28
Direct Current Circuits
CHAPTER OUTLINE
28.1
28.2
28.3
28.4
28.5
28.6
Electromotive Force
Resistors in Series and
Parallel
Kirchhoff’s Rules
RC Circuits
Electrical Meters
Household Wiring and
Electrical Safety
ANSWERS TO QUESTIONS
Q28.1
The load resistance in a circuit is the effective resistance of all of
the circuit elements excluding the emf source. In energy terms,
it can be used to determine the energy delivered to the load by
electrical transmission and there appearing as internal energy
to raise the temperature of the resistor. The internal resistance
of a battery represents the limitation on the efficiency of the
chemical reaction that takes place in the battery to supply
current to the load. The emf of the battery represents its
conversion of chemical energy into energy which it puts out by
electric transmission; the battery also creates internal energy
within itself, in an amount that can be computed from its
internal resistance. We model the internal resistance as
constant for a given battery, but it may increase greatly as the
battery ages. It may increase somewhat with increasing current
demand by the load. For a load described by Ohm’s law, the
load resistance is a precisely fixed value.
Q28.2
The potential difference between the terminals of a battery will equal the emf of the battery when
there is no current in the battery. At this time, the current though, and hence the potential drop
across the internal resistance is zero. This only happens when there is no load placed on the
battery—that includes measuring the potential difference with a voltmeter! The terminal voltage
will exceed the emf of the battery when current is driven backward through the battery, in at its
positive terminal and out at its negative terminal.
Q28.3
No. If there is one battery in a circuit, the current inside it will be from its negative terminal to its
positive terminal. Whenever a battery is delivering energy to a circuit, it will carry current in this
direction. On the other hand, when another source of emf is charging the battery in question, it will
have a current pushed through it from its positive terminal to its negative terminal.
Q28.4
Connect the resistors in series. Resistors of 5.0 kΩ, 7.5 kΩ and 2.2 kΩ connected in series present
equivalent resistance 14.7 kΩ.
Q28.5
Connect the resistors in parallel. Resistors of 5.0 kΩ, 7.5 kΩ and 2.2 kΩ connected in parallel present
equivalent resistance 1.3 kΩ.
129
130
Direct Current Circuits
Q28.6
Q28.7
In series, the current is the same through each resistor. Without knowing individual resistances,
nothing can be determined about potential differences or power.
Q28.8
In parallel, the potential difference is the same across each resistor. Without knowing individual
resistances, nothing can be determined about current or power.
Q28.9
In this configuration, the power delivered to one individual resistor is significantly less than if only
one equivalent resistor were used. This decreases the possibility of component failure, and possible
electrical disaster to some more expensive circuit component than a resistor.
Q28.10
Each of the two conductors in the extension cord itself has a small resistance. The longer the
extension cord, the larger the resistance. Taken into account in the circuit, the extension cord will
reduce the current from the power supply, and also will absorb energy itself in the form of internal
energy, leaving less power available to the light bulb.
Q28.11
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.
Q28.12
The potential difference across a resistor is positive when it is measured against the direction of the
current in the resistor.
Q28.13
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.
Q28.14
An ideal ammeter has zero resistance. An ideal voltmeter has infinite resistance. Real meters cannot
attain these values, but do approach these values to the degree that they do not alter the current or
potential difference that is being measured within the accuracy of the meter. Hooray for
experimental uncertainty!
Q28.15
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.
Chapter 28
131
Q28.16
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.
Q28.17
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.
Q28.18
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 + qIR 2 ,
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.
Q28.19
∆V 2
, the bulbs present resistances
R
2
2
2
120 V
120 V
120 V
∆V 2
= 190 Ω , and
= 72 Ω . The nominal 60 W lamp
R=
=
= 240 Ω , and
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.
At their normal operating temperatures, from P =
a
f
a
f
a
f
Q28.20
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.
Q28.21
Series, because the circuit breaker trips and opens the circuit when the current in that circuit loop
exceeds a certain preset value. The circuit breaker must be in series to sense the appropriate current
(see Fig. 28.30).
Q28.22
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 Section 26.4. 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 28.11, 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.
132
Direct Current Circuits
Q28.23
So long as you only grab one wire, and you do not touch anything that is grounded, you are safe
(see Question 28.11). If the wire breaks, let go! If you continue to hold on to the wire, there will be a
large—and rather lethal—potential difference between the wire and your feet when you hit the
ground. Since your body can have a resistance of about 10 kΩ, the current in you would be sufficient
to ruin your day.
Q28.24
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 stepdown 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.
Q28.25
As Luigi Galvani showed with his experiment with frogs’ legs, muscles contract when electric
current exists in them. If an electrician contacts a “live” wire, the muscles in his hands and fingers
will contract, making his hand clench. If he touches the wire with the front of his hand, his hand will
clench around the wire, and he may not be able to let go. Also, the back of his hand may be drier
than his palm, so an actual shock may be much weaker.
Q28.26
Grab an insulator, like a stick or baseball bat, and bat for a home run. Hit the wire away from the
person or hit them away from the wire. If you grab the person, you will learn very quickly about
electrical circuits by becoming part of one.
Q28.27
A high voltage can lead to a high current when placed in a circuit. A device cannot supply a high
current—or any current—unless connected to a load. A more accurate sign saying potentially high
current would just confuse the poor physics student who already has problems distinguishing
between electrical potential and current.
Q28.28
The two greatest factors are the potential difference between the wire and your feet, and the
conductivity of the kite string. This is why Ben Franklin’s experiment with lightning and flying a
kite was so dangerous. Several scientists died trying to reproduce Franklin’s results.
Q28.29
Suppose ε = 12 V and each lamp has R = 2 Ω. Before the switch is closed the current is
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
in the battery
= 3 A becomes larger than before and (a) lamps A and B get brighter. (d) The
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.
a fa f
a fa f
a fa f
Q28.30
a fa f
The starter motor draws a significant amount of current from the battery while it is starting the car.
This, coupled with the internal resistance of the battery, decreases the output voltage of the battery
below its the nominal 12 V emf. Then the current in the headlights decreases.
Chapter 28
Q28.31
. 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. A loop rule: the total change in altitude must be zero for any skier
completing a closed path.
SOLUTIONS TO PROBLEMS
Section 28.1
P28.1
(a)
(b)
Electromotive Force
P=
a∆V f
2
R
a11.6 V f
becomes
20.0 W =
so
R = 6.73 Ω .
2
R
FIG. P28.1
∆V = IR
a
f
so
11.6 V = I 6.73 Ω
and
I = 1.72 A
ε = IR + Ir
so
15.0 V = 11.6 V + 1.72 A r
a
f
r = 1.97 Ω .
P28.2
(a)
(b)
P28.3
∆Vterm = IR
a
becomes
10.0 V = I 5.60 Ω
so
I = 1.79 A .
∆Vterm = ε − Ir
a
f
fa
becomes
10.0 V = ε − 1.79 A 0.200 Ω
so
ε = 10.4 V .
The total resistance is R =
f
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
133
FIG. P28.3
134
P28.4
Direct Current Circuits
f
12.6 V
(b)
Let I 1 and I 2 be the currents flowing through the battery and the
headlights, respectively.
Then,
I 1 = I 2 + 35.0 A , and ε − I 1 r − I 2 r = 0
b5.00 Ω + 0.080 0 Ωg = 2.48 A .
Then, ∆V = IR = a 2.48 A fa5.00 Ωf = 12.4 V .
b
R+r
=
g a
gb
ε = I 2 + 35.0 A 0.080 0 Ω + I 2 5.00 Ω = 12.6 V
giving
I 2 = 1.93 A .
Thus,
∆V2 = 1.93 A 5.00 Ω = 9.65 V .
a
fa
FIG. P28.4
f
so
f
Resistors in Series and Parallel
a
f
g a fb
g
Therefore, a 2.00 A fR = a1.60 A fb R + 3.00 Ωg or R = 12.0 Ω .
b
∆V = I 1 R1 = 2.00 A R1 and ∆V = I 2 R1 + R 2 = 1.60 A R1 + 3.00 Ω
1
P28.6
ε
Here ε = I R + r , so I =
Section 28.2
P28.5
a
(a)
(a)
Rp =
1
1
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
34.0 V = I 17.1 Ω
a
f
FIG. P28.6
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
P28.7
f
I = 0.818 A for 10.0 Ω resistor.
For the bulb in use as intended,
P
75.0 W
=
= 0.625 A
∆V
120 V
∆V
120 V
=
= 192 Ω .
R=
0.625 A
I
I=
and
FIG. P28.7
Now, presuming the bulb resistance is unchanged,
I=
Across the bulb is
so its power is
120 V
= 0.620 A .
193.6 Ω
a
f
P = I∆V = 0.620 Aa119 V f =
∆V = IR = 192 Ω 0.620 A = 119 V
73.8 W .
Chapter 28
P28.8
120 V = IReq = I
∆V2 =
P28.9
Iρ
=
A2 A2
FG ρ
HA
e
+
1
1
A1
IJ
K
ρ
ρ
ρ
+
+
, or Iρ =
A 2 A3 A 4
a120 V f
+
1
A2
+
1
A3
+
1
A4
j
e
a120 Vf
1
A1
+
1
A2
+
1
A3
+
1
A4
135
j
= 29.5 V
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:
1
Req =
= 2.94 Ω .
1
1
1
10.0 Ω + 5.00 Ω + 25 .0 Ω
c
h
This is shown in figure (c), which in turn reduces to the circuit shown in
figure (d).
∆V
Next, we work backwards through the diagrams applying I =
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
∆V
25.0 V
=
= 1.93 A .
I=
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:
∆V = IR = 1.93 A 2.94 Ω = 5.68 V .
a
fa
f
From figure (b), we see that this potential difference is the same across ∆Vab ,
the 10 Ω resistor, and the 5.00 Ω resistor.
*P28.10
(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,
∆Vab 5.68 V
=
= 0.227 A = 227 mA .
I=
25.0 Ω
R ab
FIG. P28.9
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
resistors is
. The voltmeter displays
1.00 MΩ + Rshoes
50.0 V 1.00 MΩ
∆V = I 1.00 MΩ =
= ∆V .
1.00 MΩ + Rshoes
a
f
a
f
a
a
f
f
b
50.0 V 1.00 MΩ = ∆V 1.00 MΩ + ∆V Rshoes
1.00 MΩ 50.0 − ∆V
Rshoes =
.
∆V
(a)
We solve to obtain
(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
a
f
g
136
P28.11
Direct Current Circuits
(a)
Since all the current in the circuit must pass through the series
100 Ω resistor, P = I 2 R
2
Pmax = RI max
P
25.0 W
=
= 0.500 A
100 Ω
R
I max =
so
R eq = 100 Ω +
FG 1 + 1 IJ
H 100 100 K
−1
FIG. P28.11
Ω = 150 Ω
∆Vmax = R eq 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
P28.12
2
= 6.25 W
Using 2.00-Ω, 3.00-Ω, 4.00-Ω resistors, there are 7 series, 4 parallel, and 6 mixed combinations:
Series
2.00 Ω
3.00 Ω
4.00 Ω
5.00 Ω
P28.13
f
Parallel
0.923 Ω
1.20 Ω
1.33 Ω
1.71 Ω
6.00 Ω
7.00 Ω
9.00 Ω
Mixed
1.56 Ω
2.00 Ω
2.22 Ω
3.71 Ω
4.33 Ω
5.20 Ω
The resistors may be arranged in patterns:
The potential difference is the same across either combination.
∆V = IR = 3 I
1+
c
1
1
R
+
1
500
R
=3
500
h
FG 1 + 1 IJ = 3
H R 500 K
so
R
and
R = 1 000 Ω = 1.00 kΩ .
FIG. P28.13
*P28.14
When S is open, R1 , R 2 , R3 are in series with the battery. Thus:
6V
(1)
R1 + R 2 + R3 = −3 = 6 kΩ .
10 A
When S is closed in position 1, the parallel combination of the two R 2 ’s is in series with R1 , R3 , and the
battery. Thus:
1
6V
= 5 kΩ.
(2)
R1 + R 2 + R3 =
2
1.2 × 10 −3 A
When S is closed in position 2, R1 and R 2 are in series with the battery. R3 is shorted. Thus:
6V
= 3 kΩ.
(3)
R1 + R 2 =
2 × 10 −3 A
From (1) and (3): R3 = 3 kΩ.
Subtract (2) from (1): R 2 = 2 kΩ .
From (3): R1 = 1 kΩ.
Answers: R1 = 1.00 kΩ , R 2 = 2.00 kΩ , R3 = 3.00 kΩ .
Chapter 28
P28.15
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
137
∆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
P28.16
P3 =
b∆V g = a2.00 V f
P1 =
b∆V g = a2.00 Vf
3
2
2
3.00 Ω
R3
1
= 1.33 W in 3.00 Ω
FIG. P28.15
2
= 4.00 W in 1.00 Ω
1.00 Ω
R1
g
Denoting the two resistors as x and y,
x + y = 690, and
1
1 1
= +
150 x y
a
f
690 − x + x
1
1
1
= +
=
150 x 690 − x
x 690 − x
a
f
2
x − 690 x + 103 500 = 0
x=
690 ±
a690f
− 414 000
2
x = 470 Ω
*P28.17
2
y = 220 Ω
a f
A certain quantity of energy ∆Eint = P time is required to raise the temperature of the water to
a∆V f where a∆V f is a constant.
R
∆V f ∆t a ∆V f 2 ∆t
a
=
. Then R = 2 R .
Thus comparing coils 1 and 2, we have for the energy
2
100°C . For the power delivered to the heaters we have P = I∆V =
2
2
R1
(a)
2
R2
When connected in parallel, the coils present equivalent resistance
Rp =
a f
a f
2
2
∆V ∆t p
∆V ∆t
2 R1
1
1
=
=
=
. Now
1 R1 + 1 R 2 1 R1 + 1 2 R1
3
2 R1 3
R1
∆t p =
a ∆V f ∆t = a ∆ V f ∆ t
2
(b)
1
For the series connection, Rs = R1 + R 2 = R1 + 2 R1 = 3 R1 and
∆t s = 3 ∆t .
R1
2
3 R1
s
2 ∆t
.
3
138
P28.18
Direct Current Circuits
(a)
a
f
a
∆V = IR :
33.0 V = I1 11.0 Ω
33.0 V = I 2 22.0 Ω
I 3 = 3.00 A
I 2 = 1.50 A
P = I 2R:
P1 = 3.00 A
P1 = 99.0 W
a
f a11.0 Ωf
2
a
f
fa
2
P2 = 1.50 A 22.0 Ω
P2 = 49.5 W
f
The 11.0- Ω resistor uses more power.
FIG. P28.18(a)
a f a fa f
P = I ∆V = 4.50 33.0 = 148 W
(b)
P1 + P2 = 148 W
(c)
Rs = R1 + R 2 = 11.0 Ω + 22.0 Ω = 33.0 Ω
a
f
∆V = IR :
33.0 V = I 33.0 Ω , so I = 1.00 A
P = I 2R:
P1 = 1.00 A
P1 = 11.0 W
a
f a11.0 Ωf
2
a
f a22.0 Ωf
P2 = 1.00 A
P2 = 22.0 W
2
FIG. P28.18(c)
The 22.0- Ω resistor uses more power.
(d)
(e)
*P28.19
g a f a33.0 Ωf =
P = I a ∆V f = a1.00 A fa33.0 V f = 33.0 W
b
P1 + P2 = I 2 R1 + R2 = 1.00 A
2
33.0 W
The parallel configuration uses more power.
(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 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
(b)
Based on the reasoning above the potential differences are
ε
2ε
4ε
2ε
.
∆V1 = , ∆V2 =
, ∆V3 =
, ∆V4 =
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 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
I
2I
the resistors are I 1 = I , I 2 = I 3 = , I 4 =
.
3
3
continued on next page
Chapter 28
(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
, I2 = I3 = 0, I4 =
.
and 4, and none passes through 2 or 3. Therefore I 1 =
4
4
Section 28.3
P28.20
Kirchhoff’s Rules
a f a fa f
+15.0 − 7.00 I1 − 2.00 5.00 = 0
5.00 = 7.00 I 1
so
I 1 = 0.714 A
so
I 2 = 1.29 A
I 3 = I 1 + I 2 = 2.00 A
0.714 + I 2 = 2.00
a f
a f
+ε − 2.00 1.29 − 5.00 2.00 = 0
P28.21
139
FIG. P28.20
ε = 12.6 V
We name currents I 1 , I 2 , and I 3 as shown.
From Kirchhoff’s current rule, I 3 = I 1 + I 2 .
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 I1 = 0
a8.00fI
1
a f
= 4.00 + 6.00 I 2 .
FIG. P28.21
Solving the above linear system, we proceed to the pair of simultaneous equations:
RS8 = 4I + 4I
T8I = 4 + 6 I
1
1
2
+ 6I2
or
2
RS8 = 4I + 10I
TI = 1.33I − 0.667
1
2
2
1
and to the single equation 8 = 4I 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.
140
P28.22
Direct Current Circuits
The solution figure is shown to the right.
FIG. P28.22
P28.23
We use the results of Problem 28.21.
(a)
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 .
By the 4.00-V battery:
By the 12.0-V battery:
(b)
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
By the 8.00-Ω resistor:
.
2
By the 5.00-Ω resistor:
2
By the 1.00-Ω resistor:
2
By the 3.00-Ω resistor:
2
By the 1.00-Ω resistor:
(c)
2
−222 J + 1.88 kJ = 1.66 kJ from chemical to electrical.
687 J + 128 J + 25.6 J + 616 J + 205 J = 1.66 kJ from electrical to internal.
P28.24
We name the currents I 1 , I 2 , and I 3 as shown.
[2]
f a
f
80.0 − I a 4.00 kΩf − 60.0 − I a3.00 kΩf = 0
[3]
I 2 = I1 + I 3
[1]
(a)
a
70.0 − 60.0 − I 2 3.00 kΩ − I 1 2.00 kΩ = 0
3
2
Substituting for I 2 and solving the resulting simultaneous
equations yields
bthrough R g
I = 2.69 mA bthrough R g
I = 3.08 mA bthrough R g
∆V = −60.0 V − a3.08 mA fa3.00 kΩf =
I 1 = 0.385 mA
(b)
1
3
3
2
2
cf
Point c is at higher potential.
−69.2 V
FIG. P28.24
Chapter 28
P28.25
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:
and
a2.71RfI + a1.71RfI
a1.71RfI + a3.71RfI
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
(b)
Finally, applying Kirchhoff’s point rule at point a in Figure (a)
gives:
FIG. P28.25
I = I 4 − I 1 = 60.0 mA − 10.0 mA = +50.0 mA ,
or
P28.26
I = 50.0 mA from point a to point e .
Name the currents as shown in the figure to the right. Then w + x + z = y . Loop
equations are
−200 w − 40.0 + 80.0 x = 0
FIG. P28.26
−80.0 x + 40.0 + 360 − 20.0 y = 0
+360 − 20.0 y − 70.0 z + 80.0 = 0
Eliminate y by substitution.
Eliminate x.
Eliminate z = 17.5 − 13.5 w to obtain
R|x = 2.50w + 0.500
S|400 − 100 x − 20.0 w − 20.0 z = 0
T440 − 20.0 w − 20.0 x − 90.0 z = 0
RS350 − 270 w − 20.0 z = 0
T430 − 70.0w − 90.0z = 0
430 − 70.0 w − 1 575 + 1 215 w = 0
w=
Now
70.0
= 1.00 A upward in 200 Ω .
70.0
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,
a
fa
f
∆V = IR = 1.00 A 200 Ω = 200 V .
141
142
P28.27
Direct Current Circuits
Using Kirchhoff’s rules,
b g b g
10.0 + a1.00fI − a0.060 0fI = 0
12.0 − 0.010 0 I 1 − 0.060 0 I 3 = 0
2
and
3
I1 = I 2 + I 3
a
f a
f
10.0 + a1.00fI − b0.060 0 gI = 0
12.0 − 0.010 0 I 2 − 0.070 0 I 3 = 0
2
FIG. P28.27
3
Solving simultaneously,
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.
P28.28
a f a fb g
= a1.00fI + a1.00fI + a5.00fb I − I + I g
= a3.00fb I − I g + a5.00fb I − I + I g
∆Vab = 1.00 I 1 + 1.00 I 1 − I 2
∆Vab
∆Vab
1
2
1
1
1
2
2
Let I = 1.00 A , I 1 = x , and I 2 = y .
Then, the three equations become:
FIG. P28.28
∆Vab = 2.00 x − y , or y = 2.00 x − ∆Vab
∆Vab = −4.00 x + 6.00 y + 5.00
and ∆Vab = 8.00 − 8.00 x + 5.00 y .
Substituting the first into the last two gives:
7.00 ∆Vab = 8.00 x + 5.00 and 6.00 ∆Vab = 2.00 x + 8.00 .
Solving these simultaneously yields ∆Vab =
Then, R ab =
P28.29
27
V
∆Vab
= 17
1.00 A
I
27
V.
17
R ab =
or
27
Ω .
17
We name the currents I 1 , I 2 , and I 3 as shown.
(a)
I1 = I 2 + I 3
Counterclockwise around the top loop,
a
f a
f
12.0 V − 2.00 Ω I 3 − 4.00 Ω I 1 = 0 .
Traversing the bottom loop,
a
f a
f
8.00 V − 6.00 Ω I 2 + 2.00 Ω I 3 = 0
4 1
1
I 1 = 3.00 − I 3 , I 2 = + I 3 , and I 3 = 909 mA .
3 3
2
(b)
a
fa
f
Va − 0.909 A 2.00 Ω = Vb
Vb − Va = −1.82 V
FIG. P28.29
Chapter 28
P28.30
143
We apply Kirchhoff’s rules to the second diagram.
50.0 − 2.00 I 1 − 2.00 I 2 = 0
(1)
20.0 − 2.00 I 3 + 2.00 I 2 = 0
(2)
I1 = I 2 + I 3
(3)
Substitute (3) into (1), and solve for I 1 , I 2 , and I 3
I 1 = 20.0 A ; I 2 = 5.00 A ; I 3 = 15.0 A .
Then apply P = I 2 R to each resistor:
a
f a f a2.00 Ωf =
F 5.00 AIJ a4.00 Ωf = 25.0 W
P =G
H 2 K
a2.00 Ωf :
a4.00 Ωf :
P = I 12 2.00 Ω = 20.0 A
1
P28.32
FIG. P28.30
(Half of I 2 goes through each)
a
f a
P = I 32 2.00 Ω = 15.0 A
3
P28.31
800 W
2
a2.00 Ωf :
Section 28.4
2
f a2.00 Ωf =
2
450 W .
RC Circuits
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)
It =
(a)
I t = − I 0 e − t RC
5.10 × 10 −6 C
Q
=
= 1.96 A
I0 =
RC
1 300 Ω 2.00 × 10 −9 F
ja
e
af
FG
H
f
IJ
K
LM
MN e
ε − t RC
−10.0
30.0
=
exp
e
6
R
1.00 × 10
1.00 × 10 6 5.00 × 10 −6
je
OP
=
j PQ
4.06 µA
FIG. P28.31
af
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
= b5.10 µC g exp M
qat f = Qe
MN a1 300 Ωfe2.00 × 10 Fj PQ = 0.235 µC
−6
−9
(b)
(c)
P28.33
U=
−6
− t RC
−9
The magnitude of the maximum current is I 0 = 1.96 A .
a f
1
C ∆V
2
2
and ∆V =
Q
.
C
Q2
and when the charge decreases to half its original value, the stored energy is one2C
1
quarter its original value: U f = U 0 .
4
Therefore, U =
144
P28.34
Direct Current Circuits
so
qt
= 1 − e − t RC
Q
0.600 = 1 − e −0 .900 RC
or
e −0.900 RC = 1 − 0.600 = 0.400
thus
RC =
af
a
−0.900
= ln 0.400
RC
*P28.35
f
−0.900
= 0.982 s .
ln 0.400
a
f
We are to calculate
z
∞
e −2 t RC dt = −
0
P28.36
af
q t = Q 1 − e − t RC
z
∞
IJ
K
FG
H
RC −2 t RC 2dt
RC −2 t RC
−
=−
e
e
RC
2 0
2
e
je
∞
0
=−
j
(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
j
FG 10.0 V IJ e
H 100 × 10 Ω K
200 µA + b100 µA ge
.
(a)
− t 1.00 s
3
.
− t 1.00 s
So the switch carries downward current
P28.37
RC −∞
RC
RC
e − e0 = −
0−1 = +
.
2
2
2
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 V f = 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. P28.37(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=
and
so
1
= 3.60 Ω
1 9.00 Ω + 1 6.00 Ω
b
g b
g
RC = 3.60 × 10 −6 s
1
e − t RC =
10
t = RC ln 10 = 8. 29 µs .
FIG. P28.37(b)
Chapter 28
*P28.38
(a)
af
af
t = RC ln
FG
H
∆V0
t
= ln
RC
∆V
∆V0
= e + t RC
∆V
∆V
= e − t RC
∆V0
P28.39
3 000 V
We model the person’s body and street shoes
as shown. For the discharge to reach 100 V,
q t = Qe − t RC = C∆V t = C∆V0 e − t RC
FG ∆V IJ = 5 000 × 10 Ωe230 × 10
H ∆V K
6
0
e
−12
150 pF
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
(a)
τ = RC = 4.00 × 10 6 Ω 3.00 × 10 −6 F = 12.0 s
(b)
I=
je
3.91 s
ε − t RC
12.0
=
e
e − t 12.0 s
6
R
4.00 × 10
a f
q = 36.0 µC 1 − e − t 12 .0
∆V0 =
FIG. P28.38(a)
j
q = Cε 1 − e − t RC = 3.00 × 10 −6 12.0 1 − e − t 12.0
P28.40
5 000 M Ω
80 pF
j
(b)
e
I = 3.00 µAe − t 12 .0
Q
C
FIG. P28.39
b g
Then, if q(t ) = Q e − t RC
∆V ( t ) = ∆V0 e − t RC
and
b∆V g = e
∆V ( t )
− t RC .
0
b g
When ∆V ( t ) = 1 ∆V0 , then
2
e − t RC =
−
Section 28.5
P28.41
FG IJ
H K
t
1
= ln
= − ln 2 .
RC
2
R=
Thus,
1
2
t
C ln 2
a f
.
Electrical Meters
e
j
∆V = I g rg = I − I g R p , or R p =
I g rg
eI − I
g
a f
j eI − I j
=
I g 60.0 Ω
g
Therefore, to have I = 0.100 A = 100 mA when I g = 0.500 mA :
Rp =
a0.500 mAfa60.0 Ωf =
99.5 mA
145
0.302 Ω .
FIG. P28.41
146
P28.42
Direct Current Circuits
a
f e
j
Applying Kirchhoff’s loop rule, − I g 75.0 Ω + I − I g R p = 0 .
Therefore, if I = 1.00 A when I g = 1.50 mA ,
Rp =
a f = e1.50 × 10 Aja75.0 Ωf =
eI − I j 1.00 A − 1.50 × 10 A
−3
I g 75.0 Ω
−3
g
P28.43
Series Resistor → Voltmeter
0.113 Ω .
b
25.0 = 1.50 × 10 −3 Rs + 75.0
∆V = IR :
FIG. P28.42
g
Rs = 16.6 kΩ .
Solving,
FIG. P28.43
P28.44
(a)
In Figure (a), the
emf sees an
equivalent
resistance of
200.00 Ω .
6.000 0 V
200.00 Ω
= 0.030 000 A
I=
6.0000 V
A
V
A
V
180.00 Ω
180.00 Ω
180.00 Ω
(a)
(b)
(c)
FIG. P28.44
(b)
20.000 Ω
20.000 Ω
20.000 Ω
b
f
ga
The terminal potential difference is
∆V = IR = 0.030 000 A 180.00 Ω = 5.400 0 V .
In Figure (b),
R eq =
F 1 + 1 I
GH 180.00 Ω 20 000 Ω JK
−1
= 178.39 Ω.
The equivalent resistance across the emf is 178.39 Ω + 0.500 00 Ω + 20.000 Ω = 198.89 Ω .
The ammeter reads
and the voltmeter reads
I=
ε
R
=
6.000 0 V
= 0.030 167 A
198.89 Ω
b
ga
F 1 + 1 I
GH 180.50 Ω 20 000 Ω JK
f
∆V = IR = 0.030 167 A 178.39 Ω = 5.381 6 V .
−1
(c)
In Figure (c),
= 178.89 Ω .
Therefore, the emf sends current through
Rtot = 178.89 Ω + 20.000 Ω = 198.89 Ω.
The current through the battery is
I=
but not all of this goes through the ammeter.
6.000 0 V
= 0.030 168 A
198.89 Ω
b
ga
f
The voltmeter reads
∆V = IR = 0.030 168 A 178.89 Ω = 5.396 6 V .
The ammeter measures current
I=
∆V 5.396 6 V
=
= 0.029 898 A .
R
180.50 Ω
The connection shown in Figure (c) is better than that shown in Figure (b) for accurate
readings.
Chapter 28
P28.45
Consider the circuit diagram shown, realizing that
I g = 1.00 mA . For the 25.0 mA scale:
a24.0 mAfbR
1
g a
fa
+ R 2 + R3 = 1.00 mA 25.0 Ω
f
FG
H
FIG. P28.45
IJ
K
25.0
R1 + R 2 + R3 =
Ω.
24.0
or
(1)
(2)
For the 100 mA scale:
a49.0 mAfbR + R g = a1.00 mAfb25.0 Ω + R g
49.0b R + R g = 25.0 Ω + R .
a99.0 mAfR = a1.00 mAfb25.0 Ω + R + R g
or
99.0 R1 = 25.0 Ω + R 2 + R3 .
(3)
For the 50.0 mA scale:
1
or
1
2
3
2
3
1
2
3
Solving (1), (2), and (3) simultaneously yields
R1 = 0.260 Ω , R 2 = 0.261 Ω , R3 = 0.521 Ω .
P28.46
∆V = IR
(a)
jb
e
20.0 V = 1.00 × 10 −3 A R1 + 60.0 Ω
g
R1 = 1.994 × 10 4 Ω = 19.94 kΩ
P28.47
FIG. P28.46
e
jb
g
R 2 = 30.0 kΩ
e
jb
g
R3 = 50.0 kΩ
(b)
50.0 V = 1.00 × 10 −3 A R 2 + R1 + 60.0 Ω
(c)
100 V = 1.00 × 10 −3 A R3 + R1 + 60.0 Ω
e
ja
Ammeter:
I g r = 0.500 A − I g 0.220 Ω
or
I g r + 0.220 Ω = 0.110 V
Voltmeter:
2.00 V = I g r + 2 500 Ω
a
f
b
f
(1)
g
(2)
Solve (1) and (2) simultaneously to find:
I g = 0.756 mA and r = 145 Ω .
FIG. P28.47
Section 28.6
Household Wiring and Electrical Safety
FG ρ IJ = a1.00 Af e1.70 × 10 Ω ⋅ mja16.0 ftfb0.304 8 m ftg =
H AK
π e0.512 × 10 mj
2
P28.48
147
2
2
(a)
P =I R=I
(b)
P = I 2 R = 100 0.101 Ω = 10.1 W
−8
−3
a
f
2
0.101 W
148
P28.49
Direct Current Circuits
(a)
P = I∆V :
1 500 W
P
=
= 12.5 A .
∆V
120 V
750 W
I=
= 6.25 A .
120 V
I=
So for the Heater,
For the Toaster,
I=
And for the Grill,
(b)
1 000 W
= 8.33 A .
120 V
12.5 + 6.25 + 8.33 = 27.1 A
The current draw is greater than 25.0 amps, so this circuit breaker would not be sufficient.
P28.50
2
2
I Al
R Al = I Cu
RCu
P28.51
(a)
RCu
I Cu =
R Al
I Al =
so
ρ Cu
1.70
20.0 = 0.776 20.0 = 15.5 A
I Cu =
2.82
ρ Al
a f
a f
Suppose that the insulation between either of your fingers and the conductor adjacent is a
chunk of rubber with contact area 4 mm 2 and thickness 1 mm. Its resistance is
e
je
j
10 13 Ω ⋅ m 10 −3 m
ρ
≈
≈ 2 × 10 15 Ω .
A
4 × 10 −6 m 2
R=
The current will be driven by 120 V through total resistance (series)
2 × 10 15 Ω + 10 4 Ω + 2 × 10 15 Ω ≈ 5 × 10 15 Ω .
It is: I =
(b)
∆V
120 V
~
~ 10 −14 A .
15
R
5 × 10 Ω
Vh
, where Vh
2
is the potential of the “hot” wire. The potential difference between your finger and thumb is
∆V = IR ~ 10 −14 A 10 4 Ω ~ 10 −10 V . So the points where the rubber meets your fingers are
The resistors form a voltage divider, with the center of your hand at potential
e
je
j
at potentials of
~
Vh
+ 10 −10 V
2
and
~
Vh
− 10 −10 V .
2
Additional Problems
P28.52
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 =
Q
:
∆t
fb
E
:
∆t
∆t =
∆t =
g
E
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
149
Chapter 28
P28.53
aR + r f = FGH εP IJK R .
R + a 2 r − x fR − r = 0 .
R + a 2.40 − xfR − 1.44 = 0 ,
−a 2.40 − x f ± a 2.40 − x f
R=
ε
ε 2R
, so P = I 2 R =
or
2
R+r
R+r
ε2
2
, then R + r = xR or
Let x ≡
P
With r = 1. 20 Ω, this becomes
I=
a
f
a
2
f
2
With
(b)
For
2
− 5.76
2
ε = 9.20 V and
R=
2
2
which has solutions of
(a)
2
+4.21 ±
P = 12.8 W , x = 6.61 :
a4.21f
2
− 5.76
2
= 3.84 Ω or
ε = 9.20 V and
+1.59 ±
.
0.375 Ω .
P = 21.2 W , x ≡
a1.59f
2
ε2
= 3.99
P
− 5.76
1.59 ± −3.22
=
.
2
2
The equation for the load resistance yields a complex number, so there is no resistance
R=
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 =
P28.54
ε2
= 17.6 W .
4r
Using Kirchhoff’s loop rule for the closed loop, +12.0 − 2.00 I − 4.00 I = 0 , so I = 2.00 A
a
fa
f a fa
f
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 .
P28.55
*P28.56
I=
ε
3R
Pseries = ε I =
I=
3ε
R
Pparallel = ε I =
(a)
Req = 3 R
(b)
Req =
(c)
Nine times more power is converted in the parallel connection.
(a)
We model the generator as a constant-voltage power supply.
Connect two light bulbs across it in series. Each bulb is designed to
P
100 W
=
= 0.833 A . Each has resistance
carry current I =
∆V 120 V
∆V
120 V
=
= 144 Ω . In the 240-V circuit the equivalent
R=
0.833 A
I
resistance is 144 Ω + 144 Ω = 288 Ω . The current is
∆V 240 V
=
= 0.833 A and the generator delivers power
I=
288 Ω
R
P = I∆V = 0.833 A 240 V = 200 W .
1
R
=
3
1R + 1R + 1R
b g b g b g
a
continued on next page
f
ε2
3R
3ε 2
R
FIG. P28.56(a)
150
Direct Current Circuits
(b)
The hot pot is designed to carry
current
I=
P
∆V
=
28.8 Ω
500 W
= 4.17 A .
120 V
144 Ω
240 V
It has resistance
R=
FIG. P28.56(b)
∆V 120 V
=
= 28.8 Ω .
4.17 A
I
4.17 A
= 5 , we can place five light bulbs in parallel and the hot
0.833 A
pot in series with their combination. The current in the generator is then 4.17 A and it
In terms of current, since
a
f
delivers power P = I∆V = 4.17 A 240 V = 1 000 W .
P28.57
The current in the simple loop circuit will be I =
(a)
∆Vter = ε − Ir =
(b)
I=
(c)
P = I2R = ε 2
εR
R+r
ε
R+r
Then 2R = R + r
P28.58
.
and
∆Vter → ε as R → ∞ .
and
I→
ε
as R → 0 .
r
ε2
− 2ε 2 R
dP
=
+
3
dR
R+r
R+r
R
aR + r f
ε
R+r
a
2
and
f a
f
2
FIG. P28.57
=0
R=r .
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
j
− 3 .00 × 10 5 Ω R
f Re10.0 ×10
−6
sΩ
j O.
PQ
.
= 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
Chapter 28
P28.59
Let the two resistances be x and y.
Ps
225 W
=
= 9.00 Ω
2
2
I
5.00 A
Then,
Rs = x + y =
and
Rp =
so
x 9.00 Ω − x
= 2.00 Ω
x + 9.00 Ω − x
a
a
y = 9.00 Ω − x
f
x
Pp
xy
50.0 W
= 2 =
= 2.00 Ω
2
x+y I
5.00 A
f
a
a
y
x
f
y
x 2 − 9.00 x + 18.0 = 0 .
f
Factoring the second equation,
ax − 6.00fax − 3.00f = 0
so
x = 6.00 Ω or x = 3.00 Ω .
Then, y = 9.00 Ω − x gives
y = 3.00 Ω or y = 6.00 Ω .
FIG. P28.59
The two resistances are found to be 6.00 Ω and 3.00 Ω .
P28.60
Let the two resistances be x and y.
Then, Rs = x + y =
Pp
xy
Ps
=
=
and
R
.
p
x + y I2
I2
From the first equation, y =
becomes
e
x Ps I 2 − x
e
2
j
x + Ps I − x
j
=
Pp
I
2
Using the quadratic formula, x =
Then, y =
(a)
(b)
FG P IJ x + P P
HI K I
s p
4
s
2
Ps ± Ps 2 − 4Ps Pp
2I 2
Ps + Ps 2 − 4Ps Pp
2I 2
and
= 0.
FIG. P28.60
.
Ps − Ps 2 − 4Ps Pp
2I 2
.
c ∑ R h − bε + ε g = 0
40.0 V − a 4.00 A f a 2.00 + 0.300 + 0.300 + RfΩ − a6.00 + 6.00 f V = 0 ;
ε−I
1
2
g a
fa
2
f
2
2
2
For the limiting resistor,
b
so
R = 4.40 Ω
a f a2.00 Ωf = 32.0 W .
P = I R = a 4.00 A f a0.600 Ωf = 9.60 W .
P = a 4.00 A f a 4.40 Ωf = 70.4 W .
P = I 2 R = 4.00 A
Inside the supply,
Inside both batteries together,
(c)
y
Ps ∓ Ps 2 − 4Ps Pp
Ps
.
−
=
x
gives
y
2I 2
I2
The two resistances are
P28.61
x
Ps
− x , and the second
I2
or x 2 −
y
x
P = I ε 1 + ε 2 = 4.00 A 6.00 + 6.00 V = 48.0 W
151
152
*P28.62
Direct Current Circuits
(a)
∆V1 = ∆V2
I 1 R1 = I 2 R 2
I R
R + R1
I = I 1 + I 2 = I1 + 1 1 = I1 2
R2
R2
I1 =
I2 =
(b)
R2
I2
FIG. P28.62(a)
I 1 R1
IR1
=
= I2
R2
R1 + R 2
b
The power delivered to the pair is P = I12 R1 + I 22 R2 = I 12 R1 + I − I1
dP
we want to find I 1 such that
= 0.
dI 1
ga f
dP
= 2 I 1 R1 + 2 I − I 1 −1 R 2 = 0
dI 1
I1 =
I1
I
IR 2
R1 + R 2
b
R1
g R . For minimum power
2
2
I 1 R1 − IR 2 + I 1 R 2 = 0
IR 2
R1 + R 2
This is the same condition as that found in part (a).
P28.63
Let Rm = measured value, R = actual value,
(a)
I R = current through the resistor R
I = current measured by the ammeter.
(a)
b
g
When using circuit (a), I R R = ∆V = 20 000 I − I R or R = 20 000
But since I =
∆V
∆V
and I R =
, we have
Rm
R
I
R
=
I R Rm
FIG. P28.63
bR − R g .
m
R = 20 000
When R > Rm , we require
bR − R g ≤ 0.050 0 .
b
When using circuit (b),
But since I R =
∆V
,
Rm
When Rm > R , we require
From (2) we find
(b)
R
and
Rm
(1)
m
R
g
Therefore, Rm ≥ R 1 − 0.050 0 and from (1) we find
(b)
LM I − 1OP .
NI Q
R ≤ 1 050 Ω .
a
f
I R R = ∆V − I R 0.5 Ω .
a
f
Rm = 0.500 + R .
bR
m
−R
R
g ≤ 0.050 0 .
R ≥ 10.0 Ω .
(2)
Chapter 28
P28.64
FG
H
dE
ε −1 RC
e
.
= P = εI = ε
dt
R
Then the total energy put out by the battery is
z z
z
dE =
FG
H
fz
∞
ε2
t
− RC exp −
R
RC
0
a
dE =
IJ FG − dt IJ = −ε C expFG − t IJ
K H RC K
H RC K
IJ
K
FG
H
∞
ε2
t
exp −
dt
R
RC
t=0
∞
2
= −ε 2 C 0 − 1 = ε 2C .
0
FG
H
IJ
K
The power delivered to the resistor is
ε2
dE
2t
.
= P = ∆VR I = I 2 R = R 2 exp −
dt
RC
R
So the total internal energy appearing in the resistor is
z z
IJ z expFG − 2t IJ FG − 2dt IJ = − ε C expFG − 2t IJ
z
K H RC K H RC K 2 H RC K
1
The energy finally stored in the capacitor is U = C a ∆V f
2
dE =
FG
H
∞
ε2
RC
−
R
2
battery.
(a)
2
0
conserved ε 2 C =
P28.65
IJ
K
The battery supplies energy at a changing rate
153
e
I=
0
2
=
1 2
Cε . Thus, energy of the circuit is
2
j
L
F ja10.0 V fM1 − e
N
−10.0
e 2.00 ×10 je1.00 ×10 j O =
−6
6
PQ
9.93 µC
FG IJ
H K
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 W = 334 nW
dt GH 1.00 × 10 C V JK
P
= Iε = e3.37 × 10 A ja10.0 V f = 3.37 × 10 W = 337 nW
−5.00
6
−8
2
−6
−8
−6
battery
IJ
K
ε 2C
ε 2C
0−1 =
.
2
2
dq
∆V − t RC
=
e
dt
R
I=
(d)
=−
FG
H
ε2
2t
exp −
dt
R
RC
0
q = C∆V 1 − e − t RC
e
(c)
∞
∞
1 2
1
ε C + ε 2C and resistor and capacitor share equally in the energy from the
2
2
q = 1.00 × 10 −6
(b)
dE =
−8
−7
−7
154
P28.66
Direct Current Circuits
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 R 2 C
and the switch has just closed. The voltage is
∆V
Voltage
controlled
switch
a f LMN 32 ∆V OPQe .
1
We want to know when ∆V atf will reach ∆V .
3
1
L2 O
Therefore,
∆V = M ∆V P e
3
N3 Q
R2
− t R 2C
∆VC t =
+
R1
C
V ∆Vc
C
− t R2C
1
2
or
e − t R 2C =
or
t1 = R 2 C ln 2 .
FIG. P28.66
1
After the switch opens, the voltage is ∆V , increasing toward ∆V with time constant R1 + R 2 C :
3
b
LM 2 ∆V OPe
N3 Q
af
∆VC t = ∆V −
.
af
2
∆V
3
2
2
∆V = ∆V − ∆Ve − t b R1 + R2 gC
3
3
b
g
t 2 = R1 + R 2 C ln 2
So
(a)
g
∆VC t =
When
P28.67
b
− t R1 + R 2 C
g
or
e − t b R1 + R2 gC =
and
T = t1 + t 2 =
First determine the resistance of each light bulb: P =
a∆V f = a120 Vf
2
R=
bR
1
g
+ 2 R 2 C ln 2 .
2
R
2
60.0 W
P
a∆V f
1
.
2
= 240 Ω .
We obtain the equivalent resistance Req of the network of light
FIG. P28.67
bulbs by identifying series and parallel equivalent resistances:
Req = R1 +
1
= 240 Ω + 120 Ω = 360 Ω.
+ 1 R3
b1 R g b g
2
a∆V f = a120 Vf
2
The total power dissipated in the 360 Ω is
(b)
P=
The current through the network is given by P = I 2 Req : I =
The potential difference across R1 is
Req
360 Ω
2
= 40.0 W .
P
40.0 W 1
=
= A.
360 Ω
3
Req
∆V1 = IR1 =
FG 1 AIJ a240 Ωf =
H3 K
The potential difference ∆V23 across the parallel combination of R 2 and R3 is
∆V23 = IR 23 =
I=
1
FG 1 AIJ FG
H 3 K H b1 240 Ωg + b1 240 Ωg JK
40.0 V .
80.0 V .
Chapter 28
*P28.68
(a)
155
With the switch closed, current exists in a simple series
circuit as shown. The capacitors carry no current. For R 2
we have
P
2.40 V ⋅ A
=
= 18.5 mA .
P = I 2 R2
I=
R2
7 000 V A
The potential difference across R1 and C 1 is
jb
e
g
∆V = IR1 = 1.85 × 10 −2 A 4 000 V A = 74.1 V .
The charge on C 1
ja
e
FIG. P28.68(a)
f
Q = C1 ∆V = 3.00 × 10 −6 C V 74.1 V = 222 µC .
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
ja
e
f
Q = C 2 ∆V = 6.00 × 10 −6 C V 130 V = 778 µC .
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 C 2
e
ja
f
Q = C 2 ∆V = 6.00 × 10 −6 C V 204 V = 1 222 µC
for a change of 1 222 µC − 778 µC = 444 µC .
*P28.69
FIG. P28.68(b)
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. P28.69
211
=
a150 + 45 + 14 + 2f mA = 211 mA , nearly the same as before. The ratio is 213
(b)
The resistor with least resistance carries 150 mA. Doubling its resistance changes this current
to 75 mA and changes the total to
138
75 + 45 + 14 + 4 mA = 138 mA . The ratio is
= 0.648 , representing a much larger
213
reduction (35.2% instead of 0.9%).
a
(c)
0.991 .
f
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.
156
*P28.70
Direct Current Circuits
From the hint, the equivalent resistance of
RT +
That is,
RT +
.
1
= R eq
1 RL + 1 R eq
RL R eq
= R eq
RL + R eq
2
RT RL + RT R eq + RL R eq = RL R eq + R eq
2
− RT R eq − RT RL = 0
R eq
R eq =
a fb
2a1f
RT ± RT2 − 4 1 − RT RL
g
Only the + sign is physical:
R eq =
P28.71
1
2
FH
IK
4RT RL + RT2 + RT .
For example, if
RT = 1 Ω.
And
RL = 20 Ω , R eq = 5 Ω.
(a)
After steady-state conditions have been reached, there is no DC current through the
capacitor.
b
g
I R3 = 0 steady-state .
Thus, for R3 :
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 R 2 :
(b)
ε
IbR
1 + R2
g = R1 + R 2
=
9.00 V
a12.0 kΩ + 15.0 kΩf =
b
After the transient currents have ceased, the potential
difference across C is the same as the potential
difference across R 2 = IR 2 because there is no voltage
drop across R3 . Therefore, the charge Q on C is
b
a f
Q = C ∆V
R2
= 50.0 µC .
continued on next page
b g b
g
gb
ga
= C IR 2 = 10.0 µF 333 µA 15.0 kΩ
g
333 µA steady-state .
f
FIG. P28.71(b)
Chapter 28
(c)
When the switch is opened, the branch containing R1
is no longer part of the circuit. The capacitor discharges
through R 2 + R3 with a time constant of
b
g
g a
b
fb
g
R 2 + R3 C = 15.0 kΩ + 3.00 kΩ 10.0 µF = 0.180 s . The
initial current I i in this discharge circuit is determined
by the initial potential difference across the capacitor
applied to R 2 + R3 in series:
Ii =
a f
∆V
bR
2
b
C
+ R3
g
=
g b
b
a
g
f
f
ga
FIG. P28.71(c)
333 µA 15.0 kΩ
IR 2
=
= 278 µA .
R 2 + R3
15.0 kΩ + 3.00 kΩ
Thus, when the switch is opened, the current through R 2 changes instantaneously from
333 µA (downward) to 278 µA (downward) as shown in the graph. Thereafter, it decays
according to
I R 2 = I i e − t b R2 + R3 gC =
(d)
b278 µAge
a
− t 0 .180 s
f afor t > 0f
The charge q on the capacitor decays from Qi to
q = Qi e −
b
g
.
Qi
according to
5
t R 2 + R3 C
Qi
− t 0.180 s g
= Qi e b
5
5 = e t 0.180 s
t
180 ms
t = 0.180 s ln 5 = 290 ms
ln 5 =
a
*P28.72
(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 five resistors
on the top are in the same connection as
those in Example 28.5; the same argument
tells us that the middle resistor can be
removed without affecting the circuit. The
remaining resistors over the three parallel
branches have equivalent resistance
R eq =
(b)
fa f
FG 1 + 1 + 1 IJ
H 20 20 10 K
−1
= 5.00 Ω .
So the current through the battery is
∆V 12.0 V
=
= 2.40 A .
R eq 5.00 Ω
FIG. P28.72(a)
157
158
P28.73
Direct Current Circuits
∆V = ε e − t RC
FG ε IJ = FG 1 IJ t .
H ∆V K H RC K
F ε IJ versus t should
A plot of lnG
H ∆V K
so
ln
be a straight line with slope equal
1
to
.
RC
Using the given data values:
FIG. P28.73
(a)
A least-square fit to this data yields the graph above.
∑ xi = 282 ,
∑ xi yi = 244,
∑ = 1.86 × 10
∑ yi = 4.03 ,
x i2
4
af
,
4.87
N=8
11.1
19.4
c∑ x y h − c∑ x hc∑ y h = 0.011 8
30.8
N e∑ x j − c∑ x h
46.6
67.3
∑ x jc∑ y h − c∑ x hc∑ x y h
e
= 0.088 2
Intercept =
102.2
N e∑ x j − c∑ x h
F ε IJ = b0.011 8gt + 0.088 2
The equation of the best fit line is:
lnG
H ∆V K
Slope =
N
i i
i
i
i
i
2
i
(b)
i
2
2
i
2
i
i i
2
i
Thus, the time constant is
and the capacitance is
P28.74
(a)
1
R1 .
2
(b)
FG
H
5.55
0.109
4.93
0.228
4.34
0.355
3.72
0.509
3.09
0.695
2.47
0.919
1.83
1.219
.
R1
a
IJ
K
If R1 = 13.0 Ω and R 2 = 6.00 Ω, then R x = 2.75 Ω .
Ry
Rx
(1)
1 1
1
R1 + R x , or R x = R 2 − R1 .
2 2
4
b
c
Ry
For the second measurement, the equivalent circuit is
shown in Figure 2.
1
(2)
R ac = R 2 = R y + R x .
Thus,
2
Substitute (1) into (2) to obtain:
R2 =
0
1
1
=
= 84.7 s
slope 0.011 8
τ
84.7 s
= 8.47 µF .
C= =
R 10.0 × 10 6 Ω
R ab = R1 = R y + R y = 2 R y
Ry =
b
ln ε ∆V
∆V V
6.19
τ = RC =
For the first measurement, the equivalent circuit is as
shown in Figure 1.
so
af
ts
0
Figure 1
a
Ry
R2
Ry
Figure 2
FIG. P28.74
The antenna is inadequately grounded since this exceeds the limit of 2.00 Ω .
c
Rx
g
Chapter 28
P28.75
The total resistance between points b and c is:
2.00 kΩ 3.00 kΩ
R=
= 1. 20 kΩ .
2.00 kΩ + 3.00 kΩ
The total capacitance between points d and e is:
C = 2.00 µF + 3.00 µF = 5.00 µF .
a
fa
2.00 kΩ
f
b
3.00 kΩ
The potential difference between point d and e in this series RC
circuit at any time is:
a
f
∆V = ε 1 − e − t RC = 120.0 V 1 − e −1 000 t
6
S
a
and q 2
f
b
ga
= C a ∆V f = b3.00 µFga120.0 V f 1 − e
*P28.76
(a)
2
C1 = 2.00 µF
d
e
C2 = 3.00 µF
120 V
+
-
f
.
FIG. P28.75
Therefore, the charge on each capacitor between points d and e is:
q1 = C1 ∆V = 2.00 µF 120.0 V 1 − e −1 000 t
c
159
6
=
b240 µCg 1 − e
= b360 µC g 1 − e
−1 000 t 6
−1 000 t 6
−1 000 t 6
.
Let i represent the current in the battery and i c the current charging the capacitor. Then i − i c is
q
the current in the voltmeter. The loop rule applied to the inner loop is +ε − iR − = 0 . The loop
C
dq
dq
rule for the outer perimeter is ε − iR − i − i c r = 0 . With i c = , this becomes ε − iR − ir + r = 0 .
dt
dt
q
ε
by substitution to obtain
Between the two loop equations we eliminate i = −
R RC
b g
a
ε − R+r
fFGH Rε − RCq IJK + dqdt r = 0
dq
R+r
R+r
ε+
q+ r =0
R
RC
dt
q
r
Rr dq
−
ε+ +
=0
R+r
C R + r dt
ε−
This is the differential equation required.
(b)
To solve we follow the same steps as on page 875.
FG
H
ε rC
dq ε R + r
R+r
= −
q=−
q−
dt R RrC
RrC
R+r
z
q
0
z
t
dq
R+r
=−
dt
q − ε rC R + r
RrC 0
a
f
IJ
K
F ε rc IJ
lnG q −
H R+rK
q
=−
0
F q − ε rc aR + r f I = − R + r t q − ε rc = − ε rc e
lnG
R+r
R+r
H − ε rc aR + r f JK RrC
r
Rr
where R =
q=
Cε e1 − e
j
r+R
R+r
− t Req C
a f
− R + r RrC t
eq
The voltage across the capacitor is VC =
(c)
t
R+r
t
RrC 0
q
r
−t R C
=
ε 1 − e eq .
C r+R
e
j
r
rε
ε 1−0 =
. If the switch is then opened,
r+R
r+R
the capacitor discharges through the voltmeter. Its voltage decays exponentially according
rε − t rC
.
to
e
r+R
As t → ∞ the capacitor voltage approaches
a f
160
Direct Current Circuits
ANSWERS TO EVEN PROBLEMS
P28.2
(a) 1.79 A ; (b) 10.4 V
P28.42
0.113 Ω
P28.4
(a) 12.4 V ; (b) 9.65 V
P28.44
P28.6
(a) 17.1 Ω ; (b) 1.99 A in 4 Ω and 9 Ω;
1.17 A in 7 Ω; 0.818 A in 10 Ω
(a) 30.000 mA , 5.400 0 V ;
(b) 30.167 mA , 5.381 6 V ;
(c) 29.898 mA ; 5.396 6 V
P28.8
29.5 V
P28.46
see the solution
P28.10
(a) see the solution; (b) no
P28.48
(a) 0.101 W; (b) 10.1 W
P28.12
see the solution
P28.50
15.5 A
P28.14
R1 = 1.00 kΩ; R 2 = 2.00 kΩ ; R3 = 3.00 kΩ
P28.52
2.22 h
P28.16
470 Ω and 220 Ω
P28.54
a is 4.00 V higher
P28.18
(a) 11.0 Ω ; (b) and (d) see the solution;
(c) 220 Ω; (e) Parallel
P28.56
(a) see the solution; 833 mA; 200 W;
(b) see the solution; 4.17 A; 1.00 kW
P28.20
I 1 = 714 mA ; I 2 = 1.29 A ; ε = 12.6 V
P28.58
587 kΩ
P28.22
see the solution
P28.60
P28.24
(a) 0.385 mA in R1 ; 2.69 mA in R3 ;
3.08 mA in R 2 ; (b) c higher by 69.2 V
P28.62
(a) I 1 =
1.00 A up in 200 Ω ; 4.00 A up in 70 Ω ;
3.00 A up in 80 Ω ; 8.00 A down in 20 Ω;
200 V
P28.64
see the solution
P28.28
see the solution
P28.66
bR
P28.30
800 W to the left-hand resistor; 25.0 W to
each 4 Ω; 450 W to the right-hand resistor
P28.68
(a) 222 µC ; (b) increase by 444 µC
P28.32
(a) −61.6 mA ; (b) 0.235 µC ; (c) 1.96 A
P28.70
see the solution
P28.34
0.982 s
P28.72
(a) 5.00 Ω; (b) 2.40 A
P28.36
(a) 1.50 s; (b) 1.00 s;
(c) 200 µA + 100 µA e − t 1.00 s
P28.74
(a) R x = R 2 −
P28.76
(a) and (b) see the solution; (c)
P28.26
b
g
P28.38
(a) 3.91 s; (b) 0.782 ms
P28.40
t
C ln 2
Ps + Ps 2 − 4Ps Pp
and
2I 2
Ps − Ps 2 − 4Ps Pp
2I 2
IR 2
IR1
;
; I2 =
R
+
R
1
2
1 + R2
(b) see the solution
1
bR
g
g
+ 2 R 2 C ln 2
R1
; (b) no; R x = 2.75 Ω
4
rε − t rC
e
r+R