Current and Direct
Current Circuits
CHAPTER OUTLINE
21.1
21.2
21.3
21.4
21.5
21.6
21.7
21.8
21.9
21.10
Electric Current
Resistance and Ohm’s Law
Superconductors
A Structural Model for
Electrical Conduction
Electric Energy and Power
Sources of emf
Resistors in Series and in
Parallel
Kirchhoff’s Rules
RC Circuits
Context ConnectionThe
Atmosphere as a
Conductor
ANSWERS TO QUESTIONS
Q21.1
Consider all of the eastbound lanes on a highway, taken together.
Individual vehicles correspond to bits of charge. The number of
vehicles that pass a certain milepost, divided by the time during
which they go past, corresponds to the value of the current.
Q21.2
Geometry and resistivity. In turn, the resistivity of the material
depends on the temperature.
Q21.3
The radius of wire B is 3 times the radius of wire A, to make its cross–sectional area 3 times larger.
Q21.4
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, and a current steadily
increasing in time.
On the other hand, 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.
Q21.5
The amplitude of atomic vibrations increases with temperature. Atoms can then scatter electrons
more efficiently.
Q21.6
A current will continue to exist in a superconductor without voltage because there is no resistance
loss.
Q21.7
Because there are so many electrons in a conductor (approximately 10 28 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.
Q21.8
The 25 W bulb has a higher resistance. The 100 W bulb carries more current.
Q21.9
One ampere–hour is 3 600 coulombs. The ampere–hour rating is the quantity of charge that the
battery can lift though its nominal potential difference.
577
578
Current and Direct Current Circuits
Q21.10
In series, the current is the same through each resistor. Without knowing individual resistances,
nothing can be determined about potential differences or power.
Q21.11
In parallel, the potential difference is the same across each resistor. Without knowing individual
resistances, nothing can be determined about current or power.
Q21.12
A short circuit can develop when the last bit of insulation frays away between the two conductors in
a lamp cord. Then the two conductors touch each other, opening a low-resistance branch in parallel
with the lamp. The lamp will immediately go out, carrying no current and presenting no danger. A
very large current exists in the power supply, the house wiring, and the rest of the lamp cord up to
the contact point. Before it blows the fuse or pops the circuit breaker, the large current can quickly
raise the temperature in the short-circuit path.
Q21.13
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.
Q21.14
A wire or cable in a transmission line is thick and made of material with very low resistivity. Only
when its length is very large does its resistance become significant. To transmit power over a long
distance it is most efficient to use low current at high voltage, minimizing the I 2 R power loss in the
transmission line. Alternating current, as opposed to the direct current we study first, can be stepped
up in voltage and then down again, with high-efficiency transformers at both ends of the power
line.
Q21.15
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.
Q21.16
Car headlights are in parallel. If they were in series, both would go out when the filament of one
failed. An important safety factor would be lost.
Q21.17
Kirchhoff’s junction rule expresses conservation of electric charge. If the total current into a point
were different from the total current out, then charge would be continuously created or annihilated
at that point.
Kirchhoff’s loop rule expresses conservation of energy. For a single-loop circle with two
resistors, the loop rule reads +ε − IR1 − IR 2 = 0 . This is algebraically equivalent to qε = qIR1 + qIR2 ,
where q = I∆t is the charge passing a point in the loop in the time interval ∆t . The equivalent
equation states that the power supply injects energy into the circuit equal in amount to that which
the resistors degrade into internal energy.
Q21.18
At their normal operating temperatures, from P =
∆V 2
, the bulbs present resistances
R
2
2
2
120 V
120 V
120 V
∆V 2
R=
=
= 240 Ω , and
= 190 Ω , and
= 72 Ω . The nominal 60 W lamp
P
60 W
75 W
200 W
has greatest resistance. When they are connected in series, they all carry the same small current.
Here the highest-resistance bulb glows most brightly and the one with lowest resistance is faintest.
This is just the reverse of their order of intensity if they were connected in parallel, as they are
designed to be.
a
f
a
f
a
f
Chapter 21
579
Q21.19
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.
Q21.20
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 Example 21.10. 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 21.13, 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.
Q21.21
12 V
= 2 A.
6Ω
The potential difference across each lamp is 2 A 2 Ω = 4 V . The power of each lamp is
2 A 4 V = 8 W , totaling 24 W for the circuit. Closing the switch makes the switch and the wires
connected to it a zero-resistance branch. All of the current through A and B will go through the
switch and (b) lamp C goes out, with zero voltage across it. With less total resistance, the (c) current
12 V
= 3 A becomes larger than before and (a) lamps A and B get brighter. (d) The
in the battery
4Ω
voltage across each of A and B is 3 A 2 Ω = 6 V , larger than before. Each converts power
3 A 6 V = 18 W , totaling 36 W, which is (e) an increase.
Suppose
ε = 12 V
and each lamp has R = 2 Ω . Before the switch is closed the current is
a fa f
a fa f
a fa f
a fa f
Q21.22
. Three runs in parallel:
Two runs in series:
two runs:
. Junction of one lift and
.
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, so that if we count the outward bound skiers as negative the net
accumulation of skiers is zero. A loop rule: the total change in altitude must be zero for any skier
completing a closed path.
SOLUTIONS TO PROBLEMS
Section 21.1 Electric Current
P21.1
I=
∆Q
∆t
N=
e
ja
f
∆Q = I∆t = 30.0 × 10 −6 A 40.0 s = 1.20 × 10 −3 C
Q
1.20 × 10 −3 C
=
= 7.50 × 10 15 electrons
e 1.60 × 10 −19 C electron
580
P21.2
Current and Direct Current Circuits
The period of revolution for the sphere is T =
revolving charge is I =
P21.3
z
t
af
e
P21.4
af
(b)
Q 10τ = I 0τ 1 − e −10 =
(c)
Q ∞ = I 0τ 1 − e −∞ = I 0τ
0
j b0.999 95gI τ
e
e
0
j
q = 4t 3 + 5t + 6
e
A = 2.00 cm 2
(a)
(b)
P21.5
j a0.632fI τ
e
Q τ = I 0 τ 1 − e −1 =
af
, and the average current represented by this
j
(a)
a f
ω
q
qω
=
.
2π
T
Q t = Idt = I 0τ 1 − e − t τ
0
2π
a
1.00 m I
J
jFGH 100
cm K
f
I 1.00 s =
J=
dq
dt
2
= 2.00 × 10 −4 m 2
e
= 12t 2 + 5
t = 1.00 s
j
t =1.00 s
= 17.0 A
I
17.0 A
=
= 85.0 kA m 2
A 2.00 × 10 −4 m 2
We use I = nqAv d . Here 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
27.0 g
=
= 4.49 × 10 −23 g atom .
23
NA
6.02 × 10
Thus,
n=
density of aluminum
2.70 g cm 3
=
mass per atom
4.49 × 10 −23 g atom
n = 6.02 × 10 22 atoms cm3 = 6.02 × 10 28 atoms m3 .
I
5.00 A
=
= 1.30 × 10 −4 m s
28
−
3
nqA
6.02 × 10 m
1.60 × 10 −19 C 4.00 × 10 −6 m 2
Therefore,
vd =
or,
v d = 0.130 mm s .
e
je
je
j
Chapter 21
581
Section 21.2 Resistance and Ohm’s Law
∆V 120 V
=
= 0.500 A = 500 mA
R
240 Ω
P21.6
I=
P21.7
∆V = IR
and
F 1.00 m I = 6.00 × 10
A = a0.600 mmf G
H 1 000 mmJK
∆VA a0.900 V fe6.00 × 10 m j
I=
=
ρl
e5.60 × 10 Ω ⋅ mja1.50 mf
ρl
R=
:
A
2
2
−7
Iρl
∆V =
:
A
−7
m2
2
−8
I = 6.43 A
P21.8
(a)
Given
M = ρ d V = ρ d Al
we obtain:
A=
Thus,
(b)
V=
l=
M
ρd
M
.
ρdl
where
ρ d ≡ mass density,
Taking ρ r ≡ resistivity,
R=
e1.00 × 10 ja0.500f
e1.70 × 10 je8.92 × 10 j
=
ρr ρd
−8
π r 2l =
M
π ρdl
=
e
1.00 × 10 −3
b
g e
M
ρd
.
r = 1.40 × 10 −4 m .
ja f
π 8.92 × 10 3 1.82
diameter = 280 µm .
The diameter is twice this distance:
P21.9
ρrl
ρ ρ l2
= r d .
M ρdl
M
l = 1.82 m .
3
or
r=
A
=
−3
MR
,
Thus,
ρrl
a
f
j OP =
PP
Q
49.9 mA
j
(a)
ρ = ρ 0 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=
(c)
F π d I = 6.35 × 10
I = JA = J G
H 4 JK e
continued on next page
E
ρ
=
0.200 V m
3.15 × 10 −8 Ω ⋅ m
2
= 6.35 × 10 6 A m 2
6
L π e1.00 × 10
A m jM
MM
4
N
2
−4
m
2
582
Current and Direct Current Circuits
(d)
6.02 × 10 23 electrons
n=
e
6
26.98 g 2.70 × 10 g m
3
j
= 6.02 × 10 28 electrons m 3
e
j
6.35 × 10 6 A m 2
J
vd =
=
= 659 µm s
ne
6.02 × 10 28 electrons m 3 1.60 × 10 −19 C
e
je
∆V = El = b0. 200 V mga 2.00 mf = 0.400 V
(e)
P21.10
j
b
g
At the low temperature TC we write
RC =
∆V
= R0 1 + α TC − T0
IC
At the high temperature Th ,
Rh =
∆V ∆V
=
= R0 1 + α Th − T0 .
Ih
1A
where T0 = 20.0° C .
b
g
a∆V f a1.00 Af = 1 + e3.90 × 10 ja38.0f
a∆V f I 1 + e3.90 × 10 ja−108f
F 1.15 IJ = 1.98 A .
I = a1.00 A fG
H 0.579 K
−3
Then
−3
C
and
P21.11
C
For aluminum,
R=
ρl
A
=
b
ga
α E = 3.90 × 10 −3 ° C −1
(Table 21.1)
α = 24.0 × 10 −6 ° C −1
(Table 16.1)
ρ 0 1 + α E ∆T l 1 + α∆T
a
A 1 + α∆T
f
2
f = R b1 + α ∆T g = a1.234 ΩfF 1.39 I =
GH 1.002 4 JK
a1 + α∆T f
0
E
Section 21.3 Superconductors
No problems in this section
Section 21.4 A Structural Model for Electrical Conduction
P21.12
(a)
(b)
n is unaffected
J =
I
∝I
A
so it doubles .
continued on next page
1.71 Ω
Chapter 21
(c)
J = nev d
so v d
τ=
(d)
doubles .
mσ
nq 2
is unchanged as long as σ does not change due to a temperature change in the
conductor.
P21.13
ρ=
m
nq 2τ
τ=
so
e
je
j
je
9.11 × 10 −31
m
=
= 2.47 × 10 −14 s
ρ nq 2
1.70 × 10 −8 8.49 × 10 28 1.60 × 10 −19
e
vd =
j
qE
τ
m
e1.60 × 10 jEe2.47 × 10 j
−19
−14
so
7.84 × 10 −4 =
Therefore,
E = 0.181 V m . We could also compute the field from E = ρ J with J = nqv d .
9.11 × 10 −31
Section 21.5 Electric Energy and Power
P21.14
I=
P
∆V
=
and R =
P21.15
600 W
= 5.00 A
120 V
∆V 120 V
=
= 24.0 Ω .
5.00 A
I
b
gb
g
P = 0.800 1 500 hp 746 W hp = 8.95 × 10 5 W
b
8.95 × 10 5 = I 2 000
P = I ∆V
*P21.16
g
I = 448 A
The battery takes in energy by electric transmission
a fa f
e
j
FG 3 600 s IJ = 469 J.
H 1h K
P ∆t = ∆V I ∆t = 2.3 J C 13.5 × 10 −3 C s 4.2 h
It puts out energy by electric transmission
a∆V fIa∆tf = 1.6 J C e18 × 10
(a)
583
efficiency =
continued on next page
−3
j
C s 2.4 h
FG 3 600 s IJ = 249 J.
H 1h K
useful output 249 J
=
= 0.530
total input
469 J
584
Current and Direct Current Circuits
(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 =
P21.17
221 J
Q
=
mc 0.015 kg
kg ° C
= 15.1° C
975 J
a f R = FG ∆V IJ = FG 140 IJ = 1.361
b g R H ∆V K H 120 K
F P − P IJ a100%f = FG P − 1IJ a100%f = a1.361 − 1f100% =
∆% = G
H P K
HP K
∆V
P
=
P0
∆V0
2
2
2
2
0
0
0
*P21.18
36.1%
0
The energy taken in by electric transmission for the fluorescent lamp is
a
sI
fFGH 3 600
J = 3.96 × 10 J
1h K
F $0.08 IJ FG k IJ FG W ⋅ s IJ FG h IJ = $0.088
JG
H kWh K H 1 000 K H J K H 3 600 s K
6
P ∆t = 11 J s 100 h
cost = 3.96 × 10 6
For the incandescent bulb,
a
sI
fFGH 3 600
J = 1.44 × 10
1h K
F $0.08 IJ = $0.32
JG
H 3.6 × 10 J K
P ∆t = 40 W 100 h
cost = 1.44 × 10 7
7
J
6
saving = $0.32 − $0.088 = $0.232
P21.19
At operating temperature,
a
fa
f
(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.
b
R = R0 1 + α ∆T
∆T = 441° C
g
120 120
=
1 + 0.400 × 10 −3 ∆T
1.53 1.80
e
j
T = 20.0° C + 441° C = 461° C
Chapter 21
P21.20
The total clock power is
e270 × 10
From e =
6
jFGH
clocks 2.50
Js
clock
IJ FG 3 600 s IJ = 2.43 × 10
KH 1 h K
12
J h.
Wout
, the power input to the generating plants must be:
Qin
Q in Wout ∆ t 2. 43 × 10 12 J h
=
=
= 9.72 × 10 12 J h
∆t
0.250
e
and the rate of coal consumption is
e
Rate = 9.72 × 10 12 J h
P21.21
6
5
kg coal h = 295 metric ton h .
You pay the electric company for energy transferred in the amount E = P ∆t
(a)
(b)
(c)
*P21.22
.00 kg coal I
jFGH 133.0
J = 2.95 × 10
× 10 J K
7 d I F 86 400 s I F 1 J I
fFGH 1 week
JK GH 1 d JK GH 1 W ⋅ s JK = 48.4 MJ
F 7 d IJ FG 24 h IJ FG k IJ = 13.4 kWh
P ∆t = 40 Wa 2 weeksfG
H 1 week K H 1 d K H 1 000 K
F 7 d IJ FG 24 h IJ FG k IJ FG 0.12 $ IJ = $1.61
P ∆t = 40 Wa 2 weeksfG
H 1 week K H 1 d K H 1 000 K H kWh K
a
P ∆t = 40 W 2 weeks
h I F k I F 0.12 $ I
fFGH 601min
JK GH 1 000 JK GH kWh JK = $0.005 82
F 1 h IJ FG k IJ FG 0.12 $ IJ = $0.416
P ∆t = 5 200 Wa 40 minfG
H 60 min K H 1 000 K H kWh K
a
P ∆t = 970 W 3 min
= 0.582¢
The heater should put out constant power
P=
d
i b
gb
ga
0.250 kg 4 186 J 100° C − 20° C
Q mc T f − Ti
=
=
∆t
∆t
kg⋅° C 4 min
a
f
Then its resistance should be described by
a f a∆V faR∆V f
a∆V f = b120 J Cg
R=
P = ∆V I =
2
P
continued on next page
349 J s
2
= 41.3 Ω
f FG 1 min IJ = 349 J s
H 60 s K
585
586
Current and Direct Current Circuits
Its resistivity at 100°C is given by
b
g e
a f
j
ρ = ρ 0 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
l
l
l4
=ρ
=ρ
2
A
πr
π d2
j π4dl
e
41.3 Ω = 1.55 × 10 −6 Ω ⋅ m
l
= 2.09 × 10 +7 m
2
d
2
e
j
d 2 = 4.77 × 10 −8 m l
or
One possible choice is l = 0.900 m and d = 2.07 × 10 −4 m. If l 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 l and d less than those described by
π d2
π
4.77 × 10 −8 m l 2 = 0.5 × 10 −6 m3 ,
l = 0.5 × 10 −6 m3 . Substituting d 2 = 4.77 × 10 −8 m l gives
4
4
l = 3.65 m and d = 4.18 × 10 −4 m. Thus our answer is: Any diameter d and length l related by
d 2 = 4.77 × 10 −8 m l would have the right resistance. One possibility is length 0.900 m and diameter
e
e
j
e
j
j
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 .
P21.23
(a)
P = I ∆V
so I =
(b)
∆t =
P
=
∆V
∆U
P
=
8.00 × 10 3 W
= 667 A .
12.0 V
2.00 × 10 7 J
8.00 × 10 3 W
b
= 2.50 × 10 3 s
ge
j
and ∆x = v∆t = 20.0 m s 2.50 × 10 3 s = 50.0 km .
P21.24
Consider a 400-W blow dryer used for ten minutes daily for a year. The energy transferred to the
dryer is
b
gb
FG 1 kWh IJ ≈ 20 kWh .
H 3.6 × 10 J K
f
ga
P ∆t = 400 J s 600 s d 365 d ≈ 9 × 10 7 J
6
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
a
fb
g
Cost ≈ 20 kWh $0.10 kWh = $2 ~ $1 .
Chapter 21
Section 21.6 Sources of emf
P21.25
(a)
(b)
P=
a∆V f
2
R
a11.6 Vf
becomes
20.0 W =
so
R = 6.73 Ω .
2
R
FIG. P21.25
∆V = IR
a
so
11.6 V = I 6.73 Ω
and
I = 1.72 A
ε = IR + Ir
so
f
a
f
15.0 V = 11.6 V + 1.72 A r
r = 1.97 Ω .
P21.26
The total resistance is R =
3.00 V
= 5.00 Ω .
0.600 A
(a)
Rlamp = R − rbatteries = 5.00 Ω − 0.408 Ω = 4.59 Ω
(b)
0.408 Ω I 2
Pbatteries
=
= 0.081 6 = 8.16%
Ptotal
5.00 Ω I 2
a
a
f
f
FIG. P21.26
Section 21.7 Resistors in Series and in Parallel
P21.27
(a)
Rp =
1
b1 7.00 Ωg + b1 10.0 Ωg = 4.12 Ω
Rs = R1 + R 2 + R3 = 4.00 + 4.12 + 9.00 = 17.1 Ω
(b)
∆V = IR
a
34.0 V = I 17.1 Ω
FIG. P21.27
f
I = 1.99 A for 4.00 Ω , 9.00 Ω resistors.
a1.99 Afa4.12 Ωf = 8.18 V
8.18 V = I a7.00 Ωf
Applying ∆V = IR ,
so
I = 1.17 A for 7.00 Ω resistor
a
8.18 V = I 10.0 Ω
so
f
I = 0.818 A for 10.0 Ω resistor.
587
588
P21.28
Current and Direct Current Circuits
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
50.0 V
. The voltmeter displays
resistors is
1.00 MΩ + Rshoes
a
f
a
f
∆V = I 1.00 MΩ =
(a)
We solve to obtain
a
f
b
a
g
f
1.00 MΩ 50.0 − ∆V
.
∆V
With Rshoes → 0 , the current through the person’s body is
50.0 V
= 50.0 µA
1.00 MΩ
P21.29
f
50.0 V 1.00 MΩ = ∆V 1.00 MΩ + ∆V Rshoes
Rshoes =
(b)
a
50.0 V 1.00 MΩ
= ∆V .
1.00 MΩ + Rshoes
The current will never exceed 50 µA .
If we turn the given diagram on its side, we find that it is the same as
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 =
c
1
1
10.0 Ω
+ 5.001 Ω +
1
25.0 Ω
h = 2.94 Ω .
This is shown in figure (c), which in turn reduces to the circuit shown in
figure (d).
∆V
and
Next, we work backwards through the diagrams applying I =
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
25.0 V
∆V
I=
=
= 1.93 A .
12.94 Ω
R
In figure (c), this 1.93 A goes through the 2.94 Ω equivalent resistor to
give a potential difference of:
a
fa
f
∆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.
(b)
Therefore, ∆Vab = 5.68 V .
(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 .
25.0 Ω
Rab
FIG. P21.29
Chapter 21
P21.30
(a)
589
Since all the current in the circuit must pass through the series
100 Ω resistor, P = I 2 R
2
Pmax = RI max
so
P
I max =
R
R eq = 100 Ω +
=
FIG. P21.30
25.0 W
= 0.500 A
100 Ω
FG 1 + 1 IJ
H 100 100 K
−1
Ω = 150 Ω
∆Vmax = Req I max = 75.0 V
a
fa
f
P = I∆V = 0.500 A 75.0 V = 37.5 W total power
(b)
P1 = 25.0 W
a
fa
P2 = P3 = RI 2 = 100 Ω 0.250 A
P21.31
2
= 6.25 W
FG 1 + 1 IJ = 0.750 Ω
H 3.00 1.00 K
= a 2.00 + 0.750 + 4.00f Ω = 6.75 Ω
−1
Rp =
Rs
f
∆V 18.0 V
=
= 2.67 A
Rs 6.75 Ω
I battery =
a
P = I 2R:
P2 = 2.67 A
f a2.00 Ωf
2
P2 = 14.2 W in 2.00 Ω
a
f a4.00 Af = 28.4 W
= a 2.67 A fa 2.00 Ωf = 5.33 V,
= a 2.67 A fa 4.00 Ωf = 10.67 V
P4 = 2.67 A
∆V2
∆V4
2
in 4.00 Ω
b
∆Vp = 18.0 V − ∆V2 − ∆V4 = 2.00 V = ∆V3 = ∆V1
P3 =
b∆V g = a2.00 Vf
P1 =
b∆V g = a2.00 Vf
3
R3
2
3.00 Ω
1
R1
1.00 Ω
g
2
= 1.33 W in 3.00 Ω
FIG. P21.31
2
= 4.00 W in 1.00 Ω
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 series,
1
potential difference is shared in proportion to the resistance, so resistor 1 gets of the
3
2
battery voltage and the 2-3-4 parallel combination get of the battery voltage. This is the
3
1
potential difference across resistor 4, but resistors 2 and 3 must share this voltage. goes to
3
2
2 and to 3. The ranking by potential difference is ∆V4 > ∆V3 > ∆V1 > ∆V2 .
3
continued on next page
*P21.32
(a)
590
Current and Direct Current Circuits
(b)
Based on the reasoning above the potential differences are
∆V1 =
*P21.33
ε , ∆V
2
3
=
2ε
4ε
2ε
, ∆V3 =
, ∆V4 =
.
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 I 1 > I 4 > I 2 = I 3 .
(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 current through
2I
I
.
the resistors are I 1 = I , I 2 = I 3 = , I 4 =
3
3
(e)
Increasing resistor 3 increases the equivalent resistance of the entire circuit. The current in
the circuit, which is the current through resistor 1, decreases. This decreases the potential
difference across resistor 1, increasing the potential difference across the parallel
combination. With a larger potential difference the current through resistor 4 is increased.
With more current through 4, and less in the circuit to start with, the current through
resistors 2 and 3 must decrease. To summarize, I 4 increases and I 1 , I 2 , and I 3 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
3
now has an equivalent resistance of 4R. The current in the circuit drops to of the original
4
4
current because the resistance has increased by . All this current passes through resistors 1
3
3I
3I
.
and 4, and none passes through 2 or 3. Therefore I 1 = , I 2 = I 3 = 0 , I 4 =
4
4
(a)
We find the resistance intrinsic to the vacuum
cleaner:
P = I∆V =
a∆V f
120 V = ε
2
R
a∆V f = a120 V f
2
R=
535 W
P
2
= 26.9 Ω
with the inexpensive cord, the equivalent
resistance is 0.9 Ω + 26.9 Ω + 0.9 Ω = 28.7 Ω so the
current throughout the circuit is
I=
continued on next page
ε
RTot
0.9 Ω = r
=
120 V
= 4.18 A
28.7 Ω
0.9 Ω = r
FIG. P21.33
26.9 Ω = R
Chapter 21
and the cleaner power is
a f
Pcleaner = I ∆V
In symbols, RTot = R + 2r , I =
(b)
R + 2r =
FεRI
GH P JK
2
12
= 120 V
cleaner
r=
ε
R + 2r
FG 26.9 Ω IJ
H 525 W K
a
= I 2 R = 4.18 A
f a26.9 Ωf =
2
and Pcleaner = I 2 R =
ε 2R
a R + 2r f
2
470 W .
.
12
= 27.2 Ω
ρl ρl 4
27.2 Ω − 26.9 Ω
= 0.128 Ω =
=
A π d2
2
F 4ρl I
d=G
H π r JK
(c)
cleaner
12
F 4e1.7 × 10 Ω ⋅ mja15 mf I
JJ
=G
GH
π a0.128 Ωf
K
−8
12
= 1.60 mm or more
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:
εF
r=
R
G
2 HP
F 4ρl I
d=G
H π r JK
IJ
K
F 4e1.7 × 10 Ω ⋅ mja15 mf I
=G
GH π b0.037 9 Ωg JJK
cleaner
12
IJ
K
12
−
FG
H
R 120 V 26.9 Ω
=
2
2
532 W
−8
12
−
26.9 Ω
= 0.037 9 Ω
2
12
= 2.93 mm or more
Section 21.8 Kirchhoff’s Rules
P21.34
a f a fa f
+15.0 − 7.00 I 1 − 2.00 5.00 = 0
5.00 = 7.00 I 1
so
I 1 = 0.714 A
so
I 2 = 1.29 A
I 1 + I 2 − 2.00 A = 0
0.714 + I 2 = 2.00
+ε − 2.00(1.29) − 5.00( 2.00) = 0
ε = 12.6 V
FIG. P21.34
591
592
P21.35
Current and Direct Current Circuits
We name currents I 1 , I 2 , and I 3 as shown.
From Kirchhoff’s current rule, I 3 − I 1 − I 2 = 0 .
Applying Kirchhoff’s voltage rule to the loop containing I 2 and I 3 ,
a f a f
8.00 = a 4.00fI + a6.00fI
12.0 V − 4.00 I 3 − 6.00 I 2 − 4.00 V = 0
3
2
Applying Kirchhoff’s voltage rule to the loop containing I 1 and I 2 ,
a f
a f
− 6.00 I 2 − 4.00 V + 8.00 I 1 = 0
a8.00fI
1
FIG. P21.35
a f
= 4.00 + 6.00 I 2 .
Solving the above linear system, we proceed to the pair of simultaneous equations:
RS8 = 4I + 4I
T8I = 4 + 6 I
1
1
2
+ 6I 2
or
2
RS8 = 4I + 10 I
TI = 1.33 I − 0.667
1
2
2
1
and to the single equation 8 = 4 I 1 + 13.3 I 1 − 6.67
I1 =
and
14.7 V
= 0.846 A .
17.3 Ω
I 3 = I1 + I 2
Then
give
a
f
I 2 = 1.33 0.846 A − 0.667
I 1 = 846 mA, I 2 = 462 mA, I 3 = 1.31 A .
All currents are in the directions indicated by the arrows in the circuit diagram.
P21.36
See Figure P21.36 for solution to this problem.
FIG. P21.36
P21.37
We use the results of Problem 21.35.
(a)
By the 4.00-V battery:
By the 12.0-V battery:
(b)
By the 8.00-Ω resistor:
By the 5.00-Ω resistor:
By the 1.00-Ω resistor:
By the 3.00-Ω resistor:
By the 1.00-Ω resistor:
continued on next page
a f a fa
f
a12.0 Vfa1.31 Af120 s = 1.88 kJ .
∆U = ∆V I∆t = 4.00 V −0.462 A 120 s = −222 J .
a
f a8.00 Ωf120 s = 687 J
a0.462 Af a5.00 Ωf120 s = 128 J .
a0.462 Af a1.00 Ωf120 s = 25.6 J .
a1.31 Af a3.00 Ωf120 s = 616 J .
a1.31 Af a1.00 Ωf120 s = 205 J .
I 2 R∆t = 0.846 A
2
2
2
2
2
.
Chapter 21
(c)
We can count the energy twice, like a child counting his lunch money at ten o’clock and
again at eleven.
−222 J + 1.88 kJ = 1.66 kJ from chemical to electrically transmitted.
687 J + 128 J + 25.6 J + 616 J + 205 J = 1.66 kJ from electrically transmitted to internal.
*P21.38
(a)
The first equation represents
Kirchhoff’s loop theorem. We choose
to think of it as describing a clockwise
trip around the left-hand loop in a
circuit, see Figure (a). For the righthand loop see Figure (b). The
junctions must be between the 5.8 V
and the 370 Ω and between the
370 Ω and the 150 Ω . Then we have
Figure (c). This is consistent with the
third equation,
I1
220 Ω
370 Ω
5.8 V
Figure (a)
I2
I3
370 Ω
150 Ω
5.8 V
We substitute:
I1
220 Ω
−220 I 1 + 5.8 − 370 I 1 − 370 I 3 = 0
+370 I 1 + 370 I 3 + 150 I 3 − 3.1 = 0
I2
I3
150 Ω
3.10 V
370 Ω
Figure (c)
Next
FIG. P21.38
5.8 − 590 I 1
370
520
370 I 1 +
5.8 − 590 I 1 − 3.1 = 0
370
370 I 1 + 8.15 − 829 I 1 − 3.1 = 0
I3 =
b
I1 =
I3 =
g
11.0 mA in the 220- Ω resistor and out of
5.05 V
=
the positive pole of the 5.8-V battery
459 Ω
b
5.8 − 590 0.011 0
3.10 V
Figure (b)
I1 + I 3 − I 2 = 0
I 2 = I1 + I 3
(b)
I2
g = −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
I 2 = 11.0 − 1.87 = 9.13 mA in the 370- Ω resistor
593
594
P21.39
Current and Direct Current Circuits
Label the currents in the branches as shown in the first figure.
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:
a2.71RfI + a1.71RfI
a1.71RfI + a3.71RfI
and
1
2
= 250
1
2
= 500 .
(a)
With R = 1 000 Ω, simultaneous solution of these equations yields:
I 1 = 10.0 mA
and
I 2 = 130.0 mA .
From Figure (b),
Vc − Va = I 1 + I 2 1.71R = 240 V .
Thus, from Figure (a),
I4 =
b
ga
f
Vc − Va
240 V
=
= 60.0 mA .
4R
4 000 Ω
Finally, applying Kirchhoff’s point rule at point a in Figure (a)
gives:
(b)
FIG. P21.39
I 4 − I1 − I = 0
I = I 4 − I 1 = 60.0 mA − 10.0 mA = +50.0 mA
or
P21.40
I = 50.0 mA from point a to point e .
Using Kirchhoff’s rules,
b g b g
10.0 + a1.00 fI − a0.060 0fI = 0
12.0 − 0.010 0 I 1 − 0.060 0 I 3 = 0
2
3
and
I 1 − I 2 − I 3 = 0 or
then
12.0 − 0.010 0 I 2 − 0.070 0 I 3 = 0
and
I 2 = 0.06 I 3 − 10
a
f a
Solving simultaneously,
I1 = I 2 + I 3
f
a fb
FIG. P21.40
g
12 − 0.01 0.06 I 3 − 10 − 0.07 I 3 = 0
I 2 = 0.283 A downward in the dead battery
and
I 3 = 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.
Chapter 21
595
Section 21.9 RC Circuits
P21.41
P21.42
e
je
j
(a)
RC = 1.00 × 10 6 Ω 5.00 × 10 −6 F = 5.00 s
(b)
Q = Cε = 5.00 × 10 −6 C (30.0 V ) = 150 µC
(c)
I (t ) =
(a)
I t = − I 0 e − t RC
c
h
L
30.0
−10.0
H 1.00 × 10 IK expMMN c1.00 × 10 hc5.00 × 10
ε e − t RC = F
R
6
6
−6
OP =
h PQ
4.06 µA
FIG. P21.41
af
I0 =
Q
5.10 × 10 −6 C
= 1.96 A
=
RC
1 300 Ω 2.00 × 10 −9 F
b
ge
j
L −9.00 × 10 s OP
I at f = −a1.96 A f exp M
MN b1 300 Ωge2.00 × 10 Fj PQ = −61.6 mA
L −8.00 × 10 s OP
qat f = Qe
= b5.10 µC g exp M
MN a1 300 Ωfe2.00 × 10 Fj PQ = 0.235 µC
−6
−9
(b)
P21.43
−6
− t RC
−9
(c)
The magnitude of the maximum current is I 0 = 1.96 A .
(a)
τ = RC = 1.50 × 10 5 Ω 10.0 × 10 −6 F = 1.50 s
(b)
τ = 1.00 × 10 5 Ω 10.0 × 10 −6 F = 1.00 s
(c)
The battery carries current
10.0 V
= 200 µA .
50.0 × 10 3 Ω
The 100 kΩ carries current of magnitude
I = I 0 e − t RC =
e
je
e
j
je
j
FG 10.0 V IJ e
H 100 × 10 Ω K
200 µA + b100 µA ge
.
(a)
150 pF
af
q t = Qe − t RC = C∆V t = C∆V0 e − t RC
∆V0
= e + t RC
∆V
∆V
= e − t RC
∆V0
FG
H
∆V0
t
= ln
∆V
RC
FG ∆V IJ = 5 000 × 10 Ωe230 × 10
H ∆V K
t = RC ln
(b)
6
0
e
j
.
3 000 V
We model the person’s body and street shoes
as shown. For the discharge to reach 100 V,
af
− t 1.00 s
− t 1.00 s
So the switch carries downward current
*P21.44
3
−12
IJ
K
000 I
JK =
j FGH 3100
F ln
t = 1 × 10 6 V A 230 × 10 −12 C V ln 30 = 782 µs
80 pF
FIG. P21.44(a)
3.91 s
5 000 M Ω
596
P21.45
Current and Direct Current Circuits
(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:
10.0 V
= 2.00 A
5.00 Ω
VL = 10.0 V − 2.00 A 1.00 Ω = 8.00 V
IL =
a
fa f
F 2.00 Ω IJ a10.0 Vf = 2.00 V
=G
H 2.00 Ω + 8.00 Ω K
Likewise,
VR
or
IR =
10.0 V
= 1.00 A
10.0 Ω
a
f a
fa
FIG. P21.45(a)
f
VR = 10.0 V − 8.00 Ω 1.00 A = 2.00 V .
(b)
Therefore,
∆V = VL − VR = 8.00 − 2.00 = 6.00 V .
Redraw the circuit
R=
1
= 3.60 Ω
1 9.00 Ω + 1 6.00 Ω
b
g b
g
RC = 3.60 × 10 −6 s
P21.46
1
10
and
e − t RC =
so
t = RC ln 10 = 8.29 µs .
FIG. P21.45(b)
af
e
j
The potential difference across the capacitor
∆V t = ∆Vmax 1 − e − t RC .
Using 1 Farad = 1 s Ω ,
4.00 V = 10.0 V 1 − e
Therefore,
0.400 = 1.00 − e
or
e
Taking the natural logarithm of both sides,
−
and
R=−
fLMN
a
e
j
− 3.00 ×10 5 Ω R
a
− 3.00 s
e
f Re10.0 ×10
J = σE
so
σ=
sΩ
j O.
PQ
j
− 3 .00 × 10 5 Ω R
= 0.600 .
3.00 × 10 5 Ω
= ln 0.600
R
a
f
3.00 × 10 5 Ω
= +5.87 × 10 5 Ω = 587 kΩ .
ln 0.600
a
f
Section 21.10 Context Connection
The Atmosphere as a Conductor
P21.47
−6
J 6.00 × 10 −13 A m 2
=
= 6.00 × 10 −15 (Ω ⋅ m) −1
E
100 V m
Chapter 21
P21.48
(a)
e
je
597
j
P = ∆VI = 300 × 10 3 J C 1.00 × 10 3 C s = 3.00 × 10 8 W
A large electric generating station, fed by a trainload of coal each day, converts energy faster.
(b)
I=
P
A
P
π r2
=
e j e
j
P = I π r 2 = 1 370 W m 2 π ( 6.37 × 10 6 m) 2 = 1.75 × 10 17 W
Terrestrial solar power is immense compared to lightning and compared to all human
energy conversions.
Additional Problems
P21.49
(a)
I=
∆V
R
a∆V f = a120 Vf
2
R=
(b)
I=
P
∆V
∆t =
=
P = I∆V =
and
R=
2
25.0 W
P
so
a∆V f
2
R
a∆V f = a120 Vf
2
= 576 Ω
100 W
P
2
= 144 Ω
25.0 W
Q 1.00 C
= 0.208 A =
=
120 V
∆t
∆t
1.00 C
= 4.80 s
0.208 A
The bulb takes in charge at high potential and puts out the same amount of charge at low
potential. The charge in itself is identical.
(c)
P = 25.0 W =
∆U 1.00 J
=
∆t
∆t
∆t =
1.00 J
= 0.040 0 s
25.0 W
The bulb takes in energy by electrical transmission and puts out the same amount of energy
by heat and light.
(d)
b
gb
f
ga
∆U = P∆t = 25.0 J s 86 400 s d 30.0 d = 64.8 × 10 8 J
The electric company sells energy .
FG $0.070 0 IJ FG k IJ FG W ⋅ s IJ FG h IJ = $1.26
H kWh K H 1 000 K H J K H 3 600 s K
$0.070 0 F
Cost per joule =
G kWh IJ = $1.94 × 10 J
kWh H 3.60 × 10 J K
Cost = 64.8 × 10 6 J
−8
6
P21.50
ρ=
a f
∆V A
RA
=
l
I l
ρ ( Ω ⋅ m)
l (m)
0.540
R ( Ω)
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 (in agreement with tabulated value of 1.50 × 10 −6 Ω ⋅ m in Table 21.1)
598
P21.51
P21.52
Current and Direct Current Circuits
a
a
f
f
(a)
r
0 − 4.00 V
dV $
E=−
i=−
= 8.00 $i V m
0.500 − 0 m
dx
(b)
R=
(c)
I=
(d)
r I
6.28 A
J = $i =
A
π 1.00 × 10 −4 m
(e)
ρ J = 4.00 × 10 −8 Ω ⋅ m 2.00 × 10 8 $i A m 2 = 8.00 $i V m = E
(a)
r
dV x $
V$
E=−
i=
i
dx
L
(b)
R=
(c)
I=
ρl
A
e4.00 × 10 Ω ⋅ mja0.500 mf =
π e1.00 × 10 mj
−8
=
−4
2
0.637 Ω
∆V
4.00 V
=
= 6.28 A
0.637 Ω
R
e
r
j
e
A
= 2.00 × 10 8 $i A m 2 = 200 $i MA m 2
je
af
ρl
2
=
4ρ L
π d2
Vπ d 2
∆V
=
4ρ L
R
r
j
Note that the field is independent of the material and the area.
Independent of voltage, field, and current.
This is how Georg Simon Ohm thought of his experimental results on
all the things current depends on.
(d)
(e)
P21.53
(a)
(b)
r I
V $
i
J = $i =
A
ρL
Independent of area.
r
V$
Independent of length and area. At each point within the material,
i= E
L
vector field causes vector current density according to this relation. This is how an atom
thinks of Ohm’s law.
r
ρJ =
P = I∆V :
P
1 500 W
= 12.5 A .
120 V
So for the Heater,
I=
For the Toaster,
I=
750 W
= 6.25 A .
120 V
And for the Grill,
I=
1 000 W
= 8.33 A .
120 V
∆V
=
12.5 + 6.25 + 8.33 = 27.1 A
The current draw is greater than 25.0 amps, so this circuit would not be sufficient.
Chapter 21
P21.54
(a)
599
A thin cylindrical shell of radius r, thickness dr, and length L contributes resistance
dR =
ρ dl
=
A
F
b g GH
I
JK
ρ dr
ρ dr
=
.
2π L r
2π r L
The resistance of the whole annulus is the series summation of the contributions of the thin
shells:
FG IJ
H K
z
r
R=
r
ρ b dr
ρ
ln b
=
2π L r r
2π L
ra
a
(b)
FG IJ
H K
r
∆V
ρ
ln b
=
2π L
I
ra
In this equation
we solve for ρ =
*P21.55
.
2π L∆V
I ln rb ra
b
g
.
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
a
fb
g
∆U = q∆V = 240 C 6.00 J C = 1 440 J .
∆V 6.00 V
=
= 0.030 0 A .
200 Ω
R
The radio draws current
I=
So, its power is
P = ∆V I = 6.00 V 0.030 0 A = 0.180 W = 0.180 J s.
a f a
Then for the time the energy lasts, we have P =
We could also compute this from I =
P21.56
I=
ε
R+r
Let x ≡
, so P = I 2 R =
ε2
P
a
, then R + r
ε 2R
2
E
:
∆t
∆t =
ε = 9.20 V
R=
continued on next page
+4.21 ±
P
=
1 440 J
= 8.00 × 10 3 s .
0.180 J s
240 C
Q
=
= 8.00 × 10 3 s = 2.22 h .
I 0.030 0 A
FG ε IJ R .
HPK
2
a f
R + a 2.40 − xfR − 1.44 = 0 ,
−a 2.40 − x f ± a 2.40 − x f
R=
R 2 + 2r − x R − r 2 = 0 .
= xR or
2
which has solutions of
With
E
(R + r ) 2 =
With r = 1.20 Ω, this becomes
(a)
g
∆t =
or
(R + r ) 2
f
Q
:
∆t
fb
2
and
a4.21f
2
P = 12.8 W , x = 6.61 :
2
− 5.76
= 3.84 Ω or
0.375 Ω .
2
− 5.76
.
600
Current and Direct Current Circuits
(b)
ε = 9.20 V
For
R=
and
+1.59 ±
P = 21.2 W , x ≡
a1.59f
2
− 5.76
2
=
ε2
P
= 3.99
1.59 ± −3.22
.
2
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 =
P21.57
The current in the simple loop circuit will be I =
(a)
∆Vter = ε − Ir =
(b)
I=
ε
R+r
P = I2R = ε 2
(c)
εR
R+r
4r
ε
R+r
= 17.6 W .
.
and
∆Vter → ε as R → ∞ .
and
I→
ε
r
as R → 0 .
FIG. P21.57
dP
ε2 =0
−2ε 2 R
=
+
dR ( R + r ) 3 ( R + r ) 2
R
(R + r ) 2
Then 2 R = R + r
P21.58
ε2
and
R=r .
FG
IJ
H
K
z dE = z εR expFH − RCt IK dt
dE
ε e −1 RC .
= P =ε I =ε
dt
R
The battery supplies energy at a changing rate
∞
Then the total energy put out by the battery is
2
t =0
z
dE =
ε 2 (− RC )∞ expF −
z
R
H
0
t
RC
IK FH − dt IK = −ε C expFH − t IK
RC
RC
2
∞
0
= −ε 2 C 0 − 1 = ε 2C .
F
H
I
K
The power delivered to the resistor is
dE
ε2
2t
= P = ∆VR I = I 2 R = R 2 exp −
.
RC
dt
R
So the total internal energy appearing in the resistor is
z zε
z
dE =
ε 2 F − RC I ∞ expF −
R
H
2
Kz
0
H
2t
RC
IK FH − 2dt IK = − ε C expFH − 2t IK
2
RC
RC
2
The energy finally stored in the capacitor is U =
conserved
battery.
ε 2C = 1 ε 2C + 1 ε 2C
2
2
dE =
∞
0
=−
∞
2
0
R
ε 2C
2
FH
exp −
0 −1 =
IK
2t
dt
RC
ε 2C .
2
1
1
C ( ∆V ) 2 = Cε 2 . Thus, energy of the circuit is
2
2
and resistor and capacitor share equally in the energy from the
Chapter 21
P21.59
(a)
e
q = C∆V 1 − e − t RC
j
fLMN
ja
e
q = 1.00 × 10 −6 F 10.0 V 1 − e
(b)
I=
−10.0
−6
FG 10.0 V IJ e = 3.37 × 10 A = 33.7 nA
H 2.00 × 10 Ω K
dU d F 1 q I F q I dq F q I
=
=G J
= G JI
dt dt GH 2 C JK H C K dt H C K
dU F 9.93 × 10 C I
=
e3.37 × 10 Aj = 3.34 × 10
dt GH 1.00 × 10 C V JK
−5.00
6
PQ
9.93 µC
−8
2
−8
−6
P21.60
6
FG IJ
H K
−6
(d)
e 2.00 ×10 je1.00 ×10 j O =
dq
∆V − t RC
=
e
dt
R
I=
(c)
601
c
−7
W = 334 nW
h
Pbattery = Iε = 3.37 × 10 −8 A (10.0 V ) = 3.37 × 10 −7 W = 337 nW
Start at the point when the voltage has just reached
2
∆V
3
2
∆V and is
3
decaying towards 0 V with a time constant RBC
and the switch has just closed. The voltage is
a f LMN 32 ∆V OPQe .
1
We want to know when ∆V at f will reach ∆V .
3
1
L2 O
∆ V = M ∆V P e
Therefore,
3
N3 Q
Voltage
controlled
switch
∆V
R2B
C
− t RBC
∆VC t =
+
R1A
V ∆Vc
C
− t RBC
1
2
or
e − t RBC =
or
t1 = RBC ln 2 .
FIG. P21.60
After the switch opens, the voltage is
af
∆VC t = ∆V −
When
af
∆VC t =
LM 2 ∆V OPe
N3 Q
b
b
b
g
− t R A + RB C
.
2
∆V
3
2
2
∆V = ∆V − ∆Ve − t b R A + RB gC
3
3
So
g
t 2 = R A + RB C ln 2
g
1
∆V , increasing toward ∆V with time constant R A + RB C :
3
or
e − t b R A + RB gC =
and
T = t1 + t 2 =
1
.
2
bR
A
g
+ 2 RB C ln 2 .
602
P21.61
Current and Direct Current Circuits
(a)
With the switch closed, current exists in a simple series
circuit as shown. The capacitors carry no current. For R 2
we have
P = I 2 R2
I=
P
R2
=
2.40 V ⋅ A
= 18.5 mA .
7 000 V A
The potential difference across R1 and C 1 is
e
∆V = IR1 = 1.85 × 10
−2
jb
FIG. P21.61(a)
g
A 4 000 V A = 74.1 V .
The charge on C 1 is
ja
e
f
Q = C1 ∆V = 3.00 × 10 −6 C V 74.1 V = 222 µ C .
The potential difference across R 2 and C 2 is
jb
e
g
∆V = IR 2 = 1.85 × 10 −2 A 7 000 Ω = 130 V .
The charge on C 2 is
ja
e
f
Q = C 2 ∆V = 6.00 × 10 −6 C V 130 V = 778 µ C .
The battery emf is
b
g
b
g
IR eq = I R1 + R 2 = 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
C 2 is
e
ja
f
Q = C 2 ∆V = 6.00 × 10 −6 C V 204 V = 1 222 µ C
FIG. P21.61(b)
for a change of 1 222 µ C − 778 µ C = 444 µ C .
P21.62
∆V = ε e − t RC
FG ε IJ = FG 1 IJ t .
H ∆V K H RC K
ε
A plot of lnF
H IK versus t should
so
ln
∆V
be a straight line with slope equal
1
to
.
RC
Using the given data values:
FIG. P21.62
continued on next page
Chapter 21
(a)
A least-square fit to this data yields the graph above.
∑ xi = 282 ,
∑
∑ xi yi = 244,
∑ yi = 4.03 ,
xi2
4
= 1.86 × 10 ,
N=8
t( s )
0
4.87
11.1
19.4
30.8
46.6
67.3
102.2
c∑ x y h − c∑ x hc∑ y h = 0.011 8
N e∑ x j − c∑ x h
e∑ x jc∑ y h − c∑ x hc∑ x y h = 0.088 2
Intercept =
N e∑ x j − c∑ x h
ε
The equation of the best fit line is:
lnF
H IK = a0.011 8ft + 0.088 2
Slope =
N
i i
i
2
i
i
2
2
i
i
i
i
2
i
i i
2
P21.63
a
ln ε ∆V
0
0.109
0.228
0.355
0.509
0.695
0.919
1.219
f
i
∆V
(b)
∆V (V )
6.19
5.55
4.93
4.34
3.72
3.09
2.47
1.83
603
Thus, the time constant is
τ = RC =
and the capacitance is
C=
τ
R
=
.
1
1
=
= 84.7 s
slope 0.011 8
84.7 s
= 8.47 µF .
10.0 × 10 6 Ω
An editorial in The Physics Teacher suggested the
idea for this problem.
The battery current is
a150 + 45 + 14 + 4f 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. Then
the total current is
FIG. P21.63
211
=
a150 + 45 + 14 + 2f mA = 211 mA, nearly the same as before. The ratio is 213
(b)
0.991 .
The resistor with least resistance carries 150 mA. Doubling its resistance changes this current
to 75 mA and changes the total to
=
a75 + 45 + 14 + 4f mA = 138 mA . The ratio is 138
213
0.648 , representing a much larger
reduction (35.2% instead of 0.9%).
(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. Doubling the thermal
resistance of the ceiling will produce a much larger saving. The ceiling originally has the
smallest thermal resistance.
604
*P21.64
Current and Direct Current Circuits
(a)
For the first measurement, the equivalent circuit is as
shown in Figure 1.
Ry =
1
R1 .
2
R ac = R2 =
(1)
1
Ry + Rx .
2
(2)
Ry
Rx
Figure 1
For the second measurement, the equivalent circuit is
shown in Figure 2.
Thus,
b
c
Ry
R ab = R1 = R y + Ry = 2 R y
so
R1
a
R2
a
Ry
Ry
c
Rx
Substitute (1) into (2) to obtain:
R2 =
FG
H
IJ
K
1 1
1
R1 + R x , or R x = R 2 − R1 .
4
2 2
Figure 2
FIG. P21.64
If R1 = 13.0 Ω and R 2 = 6.00 Ω, then R x = 2.75 Ω .
(b)
The antenna is inadequately grounded since this exceeds the limit of 2.00 Ω .
ANSWERS TO EVEN PROBLEMS
P21.2
qω
2π
P21.4
(a) 17.0 A ; (b) 85.0 kA m 2
P21.6
500 mA
P21.8
(a) 1.82 m; (b) 280 µm
P21.10
1.98 A
P21.12
(a) no change; (b) doubles; (c) doubles;
(d) no change
P21.14
5.00 A; 24.0 Ω
P21.16
(a) 0.530; (b) 221 J; (c) 15.1°C
P21.18
$0.232
P21.20
295 metric ton/h
P21.22
Any diameter d and length l related by
d 2 = 4.77 × 10 −8 m l , such as length
e
j
0.900 m and diameter 0.207 mm. Yes.
P21.24
~$1
P21.26
(a) 4.59 Ω; (b) 8.16%
P21.28
(a) see the solution; (b) no
P21.30
(a) 75.0 V;
(b) 25.0 W, 6.25 W, and 6.25 W; 37.5 W
P21.32
(a) ∆V4 > ∆V3 > ∆V1 > ∆V2 ;
ε
2ε
4ε
, ∆V3 =
,
(b) ∆V1 = , ∆V2 =
3
9
9
2ε
∆V4 =
;
3
(c) I 1 > I 4 > I 2 = I 3 ;
I
2I
(d) I 1 = I , I 2 = I 3 = , I 4 = ;
3
3
(e) I 4 increases while I 1 , I 2 , and I 3
decrease;
3I
3I
(f) I 1 = , I 2 = I 3 = 0 , I 4 =
4
4
P21.34
0.714 A, 1.29 A, 12.6 V
P21.36
see the solution
Chapter 21
P21.38
(a) see the solution;
(b) The current in the 220-Ω resistor and
the 5.80-V battery is 11.0 mA out of the
positive battery pole. The current in the
370-Ω resistor is 9.13 mA. The current in
the 150-Ω resistor and the 3.10-V battery is
1.87 mA out of the negative battery pole.
P21.40
starter 171 A; battery 0.283 A
P21.42
(a) –61.6 mA; (b) 0.235 µC ; (c) 1.96 A
P21.44
(a) 3.91 s; (b) 0.782 ms
P21.46
587 kΩ
P21.48
(a) 300 MW; (b) 175 PW
P21.50
Experimental resistivity = 1.47 µΩ ⋅ m ± 4%,
in agreement with 1.50 µΩ ⋅ m
P21.52
FG IJ
H K
2π L∆V
r
ρ
ln b ; (b) ρ =
r
ra
2π L
I ln rb
a
P21.54
(a) R =
P21.56
(a) either 3.84 Ω or 0.375 Ω;
(b) impossible
P21.58
see the solution
P21.60
bR
P21.62
+ 2 RB C ln 2
(b) 84.7 s, 8.47 µF
R1
;
4
(b) R x = 2.75 Ω , the station is inadequately
grounded
(a) R x = R 2 −
P21.64
Vπ d 2
V$
4ρL
V $
; (d)
i ; (b)
i;
;
(c)
2
4 ρL
ρL
L
πd
(e) see the solution
(a)
The resistance of the spherical shell of air carrying radial electric current is
R=
jb
e
g
13
ρ L 2 × 10 Ω ⋅ m 4 000 m
=
= 157 Ω .
2
A
4π 6.37 × 10 6 m
e
b
j
gb
g
The time constant for discharge is RC = 157 V A 0.8 C V = 126 s .
(a)
ej
g
F ε IJ = 0.011 8t + 0.088 2 ;
(a) lnG
H ∆V K
A
CONTEXT 6 CONCLUSION SOLUTIONS TO PROBLEMS
CC6.1
605
q = Q e − t RC describes the discharge
q
= e − t RC
Q
e + t RC =
Q
q
F I
GH JK
F QI
t = RC lnG J
H qK
F F 4 × 10 C I I =
t = 126 sG lnG
H H 2 × 10 C JK JK
t
Q
= ln
RC
q
4
4
continued on next page
87.0 s
(b)
(c)
F F 4 × 10
GH GH 5 × 10
F F 4 × 10
t = 126 sG lnG
H H 0
t = 126 s ln
4
3
4
I I = 261 s
JK JK
C II
JK JK → ∞
C
C
The exponential decay goes on forever, as modeled mathematically.
CC6.2
(a)
I=
Q
∆t
∆t =
CC6.3
25 C
Q
=
= 1.25 × 10 −2 s ≈ 0.01 s
I 2000 C s
FG 86 400 s IJ FG 1 IJ ≈
H 1 d K H 1.25 × 10 s K
7 × 10 6
(b)
number = 1 d
(a)
For radial currents, the two layers of the atmosphere are resistors in series, presenting
equivalent resistance
−2
e
j
e
j
3
3
13
13
ρ 1 L ρ 2 L 2 × 10 Ω ⋅ m 2.5 × 10 m 0.2 × 10 Ω ⋅ m 2.5 × 10 m
+
=
+
= 108 Ω .
2
2
A
A
4π 6.37 × 10 6 m
4π 6.37 × 10 6 m
e
j
e
j
Then the time constant for discharge is
a
f
RC = 0.9 F 108 Ω = 97.1 s .
According to the argument of the context conclusion, we consider six time constants as the
time for 2 × 10 4 lightning strikes. Then the number in a day is
2 × 10 4
2 × 10 4
=
6 RC
6 97.1 s
a
(b)
FG 86 400 s IJ =
fH 1 d K
3 × 10 6 .
For radial currents through the atmosphere, the air in the two hemispheres presents
resistances in parallel, with equivalent resistance
F 2π e6.37 × 10 mj
I
2π e6.37 × 10 mj
1
G
J
R =
=
+
FG ρ L IJ + FG ρ L IJ GH 2 × 10 Ω ⋅ me5 × 10 mj 0.2 × 10 Ω ⋅ me5 × 10 mj JK
HAK H AK
2 × 10
2 × 10
FG 86 400 s IJ = 9 × 10
number of daily strokes =
=
6 RC
6a35.7 Ωfa0.9 F f H 1 d K
2
6
eq
1
−1
2
−1
3
13
4
4
6
3
13
6
ANSWERS TO EVEN CONTEXT 6 CONCLUSION PROBLEMS
CC6.2
(a) 0.01 s; (b) 7 × 10 6
2
−1
= 35.7 Ω