746 W/1 hp)
total power input (120 V) I

I = 1 860 J/s = 2 070 J/s = 17.3 A
0.9(120 V) 120 J/C

(b) energy input = Pinput ∆t = (2 070 J/s) 3 (3 600 s) = 2.24 × 107 J

(c) cost = 2.24 × 107 J ⎛ S/ 0.16⎞ ⎛ k J h⎞ = $ 0.995
⎝ 1 kWh ⎠ ⎝ 103 W s 3 600 s⎠

108 Chapter 27

P (∆V )2 R ⎛ ∆V ⎞ 2 140 2
P0 R ⎝⎜ ∆V0 ⎟⎠ 120
P27.27 = (∆V0 )2 = = ⎛ ⎞ = 1.361
⎝ ⎠

∆% = ⎛P − P0 ⎞ (100%) = ⎛P − ⎞ (100%) = (1.361− 1)100% = 36.1%
P0 ⎟⎠ ⎝⎜ P0 1⎟⎠
⎝⎜

P27.28 The battery takes in energy by electric transmission

)P ⎛ 3 600 s⎞
∆t = (∆V ) I (∆t ) = 2.3 J (C 13.5 × 10−3 C s 4.2 h ⎝ 1h ⎠ = 469 J

It puts out energy by electric transmission

)(∆V ) I (∆t ) = 1.6 ⎛ 3 600 s⎞
J (C 18 × 10−3 C s 2.4 h ⎝ 1h ⎠ = 249 J

(a) efficiency = useful output = 249 J = 0.530
total input 469 J

(b) The only place for the missing energy to go is into internal energy:
469 J = 249 J + ∆Eint

∆Eint = 221 J

(c) We imagine toasting the battery over a fire with 221 J of heat input:

Q = mc∆T

∆T = Q = 221 J kg°C = 15.1°C
mc 0.015 kg 975 J

P27.29 P = I (∆V ) = (∆V )2 = 500 W R = (110 V)2 = 24.2 Ω
(500 W)
R

( )so 2
(a) R = ρ RA (24.2 Ω)π 2.50 × 10−4 m
A = ρ = = 3.17 m
1.50 × 10−6 Ω ⋅ m

( )(b) R = R0 [1+α∆T ] = 24.2 Ω⎣⎡1+ 0.400 ×10−3 (1180)⎤⎦ = 35.6 Ω

P = (∆V )2 = (110)2 = 340 W

R 35.6

( ( ) )P27.30
R= ρ = 1.50 × 10−6 Ω ⋅ m 25.0 m
A π 0.200 × 10−3 m 2 = 298 Ω

∆V = IR = (0.500 A)(298 Ω) = 149 V

(a) E = ∆V = 149 V = 5.97 V m
25.0 m

(b) P = (∆V ) I = (149 V)(0.500 A) = 74.6 W

( )(c) R = R0 ⎣⎡1+α (T − T0 )⎦⎤ = 298 Ω⎣⎡1+ 0.400 ×10−3 °C 320°C⎦⎤ = 337 Ω

I = ∆V = (149 V) = 0.443 A
R (337 Ω)

P = (∆V ) I = (149 V)(0.443 A) = 66.1 W

Current and Resistance 109

P27.31 (a) ∆U = q (∆V ) = It (∆V ) = (55.0 A ⋅ h) (12.0 V) ⎛⎝⎜11AC⋅ ⎞⎠⎟ ⎜⎛⎝1 1J ⎞⎠⎟ ⎛⎜⎝1 W⋅ s ⎠⎞⎟
V⋅ 1J
s C

= 660 W ⋅ h = 0.660 kWh

(b) Cost = 0.660 kWh ⎛ $0.060 0 ⎞ = 3.96¢
⎜⎝ 1 kWh ⎟⎠

*P27.32 (a) The resistance of 1 m of 12-gauge copper wire is

( )R = ρ
( )A
= π ρ 2)2 = 4ρ = 4 1.7 × 10−8 Ω ⋅ m 1 m = 5.14 × 10−3 Ω
π d2 π 0.205 3 × 10−2 m
(d 2

The rate of internal energy production is P = I ∆V = I 2R = (20 A)2 5.14 × 10−3 Ω = 2.05 W .

(b) PAl = I2R = I 2 4ρAl
π d2

PAl = ρAl PAl = 2.82 × 10−8 Ω ⋅ m 2.05 W = 3.41 W
P ρCu Cu 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

P∆t = 11 J s (100 h)⎝⎛ 3 600 s⎞ = 3.96 × 106 J
1h ⎠

cost = 3.96 × 106 J⎝⎛⎜ $k0W.0h8⎞⎠⎟ ⎛ k⎞ ⎛ W ⋅s⎞ ⎛h ⎞ = $0.088
⎝⎜ 1 000⎟⎠ ⎝ J⎠ ⎜⎝ 3 600 s ⎠⎟

For the incandescent bulb,

P∆t = 40 W (100 h)⎛⎝ 3 600 s⎞ = 1.44 × 107 J
1h ⎠

cost = 1.44 × 107 J ⎛ $0.08 ⎞ = $0.32
⎜⎝ 3.6 × 106 J ⎠⎟

saving = $0.32 − $0.088 = $0.232

P27.34 The total clock power is

( )270 × 106 ⎛ 2.50 Js ⎞ ⎛ 3 600 s⎞ = 2.43 × 1012 Jh
clocks ⎝ clock ⎠ ⎝ 1h ⎠

From e = Wout , the power input to the generating plants must be:
Qin

Qin = Wout ∆ t = 2.43 × 1012 J h = 9.72 × 1012 J h
∆t e 0.250

and the rate of coal consumption is

( )Rate = ⎛ 1.00 kg coal ⎞
9.72 × 1012 J h ⎝ 33.0 × 106 J ⎠ = 2.95 × 105 kg coal h = 295 metric ton h

110 Chapter 27

P27.35 P = I (∆V ) = (1.70 A)(110 V) = 187 W

Energy used in a 24-hour day = (0.187 kW)(24.0 h) = 4.49 kWh.

Therefore daily cost = 4.49 kWh ⎛ $0.060 0 ⎟⎠⎞ = $0.269 = 26.9¢ .
⎜⎝ kWh

P27.36 P = I ∆V = (2.00 A)(120 V) = 240 W

∆Eint = (0.500 kg)(4 186 J kg ⋅°C)(77.0°C) = 161 kJ

∆t = ∆Eint = 1.61 × 105 J = 672 s
240 W
P

P27.37 At operating temperature,

(a) P = I ∆V = (1.53 A)(120 V) = 184 W

(b) Use the change in resistance to find the final operating temperature of the toaster.

R = R0 (1 + α∆T ) ( )120 120 ⎡⎣1 ∆T ⎦⎤
1.80
1.53
= + 0.400 × 10−3

∆T = 441°C T = 20.0°C + 441°C = 461°C

P27.38 You pay the electric company for energy transferred in the amount E = P ∆t.

(a) P ∆t = 40 W(2 weeks)⎝⎛ 1 7d ⎞ ⎛ 86 400 s⎞ ⎛ 1 1J ⎞ = 48.4 MJ
week ⎠ ⎝ 1d ⎠ ⎝ W ⋅s⎠

P ∆t = 40 W(2 weeks)⎝⎛ 1 7d ⎞ ⎛ 24 h⎞ ⎛ 1 k⎞ = 13.4 kWh
week ⎠ ⎝ 1d ⎠ ⎝⎜ 000 ⎠⎟

P ∆t = 40 W(2 weeks)⎝⎛ 1 7d ⎞ ⎛ 24 h⎞ ⎛ 1 k⎞ ⎛ 0.12 $ ⎞⎠⎟ = $1.61
week ⎠ ⎝ 1d ⎠ ⎜⎝ 000 ⎠⎟ ⎜⎝ kWh

(b) P ∆t = 970 W (3 min)⎝⎛ 1h ⎞ ⎛ k⎞ ⎛ 0k.1W2h$ ⎟⎠⎞ = $0.005 82 = 0.582¢
60 min⎠ ⎜⎝ 1 000 ⎠⎟ ⎜⎝

(c) P ∆t = 5 200 W (40 min)⎝⎛ 1h ⎞ ⎛ 1 k ⎞ ⎛ 0k.1W2h$ ⎠⎞⎟ = $0.416
60 min⎠ ⎜⎝ 000 ⎟⎠ ⎝⎜

P27.39 Consider a 400-W blow dryer used for ten minutes daily for a year. The energy transferred to the
dryer is

P ∆t = (400 J s)(600 s d)(365 d) ≈ 9 × 107 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 so P = I ∆V = (∆V )2
(b) R
R
R = (∆V )2 = (120 V)2 = 576 Ω and
P 25.0 W R = (∆V )2 = (120 V)2 = 144 Ω
P 100 W

I = P = 25.0 W = 0.208 A = Q = 1.00 C
∆V 120 V ∆t ∆t

∆t = 1.00 C = 4.80 s
0.208 A

The charge itself is the same. It comes out at a location that is at lower potential.

(c) P = 25.0 W = ∆U = 1.00 J ∆t = 1.00 J = 0.040 0 s
∆t ∆t 25.0 W

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 × 106 J

The electric company sells energy .

Cost = 64.8 × 106 J ⎛ $0.070 0⎞ ⎛ k⎞ ⎛ W ⋅s⎞ ⎛h ⎞ = $1.26
⎜⎝ kWh ⎠⎟ ⎜⎝ 1 000⎟⎠ ⎜⎝ J ⎟⎠ ⎜⎝ 3 600 s ⎠⎟

Cost per joule = $0.070 0 ⎛ kWh ⎞ = $1.94 × 10−8 J
kWh ⎝⎜ 3.60 × 106 J ⎟⎠

*P27.41 The heater should put out constant power

( )P = Q = mc Tf − Ti = (0.250 kg)(4 186 J)(100°C − 20°C) ⎛ 1 min⎞ = 349 J s
∆t ∆t
kg ⋅°C(4 min) ⎝ 60 s ⎠

Then its resistance should be described by R = (∆V )2 = (120 J C)2 = 41.3 Ω

P = (∆V ) I = (∆V )(∆V ) P 349 J s

R

Its resistivity at 100°C is given by

( ) ( ) ( )ρ = ρ0 ⎡⎣1 + α T − T0 ⎤⎦ = 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 =ρ =ρ 4
A π r2 π d2

( )41.3 Ω = 1.55 × 10−6 Ω ⋅ m 4
π d2

= 2.09 × 10+7 m ( )or d 2 = 4.77 × 10−8 m
d2

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 cm3, we want and d less

π d2

( )than those described by
4 = 0.5 × 10−6 m3. Substituting d 2 = 4.77 × 10−8 m gives

( )π 4.77 × 10−8 m 2 = 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 cm3.

112 Chapter 27

P27.42 The original stored energy is Ui = 1 Q ∆Vi = 1 Q2 .
2 2 C

(a) When the switch is closed, charge Q distributes itself over the plates of C and 3C in

parallel, presenting equivalent capacitance 4C. Then the final potential difference is

∆Vf = Q for both.
4C

(b) The smaller capacitor then carries charge C ∆V f = QC = Q . The larger capacitor
4C 4

carries charge 3C Q = 3Q .
4C 4

( )(c) 1C 2 1 ⎛ Q ⎞2 = Q2
The smaller capacitor stores final energy 2 ∆Vf = 2 C ⎝⎜ 4C ⎟⎠ 32C . The larger

capacitor possesses energy 1 ⎛ Q ⎞2 = 3Q2 .
2 3C ⎝⎜ 4C ⎠⎟ 32C

(d) The total final energy is Q2 + 3Q2 = Q2 . The loss of potential energy is the energy
32C 32C 8C
Q2 Q2 3Q2
appearing as internal energy in the resistor: 2C = 8C + ∆ Eint ∆ Eint = 8C

P27.43 We begin with the differential equation α = 1 dρ
ρ dT

(a) Separating variables, ∫ ∫ρ dρ = T αdT
ρρ0
T0

ρ( )⎛ ⎞ = α ρ = ρ0eα(T−T0 ) .
ρ0 ⎠⎟
ln ⎝⎜
T − T0 and

(b) From the series expansion ex ≈ 1+ x, (x << 1), we have

( )ρ ≈ ρ0 ⎡⎣1 + α T − T0 ⎤⎦ .

P27.44 We find the drift velocity from I = nqvd A = nqvdπ r2

= I = 1 000 A = 2.35 × 10−4
nqπ r2 8.46 × 1028 m−3 1.60 × 10−19
( ) ( )vd 2 m s
C π 10−2 m

v= x t = x = 200 × 103 m s = 8.50 × 108 s = 27.0 yr
t v 2.35 × 10−4 m

( )*P27.45 From ρ = RA = ∆V A we compute (m) R (Ω) ρ (Ω ⋅ m)
I 0.540 10.4 1.41 × 10−6
1.028 21.1 1.50 × 10−6
1.543 31.8 1.50 × 10−6

ρ = 1.47 × 10−6 Ω ⋅ 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 113

P27.46 2 wires → = 100 m

)(R = 0.108 Ω 100 m = 0.036 0 Ω
300 m

(a) (∆V ) = (∆V ) − IR = 120 − (110)(0.036 0) = 116 V
home line

(b) P = I (∆V ) = (110 A)(116 V) = 12.8 kW

(c) Pwires = I 2R = (110 A)2 (0.036 0 Ω) = 436 W

)(E = − dV ˆi = −
*P27.47 (a) 0 − 4.00 V 8.00ˆi V m
(b) =
(c) )(dx 0.500 − 0 m

( ) )(R = ρ = 4.00 × 10−8 Ω ⋅ m 0.500 m = 0.637 Ω
( )A π 1.00 × 10−4 m 2

I = ∆V = 4.00 V = 6.28 A
R 0.637 Ω

= I = 6.28 A = 2.00 × 108 A m2 = 200 MA m2
A 1.00 × 10−4
( )(d) J π 2 The field and the current

m

are both in the x direction.

( )( )(e) ρJ = 4.00 × 10−8 Ω ⋅ m 2.00 × 108 A m2 = 8.00 V m = E

*P27.48 (a) E = − dV (x) ˆi = V ˆi
dx L

(b) R= ρ = 4ρL
A π d2

(c) I = ∆V = Vπ d 2
R 4ρ L

(d) J = I = V The field and the current are both in the x direction.
A ρL

(e) ρ J = V = E
L

114 Chapter 27

P27.49 (a) P = I ∆V

so I = P = 8.00 × 103 W = 667 A

∆V 12.0 V

(b) ∆t = ∆U = 2.00 × 107 J = 2.50 × 103 s
8.00 × 103 W
P

and ∆ x = v∆t = (20.0 m s)(2.50 × 103 s) = 50.0 km

*P27.50 (a) We begin with R = ρ = ρ0 ⎡⎣1+ α (T − T0 )⎤⎦ 0 ⎣⎡1+ α′ (T − T0 )⎦⎤ ,
which reduces to A A0 ⎡⎣1+ 2α′ (T − T0 )⎦⎤

(b) For copper: R= R0 ⎣⎡1+ α (T − T0 )⎤⎦ ⎣⎡1+ α ′(T − T0 )⎦⎤
⎣⎡1+ 2α ′(T − T0 )⎤⎦

ρ0 = 1.70 × 10−8 Ω ⋅ m, α = 3.90 × 10−3°C−1 , and

α′ = 17.0 × 10−6°C−1

( ( ) )R0
= ρ0 0 = 1.70 × 10−8 (2.00)
A0
π 0.100 × 10−3 2 = 1.08 Ω

The simple formula for R gives:

( )R = (1.08 Ω)⎣⎡1+ 3.90 ×10−3°C−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 115

P27.51 Let a be the temperature coefficient at 20.0°C, and α ′ be the temperature coefficient at 0°C.

( )Then ρ = ρ0 ⎡⎣1+ α (T − 20.0°C)⎦⎤, and ρ = ρ′ ⎡⎣1 + α ′ T − 0°C ⎤⎦ must both give the correct

resistivity at any temperature T. That is, we must have:

ρ0 [1+ α (T − 20.0°C)] = ρ′[1+ α ′(T − 0°C)] (1)

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 = ρ0 [1− α (20.0°C)][1+ α ′(20.0°C)]

Therefore 1 + α ′ ( 20.0°C) = 1 − α ( 1

20.0°C)

which simplifies to α ′ = [1 − α 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
Copper 4.1 × 10−3 °C
Gold 4.2 × 10−3 °C
Aluminum 3.6 × 10−3 °C
Tungsten 4.2 × 10−3 °C
Iron 4.9 × 10−3 °C
Platinum 5.6 × 10−3 °C
Lead 4.25 × 10−3 °C
Nichrome 4.2 × 10−3 °C
Carbon 0.4 × 10−3 °C
Germanium −0.5 × 10−3 °C
Silicon −24 × 10−3 °C
−30 × 10−3 °C

116 Chapter 27

P27.52 (a) A thin cylindrical shell of radius r, thickness dr, and length L contributes resistance

dR = ρd = ρdr L = ⎛ ρ ⎞ dr
A ⎜⎝ 2π L ⎟⎠ r
(2π r)

The resistance of the whole annulus is the series summation of the contributions of the thin
shells:

∫R = ρ rb dr = ρ L ln ⎛ rb ⎞
2π ⎝⎜ ra ⎠⎟
2π L ra r

(b) In this equation ∆V = ρ ln ⎛ rb ⎞
⎜ ⎟
I 2π L ⎝ ra ⎠

we solve for ρ = 2π L∆V )

I ln(rb ra

*P27.53 The original resistance is Ri = rLi/Ai.

The new length is L = Li + d L = Li(1 + d ).

Constancy of volume implies AL = AiLi so A = Ai Li = Ai Li = Ai
L Li (1+ δ) (1+ δ)

The new resistance is R= ρL = ρLi (1 + δ ) = Ri (1 + δ )2 = Ri (1 + 2δ + δ 2 ).
A Ai / (1 + δ )

The result is exact if the assumptions are precisely true. Our derivation contains no approxima-
tion steps where delta is assumed to be small.

P27.54 Each speaker receives 60.0 W of power. Using P = I 2R, we then have
I = P = 60.0 W = 3.87 A

R 4.00 Ω

The system is not adequately protected since the fuse should be set to melt at 3.87 A, or less .

P27.55 (a) ∆V = −E ⋅ or dV = −E ⋅ dx
∆V = −IR = −E ⋅

I = dq = E ⋅ = A E ⋅ = A E = −σ A dV = σ A dV
dt R ρ ρ dx dx

(b) Current flows in the direction of decreasing voltage. Energy flows by heat in the direction
of decreasing temperature.

P27.56 From the geometry of the longitudinal section of the resistor shown in
the figure, we see that

(b − r) = (b − a)

yh

From this, the radius at a distance y from the base is r = (a − b) y + b FIG. P27.56

h

For a disk-shaped element of volume dR = ρdy : R = ρ h ⎡⎣(a − dy h) + b ⎦⎤2
πr2 π
∫ b)(y

0

Using the integral formula ∫ du = − 1 , R= ρ h
π ab
(au + b)2 a (au + b)

Current and Resistance 117

ρdx = ρdx y2 − y1
A wy L
P27.57 ∫ ∫R = where y = y1 + x

∫R = ρ L dx = ρL ln ⎢⎣⎡y1 + y2 − y1 x⎤⎥⎦ L
L 0
w0 y1 + [(y2 − y1 ) L] x w (y2 − y1 )
FIG. P27.57

R= w ( ρL y1 ) ln ⎛ y2 ⎞
y2 − ⎜ y1 ⎟
⎝ ⎠

*P27.58 A spherical layer within the shell, with radius r and thickness dr, has resistance

dR = ρdr
4πr2

The whole resistance is the absolute value of the quantity

b rb ρdr =ρ r −1 rb −ρ ⎛ 1 1⎞ ρ ⎛1 1⎞
ra 4π r2 4π −1 ra 4π ⎝⎜ ra rb ⎟⎠ 4π ⎝⎜ ra rb ⎠⎟
dR =

a
∫ ∫R = = − + = −

*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 resis-
tive object.

P27.60 (a) The resistance of the dielectric block is R = ρ = d .
A σA

The capacitance of the capacitor is C = κ∈0 A .
d

Then RC = d κ ∈0 A = κ ∈0 is a characteristic of the material only.
σA d σ

(b) R = κ ∈0 = ρ κ ∈0 = 75 ×1016 Ω ⋅ m (3.78)8.85 ×10−12 C2 = 1.79 ×1015 Ω
σC C 14 ×10−9
F N⋅m2

P27.61 (a) Think of the device as two capacitors in parallel. The one on the left has κ1 = 1 ,
A1 = ⎝⎛⎜ 2 + x⎞⎠⎟ . The equivalent capacitance is

κ1 ∈0 A1 + κ 2 ∈0 A2 = ∈0 ⎛ 2 + x⎞⎠ + κ ∈0 ⎛ 2 − x⎞⎠ = ∈0 ( + 2x +κ − 2κ x)
d dd ⎝ d ⎝
2d

(b) The charge on the capacitor is Q = C∆V

Q = ∈0 ∆V ( + 2x +κ − 2κ x)

2d
The current is

I = dQ = dQ dx = ∈0 ∆V (0 + 2 + 0 − 2κ ) v = − ∈0 ∆V v (κ − 1)
dt dx dt 2d d

The negative value indicates that the current drains charge from the capacitor. Positive

current is clockwise ∈0 ∆V v (κ −1) .

d

118 Chapter 27

P27.62 I = I0 ⎡ ⎛ e∆V ⎞ ⎤ and R = ∆V
⎢exp ⎜ kBT ⎟ − 1⎥ I
⎣ ⎝ ⎠
⎦

with I0 = 1.00 × 10−9 A, e = 1.60 × 10−19 C, and kB = 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 5.38
0.480 0.429 1.12
0.520 2.25 0.232
0.560 11.8 0.047 6
0.600 61.6 0.009 7

(ii) For T = 300 K: FIG. P27.62(i)
FIG. P27.62(ii)
∆V (V) I (A) R (Ω) FIG. P27.62(iii)

0.400 0.005 77.3
0.440 0.024 18.1
0.480 0.114
0.520 0.534 4.22
0.560 2.51 0.973
0.600 11.8 0.223
0.051

(iii) For T = 320 K:

∆V (V) I (A) R (Ω)

0.400 0.002 0 203

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

Current and Resistance 119

P27.63 The volume of the gram of gold is given by ρ = m
V

m 10−3 kg
ρ × 103 kg
( )V
= = 19.3 m3 = 5.18 × 10−8 m3 = A 2.40 × 103 m

A = 2.16 × 10−11 m2

( )R = ρ
A
2.44 × 10−8 Ω ⋅ m 2.4 × 103 m = 2.71 × 106 Ω
= 2.16 × 10−11 m2

P27.64 The resistance of one wire is ⎛ 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 P = (∆V ) I = (50.0 × 103 V)(1 000 A) = 50.0 MW .

P27.65 R = R0 ⎣⎡1+ α (T − T0 )⎤⎦ so T = T0 + 1 ⎡R − ⎤ = T0 + 1 ⎡ I0 − 1⎥⎦⎤
α ⎢ 1⎥ α ⎢⎣ I
⎣ R0 ⎦

In this case, I = I0 , so T = T0 + 1 (9) = 20° + 9 °C = 2 020°C
10 α 0.004 50

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 (b) Current is the same, current density is smaller. 5.00 A, 24.9 kA/m2,

0.800 cm
P27.10 0.130 mm/s
P27.12 500 mA
P27.14 (a) ~1018 Ω (b) ~10−7 Ω (c) ~100 aA, ~1 GA
P27.16 (a) no change (b) doubles (c) doubles (d) 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

120 Chapter 27

P27.26 (a) 17.3 A (b) 22.4 MJ (c) $0.995

P27.28 (a) 0.530 (b) 221 J (c) 15.1°C

P27.30 (a) 5.97 V/m (b) 74.6 W (c) 66.1 W

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 (b) $0.005 82 (c) $0.416

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 (c) Q2/32C and 3Q2/32C (d) 3Q2/8C

P27.44 8.50 × 108 s = 27.0 yr

P27.46 (a) 116 V (b) 12.8 kW (c) 436 W

P27.48 (a) E = V/L in the x direction (b) R = 4rL/pd2 (c) I = Vp d2/4rL (d) J = V/rL
(e) See the solution.

P27.50 (a) See the solution. (b) 1.418 Ω nearly agrees with 1.420 Ω.

P27.52 (a) R = ρ ln rb (b) ρ = 2π L∆V
P27.54 2π L ra
I ln (rb ra )

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

28

Direct Current Circuits

CHAPTER OUTLINE ANSWERS TO QUESTIONS
28.1 Electromotive Force
28.2 Resistors in Series and Parallel Q28.1 No. If there is one battery in a circuit, the current inside
28.3 Kirchhoff’s Rules it will be from its negative terminal to its positive termi-
28.4 RC Circuits nal. Whenever a battery is delivering energy to a circuit,
28.5 Electrical Meters it will carry current in this direction. On the other hand,
28.6 Household Wiring and Electrical when another source of emf is charging the battery in
question, it will have a current pushed through it from its
Safety positive terminal to its negative terminal.

*Q28.3 *Q28.2 The terminal potential difference is e – Ir where I is the

current in the battery in the direction from its negative
to its positive pole. So the answer to (i) is (d) and the
answer to (ii) is (b). The current might be zero or an
outside agent might push current backward through the
battery from positive to negative terminal.

*Q28.4 Answers (b) and (d), as described by Kirchhoff’s junction rule.

*Q28.5 Answer (a).

Q28.6 The whole wire is very nearly at one uniform potential. There is essentially zero difference in
potential between the bird’s feet. Then negligible current goes through the bird. The resistance
through the bird’s body between its feet is much larger than the resistance through the wire
between the same two points.

*Q28.7 Answer (b). Each headlight’s terminals are connected to the positive and negative terminals of
the battery. Each headlight can operate if the other is burned out.

121

122 Chapter 28

Q28.8 Answer their question with a challenge. If the student is just looking at a diagram, provide the
materials to build the circuit. If you are looking at a circuit where the second bulb really is
fainter, get the student to unscrew them both and interchange them. But check that the student’s
understanding of potential has not been impaired: if you patch past the first bulb to short it out,
the second gets brighter.

*Q28.9 Answer (a). When the breaker trips to off, current does not go through the device.

*Q28.10 (i) For both batteries to be delivering electric energy, currents are in the direction g to a to b, h to
d to c, and so e to f. Points f, g, and h are all at zero potential. Points b, c, and e are at the same
higher voltage, d still higher, and a highest of all. The ranking is a > d > b = c = e > f = g = h.
(ii) The current in ef must be the sum of the other two currents. The ranking is e = f > g = a = b >
h = d = c.

*Q28.11 Closing the switch removes lamp C from the circuit, decreasing the resistance seen by the battery,
and so increasing the current in the battery. (i) Answer (a). (ii) Answer (d). (iii) Answer (a).
(iv) Answer (a). (v) Answer (d). (vi) Answer (a).

*Q28.12 Closing the switch lights lamp C. The action increases the battery current so it decreases the
terminal voltage of the battery. (i) Answer (b). (ii) Answer (a). (iii) Answer (a).
(iv) Answer (b). (v) Answer (a). (vi) Answer (a).

Q28.13 Two runs in series: . Three runs in parallel: . Junction

of one lift and two runs: .

Gustav Robert Kirchhoff, Professor of Physics at Heidelberg and Berlin, was master of the
obvious. A junction rule: The number of skiers coming into any junction must be equal to the
number of skiers leaving. A loop rule: the total change in altitude must be zero for any skier
completing a closed path.

Q28.14 The bulb will light up for a while immediately after the switch is closed. As the capacitor charges,
the bulb gets progressively dimmer. When the capacitor is fully charged the current in the circuit
is zero and the bulb does not glow at all. If the value of RC is small, this whole process might
occupy a very short time interval.

Q28.15 The hospital maintenance worker is right. A hospital room is full of electrical grounds,
including the bed frame. If your grandmother touched the faulty knob and the bed frame at the
same time, she could receive quite a jolt, as there would be a potential difference of 120 V
across her. If the 120 V is DC, the shock could send her into ventricular fibrillation, and the
hospital staff could use the defibrillator you read about in Chapter 26. If the 120 V is AC,
which is most likely, the current could produce external and internal burns along the path of
conduction. Likely no one got a shock from the radio back at home because her bedroom
contained no electrical grounds—no conductors connected to zero volts. Just like the bird
in Question 28.6, granny could touch the “hot” knob without getting a shock so long as there was
no path to ground to supply a potential difference across her. A new appliance in the bedroom or a
flood could make the radio lethal. Repair it or discard it. Enjoy the news from Lake Wobegon on
the new plastic radio.

Q28.16 Both 120-V and 240-V lines can deliver injurious or lethal shocks, but there is a somewhat better
safety factor with the lower voltage. To say it a different way, the insulation on a 120-V line can
be thinner. On the other hand, a 240-V device carries less current to operate a device with the
same power, so the conductor itself can be thinner. Finally, as we will see in Chapter 33, the last
step-down transformer can also be somewhat smaller if it has to go down only to 240 volts from
the high voltage of the main power line.

Direct Current Circuits 123

SOLUTIONS TO PROBLEMS

Section 28.1 Electromotive Force

P28.1 (a) P = (∆V )2

R 20.0 W = (11.6 V)2
becomes
R
so R = 6.73 Ω FIG. P28.1

(b) ∆V = IR 11.6 V = I (6.73 Ω)
so
and I = 1.72 A

so ε = IR + Ir

15.0 V = 11.6 V + (1.72 A)r

r = 1.97 Ω

P28.2 The total resistance is R = 3.00 V = 5.00 Ω.
P28.3 0.600 A

(a) Rlamp = R − rbatteries = 5.00 Ω − 0.408 Ω = 4.59 Ω

(b) Pbatteries = (0.408 Ω) I 2 = 0.081 6 = 8.16%
Ptotal (5.00 Ω) I 2

FIG. P28.2

(a) Here ε = I (R + r) , so I = ε r = (5.00 12.6 V 0 Ω) = 2.48 A.
Ω + 0.080
R+

Then, ∆V = IR = (2.48 A)(5.00 Ω) = 12.4 V .

(b) Let I1 and I2 be the currents flowing through the battery and the FIG. P28.3
headlights, respectively.

Then, εI1 = I2 + 35.0 A , and − I1r − I2r = 0
so ε = (I2 + 35.0 A)(0.080 0 Ω) + I2 (5.00 Ω) = 12.6 V

giving I2 = 1.93 A

Thus, ∆V2 = (1.93 A)(5.00 Ω) = 9.65 V

*P28.4 (a) At maximum power transfer, r = R. Equal powers are delivered to r and R. The efficiency
is 50.0% .

(b) For maximum fractional energy transfer to R, we want zero energy absorbed by r, so we
want r = 0 .

(c) High efficiency. The electric company’s economic interest is to minimize internal energy
production in its power lines, so that it can sell a large fraction of the energy output of its
generators to the customers.

(d) High power transfer. Energy by electric transmission is so cheap compared to the sound
system that she does not spend extra money to buy an efficient amplifier.

124 Chapter 28

Section 28.2 Resistors in Series and Parallel

P28.5 (a) Rp = (1 7.00 Ω) 1 (1 10.0 Ω) = 4.12 Ω
+

Rs = R1 + R2 + R3 = 4.00 + 4.12 + 9.00 = 17.1 Ω FIG. P28.5
FIG. P28.6
(b) ∆V = IR

34.0 V = I (17.1 Ω)

I = 1.99 A for 4.00 Ω, 9.00 Ω resistors

Applying ∆V = IR, (1.99 A)(4.12 Ω) = 8.18 V
8.18 V = I (7.00 Ω)

so I = 1.17 A for 7.00 Ω resistor

8.18 V = I (10.0 Ω)

so I = 0.818 A for 10.0 Ω resistor

*P28.6 (a) The conductors in the cord have resistance. There is a potential
difference across each when current is flowing. The potential
difference applied to the light bulb is less than 120 V, so it will
carry less current than it is designed to, and will operate at
lower power than 75 W.

(b) If the temperature of the bulb does not change much between
the design operating point and the actual operating point, we can
take the resistance of the filament as constant.
For the bulb in use as intended,

I = P = 75.0 W = 0.625 A

∆V 120 V

and R = ∆V = 120 V = 192 Ω
I 0.625 A

Now, presuming the bulb resistance is unchanged,

I = 120 V = 0.620 A
193.6 Ω

Across the bulb is ∆V = IR = 192 Ω(0.620 A) = 119 V

so its power is P = I ∆V = 0.620 A(119 V) = 73.8 W

Direct Current Circuits 125

P28.7 If we turn the given diagram on its side, we find that it is the same as
P28.8 figure (a). The 20.0 Ω and 5.00 Ω resistors are in series, so the first
reduction is shown in (b). In addition, since the 10.0 Ω, 5.00 Ω, and
25.0 Ω resistors are then in parallel, we can solve for their equivalent
resistance as:

Req = ⎛1 1 1 ⎞ = 2.94 Ω
⎜⎝ 10.0 1 25.0 Ω ⎟⎠
Ω + 5.00 Ω +

This is shown in figure (c), which in turn reduces to the circuit shown in
figure (d).
Next, we work backwards through the diagrams applying I = ∆V and

R
∆V = IR alternately to every resistor, real and equivalent. The 12.94 Ω
resistor is connected across 25.0 V, so the current through the battery in
every diagram is

I = ∆V = 25.0 V = 1.93 A
R 12.94 Ω

In figure (c), this 1.93 A goes through the 2.94 Ω equivalent resistor to
give a potential difference of:

∆V = IR = (1.93 A)(2.94 Ω) = 5.68 V

From figure (b), we see that this potential difference is the same across
∆Vab , the 10 Ω resistor, and the 5.00 Ω resistor.

Thus we have first found the answer to part (b), which is ∆Vab = 5.68 V . FIG. P28.7

(a) Since the current through the 20.0 Ω resistor is also the current
through the 25.0 Ω line ab,
I = ∆Vab = 5.68 V = 0.227 A = 227 mA
Rab 25.0 Ω

We assume that the metal wand makes low-resistance contact with the person’s hand and that the

resistance through the person’s body is negligible compared to the resistance Rshoes of the shoe
soles. The equivalent resistance seen by the power supply is 1.00 MΩ + Rshoes. The current

through both resistors is 50.0 V . The voltmeter displays

1.00 MΩ + Rshoes

(a) We solve to obtain ∆V = I (1.00 MΩ) = 50.0 V(1.00 MΩ) = ∆V

1.00 MΩ + Rshoes

50.0 V(1.00 MΩ) = ∆V (1.00 MΩ) + ∆V ( Rshoes )

Rshoes = 1.00 MΩ (50.0 − ∆V )

∆V

(b) With Rshoes → 0, the current through the person’s body is

50.0 V = 50.0 µA The current will never exceed 50 µA.
1.00 MΩ

126 Chapter 28

P28.9 (a) Since all the current in the circuit must pass through the series 100 Ω

resistor, P = I 2R

Pmax = RI 2
max

FIG. P28.9

so Imax = P= 25.0 W = 0.500 A
100 Ω
R

Req = 100 Ω + ⎛ 1 + 1 ⎞ −1 Ω = 150 Ω
⎝ 100 100 ⎠

∆Vmax = Req Imax = 75.0 V

(b) P1 = I ∆V = (0.500 A)(75.0 V) = 37.5 W total power

P1 = 25.0 W
P2 = P3 = RI 2 (100 Ω)(0.250 A)2 = 6.25 W

P28.10 Using 2.00-Ω, 3.00-Ω, and 4.00-Ω resistors,
there are 7 series, 4 parallel, and 6 mixed
combinations:

Series 6.00 Ω Parallel Mixed
2.00 Ω 7.00 Ω 0.923 Ω 1.56 Ω
3.00 Ω 9.00 Ω 1.20 Ω 2.00 Ω
4.00 Ω 1.33 Ω 2.22 Ω
5.00 Ω 1.71 Ω 3.71 Ω
4.33 Ω
5.20 Ω

FIG. P28.10

P28.11 When S is open, R1, R2 , R3 are in series with the battery. Thus:

R1 + R2 + R3 = 6V = 6 kΩ (1)
10−3 A

When S is closed in position 1, the parallel combination of the two R2’s is in series with R1, R3 ,
and the battery. Thus:

R1 + 1 R2 + R3 = 6V A = 5 kΩ (2)
2 1.2 × 10−3

When S is closed in position 2, R1 and R2 are in series with the battery. R3 is shorted. Thus:

R1 + R2 = 6V A = 3 kΩ (3)
2 × 10−3

From (1) and (3): R3 = 3 kΩ.
Subtract (2) from (1): R2 = 2 kΩ.
From (3): R1 = 1 kΩ.
Answers: R1 = 1.00 kΩ, R2 = 2.00 kΩ, R3 = 3.00 kΩ .

Direct Current Circuits 127

P28.12 Denoting the two resistors as x and y,

x + y = 690, and 1 = 1 + 1
150 x y

1 = 1 + 1 = (690 − x) + x
150 x 690 − x x(690 − x)
x2 − 690x + 103 500 = 0

690 ± (690)2 − 414 000

x=
2

x = 470 Ω y = 220 Ω

*P28.13 The resistance between a and b decreases . Closing the switch opens a new path with resistance
of only 20 Ω. The original resistance is R + 1 = R + 50 Ω.

11
+

90 + 10 10 + 90

The final resistance is R + 1 + 1 = R + 18 Ω.

11 11
++

90 10 10 90

We require R + 50 Ω = 2(R + 18 Ω) so R = 14.0 Ω

*P28.14 (a) The resistors 2, 3, and 4 can be combined to a single 2R resistor. This is in series with
resistor 1, with resistance R, so the equivalent resistance of the whole circuit is 3R. In

1
series, potential difference is shared in proportion to the resistance, so resistor 1 gets 3
of the battery voltage and the 2-3-4 parallel combination gets 2 of the battery voltage.

3
This is the potential difference across resistor 4, but resistors 2 and 3 must share this

voltage. In this branch 1 goes to 2 and 2 to 3. The ranking by potential difference
33

is ∆V4 > ∆V3 > ∆V1 > ∆V2 .

(b) Based on the reasoning above the potential differences

are ∆V1 = ε , ∆V2 = 2ε , ∆V3 = 4ε , ∆V4 = 2ε .

3 9 9 3

(c) All the current goes through resistor 1, so it gets the most. The current then splits at the
parallel combination. Resistor 4 gets more than half, because the resistance in that branch
is less than in the other branch. Resistors 2 and 3 have equal currents because they are in

series. The ranking by current is I1 > I4 > I2 = I3 .

(d) Resistor 1 has a current of I. Because the resistance of 2 and 3 in series is twice that of
resistor 4, twice as much current goes through 4 as through 2 and 3. The currents through

the resistors are I1 = I, I2 = I3 = I, I4 = 2I .
3 3

(e) Increasing resistor 3 increases the equivalent resistance of the entire circuit. The current in
the battery, which is the current through resistor 1, decreases. This decreases the potential
difference across resistor 1, increasing the potential difference across the parallel combi-
nation. With a larger potential difference the current through resistor 4 is increased. With

continued on next page

128 Chapter 28

more current through 4, and less in the circuit to start with, the current through resistors
2 and 3 must decrease. To summarize, I4 increases and I1, I2 , and I3 decrease .

(f) If resistor 3 has an infinite resistance it blocks any current from passing through that

branch, and the circuit effectively is just resistor 1 and resistor 4 in series with the

battery. The circuit now has an equivalent resistance of 4R. The current in the

circuit drops to 3 of the original current because the resistance has increased by 4.
4 3

All this current passes through resistors 1 and 4, and none passes through 2 or 3.

Therefore I1 = 3I , I2 = I3 = 0, I4 = 3I .
4 4

P28.15 Rp = ⎛ 1 + 1 ⎞ −1 = 0.750 Ω
⎝ 3.00 1.00 ⎠

Rs = (2.00 + 0.750 + 4.00) Ω = 6.75 Ω

I battery = ∆V = 18.0 V = 2.67 A
Rs 6.75 Ω

P2 = I 2R: P2 = (2.67 A)2 (2.00 Ω)

P2 = 14.2 W in 2.00 Ω

P4 = (2.67 A)2 (4.00 A) = 28.4 W in 4.00 Ω

∆V2 = (2.67 A)(2.00 Ω) = 5.33 V, FIG. P28.15
∆V4 = (2.67 A)(4.00 Ω) = 10.67 V

∆Vp = 18.0 V − ∆V2 − ∆V4 = 2.00 V (= ∆V3 = ∆V1 )

P3 = (∆V3 )2 = (2.00 V)2 = 1.33 W in 3.00 Ω

R3 3.00 Ω

P1 = (∆V1 ) = (2.00 V)2 = 4.00 W in 1.00 Ω

R1 1.00 Ω

Section 28.3 Kirchhoff’s Rules

P28.16 +15.0 − (7.00) I1 − (2.00)(5.00) = 0

5.00 = 7.00 I1 so I1 = 0.714 A

I3 = I1 + I2 = 2.00 A I2 = 1.29 A

0.714 + I2 = 2.00 so ε = 12.6 V FIG. P28.16

+ε − 2.00(1.29) − 5.00(2.00) = 0

Direct Current Circuits 129

P28.17 We name currents I1, I2 , and I3 as shown.
From Kirchhoff’s current rule, I3 = I1 + I2.
Applying Kirchhoff’s voltage rule to the loop containing I2 and I3,

12.0 V − (4.00) I3 − (6.00) I2 − 4.00 V = 0
8.00 = (4.00) I3 + (6.00) I2

Applying Kirchhoff’s voltage rule to the loop containing I1 and I2 , FIG. P28.17

− (6.00) I2 − 4.00 V + (8.00) I1 = 0 (8.00) I1 = 4.00 + (6.00) I2

Solving the above linear system, we proceed to the pair of simultaneous equations:

⎧8 = 4 I1 + 4I2 + 6I2 or ⎧⎨⎩8I2==41I.13+3 10 I 2
⎨⎩8I1 =4 + 6I2 I1 −
0.667

and to the single equation 8 = 4I1 + 13.3I1 − 6.67

I1 = 14.7 V = 0.846 A Then I2 = 1.33(0.846 A) − 0.667
17.3 Ω

and I3 = I1 + I2 give I1 = 846 mA, I2 = 462 mA, I3 = 1.31 A

All currents are in the directions indicated by the arrows in the circuit diagram.

P28.18 The solution figure is shown to the right.

FIG. P28.18

*P28.19 We use the results of Problem 28.17.

(a) By the 4.00-V battery: ∆U = (∆V ) I ∆t = (4.00 V)(−0.462 A)120 s = −222 J
By the 12.0-V battery: (12.0 V)(1.31 A)120 s = 1.88 kJ

(b) By the 8.00-Ω resistor: I 2R∆t = (0.846 A)2 (8.00 Ω)120 s = 687 J
By the 5.00-Ω resistor: (0.462 A)2 (5.00 Ω)120 s = 128 J

By the 1.00-Ω resistor: (0.462 A)2 (1.00 Ω)120 s = 25.6 J

By the 3.00-Ω resistor: (1.31 A)2 (3.00 Ω)120 s = 616 J

By the 1.00-Ω resistor: (1.31 A)2 (1.00 Ω)120 s = 205 J

(c) −222 J + 1.88 kJ = 1.66 kJ from chemical to electrically transmitted. Like

a child counting his lunch money twice, we can count the same energy again,
687 J + 128 J + 25.6 J + 616 J + 205 J = 1.66 kJ, as it is transformed from electrically
transmitted to internal. The net energy transformation is from chemical to internal .

130 Chapter 28

*P28.20 (a) The first equation represents Kirchhoff’s loop I1 I2
theorem. We choose to think of it as describing
a clockwise trip around the left-hand loop in a 220 Ω 5.8 V 370 Ω
circuit; see Figure (a). For the right-hand loop
see Figure (b). The junctions must be between Figure (a)
the 5.8 V and the 370 Ω and between the
370 Ω and the 150 Ω. Then we have I2 I3 3.10 V
Figure (c). This is consistent with the third
equation, 370 Ω 150 Ω

I1 + I3 − I2 = 0 Figure (b)
I2 = I1 + I3
I3
5.8 V

(b) We substitute: I1 220 Ω 150 Ω
I2 370 Ω 3.10 V
−220I1 + 5.8 − 370I1 − 370I3 = 0
+370I1 + 370I3 + 150I3 − 3.1 = 0 Figure (c)

FIG. P28.20

Next

I3 = 5.8 − 590I1
370

370 I1 + 520 (5.8 − 590 I1 ) − 3.1 = 0
370

370I1 + 8.15 − 829I1 − 3.1 = 0

I1 = 5.05 V = 11.0 mA in the 220-Ω resistor and out of
459 Ω the positive pole of the 5.8-V battery

I3 = 5.8 − 590 ( 0.011 0) = −1.87 mA

370

The current is 1.87 mA in the 150-Ω resistor and out of the

negative pole of the 3.1-V battery.

I2 = 11.0 − 1.87 = 9.13 mA in the 370-Ω resistor

*P28.21 Let I represent the current in the ammeter and the top 6-Ω resistor. The bottom 6-Ω resistor has
6

the same potential difference across it, so it carries an equal current. For the top loop we have

6 V – 10 Ω I – 6 Ω I = 0
10 6

For the bottom loop, 4.5 – 5 I5 – 6 I6 = 0.

For the junctions on the left side, taken together, + I10 + I5 – I6 – I6 = 0.

We eliminate I10 = 0.6 – 0.6 I6 and I5 = 0.9 – 1.2 I6 by substitution:
0.6 – 0.6 I6 + 0.9 – 1.2 I6 – 2 I6 = 0 I6 = 1.5/3.8 = 0.395 A

The loop theorem for the little loop containing the voltmeter gives
+ 6 V –∆V – 4.5 V = 0 ∆V = 1.50 V

Direct Current Circuits 131

P28.22 Label the currents in the branches as shown in the first figure. Figure (a)
Reduce the circuit by combining the two parallel resistors as
shown in the second figure.

Apply Kirchhoff’s loop rule to both loops in Figure (b) to obtain:

(2.71R) I1 + (1.71R) I2 = 250
and (1.71R) I1 + (3.71R) I2 = 500

With R = 1 000 Ω, simultaneous solution of these equations
yields:

I1 = 10.0 mA

and I2 = 130.0 mA Figure (b)
From Figure (b), FIG. P28.22
Vc − Va = ( I1 + I2 )(1.71R) = 240 V

Thus, from Figure (a), I4 = Vc − Va = 240 V = 60.0 mA
4R 4 000 Ω

Finally, applying Kirchhoff’s point rule at point a in Figure (a) gives:

I = I4 − I1 = 60.0 mA − 10.0 mA = +50.0 mA
or I = 50.0 mA from point a to point e

P28.23 Name the currents as shown in the figure to the right. Then w + x + z = y.
Loop equations are

−200w − 40.0 + 80.0x = 0 FIG. P28.23
−80.0x + 40.0 + 360 − 20.0y = 0
+360 − 20.0y − 70.0z + 80.0 = 0 ⎧x = 2.50w + 0.500
⎪⎨400 − 100x − 20.0w − 20.0z = 0
Eliminate y by substitution. ⎩⎪440 − 20.0w − 20.0x − 90.0z = 0

Eliminate x. ⎧350 − 270w − 20.0z = 0
Eliminate z = 17.5 − 13.5w to obtain ⎩⎨430 − 70.0w − 90.0z = 0
430 − 70.0w − 1 575 + 1 215w = 0

w = 70.0 = 1.00 A upward in 200 Ω
70.0

Now z = 4.00 A upward in 70.0 Ω

x = 3.00 A upward in 80.0 Ω

y = 8.00 A downward in 20.0 Ω

and for the 200 Ω, ∆V = IR = (1.00 A)(200 Ω) = 200 V

132 Chapter 28

P28.24 Using Kirchhoff’s rules, FIG. P28.24

12.0 − (0.010 0) I1 − (0.060 0) I3 = 0

10.0 + (1.00) I2 − (0.060 0) I3 = 0

and I1 = I2 + I3

12.0 − (0.010 0) I2 − (0.070 0) I3 = 0

10.0 + (1.00) I2 − (0.060 0) I3 = 0

Solving simultaneously,

I2 = 0.283 A downward in the dead battery

and I3 = 171 A downward in the starter

The currents are forward in the live battery and in the starter, relative to normal starting
operation. The current is backward in the dead battery, tending to charge it up.

P28.25 We name the currents I1, I2 , and I3 as shown.

(a) I1 = I2 + I3
Counterclockwise around the top loop,

12.0 V − (2.00 Ω) I3 − (4.00 Ω) I1 = 0

Traversing the bottom loop,

8.00 V − (6.00 Ω) I2 + (2.00 Ω) I3 = 0

1 4 1 FIG. P28.25
2 3 3
I1 = 3.00 − I3 , I2 = + I3 , and I3 = 909 mA

(b) Va − (0.909 A)(2.00 Ω) = Vb

Vb − Va = −1.82 V

P28.26 ∆Vab = (1.00) I1 + (1.00)( I1 − I2 )
∆Vab = (1.00) I1 + (1.00) I2 + (5.00)( I − I1 + I2 )
∆Vab = (3.00)(I − I1 ) + (5.00)(I − I1 + I2 )

Let I = 1.00 A, I1 = x, and I2 = y.

Then, the three equations become: FIG. P28.26

∆Vab = 2.00x − y, or y = 2.00x − ∆Vab

∆Vab = −4.00x + 6.00y + 5.00

and ∆Vab = 8.00 − 8.00x + 5.00y
Substituting the first into the last two gives:

7.00∆Vab = 8.00x + 5.00 and 6.00∆Vab = 2.00x + 8.00

Solving these simultaneously yields ∆Vab = 27 V.
17

Then, Rab = ∆Vab = V27 or Rab = 27 Ω
I 17
17

1.00 A

Direct Current Circuits 133

Section 28.4 RC Circuits

) )( (P28.27 (a) RC = 1.00 × 106 Ω 5.00 × 10−6 F = 5.00 s

(b) Q = Cε = (5.00 × 10−6 C)(30.0 V) = 150 µC

εI (t )= e−t RC = ⎛ 30.0 ⎞ ⎡ −10.0 ⎤
⎝ 1.00 × 106 ⎠ exp ⎢ × 106 5.00 ⎥
( )( )(c) ⎦⎥
R ⎢⎣ 1.00 × 10−6

= 4.06 µA FIG. P28.27

P28.28 The potential difference across the capacitor ( )∆V (t ) = ∆Vmax 1 − e−t RC
Using 1 Farad = 1 s Ω ,
Therefore, 4.00 V = (10.0 V) ⎣⎢⎡1− e−(3.00 s) (⎣⎡R 10.0×10−6 s Ω)⎤⎦ ⎤
Or ⎦⎥
Taking the natural logarithm of both sides,
0.400 = 1.00 − e (− 3.00×105 Ω) R
and
e (− 3.00×105 Ω) R = 0.600

− 3.00 × 105 Ω = ln(0.600)

R

R = − 3.00 × 105 Ω = +5.87 × 105 Ω = 587 kΩ
0.600
ln ( )

P28.29 (a) I (t ) = −I0e−t RC

= Q = 5.10 × 10−6 C = 1.96 A
RC
(1 300 Ω) 2.00 × 10−9
( )I0 F

(t ) = − (1.96 A ) exp ⎡ −9.00 × 10−6 s ⎤ −61.6 mA
⎥=
300 Ω) 2.00 × 10−9 ⎥⎦
)(I ⎢ (1 F
⎣⎢

q (t ) = Qe−t RC (5.10 µC) ⎡ −8.00 × 10−6 s ⎤
⎥=
⎢⎢⎣ (1 300 Ω) 2.00 × 10−9 ⎥⎦
( )(b) = exp F 0.235 µC

(c) The magnitude of the maximum current is I0 = 1.96 A .

P28.30 We are to calculate

e−2t∫ ∫ϱRC dt= − RC ϱ e−2t RC ⎛ − 2dt ⎞ = − RC e−2t RC ϱ = − RC ⎣⎡ e− ϱ − e0 ⎤⎦ = − RC [0 − 1] = + RC
2 0 ⎝ RC ⎠ 2 0 2 2 2
0

134 Chapter 28

P28.31 (a) Call the potential at the left junction VL and at the right VR. After a
“long” time, the capacitor is fully charged.

VL = 8.00 V because of voltage divider:

IL = 10.0 V = 2.00 A
5.00 Ω
FIG. P28.31(a)
VL = 10.0 V − (2.00 A)(1.00 Ω) = 8.00 V FIG. P28.31(b)

Likewise, VR = ⎛ 2.00 Ω ⎞ (10.0 V) = 2.00 V
⎝ 2.00 Ω + 8.00 Ω⎠

or IR = 10.0 V = 1.00 A
10.0 Ω

VR = (10.0 V) − (8.00 Ω)(1.00 A) = 2.00 V

Therefore, ∆V = VL − VR = 8.00 − 2.00 = 6.00 V

(b) Redraw the circuit R = (1 9.00 Ω) 1 (1 6.00 Ω) = 3.60 Ω
and +

RC = 3.60 × 10−6 s

e−t RC = 1
10

so t = RC ln10 = 8.29 µs

) )( (P28.32 (a) τ = RC = 1.50 × 105 Ω 10.0 × 10−6 F = 1.50 s
) )( ((b) τ = 1.00 × 105 Ω 10.0 × 10−6 F = 1.00 s

(c) The battery carries current 10.0 V Ω = 200 µA
The 100 kΩ carries current of magnitude 50.0 × 103
So the switch carries downward current
I = I0e−t RC = ⎛ 10.0 V ⎞ e−t 1.00 s
⎝ 100 × 103 Ω⎠

)(200 µA + 100 µA e−t 1.00 s

Section 28.5 Electrical Meters I g rg = Ig (60.0 Ω)
I − Ig
( )P28.33 ∆V = Igrg = I − Ig Rp , or I − Ig
( ) ( )Rp =

Therefore, to have I = 0.100 A = 100 mA when Ig = 0.500 mA:

Rp = (0.500 mA)(60.0 Ω) = 0.302 Ω

99.5 mA

FIG. P28.33

Direct Current Circuits 135

P28.34 Ammeter: ( )Igr = 0.500 A − Ig (0.220 Ω) (1)
or (2)
Voltmeter: Ig (r + 0.220 Ω) = 0.110 V

2.00 V = Ig (r + 2 500 Ω)

Solve (1) and (2) simultaneously to find:

Ig = 0.756 mA and r = 145 Ω FIG. P28.34

P28.35 Series Resistor → Voltmeter

∆V = IR: 25.0 = 1.50 × 10−3 (Rs + 75.0)

Solving, Rs = 16.6 kΩ

*P28.36 (a) In Figure (a), the 20.000 Ω 20.000 Ω FIG. P28.35
emf sees an 6.0000 V
equivalent VA 20.000 Ω
resistance 180.00 Ω
of 200.00 Ω. (a) 180.00 Ω VA
(b)
I = 6.000 0 V 180.00 Ω
200.00 Ω FIG. P28.36 (c)

= 0.030 000 A

The terminal potential difference is ∆V = IR = (0.030 000 A)(180.00 Ω) = 5.400 0 V

(b) In Figure (b), Req = ⎛1 Ω + 20 1 ⎞ −1 = 178.39 Ω
⎜⎝ 180.00 000 Ω ⎟⎠
The equivalent resistance across
the emf is 178.39 Ω + 0.500 00 Ω + 20.000 Ω = 198.89 Ω
The ammeter reads
I = ε = 6.000 0 V = 0.030 167 A
and the voltmeter reads
R 198.89 Ω
(c) In Figure (c),
∆V = IR = (0.030 167 A)(178.39 Ω) = 5.381 6 V
Therefore, the emf sends current
through ⎛ 1 1 ⎞ −1
⎝⎜ 180.50 000 Ω ⎠⎟
The current through the battery is Ω + 20 = 178.89 Ω
but not all of this goes through the
ammeter. Rtot = 178.89 Ω + 20.000 Ω = 198.89 Ω
The voltmeter reads I = 6.000 0 V = 0.030 168 A

The ammeter measures current 198.89 Ω

∆V = IR = (0.030 168 A)(178.89 Ω) = 5.396 6 V

I = ∆V = 5.396 6 V = 0.029 898 A .
R 180.50 Ω
continued on next page

136 Chapter 28

(d) Both circuits are good enough for some measurements. The connection in Figure (c) gives data
leading to value of resistance that is too high by only about 0.3%. Its value is more accurate than
the value, 0.9% too low, implied by the data from the circuit in part (b).

Section 28.6 Household Wiring and Electrical Safety

P28.37 (a) P = I ∆V: So for the Heater, I = P = 1 500 W = 12.5 A
For the Toaster,
And for the Grill, ∆V 120 V
I = 750 W = 6.25 A

120 V
I = 1 000 W = 8.33 A

120 V

(b) 12.5 + 6.25 + 8.33 = 27.1 A

The current draw is greater than 25.0 amps, so a circuit with this circuit breaker would not
be sufficient.

P28.38 (a) Suppose that the insulation between either of your fingers and the conductor adjacent is a
chunk of rubber with contact area 4 mm2and thickness 1 mm. Its resistance is

( )( )R = ρ ≈
A
1013 Ω ⋅ m 10−3 m ≈ 2 × 1015 Ω
4 × 10−6 m2

The current will be driven by 120 V through total resistance (series)

2 × 1015 Ω + 104 Ω + 2 × 1015 Ω ≈ 5 × 1015 Ω

It is: I = ∆V ~ 120 V ~10−14 A .
R 5 × 1015 Ω
Vh
(b) The resistors form a voltage divider, with the center of your hand at potential 2 , where

Vh is the potential of the “hot” wire. The potential difference between your finger and

( )( )thumb is ∆V = IR ~ 10−14 A 104 Ω ~ 10−10 V. So the points where the rubber meets your

fingers are at potentials of

~Vh + 10−10 V and ~Vh − 10−10 V
22

Additional Problems

*P28.39 Several seconds is many time constants, so the capacitor is fully charged and (d) the current in its
branch is zero .

Center loop: +8 V + 3 Ω I2 – 5 Ω I1 = 0

Right loop: +4 V – 3 Ω I2 – 5 Ω I3 = 0

Top junction: + I + I – I = 0
1 2 3

Now we will eliminate I1 = 1.6 + 0.6I2 and I3 = 0.8 – 0.6I2

by substitution: 1.6 + 0.6I2 + I2 – 0.8 + 0.6I2 = 0 Then I2 = –0.8/2.2 = –0.3636

So (b) the current in 3 Ω is 0.364 A down .

Now (a) I3 = 0.8 – 0.6(–0.364) = 1.02 A down in 4 V and in 5 Ω .

(c) I = 1.6 + 0.6(–0.364) = 1.38 A up in the 8 V battery
1

(e) For the left loop +3 V – Q/6 m F + 8 V = 0 so Q = 6 mF 11 V = 66.0 mC

Direct Current Circuits 137

*P28.40 The current in the battery is 15 V = 1.15 A.

10 Ω + 1

1 Ω + 1 Ω
5 8

The voltage across 5 Ω is 15 V – 10 Ω 1.15 A = 3.53 V.

(a) The current in it is 3.53 V/5 Ω = 0.706 A.

(b) P = 3.53 V 0.706 A = 2.49 W

(c) Only the circuit in Figure P28.40c requires the use of Kirchhoff’s rules for solution. In the
other circuits the 5-Ω and 8-Ω resistors are still in parallel with each other.

(d) The power is lowest in Figure P28.40c. The circuits in Figures P28.40b and P28.40d have
in effect 30-V batteries driving the current.

P28.41 The set of four batteries boosts the electric potential of each bit of charge that goes through them
by 4 × 1.50 V = 6.00 V. The chemical energy they store is

The radio draws current ∆U = q∆V = (240 C)(6.00 J C) = 1 440 J
So, its power is
I = ∆V = 6.00 V = 0.030 0 A
R 200 Ω

P = (∆V ) I = (6.00 V)(0.030 0 A) = 0.180 W = 0.180 J s

Then for the time the energy lasts, ∆t = E = 1 440 J = 8.00 × 103 s
0.180 J s
we have P = E : P

∆t ∆ t = Q = 240 C = 8.00 × 103 s = 2.22 h
I 0.030 0 A
We could also compute this from I = Q :
∆t

*P28.42 I= ε , so P = I2R = ε2R or (R + r )2 = ⎛ ε2 ⎞ R
⎜⎝ ⎟⎠
R+r (R + r)2 P

Let x ≡ ε2 , then (R + r)2 = xR or R2 + (2r − x) R − r2 = 0

P

With r = 1.20 Ω, this becomes R2 + (2.40 − x) R − 1.44 = 0

which has solutions of R = − (2.40 − x) ± (2.40 − x)2 − 5.76

(a) With ε = 9.20 V 2

and P = 12.8 W, x = 6.61:

R = +4.21 ± (4.21)2 − 5.76 = 3.84 Ω or 0.375 Ω . Either external

2
resistance extracts the same power from the battery.

(b) For ε = 9.20 V and P = 21.2 W, x ≡ ε2 = 3.99

P

R = +1.59 ± (1.59)2 − 5.76 = 1.59 ± −3.22

22

The equation for the load resistance yields a complex number, so there is no resistance

that will extract 21.2 W from this battery. The maximum power output occurs when

R=r = 1.20 Ω, and that maximum is Pmax = ε2 = 17.6 W.

4r

138 Chapter 28

P28.43 Using Kirchhoff’s loop rule for the closed loop, +12.0 − 2.00I − 4.00I = 0, so I = 2.00 A

Vb − Va = +4.00 V − (2.00 A)(4.00 Ω) − (0)(10.0 Ω) = −4.00 V

Thus, ∆Vab = 4.00 V and point a is at the higher potential .

I= ε P =series ε ε2
P28.44 (a) Req = 3R I=
3R 3R

I = 3ε ε εP =parallel
(b) Req = (1 R) + 1 + (1 R) = R I= 3 2
3 R R
(1 R)

(c) Nine times more power is converted in the parallel connection.

*P28.45 The charging current is given by 14.7 V – 13.2 V – I 0.85 Ω = 0 I = 1.76 A
The energy delivered by the 14.7 V supply is ∆VIt = 14.7 V 1.76 A 1.80 h (3 600 s/h) = 168 000 J
The energy stored in the battery is 13.2 V 1.76 A 1.8 (3 600 s) = 151 000 J
The same energy is released by the emf of the battery: 13.2 V (I ) 7.3 (3 600 s) = 151 000 J
so the discharge current is I = 0.435 A
The load resistor is given by 13.2 V – (0.435 A) R – (0.435 A) 0.85 Ω = 0
R = (12.8 V)/0.435 A = 29.5 Ω
The energy delivered to the load is ∆VIt = I 2R t = (0.435 A)2 29.5 Ω (7.3) 3 600 s = 147 000 J
The efficiency is 147 000 J/168 000 J = 0.873

P28.46 (a) ε − I (∑ R) − (ε1 + ε2 ) = 0

40.0 V − (4.00 A)[(2.00 + 0.300 + 0.300 + R)Ω] − (6.00 + 6.00) V = 0; so R = 4.40 Ω

(b) Inside the supply, P = I 2R = (4.00 A)2 (2.00 Ω) = 32.0 W

Inside both batteries together, P = I 2R = (4.00 A)2 (0.600 Ω) = 9.60 W

For the limiting resistor, P = (4.00 A)2 (4.40 Ω) = 70.4 W

(c) P = I (ε1 + ε2 ) = (4.00 A)[(6.00 + 6.00) V] = 48.0 W

Direct Current Circuits 139

P28.47 Let the two resistances be x and y. xy
x
Then, Rs = x+y= Ps = 225 W = 9.00 Ω y = 9.00 Ω − x y

I2 (5.00 A)2 FIG. P28.47

and Rp = xy = Pp = 50.0 W = 2.00 Ω
x+y
I2 (5.00 A)2

so x (9.00 Ω − x) = 2.00 Ω x2 − 9.00x + 18.0 = 0
x + (9.00 Ω − x)
(x − 6.00)(x − 3.00) = 0
Factoring the second equation,
x = 6.00 Ω or x = 3.00 Ω
so y = 3.00 Ω or y = 6.00 Ω

Then, y = 9.00 Ω − x gives

There is only one physical answer: The two resistances are 6.00 Ω and 3.00 Ω .

P28.48 Let the two resistances be x and y. xy
x
Then, Rs = x + y = Ps and Rp = xy = Pp . y
x+y
I2 I2 FIG. P28.48

From the first equation, y= Ps − x, and the second

I2

(( ))becomes PsP
x Ps I 2 − x = Pp or x2 − ⎛ Ps ⎞ x + p = 0.
x + Ps I 2 − x ⎝ ⎠ I4
I2 I2

Using the quadratic formula, x = Ps ± P2 − 4PsP p .
s

2I 2

Then, y= Ps −x gives y = Ps ∓ P2 − 4PsP p .
s
I2
2I 2

The two resistances are Ps + P2 − 4PsP p and Ps − P2 − 4PsP p .
s s

2I 2 2I 2

P28.49 (a) ∆V1 = ∆V2 I1R1 = I2 R2

I = I1 + I2 = I1 + I1 R1 = I1 R2 + R1 R1 I1
R2 R2 I

I1 = IR2 I2 = I1 R1 = IR1 = I2 R2 I2
R1 + R2 R2 R1 + R2
FIG. P28.49(a)

(b) The power delivered to the pair is P = I12 R1 + I 2 R2 = I12 R1 +(I − )I1 2 R2 . For minimum
2

power we want to find I1 such that dP = 0.

dI1

dP = 2I1R1 + 2( I − I1 )(−1) R2 =0 I1R1 − IR2 + I1R2 = 0

dI1

I1 = IR2 This is the same condition as that found in part (a).
R1 + R2

140 Chapter 28

*P28.50 (a) When the capacitor is fully charged, no current exists in its branch. The current in the left
resistors is 5 V/83 Ω = 0.060 2 A. The current in the right resistors is 5 V/(2 Ω + R).
Relative to the negative side of the battery, the left capacitor plate is at voltage
80 Ω (0.060 2 A) = 4.82 V. The right plate is at R (5 V)/(2 Ω + R).
The voltage across the capacitor is 4.82 V – R (5 V)/(2 Ω + R).
The charge on the capacitor is
Q = 3 mF [4.82 V – R (5 V)/(2 Ω + R)] = (28.9 Ω – 0.542 R) mC/(2 Ω + R)

(b) With R = 10 Ω, Q = (28.9 – 5.42) mC/(2 + 10) = 1.96 mC

(c) Yes. Q = 0 when 28.9 Ω – 0.542 R = 0 R = 53.3 Ω

(d) The maximum charge occurs for R = 0. It is 28.9/2 = 14.5 m C .

(e) Yes. Taking R = ∞ corresponds to disconnecting a wire to remove the branch containing R.

In this case |Q|= 0.542 R/R = 0.542 m C .

P28.51 Let Rm = measured value, R = actual value,
IR = current through the resistor R
I = current measured by the ammeter

(a) When using circuit (a), IR R = ∆V = 20 000(I − IR ) or

R = 20 000 ⎡I − ⎤
⎢ 1⎥
⎣ I ⎦
R FIG. P28.51

But since I = ∆V and IR = ∆V , we have I=R
Rm R IR Rm

and R = 20 000 (R − Rm ) (1)

When R > Rm , we require Rm

(R − Rm ) ≤ 0.050 0

R

Therefore, Rm ≥ R(1− 0.050 0) and from (1) we find R ≤ 1 050 Ω

(b) When using circuit (b), IR R = ∆V − IR (0.5 Ω)

But since IR = ∆V , Rm = (0.500 + R) (2)
Rm
(Rm − R) ≤ 0.050 0
When Rm > R, we require
R
From (2) we find R ≥ 10.0 Ω

Direct Current Circuits 141

P28.52 The battery supplies energy at a changing rate dE =P = εI = ε ε⎛ e−1 RC ⎞
dt ⎠
⎝ R

Then the total energy put out by the battery is ∫ ∫ εϱ 2 exp ⎛ − t ⎞ dt
⎝ RC ⎠
dE = R

t=0

∫ dE = ε2 ϱ exp ⎛ − t⎞ ⎛ − dt ⎞ = − ε 2C exp ⎛ − t⎞ ϱ = −ε 2C[0 − 1] = ε 2C
⎝ RC ⎠ ⎝ RC ⎠ ⎝ RC ⎠ 0
R (−RC) ∫
0

The power delivered to the resistor is dE =P = ∆VR I = I2R = ε2 exp⎝⎛ − 2t ⎞
dt RC ⎠
R R2

So the total internal energy appearing in the resistor is ∫ dE = ∞ ε2 exp ⎛ − 2t ⎞ dt
⎝ RC ⎠
∫ R

0

ε2 ⎛ RC ⎞ ϱ ⎛ 2t ⎞ ⎛ 2dt ⎞ ε 2C ⎛ 2t ⎞ ϱ ε 2C [0 − 1] ε 2C
⎝ 2⎠ 0 ⎝ RC ⎠ ⎝ RC ⎠ ⎝ RC ⎠ 0
∫ ∫dE = R − exp − − = − 2 exp − = − 2 = 2

The energy finally stored in the capacitor is U = 1 C (∆V )2 = 1 Cε 2. Thus, energy of the circuit

22

is conserved ε 2C = 1 ε 2C + 1 ε 2C and resistor and capacitor share equally in the energy from
22
the battery.

P28.53 (a) ( )q = C∆V 1 − e−t RC
(b)
( )q = ⎢⎡⎣1− e−10.0 ( )( )⎣⎡ 2.00×106 1.00×10−6 ⎦⎤ ⎤
1.00 × 10−6 F )(10.0 V ⎦⎥ = 9.93 µC

I = dq = ⎛ ∆V ⎞ e−t RC
dt ⎝ R ⎠

I = ⎛ 10.0 V ⎞ e−5.00 = 3.37 × 10−8 A = 33.7 nA
⎝ 2.00 × 106 Ω⎠

(c) dU = d ⎛1 q2 ⎞ = ⎛ q⎞ dq = ⎛ q ⎞ I
dt dt ⎝⎜ 2 C ⎠⎟ ⎝C⎠ dt ⎝ C ⎠

⎛ 9.93 × 10−6 C ⎞
⎜⎝ 1.00 × 10−6 C V ⎠⎟
( )dU

dt
= 3.37 × 10−8 A = 3.34 × 10−7 W = 334 nW

ε ( )(d) Pbattery = I = 3.37 × 10−8 A (10.0 V) = 3.37 × 10−7 W = 337 nW

The battery power could also be computed as the sum of the instantaneous powers
delivered to the resistor and to the capacitor:

I 2R + dU/dt = (33.7 × 10−9 A)2 2 × 106 Ω + 334 nW = 337 nW

142 Chapter 28

*P28.54 (a) We find the resistance intrinsic to the vacuum 0.9 Ω = r
cleaner:

P = I ∆V = (∆V )2 120 V = e 26.9 Ω = R

R 0.9 Ω = r

R = (∆V )2 = (120 V)2 = 26.9 Ω
P
535 W

with the inexpensive cord, the equivalent FIG. P28.54
resistance is 0.9 Ω + 26.9 Ω + 0.9 Ω = 28.7 Ω
so the current throughout the circuit is

I = ε = 120 V = 4.18 A

RTot 28.7 Ω

and the cleaner power is

Pcleaner = I (∆V )cleaner = I 2 R = (4.18 A)2 (26.9 Ω) = 470 W

I= ε εand 2R
In symbols, RTot = R + 2r, Pcleaner = I2R =
R + 2r (R + 2r)2

ε(b) ⎛ 2R ⎞1 2 26.9 Ω ⎞1 2
⎝⎜ Pcleaner ⎠⎟ 525 W ⎠
R + 2r = = 120 V ⎛ = 27.2 Ω
⎝

r= 27.2 Ω − 26.9 Ω = 0.128 Ω = ρ = ρ4
2 A π d2

( )d m)⎞1 2
= ⎛ 4ρ ⎞1 2 = ⎛ 4 1.7 × 10−8 Ω ⋅ m (15 = 1.60 mm or more
⎝ πr ⎠ ⎝⎜ π (0.128 Ω) ⎠⎟

(c) Unless the extension cord is a superconductor, it is impossible to attain cleaner power
535 W. To move from 525 W to 532 W will require a lot more copper, as we show here:

εr = ⎛ R ⎞12 − R = 120 V⎛ 26.9 Ω ⎞ 1 2 − 26.9 Ω = 0.037 9 Ω
2 2 ⎝ 532 W ⎠ 2
2 P⎜⎝ cleaner ⎠⎟

( )d m)⎞1 2
= ⎛ 4ρ ⎞1 2 = ⎛4 1.7 × 10−8 Ω ⋅ m (15 = 2.93 mm or more
⎝ πr ⎠ ⎝⎜ ⎟⎠
π (0.037 9 Ω)

Direct Current Circuits 143

P28.55 (a) First determine the resistance of each light bulb: P = (∆V )2

R

R = (∆V )2 = (120 V)2 = 240 Ω
P 60.0 W

We obtain the equivalent resistance Req of the network of light FIG. P28.55
bulbs by identifying series and parallel equivalent resistances:

Req = R1 + (1 1 R3 ) = 240 Ω + 120 Ω = 360 Ω

R2 ) + (1

The total power dissipated in the 360 Ω is P = (∆V )2 = (120 V)2 = 40.0 W

Req 360 Ω

(b) The current through the network is given by P = I 2 Req: I = P = 40.0 W = 1 A

Req 360 Ω 3

The potential difference across R1 is ∆V1 = IR1 = ⎛ 1 A⎠⎞ (240 Ω) = 80.0 V
⎝ 3

The potential difference ∆V23 across the parallel combination of R2 and R3 is

∆V23 = IR23 = ⎛ 1 A⎠⎞ ⎛ (1 240 1 240 ⎞ = 40.0 V
⎝ 3 ⎝⎜
Ω) + (1 Ω)⎟⎠

P28.56 (a) With the switch closed, current exists in a simple series
circuit as shown. The capacitors carry no current.
For R2 we have

P = I 2 R2 I = P = 2.40 V ⋅ A = 18.5 mA

R2 7 000 V A

The potential difference across R1 and C1 is FIG. P28.56(a)

)(∆V = IR1 = 1.85 × 10−2 A (4 000 V A) = 74.1 V

The charge on C1

)(Q = C1∆V = 3.00 × 10−6 C V (74.1 V) = 222 µC

The potential difference across R2 and C2 is

)(∆V = IR2 = 1.85 × 10−2 A (7 000 Ω) = 130 V

The charge on C2

)(Q = C2∆V = 6.00 × 10−6 C V (130 V) = 778 µC

The battery emf is

IReq = I ( R1 + R2 ) = 1.85 × 10−2 A(4 000 + 7 000) V A = 204 V

(b) In equilibrium after the switch has been opened, no current
exists. The potential difference across each resistor is zero.
The full 204 V appears across both capacitors. The new
charge on C2 is

)(Q = C2∆V = 6.00 × 10−6 C V (204 V) = 1 222 µC FIG. P28.56(b)

for a change of 1 222 µC − 778 µC = 444 µC .

144 Chapter 28

*P28.57 (a) The emf of the battery is 9.30 V . Its internal resistance is given by

9.30 V – 3.70 A r = 0 r = 2.51 Ω

(b) Total emf = 20(9.30 V) = 186 V . The maximum current is given by

20(9.30 V) – 20(2.51 Ω) I – 0 I = 0 I = 3.70 A

(c) For the circuit 20(9.30 V) – 20(2.51 Ω) I – 120 Ω I = 0 I = 186 V/170 Ω = 1.09 A

(d) P = I 2 R = (1.09 A)2 120 Ω = 143 W . This is a potentially deadly situation.

(e) The potential difference across his body is 120 Ω (0.005 00 A) = 0.600 V.

This must be the terminal potential difference of the bank of batteries:

186 V – Itot 20(2.51 Ω) = 0.6 V Itot = 185.4 V/50.3 Ω = 3.688 A
For the copper wire we then have 0.6 V = (3.688 A – 0.005 A) R R = 0.163 Ω

(f) For the experimenter’s body, P = I∆V = 0.005 A 0.6 V = 3.00 mW .
(g) For the wire P = I∆V = 3.683 A 0.6 V = 2.21 W .

(h) The power output of the emf depends on the resistance connected to it. A question about
“the rest of the power” is not meaningful when it compares circuits with different currents.
The net emf produces more current in the circuit where the copper wire is used. The net
emf delivers more power when the copper wire is used, 687 W rather than 203 W without
the wire. Nearly all of this power results in extra internal energy in the internal resistance
of the batteries, which rapidly rise to a high temperature. The circuit with the copper wire is
unsafe because the batteries overheat. The circuit without the copper wire is unsafe because
it delivers an electric shock to the experimenter.

*P28.58 The battery current is

(150 + 45 + 14 + 4) mA = 213 mA

(a) The resistor with highest resistance is that

carrying 4 mA. Doubling its resistance

will reduce the current it carries to 2 mA. FIG. P28.58

Then the total current is nearly the same as before. The ratio is 211 = 0.991 .
213
(150 + 45 + 14 + 2) mA = 211 mA,

(b) The resistor with least resistance carries 150 mA. Doubling its resistance changes this

current to 75 mA and changes the total to (75 + 45 + 14 + 4) mA = 138 mA. The ratio

is 138 = 0.648 , representing a much larger reduction (35.2% instead of 0.9%).
213

(c) This problem is precisely analogous. As a battery maintained a potential difference in
parts (a) and (b), a furnace maintains a temperature difference here. Energy flow by heat is
analogous to current and takes place through thermal resistances in parallel. Each resistance
can have its “R-value” increased by adding insulation. Doubling the thermal resistance of
the attic door will produce only a negligible (0.9%) saving in fuel. The ceiling originally

has the smallest thermal resistance. Doubling the thermal resistance of the ceiling will
produce a much larger saving.

Direct Current Circuits 145

P28.59 From the hint, the equivalent resistance of .

That is, RT + RL 1 Req = Req FIG. P28.60(a)
1 +1

RT + RL Req = Req
RL + Req

RT RL + RT Req + RL Req = RL Req + Re2q

Re2q − RT Req − RT RL = 0

Req = RT ± RT2 − 4(1)(−RT RL )

2(1)

Only the + sign is physical:

( )Req 1
= 2 4RT RL + RT2 + RT

For example, if RT = 1 Ω

And RL = 20 Ω, Req = 5 Ω

P28.60 (a) First let us flatten the circuit on a 2-D
plane as shown; then reorganize it to a
format easier to read. Notice that the two
resistors shown in the top horizontal
branch carry the same current as the
resistors in the horizontal branch
second from the top. The center
junctions in these two branches are at
the same potential. The vertical resistor
between these two junctions has no
potential difference across it and
carries no current. This middle resistor
can be removed without affecting the
circuit. The remaining resistors over
the three parallel branches have
equivalent resistance

Req = ⎛ 1 + 1 + 1 ⎞ −1 = 5.00 Ω
⎝ 20 20 10 ⎠

(b) So the current through the battery is
∆V = 12.0 V = 2.40 A
Req 5.00 Ω

146 Chapter 28

P28.61 (a) After steady-state conditions have been reached, there is no DC current through the capacitor.

Thus, for R3: IR3 = 0 (steady-state)

For the other two resistors, the steady-state current is simply determined by the 9.00-V emf
across the 12-kΩ and 15-kΩ resistors in series:

For R1 and R2: εI(R1+R2 ) = = (12.0 9.00 V kΩ) = 333 µA (steady-state)
R1 + R2 kΩ + 15.0

(b) After the transient currents have ceased, the potential
difference across C is the same as the potential

difference across R2 (= IR2 ) because there is no voltage

drop across R3. Therefore, the charge Q on C is

Q = C (∆V )R2 = C (IR2 ) = (10.0 µF)(333 µA)(15.0 kΩ)

= 50.0 µC

FIG. P28.61(b)

(c) When the switch is opened, the branch containing R1
is no longer part of the circuit. The capacitor discharges

through (R2 + R3 ) with a time constant of
(R2 + R3 )C = (15.0 kΩ + 3.00 kΩ)(10.0 µF) = 0.180 s.

The initial current Ii in this discharge circuit is
determined by the initial potential difference across the

capacitor applied to (R2 + R3 ) in series:

(∆V )C (333 µA)(15.0 kΩ) FIG. P28.61(c)

Ii = (R2 + R3 ) = IR2 = (15.0 kΩ + 3.00 kΩ) = 278 µA

(R2 + R3 )

Thus, when the switch is opened, the current through R2 changes instantaneously from
333 µA (downward) to 278 µA (downward) as shown in the graph. Thereafter, it decays

according to

I e−t (R2 +R3 )C
i
( )IR2
= = 278 µA e−t (0.180 s) (for t > 0)

(d) The charge q on the capacitor decays from Qi to Qi according to
5

q = Qi e−t (R2 +R3)C

Qi = Qi e(−t 0.180 s)
5

5 = et 0.180 s

ln 5 = t
180 ms

t = (0.180 s)(ln 5) = 290 ms

Direct Current Circuits 147

P28.62 ε∆V = e−t RC

so ln ⎛ ε ⎞ = ⎛ 1 ⎞ t
⎝ ⎠ ⎝ RC ⎠
∆V

A plot of ln ⎛ ε ⎞ versus t
⎝ ⎠
∆V

should be a straight line with

slope equal to 1 .
RC

Using the given data FIG. P28.62
values:

(a) A least-square fit to this data yields the graph above. t (s) ∆V (V ) ln (ε ∆V )

∑ xi = 282, ∑ xi2 = 1.86 × 104 , 0 6.19 0
4.87 5.55 0.109
∑ xi yi = 244, ∑ yi = 4.03, N =8 11.1 4.93 0.228
19.4 4.34 0.355
Slope = N (∑ xiyi ) − (∑ xi )(∑ yi ) = 0.011 8 30.8 3.72 0.509
N (∑ xi2 ) − (∑ xi )2 46.6 3.09 0.695
67.3 2.47 0.919
Intercept = (∑ xi2 )(∑ yi ) − (∑ xi )(∑ xiyi ) = 0.088 2 102.2 1.83 1.219
N (∑ xi2 ) − (∑ xi )2

The equation of the best fit line is: ln ⎛ ε ⎞ = (0.011 8)t + 0.088 2
⎝ ⎠
∆V

(b) Thus, the time constant is τ = RC = 1 = 1 = 84.7 s
and the capacitance is slope 0.011 8

C = τ = 84.7 s Ω = 8.47 µF
R 10.0 × 106

P28.63 (a) For the first measurement, the equivalent circuit is as R1 b
shown in Figure 1. ac Ry
Ry Rx
Rab = R1 = Ry + Ry = 2Ry c
Figure 1 Rx
so Ry = 1 R1 (1)
2 a R2

For the second measurement, the equivalent circuit is Ry Ry
shown in Figure 2.
Figure 2
Thus, Rac = R2 = 1 Ry + Rx (2)
2

Substitute (1) into (2) to obtain:

R2 = 1 ⎛ 1 R1 ⎞ + Rx , or Rx = R2 − 1 R1
2 ⎝ 2 ⎠ 4

(b) If R1 = 13.0 Ω and R2 = 6.00 Ω, then Rx = 2.75 Ω . FIG. P28.63

The antenna is inadequately grounded since this exceeds the limit of 2.00 Ω

148 Chapter 28

P28.64 Start at the point when the voltage has just reached R1 +
∆V
2 ∆V and the switch has just closed. The voltage Voltage
3 controlled R2
is 2 ∆V and is decaying towards 0 V with a switch
C V ∆Vc
3
time constant R2C

∆VC (t ) = ⎡ 2 ∆V ⎤ e−t R2C
⎢⎣ 3 ⎥⎦

We want to know when ∆VC (t ) will reach 1 ∆V.
3

Therefore, 1 ∆V = ⎡2 ∆V ⎤ e− t R2C
3 ⎣⎢ 3 ⎥⎦

or e−t R2C = 1 FIG. P28.64
2

or t1 = R2C ln 2
After the switch opens, the voltage is 1 ∆V , increasing toward ∆V with time constant

3

(R1 + R2 )C:

∆VC (t) = ∆V − ⎡ 2 ∆V ⎤ e−t (R1+R2 )C
⎢⎣ 3 ⎦⎥

When ∆VC (t) = 2 ∆V
3

2 ∆V = ∆V − 2 ∆Ve−t (R1+R2)C or e = 1−t (R1+R2 )C
33 2

So t2 = ( R1 + R2 )C ln 2 and T = t1 + t2 = ( R1 + 2R2 )C ln 2

P28.65 A certain quantity of energy ∆Eint = P (time) is required to raise the temperature of the

water to100°C. For the power delivered to the heaters we have P = I ∆V = (∆V )2

R

where (∆V ) is a constant. Thus, comparing coils 1 and 2, we have for the energy

(∆V )2 ∆t = (∆V )2 2∆t Then R2 = 2R1.

R1 .
R2

(a) When connected in parallel, the coils present equivalent resistance

Rp = 1 R2 = 1 2 R1 = 2R1 . Now (∆V )2 ∆t = (∆V )2 ∆t p ∆t p = 2∆t
1 R1 +1 1 R1 +1 3 3
R1 2R1 3

(b) For the series connection, Rs = R1 + R2 = R1 + 2R1 = 3R1 and (∆V )2 ∆t = (∆V )2 ∆ts

R1 3R1

∆ts = 3∆t

Direct Current Circuits 149

P28.66 (a) We model the person’s body and street shoes 3 000 V 80 pF 5 000 MΩ
as shown. For the discharge to reach 100 V, 150 pF

q (t ) = Qe−t RC = C∆V (t ) = C∆V0e−t RC

∆V = e−t RC ∆V0 = e+t RC FIG. P28.66(a)
∆V0 ∆V

t = ln ⎛ ∆V0 ⎞
RC ⎝ ∆V ⎠

( )t ⎛ ∆V0 ⎞ ⎛ 3 000 ⎞
= RC ln ⎝ ∆V ⎠ = 5 000 × 106 Ω 230 × 10−12 F ln ⎝ 100 ⎠ = 3.91 s

) )( ((b) t = 1 × 106 V A 230 × 10−12 C V ln 30 = 782 µs

ANSWERS TO EVEN PROBLEMS

P28.2 (a) 4.59 Ω (b) 8.16%

P28.4 (a) 50.0% (b) r = 0 (c) High efficiency. The electric company’s economic interest is to mini-
mize internal energy production in its power lines, so that it can sell a large fraction of the energy
output of its generators to the customers. (d) High power transfer. Energy by electric transmis-
sion is so cheap compared to the sound system that she does not spend extra money to buy an
efficient amplifier.

P28.6 (a) The 120-V potential difference is applied across the series combination of the two conductors in the
extension cord and the light bulb. The potential difference across the light bulb is less than 120 V and
its power is less than 75 W. (b) We assume the bulb has constant resistance—that is, that its
temperature does not change much from the design operating point. See the solution. 73.8 W

P28.8 (a) See the solution. (b) no

P28.10 See the solution.

P28.12 470 Ω and 220 Ω

P28.14 (a) ∆V4 > ∆V3 > ∆V1 > ∆V2 (b) ∆V1 = e/3, ∆V2 = 2e/9, ∆V3 = 4e/9, ∆V4 = 2e/3
P28.16
(c) I1 > I4 > I2 = I3 (d) I1 = I, I2 = I3 = I/3, I4 = 2I/3 (e) Increasing the value of resistor 3
increases the equivalent resistance of the entire circuit. The current in the battery, which

is also the current in resistor 1, therefore decreases. Then the potential difference across

resistor 1 decreases and the potential difference across the parallel combination increases.

Driven by a larger potential difference, the current in resistor 4 increases. This effect makes

the current in resistors 2 and 3 decrease. In summary, I increases while I1, I2, and I decrease.
4 3
(f) I1 = 3I/4, I2 = I3 = 0, I4 = 3I/4

I1 = 714 mA I2 = 1.29 A ε = 12.6 V

P28.18 See the solution.

P28.20 (a) See the solution. (b) The current in the 220-Ω resistor and the 5.80-V battery is 11.0 mA
out of the positive battery pole. The current in the 370-Ω resistor is 9.13 mA. The current in
the 150-Ω resistor and the 3.10-V battery is 1.87 mA out of the negative battery pole.

150 Chapter 28

P28.22 50.0 mA from a to e

P28.24 starter 171 A downward in the diagram; battery 0.283 A downward

P28.26 See the solution.

P28.28 587 kΩ

P28.30 See the solution.

P28.32 (a) 1.50 s (b) 1.00 s (c) 200 µA + (100 )µA e−t 1.00 s

P28.34 145 Ω, 0.756 mA

P28.36 (a) 30.000 mA, 5.400 0 V (b) 30.167 mA, 5.381 6 V (c) 29.898 mA, 5.396 6 V (d) Both
circuits are good enough for some measurements. The circuit in part (c) gives data leading to a
value of resistance that is too high by only about 0.3%. Its value is more accurate than the value,
0.9% too low, implied by the data from the circuit in part (b).

P28.38 (a) ~10−14 A (b) Vh/2 + ~10−10 V and Vh/2 − ~10−10 V, where Vh is the potential of the live wire, ~102 V

P28.40 (a) 0.706 A (b) 2.49 W (c) Only the circuit in Figure P28.40c requires the use of Kirchhoff’s
rules for solution. In the other circuits the 5-Ω and 8-Ω resistors are still in parallel with each
other. (d) The power is lowest in Figure P28.40c. The circuits in Figures P28.40b and P28.40d
have in effect 30-V batteries driving the current.

P28.42 (a) either 3.84 Ω or 0.375 Ω (b) No load resistor can extract more than 17.6 W from this battery.

P28.44 (a) e 2/3R (b) 3e 2/R (c) in the parallel connection

P28.46 (a) 4.40 Ω (b) 32.0 W, 9.60 W, 70.4 W (c) 48.0 W

P28.48 Ps + P2 − 4PsP p and Ps − P2 − 4PsP p
P28.50 s s

2I 2 2I 2

(a) 15.0 µC 160 Ω − 3R (b) 1.96 mC (c) Yes; 53.3 Ω (d) 14.5 mC for R = 0 (e) Yes; it
166 Ω + 83R

corresponds to disconnecting the wire; 0.542 mC

P28.52 See the solution.

P28.54 (a) 470 W (b) 1.60 mm or more (c) 2.93 mm or more

P28.56 (a) 222 µC (b) increase by 444 µC

P28.58 (a) 0.991 (b) 0.648 (c) The energy flows are precisely analogous to the currents in
parts (a) and (b). The ceiling has the smallest R-value of the thermal resistors in parallel, so
increasing its thermal resistance will produce the biggest reduction in the total energy flow.

P28.60 (a) 5.00 Ω (b) 2.40 A

P28.62 (a) ln(e /∆V) = 0.011 8 t + 0.088 2 (b) 84.7 s, 8.47 mF
P28.64 (R1 + 2R2 )C ln 2

P28.66 (a) 3.91 s (b) 0.782 ms