19
CURRENT, RESISTANCE, AND DIRECT-CURRENT CIRCUITS
Answers to Multiple-Choice Problems
1. B
2. B
3. D 4. B, D 5. A 6. A 7. C
8. A 9. C
10. C
11. C
12. C
13. C
14. A 15. E
Solutions to Problems
19.1. Set Up: 1A 5 1 C / s. An electron has charge of magnitude e 5 1.60 3 10219 C. The current tells us how much
1
2
charge flows in a given time. We can find the number of electrons that correspond to a certain amount of charge.
1
5 9.38 3 1018 electrons / s
Solve: (a) 1.50 A 5 1 1.50 C / s 2
219
1.60 3 10 C / electron
(b) DQ 5 IDt 5 1 1.50 A 2 1 300 s 2 5 450 C
DQ
7.50 C
5
5 5.00 s
(c) Dt 5
I
1.50 A
Reflect: The amount of charge associated with typical currents is quite large. In Chapter 17 we found that the force
between objects that have net charge 1.0 C is immense. The same amount of charge enters and leaves any part of the
circuit each second and each circuit element remains neutral.
DQ
Dt
Solve: DQ 5 IDt 5 1 25,000 A 2 1 40 3 1026 s 2 5 1.0 C
19.2. Set Up: I 5
19.3. Set Up: The number of ions that enter gives the charge that enters the axon in the specified time. I 5
Solve: DQ 5 1 5.6 3 1011 ions 2 1 1.60 3 10219 C / ion 2 5 9.0 3 1028 C
I5
DQ
.
Dt
DQ
9.0 3 1028 C
5
5 9.0 mA
Dt
10 3 1023 s
19.4. Set Up: Negative charge moving from A to B is equivalent to an equal magnitude of positive charge going from
DQ
. The current direction is the direction of flow of positive charge.
Dt
Solve: The total positive charge moving from B to A is DQ 5 1 5.11 3 1018 1 2 3 3.24 3 1018 4 2 1 1.60 3 10219 C 2 5
DQ
1.85 C
1.85 C. I 5
5
5 62 mA. Positive charge flows from B to A so the current is in this direction.
Dt
30 s
B to A. I 5
19.5. Set Up: DQ 5 IDt gives the amount of charge that passes a point in the wire in time Dt. From DQ we can find
the number of electrons, since an electron has charge 2e.
1
5 3.44 3 1016 electrons / s
Solve: 1 5.50 3 1023 C / s 2
1.60 3 10219 C / electron
Reflect: Typical currents correspond to the flow of a large number of electrons.
1
2
19-1
19-2
Chapter 19
19.6. Set Up: V 5 IR. For copper, Table 19.1 gives that r 5 1.72 3 1028 V # m and for silver,
r 5 1.47 3 1028 V # m.
1 1.72 3 1028 V # m 2 1 2.00 m 2
rL
5
5 1.65 3 1022 V.
Solve: (a) R 5
A
p 1 0.814 3 1023 m 2 2
V 5 1 12.5 3 1023 A 2 1 1.65 3 1022 V 2 5 2.06 3 1024 V.
(b) V 5
IrL V
Vc
Vs
IL
. 5
5 .
5 constant, so
rs
rc
A r
A
Vs 5 Vc
1 2
1
2
rs
1.47 3 1028 V # m
5 1 2.06 3 1024 2
5 1.76 3 1024 V.
rc
1.72 3 1028 V # m
rL
rL
. From Table 19.1, copper has r 5 1.72 3 1028 V # m.
5
A
p 1 d/2 2 2
p 1 d / 2 2 2R
p 1 1.628 3 1023 m 2 2 1 1.00 V 2
Solve: L 5
5
5 121 m
r
4 1 1.72 3 1028 V # m 2
19.7. Set Up: R 5
rL
. For copper, r 5 1.72 3 1028 V # m. A 5 pr 2.
A
1 1.72 3 1028 V # m 2 1 24.0 m 2
Solve: R 5
5 0.125 V
p 1 1.025 3 1023 m 2 2
19.8. Set Up: R 5
19.9. Set Up: R 5
rL
. The length of the wire in the spring is the circumference pd of each coil times the number
A
of coils.
Solve: L 5 1 75 2 pd 5 1 75 2 p 1 3.50 3 1022 m 2 5 8.25 m.
A 5 pr 2 5 pd 2 / 4 5 p 1 3.25 3 1023 m 2 2 / 4 5 8.30 3 1026 m2.
r5
1 1.74 V 2 1 8.30 3 1026 m2 2
RA
5
5 1.75 3 1026 V # m.
L
8.25 m
Reflect: The value of r we calculated is about a factor of 100 times larger than r for copper. The metal of the spring
is not a very good conductor.
rL
rL
. For aluminum, ral 5 2.63 3 1028 V # m. For copper, rc 5 1.72 3 1028 V # m.
5
A
pd 2 / 4
r
r al
rc
rc
Rp
1.72 3 1028 V # m
5 1 3.26 mm 2
5 2.64 mm.
5 constant, so 2 5 2 . dc 5 dal
Solve: 2 5
4L
Å ral
Å 2.63 3 1028 V # m
d
dal
dc
19.10. Set Up: R 5
rL
. For aluminum, ral 5 2.63 3 1028 V # m. For copper, rc 5 1.72 3 1028 V # m.
A
1 2.63 3 1028 V # m 2 1 3.80 m 2
ralL
5
5 2.00 3 1024 V
Solve: (a) R 5
1 1.00 3 1022 m 2 1 5.00 3 1022 m 2
A
1 2.00 3 1024 V 2 p 1 0.750 3 1023 m 2 2
rcL
Rpr 2
.
5
5 0.0205 m 5 2.05 cm
L
5
(b) R 5
rc
pr 2
1.72 3 1028 V
19.11. Set Up: R 5
rL
rL
5
. L new 5 3L; dnew 5 2d
A
p 1 d/2 2 2
rL
rL new
r3L
3
5
5
5 3R
Solve: Rnew 5
4 p 1 d/2 2 2 4
p 1 dnew / 2 2 2
p 1 2d / 2 2 2
19.12. Set Up: R 5
Current, Resistance, and Direct-Current Circuits
19-3
rL
. L new 5 3L. The volume of the wire remains the same when it is stretched.
A
L
A
A5 .
Solve: Volume 5 LA so LA 5 L new Anew . Anew 5
L new
3
19.13. Set Up: R 5
r 1 3L 2
rL
rL new
5
5 9R.
59
Anew
A
A/3
Reflect: When the length increases the resistance increases and when the area decreases the resistance increases.
Rnew 5
19.14. Set Up: RT 5 R0 3 1 1 a 1 T 2 T0 2 4 . T 5 0.0°C. T 5 11.5°C and R0 5 100.00 V.
Solve: RT 5 1 100.00 V 2 3 1 1 1 0.00040 1 C° 2 21 2 1 0.0°C 2 11.5°C 2 4 5 99.54 V
19.15. Set Up: RT 5 R0 3 1 1 a 1 T 2 T0 2 4
RT 2 R0
1.512 V 2 1.484 V
5
5 1.35 3 1023 1 C° 2 21
1 1.484 V 2 1 34.0°C 2 20.0°C 2
R0 1 T 2 T0 2
Reflect: a is positive and the resistance increases when the temperature increases.
Solve: a 5
19.16. Set Up: RT 5 R0 3 1 1 a 1 T 2 T0 2 4
Solve: T 2 T0 5
1 RT / R0 2 2 1
1 215.8 V / 217.3 V 2 2 1
5
5 13.8 C°. T 5 13.8 C° 1 4.0°C 5 17.8°C.
a
20.00050 1 C° 2 21
19.17. Set Up: Ohm’s law says V 5 IR, where R is a constant.
Solve: (a) Ohm’s law says a graph of V versus I is a straight line. That is the case here, so this object obeys
Ohm’s law.
(b) V 5 IR says R is the slope of a graph of V versus I.
0.60 V 2 0.30 V
5 150 V
(c) R 5
4.0 3 1023 A 2 2.0 3 1023 A
Reflect: We can always define R 5 V / I, but only when R is independent of I does the object obey Ohm’s law.
Vab
is a constant.
I
Solve: (a) The graph is given in Figure 19.18.
19.18. Set Up: Ohm’s law says R 5
16.0
12.0
Vab 1 V2
8.0
4.0
0.0
0.00
Figure 19.18
1.00
2.00
I 1 A2
3.00
4.00
(b) The graph of Vab versus I is a straight line so NichromeTM obeys Ohm’s law.
15.52 V 2 1.94 V
5 3.88 V.
(c) R is the slope of the graph in part (a). R 5
4.00 A 2 0.50 A
19-4
Chapter 19
Vab
is a constant.
I
Solve: (a) The graph is given in Figure 19.19a.
(b) No. The graph of Vabversus I is not a straight line so ThyriteTM does not obey Ohm’s law.
(c) The graph of R versus I is given in Figure 19.19b. R is not constant; it decreases as I increases.
19.19. Set Up: Ohm’s law says R 5
4.80
4.40
4.00
Vab 1 V2 3.60
3.20
2.80
2.40
0.00
6.00
5.00
R
4.00
1Ohms2
3.00
1.00
2.00
I 1 A2
3.00
(a)
2.00
1.00
0.00
4.00
1.00
2.00
I 1 A2
3.00
4.00
(b)
Figure 19.19
Reflect: Not all materials obey Ohm’s law.
19.20. Set Up: Since the battery has negligible internal resistance, E 5 IR.
E
1.50 V
5
5 83.3 V
I
18.0 3 1023 A
12.0 V
E
5 0.144 A
(b) I 5 5
R
83.3 V
(c) E 5 IR 5 1 0.453 A 2 1 83.3 V 2 5 37.7 V
Solve: (a) R 5
19.21. Set Up: V 5 IR, where I is the current through the battery and V is the potential difference applied to the
person.
Solve: V 5 IR 5 1 5.0 3 1023 A 2 1 1000 V 2 5 5.0 V. This is well within the range of common household voltages.
19.22. Set Up: Current is measured in A, potential difference in V. Problem 19.21 says the resistance of a wet
human is 1 kV. V 5 IR.
Solve: (a) Volts are not current; he probably meant a potential difference of 12 V.
V
6V
5 16 mA. Problem 19.21 says currents above 10 mA are dangerous, so this current is dangerous.
(b) I 5 5
R
1 kV
19.23. Set Up: 1 V 5 1 V / A and 1 V 5 1 N # m / C. 1 N 5 1 kg # m / s2
Solve: 1 V 5 1 N # m / C 5 1 1 kg # m / s2 2 # m / C 5 1 1 kg # m2 / s2 2 / A # s 5 1 kg # m2 / A # s3
1 V 5 1 V / A 5 1 kg # m2 / A2 # s3
rL
rL
5
. V 5 IR. Table 19.1 gives r 5 2.44 3 1028 V # m for gold.
A
p 1 d/2 2 2
4 1 2.44 3 1028 V # m 2 1 6.40 m 2
4rL
5
5 0.282 V
Solve: (a) R 5
pd 2
p 1 0.840 3 1023 m 2 2
(b) V 5 IR 5 1 1.15 A 2 1 0.282 V 2 5 0.324 V
19.24. Set Up: R 5
Vp 1 d / 2 2 2
rL
V
VA
Vpd 2
5
5
. V 5 IR so I 5 5
.
A
R
rL
rL
4rL
2
2
Vpd
Vpd
5
5 12 I
Solve: (a) L new 5 2L. Inew 5
4rL new
4r 1 2L 2
Vp 1 2d 2 2
Vpdnew2
(b) dnew 5 2d. Inew 5
5
5 4I
4rL
4rL
19.25. Set Up: R 5
Current, Resistance, and Direct-Current Circuits
19-5
Vp 1 2d 2 2
Vpdnew2
5
5 2I
4rL new
4r 1 2L 2
Reflect: I increases when R decreases. R decreases when L decreases or d increases.
(c) L new 5 2L; dnew 5 2d. Inew 5
19.26. Set Up: RT 5 R0 3 1 1 a 1 T 2 T0 2 4 . I0 5
15.0 A. IT 5
I0
V
. T 5 0°C, T 5 100°C and I0 5
so IT 5
R0
1 1 a 1 T 2 T0 2
15.0 A
5 12.5 A.
1 1 1 0.0020 1 C° 2 21 2 1 100 C° 2
19.27. Set Up: When the switch is open there is no current and the terminal voltage of the battery equals its emf, E.
When the switch is closed, current I flows and the terminal voltage V of the battery is V 5 E 2 Ir. The current I is
the same at all points of the circuit.
V2E
3.08 V 2 2.97 V
5
5 0.067 V.
Solve: E 5 3.08 V. V 5 E 2 Ir, with V 5 2.97 V and I 5 1.65 A, so r 5
I
1.65 A
V
2.97 V
V 2 IR 5 0 and R 5 5
5 1.80 V.
I
1.65 A
Reflect: When current flows through the battery, the terminal voltage is less than the emf because of the voltage
across the internal resistance.
19.28. Set Up: When the switch is open, there is no current. The voltmeter reads the voltage across the circuit
element it is connected across. The emf of the battery is 12.0 V.
Solve: (a) I 5 0, so the terminal voltage of the battery equals the emf. The voltmeter reads 12.0 V.
(b) I 5 0, so the voltage V 5 IR across the resistor is zero. The voltmeter reads zero.
(c) The sum of the potential changes around the circuit must be zero, so if there is a potential rise of 12.0 V across the
battery, there must be a potential drop of 12.0 V across the switch. The voltmeter reads 12.0 V.
E
12.0 V
5
5 2.40 A. (a) The terminal voltage of the battery is
(d) E 2 Ir 2 IR 5 0 and I 5
r1R
4.70 V 1 0.30 V
E 2 Ir 5 12.0 V 2 1 2.40 A 2 1 0.30 V 2 5 11.3 V. (b) The voltage across the resistor is IR 5 1 2.40 A 2 1 4.70 V 2 5
11.3 V. (c) The meter is now a path of zero resistance and the voltage across it is zero.
19.29. Set Up: The terminal voltage of the battery is given by Vab 5 E 2 Ir. The terminal voltage also equals the
voltage across the resistor, so Vab 5 IR. We have two unknowns 1 E, r 2 so we need two equations.
Vab
2.50 V
5
5 1.67 3 102 9A.
Solve: (a) With the 1500 MV resistor, I 5
R
1500 3 106 V
Vab 5 E 2 Ir gives 2.50 V 5 E 2 1 1.67 3 1029 A 2 r.
With the 5.00 V resistor, I 5
Vab
1.75 V
5
5 0.350 A.
R
5.00 V
Vab 5 E 2 Ir gives 1.75 V 5 E 2 1 0.350 A 2 r.
Subtracting the second equation from the first gives 0.75 V 5 1 0.350 A 2 r and r 5 2.14 V.
E 5 1.75 V 1 1 0.350 A 2 r 5 2.50 V
2.50 V
E
5
5 0.274 A
(b) Now R is 7.00 V. I 5
R1r
7.00 V 1 2.14 V
Vab 5 E 2 Ir 5 2.50 V 2 1 0.274 A 2 1 2.14 V 2 5 1.91 V. Or, Vab 5 IR 5 1.91 V.
Reflect: The smaller R is, the larger the current and the smaller the terminal voltage. With the 1500 MV resistor the
current is very small and the terminal voltage differs only very little from the emf of the battery.
19.30. Set Up: When a battery is short-circuited, E 2 Ir 5 0.
Solve: (a) r 5
E
1.50 V
5
5 0.0600 V
I
25.0 A
1.50 V
5 0.150 V
10.0 A
6.00 V
(c) r 5
5 6.00 3 1023 V 5 6.00 mV
1000 A
(b) r 5
19-6
Chapter 19
19.31. Set Up: When current passes through a battery in the direction from the 2 terminal toward the 1 terminal,
the terminal voltage Vab of the battery is Vab 5 E 2 Ir.
E 2 Vab
24.0 V 2 21.2 V
5
5 0.700 V.
Solve: (a) Vab 5 E 2 Ir gives r 5
I
4.00 A
Vab
21.2 V
5
5 5.30 V.
(b) Vab 2 IR 5 0 so R 5
I
4.00 A
24.0 V
5 4.00 A, which agrees with the value
Reflect: The total resistance in the circuit is R 1 r 5 6.00 V. I 5
6.00 V
specified in the problem.
E
. V 5 E 2 Ir. The graph shows I 5 2.5 A when R 5 0.
R 1 r ab
E
5.00 V
5 2.00 V
Solve: (a) When R 5 0, r 5 5
I
2.5 A
E
ER
5
. When R 5 0, Vab 5 0. As R S `, Vab S E. A qualitative sketch of Vab versus R
(b) Vab 5 IR 5
R1r
1 1 r/R
is sketched in Figure 19.32a.
19.32. Set Up: I 5
I
Vab
Vab
E
E
R
(a)
R
R
(b)
(c)
Figure 19.32
E
(c) If r 5 0 then I 5 . The graph of I versus R is sketched in Figure 19.32b. I S ` when R S 0 and I S 0 when
R
R S `. Vab 5 E; Vab is constant, independent of R. The graph of Vab versus R is sketched in Figure 19.32c.
19.33. Set Up: For a resistor, P 5 VI and V 5 IR.
P
327 W
5
5 21.8 A
V
15.0 V
15.0 V
V
5 0.688 V
(b) R 5 5
I
21.8 A
Solve: (a) I 5
19.34. Set Up: For a resistor, P 5 I 2R 5
V2
5 VI and V 5 IR.
R
P
5.0 W
5
5 0.0183 A. V 5 IR 5 1 0.0183 A 2 1 15 3 103 V 2 5 275 V.
Å R Å 15 3 103 V
1 120 V 2 2
V2
(b) P 5
5
5 1.6 W
R
9.0 3 103 V
Solve: (a) I 5
19.35. Set Up: P 5 VI. Energy 5 Pt.
Solve: (a) P 5 1 9.0 V 2 1 0.13 A 2 5 1.17 W
(b) Energy 5 1 1.17 W 2 1 1.5 h 2 1 3600 s / h 2 5 6320 J
19.36. Set Up: Power is energy per unit time. The power delivered by a voltage source is P 5 Vab I.
Solve: (a) P 5 1 25 V 2 1 12 A 2 5 300 W.
(b) Energy 5 Pt 5 1 300 W 2 1 3.0 3 1023 s 2 5 0.90 J
Current, Resistance, and Direct-Current Circuits
19-7
V2
R
1 120 V 2 2
V2
Solve: (a) R 5
5
5 240 V
P
60 W
(b) P increases when R decreases, so the 100 W bulb has less resistance.
19.37. Set Up: P 5
R5
1 120 V 2 2
V2
5
5 144 V
P
100 W
Reflect: P 5 I 2R. The bulb with smaller R draws more current and this more than compensates for smaller R in I 2R.
19.38. Set Up: V 5 IR and P 5
V2
.
R
120 V
V
5
5 0.12 A. Yes, this is very dangerous.
R
1.0 3 103 V
1 120 V 2 2
V2
5
5 14.4 W. We can also calculate P as I 2 R or VI. These expressions also give 14.4 W.
(b) P 5
R
1.0 3 103 V
Solve: (a) I 5
19.39. Set Up: P 5 VI and energy is the product of power and time.
Solve: P 5 1 500 V 2 1 80 3 1023 A 2 5 40 W.
Energy 5 Pt 5 1 40 W 2 1 10 3 1023 s 2 5 0.40 J.
Reflect: The energy delivered depends not only on the voltage and current but also on the length of the pulse.
V2
5 VI. V 5 IR. The heater consumes 540 W when V 5 120 V. Energy 5 Pt.
R
1 120 V 2 2
V2
V2
Solve: (a) P 5
so R 5
5
5 26.7 V
R
P
540 W
540 W
P
(b) P 5 VI so I 5 5
5 4.50 A
V
120 V
(c) Energy 5 1 0.540 kW 2 1 1 h 2 5 0.540 kWh. The cost is 1 0.540 kWh 2 1 7.2 cents / kWh 2 5 3.9 cents.
1 110 V 2 2
V2
5
5 453 W. P is smaller by a factor of 1 110 / 120 2 2.
(d) Assuming that R remains 26.7 V, P 5
R
26.7 V
19.40. Set Up: P 5 I 2R 5
19.41. Set Up: Energy 5 Pt. P 5 VI.
Solve: (a) The energy supplied is VIt 5 1 12.6 V 2 1 60 A 2 1 3600 s 2 5 2.7 3 106 J.
energy
2.7 3 106 J
5 6.0 3 103 s 5 100 minutes
5
(b) t 5
P
0.45 3 103 J / s
Reflect: The battery stores a large amount of energy and it takes time to return this energy to the battery in order to
recharge it.
19.42. Set Up: For an emf, P 5 VI. For a resistor, P 5 I 2R.
Solve: (a) 12.0 V 2 I 1 1.0 V 1 5.0 V 2 5 0 and I 5 2.0 A. The rate at which chemical energy is being converted to
electrical energy within the battery is PE 5 VI 5 1 12.0 V 2 1 2.0 A 2 5 24.0 W.
(b) Pr 5 I 2r 5 1 2.0 A 2 2 1 1.0 V 2 5 4.0 W
(c) PR 5 I 2R 5 1 2.0 A 2 2 1 5.0 V 2 5 20.0 W
Note that PE 5 Pr 1 PR .
19-8
Chapter 19
V2
.
R
Solve: (a) In 3.0 yr the bulbs are on for 1 3.0 yr 2 1 365.24 days / yr 2 1 4.0 h / day 2 5 4.38 3 103 h
compact bulb
The energy used is 1 23 W 2 1 4.38 3 103 h 2 5 1.01 3 105 Wh 5 101 kWh. The cost of this energy is 1 $0.080 / kWh 2
1 101 kWh 2 5 $8.08. One bulb will last longer than this. The bulb cost is $11.00, so the total cost is $19.08.
incandescent
The energy used is 1 100 W 2 1 4.38 3 103 h 2 5 4.38 3 105 Wh 5 438 kWh. The cost of this energy is
1 $0.080 / kWh 2 1 438 kWh 2 5 $35.04. Six bulbs will be used during this time and the bulb cost will be $4.50. The
total cost will be $39.54.
(b) The compact bulb will save $39.54 2 $19.08 5 $20.46.
1 120 V 2 2
V2
5
5 626 V
(c) R 5
P
23 W
Reflect: The initial cost of the bulb is much greater for the compact fluorescent bulb but the savings soon repay the
cost of the bulb. The compact bulb should last for over six years, so over a 6 year period the savings per year will be
even greater.
19.43. Set Up: A kWh is power of 1 kW for a time of 1 h. P 5
19.44. Set Up: V 5 IR. P 5 I 2R. The total resistance is the resistance of the person plus the internal resistance of
the power supply.
V
14 3 103 V
Solve: (a) I 5
5
5 1.17 A
Rtot
10 3 103 V 1 2000 V
(b) P 5 I 2R 5 1 1.17 A 2 2 1 10 3 103 V 2 5 1.37 3 104 J 5 13.7 kJ
V
14 3 103 V
(c) Rtot 5 5
5 14 3 106 V. The resistance of the power supply would need to be
I
1.00 3 1023 A
14 3 106 V 2 10 3 103 V 5 14 3 106 V 5 14 MV.
rL
. V 5 IR. P 5 I 2R.
A
1 5.0 V # m 2 1 1.6 m 2
5 1.0 3 103 V
Solve: (a) R 5
p 1 0.050 m 2 2
(b) V 5 IR 5 1 100 3 1023 A 2 1 1.0 3 103 V 2 5 100 V
(c) P 5 I 2R 5 1 100 3 1023 A 2 2 1 1.0 3 103 V 2 5 10 W
19.45. Set Up: R 5
1
1
1
5
1
1 c For resistors in series, Req 5 R1 1 R2 1 c
Req
R1
R2
These rules may have to be applied in several ways.
19.46. Set Up: For resistors in parallel,
1
1
1
1
1
1
1
1
5
so Req 5 2.9 V.
Req
25 V
12 V
5.0 V
45 V
1
1
1
5
1
(b) The 75 V and 25 V resistors are in parallel.
so their equivalent is Rp 5 18.8 V. The 9.0 V,
Rp
75 V
25 V
18.8 V, and 18 V resistors are in series, so Req 5 9.0 V 1 18.8 V 1 18 V 5 45.8 V.
(c) The 13 V and 15 V resistors are in series, so their equivalent is Rs 5 13 V 1 15 V 5 28 V. The 32 V, 28 V, and
1
1
1
1
5
1
1
14 V resistors are in parallel, so
and their equivalent is Rp1 5 7.2 V. The 72 V and 45 V
Rp1
32 V
28 V
14 V
1
1
1
1
5
resistors are in parallel so
and their equivalent is Rp2 5 27.7 V. This gives a network with 19 V,
Rp2
72 V
45 V
7.2 V, 27.7 V and 24 V in series. The equivalent resistance is Req 5 19 V 1 7.2 V 1 27.7 V 1 24 V 5 77.9 V.
(d) The 12 V, 13 V and 14 V resistors are in series and their equivalent is Rs 5 12 V 1 13 V 1 14 V 5 39 V. Rs
1
1
1
5
, so Rp 5 14.1 V. Then
1
and the 22 V resistor are in parallel and their equivalent Rp is given by
Rp
39 V
22 V
Rp and the 11 V resistor are in series and Req 5 14.1 V 1 11 V 5 25.1 V.
Solve: (a) All four resistors are in parallel.
Current, Resistance, and Direct-Current Circuits
19-9
19.47. Set Up: A network of N of the resistors in series has resistance N 1 10.0 V 2 and a network of N of the
resistors in parallel has resistance 1 10.0 V 2 / N.
Solve: (a) A parallel combination of two resistors in series with three others (Figure 19.47a).
(b) Ten in parallel.
(c) Three in parallel.
(d) Two in parallel in series with four in parallel (Figure 19.47b).
10.0 V
10.0 V
10.0 V
10.0 V
10.0 V
(a)
10.0 V
10.0 V
10.0 V
10.0 V
10.0 V
10.0 V
(b)
Figure 19.47
Reflect: There are other networks that also have the required resistance. An important additional consideration is the
power dissipated by each resistor, whether the power dissipated by any resistor in the network exceeds the maximum
power rating of the resistor.
rL
so the resistance is proportional to the length of the bar. The bars can be represented by the
A
circuit shown in Figure 19.48.
19.48. Set Up: R 5
/
R3
/
/
2R 3
2R 3
/
R3
Figure 19.48
Solve: The two resistors in parallel have an equivalent resistance
equivalent resistance R 5
1 R/3 2
R
5 . Then the three resistors in series have
2
6
2R
R
2R
9R
3R
1 1
5
5
.
3
6
3
6
2
19.49. Set Up: For a parallel connection the full line voltage appears across each resistor. The equivalent resistance
is
R1R2
1
1
1
5
1
.
and Req 5
Req
R1
R2
R1 1 R2
19-10
Chapter 19
1 40.0 V 2 1 90.0 V 2
5 27.7 V
40.0 V 1 90.0 V
V
120 V
5
5 4.33 A
(b) I 5
Req
27.7 V
V
120 V
V
120 V
5
5
5 3.00 A. For the 90.0 V resistor, I2 5
5 1.33 A.
(c) For the 40.0 V resistor, I1 5
R1
40.0 V
R2
90.0 V
Reflect: If either resistor is removed, the voltage and current for the other resistor is unchanged.
Solve: (a) Req 5
19.50. Set Up: Let R1 5 1.60 V, R2 5 2.40 V, R3 5 4.80 V. For a parallel network the full battery voltage
appears across each resistor. The equivalent resistance Req is given by
V2
1
1
1
1
5
1
1 .P5
.
Req
R1
R2
R3
R
1
1
1
1
5
1
1
and Req 5 0.80 V.
Req
1.60 V
2.40 V
4.80 V
V
28.0 V
V
28.0 V
V
28.0 V
5
5 17.5 A. I2 5
5
5 11.7 A. I3 5
5
5 5.8 A.
(b) I1 5
R1
1.60 V
R2
2.40 V
R3
4.80 V
V
28.0 V
5
5 35.0 A. This is also equal to I1 1 I2 1 I3 .
(c) I 5
Req
0.80 V
(d) The voltage is 28.0 V across each resistor.
1 28.0 V 2 2
1 28.0 V 2 2
1 28.0 V 2 2
V2
V2
V2
5
5
5
5 490 W. P2 5
5 327 W. P3 5
5 163 W.
(e) P1 5
R1
1.60 V
R2
2.40 V
R3
4.80 V
2
V
(f) Since P 5
and V is the same for all three resistors, the resistor with the smallest R dissipates the greatest
R
power.
Solve: (a)
19.51. Set Up: Let R1 5 1.60 V, R2 5 2.40 V, R3 5 4.80 V. For a series network, the current is the same in each
resistor and the sum of voltages for each resistor equals the battery voltage. The equivalent resistance is Req 5
R1 1 R2 1 R3 . P 5 I 2R.
Solve: (a) Req 5 1.60 V 1 2.40 V 1 4.80 V 5 8.80 V
V
28.0 V
5
5 3.18 A
(b) I 5
Req
8.80 V
(c) I 5 3.18 A, the same as for each resistor.
(d) V1 5 IR1 5 1 3.18 A 2 1 1.60 V 2 5 5.09 V. V2 5 IR2 5 1 3.18 A 2 1 2.40 V 2 5 7.63 V. V3 5 IR3 5
1 3.18 A 2 1 4.80 V 2 5 15.3 V.
Note that V1 1 V2 1 V3 5 28.0 V.
(e) P1 5 I 2R1 5 1 3.18 A 2 2 1 1.60 V 2 5 16.2 W. P2 5 I 2R2 5 1 3.18 A 2 2 1 2.40 V 2 5 24.3 W. P3 5 I 2R3 5
1 3.18 A 2 2 1 4.80 V 2 5 48.5 W.
(f) Since P 5 I 2R and the current is the same for each resistor, the resistor with the greatest R dissipates the greatest
power.
Reflect: When resistors are connected in parallel, the resistor with the smallest R dissipates the greatest power.
19.52. Set Up: For resistors in parallel, the voltages are the same and the currents add.
Req 5
1
1
1
5
1
so
Req
R1
R2
R1R2
, For resistors in series, the currents are the same and the voltages add. Req 5 R1 1 R2 .
R1 1 R2
Current, Resistance, and Direct-Current Circuits
19-11
Solve: The rules for combining resistors in series and parallel lead to the sequences of equivalent circuits shown in
60.0 V
Figure 19.52. Req 5 5.00 V. In Figure 19.52c, I 5
5 12.0 A. This is the current through each of the resistors
5.00 V
in Figure 19.52b.
R1
R3
R12
R2
E
E
E
Req
R34
R4
(a)
(c)
(b)
Figure 19.52
V12 5 IR12 5 1 12.0 A 2 1 2.00 V 2 5 24.0 V. V34 5 IR34 5 1 12.0 A 2 1 3.00 V 2 5 36.0 V. Note that V12 1 V34 5
V12
V12
24.0 V
24.0 V
5 8.00 A and I2 5
60.0 V. V12 is the voltage across R1 and across R2 , so I1 5
5
5
5 4.00 A.
R1
3.00 V
R2
6.00 V
V34
V34
36.0 V
36.0 V
V34 is the voltage across R3 and across R4 , so I3 5
5
5
5 9.00 A.
5 3.00 A and I4 5
R3
12.0 V
R4
4.00 V
19.53. Set Up: For resistors in parallel, the voltages are the same and the currents add.
1
1
1
5
1
so
Req
R1
R2
R1R2
. For resistors in series, the currents are the same and the voltages add. Req 5 R1 1 R2 .
R1 1 R2
Solve: The rules for combining resistors in series and parallel lead to the sequence of equivalent circuits shown in
48.0 V
5 16.0 A. In Figure 19.53b, the voltage across each
Figure 19.53. Req 5 3.00 V. In Figure 19.53c, I 5
3.00 V
48.0 V
48.0 V
5 12.0 A and I34 5
5 4.00 A. Note that I12 1 I34 5 16.0 A. Then in
resistor is 48.0 V, so I12 5
4.00 V
12.0 V
Figure 19.53a, I1 5 I2 5 12.0 A and I3 5 I4 5 4.00 A.
Req 5
E 5 48.0 V
R1 5 1.00 V
E 5 48.0 V
E 5 48.0 V
R2 5 3.00 V
R12 5 4.00 V
Req 5 3.00 V
R3 5 7.00 V
R4 5 5.00 V
R34 5 12.0 V
(b)
(a)
(c)
Figure 19.53
19.54. Set Up: We can use the power rating in P 5 V 2 / R to find the resistance of a bulb. Then we can find the
voltage across each bulb when they are connected in series and in parallel.
1 120 V 2 2
V2
5
5 144 V
Solve: R 5
P
100 W
V
120 V
5
5 0.417 A. The voltage across each bulb is V1 5 IR 5 60 V.
(a) Req 5 2R 5 288 V. I 5
Req
288 V
1 60 V 2 2
V2
5
5 25 W
R
144 V
(b) In parallel, the voltage across each bulb is 120 V and the power dissipated by each is 100 W.
P5
19.55. Set Up: The power rating of each lightbulb is the power consumed if a voltage of 120 V is applied across
the bulb. In series, the individual voltages are less than the line voltage, and each bulb consumes less than its rated
19-12
Chapter 19
V2
5 I 2R. Use the rated power to find the resistance of each bulb. In parallel, the full line voltage is
R
applied to each bulb.
1 120 V 2 2
1 120 V 2 2
V2
V2
5
5
5 240 V, R75 5
5 192 V and
Solve: The resistance of each bulb is: R60 5
P60
60 W
P75
75 W
2
2
1 120 V 2
V
R100 5
5
5 144 V.
P100
100 W
(a) In series, Req 5 R60 1 R75 1 R100 5 576 V. The current in each bulb is the current in the equivalent circuit,
V
120 V
5 0.208 A. The power consumed by each bulb is P60 5 I 2R60 5 1 0.208 A 2 2 1 240 V 2 5 10.4 W,
I5
5
Req
576 V
P75 5 1 0.208 A 2 2 1 192 V 2 5 8.31 W and P100 5 1 0.208 A 2 2 1 144 V 2 5 6.23 W.
(b) In parallel, V 5 120 V for each bulb, so P60 5 60 W, P75 5 75 W and P100 5 100 W.
(c) P is proportional to V 2 and now V 5 240 V for each bulb, so they would each consume 4 times their rated power:
P60 5 240 W, P75 5 300 W and P100 5 400 W. The power rating for each bulb would be exceeded and they would
quickly burn out.
Reflect: In series, the bulb with the greatest resistance dissipates the greatest power. When the bulbs are connected
individually across the 120 V outlet, the bulb with the least resistance dissipates the greatest power.
power. P 5
19.56. Set Up: For resistors in series the equivalent resistance is Req 5 R1 1 R2 1 R3 . The current is the same in
each. P 5 I 2R so I 5
P
2.00 W
5
5 0.316 A.
Å R20 Å 20.0 V
Req 5 5.00 V 1 10.0 V 1 20.0 V 5 35.0 V.
V 5 IReq 5 1 0.316 A 2 1 35.0 V 2 5 11.1 V.
(b) The 20.0 V resistor would dissipate more than 2.00 W and would fail.
19.57. Set Up: Assume the unknown currents have the directions shown in Figure 19.57. We have used the
junction rule to write the current through the 10.0 V battery as I1 1 I2 . There are two unknowns, I1 and I2 , so we will
need two equations. Three possible circuit loops are shown in the figure.
1 22
112
10.0 V
+
I1
I2
20.0 V
30.0 V
+
5.00 V
1 32
Figure 19.57
Solve: (a) Apply the loop rule to loop (1), going around the loop in the direction shown: 1 10.0 V 2 1 30.0 V 2 I1 5
0 and I1 5 0.333 A.
(b) Apply the loop rule to loop (3): 110.0 V 2 1 20.0 V 2 I2 2 5.00 V 5 0 and I2 5 0.250 A.
(c) I1 1 I2 5 0.333 A 1 0.250 A 5 0.583 A
Reflect: For loop (2) we get
15.00 V 1 I2 1 20.0 V 2 2 I1 1 30.0 V 2 5 5.00 V 1 1 0.250 A 2 1 20.0 V 2 2 1 0.333 A 2 1 30.0 V 2 5
5.00 V 1 5.00 V 2 10.0 V 5 0,
so that with the currents we have calculated the loop rule is satisfied for this third loop.
Current, Resistance, and Direct-Current Circuits
19-13
19.58. Set Up: The circuit diagram is given in Figure 19.58. The junction rule has been used to find the magnitude
and direction of the current in the middle branch of the circuit. There are no remaining unknown currents.
20.0 V
+
1.00 V
6.00 V
112
1.00 A
4.00 V
1.00 V
+
E1
1.00 A
2.00 A
1.00 V
+
2.00 V
E2
Figure 19.58
Solve: The loop rule applied to loop (1) gives:
120.0 V 2 1 1.00 A 2 1 1.00 V 2 1 1 1.00 A 2 1 4.00 V 2 1 1 1.00 A 2 1 1.00 V 2 2 E1 2 1 1.00 A 2 1 6.00 V 2 5 0
E1 5 20.0 V 2 1.00 V 1 4.00 V 1 1.00 V 2 6.00 V 5 18.0 V.
The loop rule applied to loop (2) gives:
120.0 V 2 1 1.00 A 2 1 1.00 V 2 2 1 2.00 A 2 1 1.00 V 2 2 E2 2 1 2.00 A 2 1 2.00 V 2 2 1 1.00 A 2 1 6.00 V 2 5 0
E2 5 20.0 V 2 1.00 V 2 2.00 V 2 4.00 V 2 6.00 V 5 7.0 V.
19.59. Set Up: The circuit is sketched in Figure 19.59. When we go around loop (1) in the direction shown there is
a potential rise across the 200.0 V battery, so there must be a drop across R and the current in R must be in the
direction shown in the figure. Similar analysis of loops (2) and (3) tell us that currents I2 and I5 must be in the
directions shown. The junction rule has been used to label the currents in all the other branches of the circuit.
I2
R
122
I1
I1 1 I 2
+
200.0 V
160 V
40.0 V
+
132
I5
20.0 V
I2 1 I5
112
142
Figure 19.59
Solve: (a) Apply the loop rule to loop (1): 1200.0 V 2 I1R 5 0. R 5
(b) Loop (2): 1160.0 V 2 I2 1 40.0 V 2 5 0. I2 5
160.0 V
5 4.00 A
40.0 V
1200.0 V
1200.0 V
5 20.0 V
5
I1
10.0 A
19-14
Chapter 19
Loop (3): 1160.0 V 2 I5 1 20.0 V 2 5 0. I5 5
160.0 V
5 8.00 A
20.0 V
A2 reads I2 5 4.00 A. A3 reads I2 1 I5 5 12.0 A. A4 reads I1 1 I2 5 14.0 A. A5 reads I5 5 8.00 A.
Reflect: The sum of potential changes around loop (4) is
1200.0 V 2 I1R 1 I2 1 40.0 V 2 2 I5 1 20.0 V 2 5 200.0 V 2 1 10.0 A 2 1 20.0 V 2 1 1 4.00 A 2 1 40.0 V 2 2
1 8.00 A 2 1 20.0 V 2 5 200.0 V 2 200.0 V 1 160.0 V 2 160.0 V 5 0.
The loop rule is satisfied for loop (4) and this is a good check of our calculations.
19.60. Set Up: The circuit diagram is given in Figure 19.60. The junction rule has been used to find the magnitude
and direction of the current in the upper branch of the circuit. There are no remaining unknown currents.
112
28.0 V
+
R
2.00 A
+
6.00 V
E
4.00 A
122
3.00 V
6.00 A
Figure 19.60
Solve: (a) The junction rule gives that the current in R is 2.00 A to the left.
(b) The loop rule applied to loop (1) gives:
2 1 2.00 A 2 R 1 28.0 V 2 1 6.00 A 2 1 3.00 V 2 5 0. R 5
(c) The loop rule applied to loop (2) gives:
28.0 V 2 18.0 V
5 5.00 V.
2.00 A
2 1 4.00 A 2 1 6.00 V 2 1 E 2 1 6.00 A 2 1 3.00 V 2 5 0. E 5 24.0 V 1 18.0 V 5 42.0 V.
Current, Resistance, and Direct-Current Circuits
19-15
19.61. Set Up: We can use the power consumption in the 5.00 V resistor to find the current through it. The circuit,
unknown currents, and a circuit loop are all shown in Figure 19.61.
15.0 V
+
7.00 V
I2
112
I1
5.00 V
+
E
2.00 V
Figure 19.61
Solve: P 5 I 2R so I1 5
20.0 W
5 2.00 A. The loop rule for loop (1) gives
Å 5.00 V
115.0 V 2 I2 1 7.00 V 2 2 1 2.00 A 2 1 5.00 V 2 5 0.
I2 5
15.0 V 2 10.0 V
5 0.714 A. The ammeter reads 0.714 A.
7.00 V
19.62. Set Up: P 5 I 2R so the power consumption of the 6.0 V resistor allows us to calculate the current through
it. Unknown currents I1 , I2 and I3 are shown in Figure 19.62. The junction rule says that I1 5 I2 1 I3 . In Figure 19.62
the two 20.0 V resistors in parallel have been replaced by their equivalent (10.0 V).
17 V
10.0 V
I3
I2
I1
+
25 V
E
122
13 V
Figure 19.62
6.0 V
+
19 V
112
3.0 V
1.0 V
19-16
Chapter 19
Solve: (a) P 5 I 2R gives I1 5
24 J / s
P
5
5 2.0 A. The loop rule applied to loop (1) gives:
Å R Å 6.0 V
2 1 2.0 A 2 1 3.0 V 2 2 1 2.0 A 2 1 6.0 V 2 1 25 V 2 I2 1 10.0 V 1 19.0 V 1 1.0 V 2 5 0.
I2 5
25 V 2 18 V
5 0.233 A.
30.0 V
(b) I3 5 I1 2 I2 5 2.0 A 2 0.233 A 5 1.77 A. The loop rule applied to loop (2) gives:
2 1 2.0 A 2 1 3.0 V 1 6.0 V 2 1 25 V 2 1 1.77 A 2 1 17 V 2 2 E 2 1 1.77 A 2 1 13 V 2 5 0.
E 5 25 V 2 18 V 2 53.1 V 5 246.1 V. The emf is 46.1 V and the polarity of the battery is opposite to what is
shown in the figure in the problem; the 1 terminal is adjacent to the 13 V resistor.
19.63. Set Up: The time constant is t 5 RC.
Solve: C 5
t
2.00 s
5
5 4.00 3 1023 F 5 4.00 mF
R
500.0 V
1 21 2
19.64. Set Up: 1 V 5 1 V / A. 1 F 5 1 C / V.
Solve: RC has units of V # F 5
V C
C
C
5 5
5 s.
A V
A
C/s
q
5 iR.
C
Solve: (a) The initial value of q is q 5 60.0 nC.
19.65. Set Up: t 5 RC. The loop rule gives
i5
q
60.0 3 1029 C
5
5 3.00 3 1024 A 5 0.300 mA.
RC
1 8.00 3 105 V 2 1 0.250 3 1029 F 2
(b) t 5 RC 5 1 8.00 3 105 V 2 1 0.250 3 1029 F 2 5 2.00 3 1024 s 5 0.200 ms
Reflect: The larger the value of R, the smaller the initial current and the longer it takes the charge to decrease to 1 / e
of its initial value.
19.66. Set Up: The loop rule gives E 5 iR 1
q
. Just after the circuit is completed, q 5 0. After a long time the
C
capacitor is fully charged and i 5 0.
Solve: (a) q 5 0 so the voltage across the capacitor is zero.
(b) q 5 0 so iR 5 E; the voltage drop across the resistor is 500 V.
(c) q 5 0
E
500 V
5 0.111 A
(d) i 5 5
R
4500 V
q
(e) (a) i 5 0 so 5 E and the voltage across the capacitor is 500 V.
C
(b) i 5 0 so the voltage across the resistor is zero.
(c) q 5 EC 5 1 500 V 2 1 6.00 3 1026 F 2 5 3.00 3 1023 C 5 3.00 mC.
(d) i 5 0
19.67. Set Up: The time constant is t 5 RC. Kirchhoff’s loop rule applied to the charging circuit gives E 5 iR 1
at all times. The current and the charge on the capacitor as functions of time are given in Equations (19.17).
Solve: (a) t 5 RC 5 1 75.0 V 2 1 6.00 3 1026 F 2 5 4.5 3 1024 s 5 450 ms
Qfinal
. Qfinal 5 EC 5 1 12.0 V 2 1 6.00 mF 2 5 72.0 mC
(b) q 5 Qfinal when i S 0, so E 5
C
q
C
Current, Resistance, and Direct-Current Circuits
19-17
(c) The current has its largest value I0 when q 5 0, so E 5 I0R.
I0 5
12.0 V
E
5
5 0.160 A
R
75.0 V
The graphs are of the same form as in Figures 19.33a and b in the textbook.
Reflect: The maximum charge on the capacitor does not depend on the resistance. The initial current is the same as
the current produced when the resistor alone is connected to the battery.
19.68. Set Up: The time constant is t 5 RC, so q 5 Qfinal 1 1 2 e 2t/RC 2 and i 5 I0e 2t/RC .
t
t
Solve: (a) q 5 0.95Qfinal gives 0.95 5 1 2 e 2t/RC. e 2t/t 5 0.05. 2 5 ln 1 0.05 2 . 5 3.0. Three time constants
t
t
are required.
(b) e 2t/t 5 0.05 so i 5 0.05I0 .
19.69. Set Up: q 5 C / V. After a long time the current has dropped to zero, there is no potential drop across the
resistor and the full battery voltage is applied to the capacitor network. Just after the switch is closed the charge and
voltage across each capacitor is zero and the battery voltage equals the voltage drop across the resistor. The time
constant is t 5 RC, where C is the equivalent capacitance of the network. i 5 I0e 2t/RC.
Solve: (a) The 20.0 pF, 30.0 pF and 40.0 pF capacitors are in series and their equivalent capacitance Cs is given by
1
1
1
1
5
1
1
.
Cs
20.0 pF
30.0 pF
40.0 pF
Cs 5 9.23 mF. After a long time the voltage across the 10.0 pF capacitor is 50.0 V. The charge on this capacitor is
q10 5 C10V10 5 1 10.0 pF 2 1 50.0 V 2 5 500 pC. The voltage across Cs is 50.0 V and
q5 5 C5V5 5 1 9.23 pF 2 1 50.0 V 2 5 461 pC.
This is the charge on each of the capacitors in series, so q20 5 q30 5 q40 5 461 pC.
461 pC
461 pC
q20
q30
5
5 23.1 V. V30 5
5
5 15.4 V.
(b) V10 5 50.0 V. V20 5
C20
20.0 pF
C30
30.0 pF
V40 5
461 pC
q40
5
5 11.5 V.
C40
40.0 pF
Note that V20 1 V30 1 V40 5 50.0 V, as it should.
50.0 V
E
5 2.50 A
(c) I 5 5
R
20.0 V
(d) The 10.0 pF capacitor and Cs are in parallel so their equivalent is Ceq 5 10.0 pF 1 9.23 pF 5 19.2 pF.
t 5 RC 5 1 20.0 V 2 1 19.2 pF 2 5 384 ps
Reflect: The charges on the capacitors start at zero and increase to their final values. The current starts at its maximum value and then decays to zero.
19.70. Set Up: The time constant for the circuit is t 5 RC. While charging, q 5 Qfinal 1 1 2 e 2t/RC 2 and
i 5 I0e 2t/RC. In the discharging circuit both the charge and current decrease to 1 / e of their initial values in time t 5 t.
Solve: t 5 RC 5 1 125 V 2 1 1.50 mF 2 5 188 ms
Qfinal
1
1
(a) q 5 Qfinal 1 1 2 e 2t/RC 2 . q 5
gives 5 1 2 e 2t/RC. e 2t/RC 5 1 2 and
e
e
e
t 5 2RC 1 ln 3 1 2 1 / e 4 2 5 2 1 188 ms 2 1 20.458 2 5 186.1 ms
(b) i 5 I0e 2t/RC 5 I0e 2186.1 ms2 / 1188 ms2 5 0.633I0
(c) t 5 t 5 188 ms
19.71. Set Up: Equations (19.17) give i and q as functions of time during charging. During discharging, both i and
q decay exponentially, as in Eq. (19.19) for i.
19-18
Chapter 19
Solve: (a) The graphs are the same as in Figure 19.33a and b in the textbook and are given in Figure 19.71a.
(b) The graphs of i and q versus t for a discharging capacitor are sketched in Figure 19.71b.
q
i
Qmax
Imax
t
t
(a)
q
i
Qmax
Imax
t
t
(b)
Figure 19.71
19.72. Set Up: There is a single current path so the current is the same at all points in the circuit. Assume the
current is counterclockwise.
Solve: (a) Apply the loop rule, traveling around the circuit in the direction of the current:
116.0 V 2 I 1 1.6 V 1 5.0 V 1 1.4 V 1 9.0 V 2 2 8.0 V 5 0. I 5
16.0 V 2 8.0 V
5 0.471 A.
17.0 V
Our calculated I is positive so I is counterclockwise, as we assumed.
(b) Vb 1 16.0 V 2 I 1 1.6 V 2 5 Va . Vab 5 16.0 V 2 1 0.471 A 2 1 1.6 V 2 5 15.2 V
19.73. Set Up: The power for a resistor R is PR 5 I 2R. The power for an emf is PE 5 EI.
Solve: (a) The loop rule for a counterclockwise direction gives
112 V 2 4 V 2 I 1 2 V 1 3 V 1 4 V 1 7 V 2 5 0. I 5
8V
5 0.50 A.
16 V
P 5 I 2 1 3.00 V 1 7.00 V 2 5 1 0.50 A 2 2 1 10.0 V 2 5 2.5 W.
(b) The net power output of the battery is
P 5 EI 2 I 2r 5 1 12.0 V 2 1 0.50 A 2 2 1 0.50 A 2 2 1 2 V 2 5 6.0 W 2 0.5 W 5 5.5 W.
(c) In the emf of this battery electrical energy is being consumed at the rate EI and in the internal resistance of the
battery electrical energy is being consumed at the rate I 2r.
P 5 1 4 V 2 1 0.50 A 2 1 1 0.50 A 2 2 1 4 V 2 5 2.0 W 1 1.0 W 5 3.0 W
(d) Chemical energy is being converted to electrical energy at a rate of 1 12.0 V 2 1 0.50 A 2 5 6.0 W in the 12.0 V
battery and electrical energy is being converted to chemical energy at a rate of 1 4.0 V 2 1 0.50 A 2 5 2.0 W in the 4.0 V
battery. The net rate of conversion of chemical energy to electrical energy is 6.0 W 2 2.0 W 5 4.0 W. The overall
rate of dissipation of electrical energy (conversion to heat) is I 2Rtotal 5 1 0.50 A 2 2 1 16 V 2 5 4.0 W. The two rates
are equal, as required by conservation of energy.
Reflect: The power output of the 12 V battery is less than EI because of energy dissipated in the internal resistance
of the battery. When current flows through a battery from the positive to the negative terminal, the positive charge
Current, Resistance, and Direct-Current Circuits
19-19
passing through the battery in the direction of the current loses electrical energy. In the battery, electrical energy is
removed from the circuit. The power EI in this case corresponds to the rate at which chemical energy is being stored.
DQ
. In 1.00 s the electron passes a point on the orbit 6.0 3 1015 times. The charge of an
Dt
electron has magnitude e.
Solve: The magnitude of the average current is
19.74. Set Up: I 5
1 6.0 3 1015 2 e
1 6.0 3 1015 2 1 1.60 3 10219 C 2
5
5 9.6 3 1024 A 5 0.96 mA.
1.00 s
1.00 s
The direction of the current is opposite to the direction of circulation of the electron, since the electron has negative
charge.
I5
19.75. Set Up: RT 5 R0 1 1 1 a 3 T 2 T0 4 2 . R 5
V
. P 5 VI. When the temperature increases the resistance
I
increases and the current decreases.
V
V
Solve: (a) 5 1 1 1 a 3 T 2 T0 4 2 . I0 5 IT 1 1 1 a 3 T 2 T0 4 2 .
IT
I0
T 2 T0 5
I0 2 IT
1.35 A 2 1.23 A
5
5 217 C°. T 5 20°C 1 217 C° 5 237°C
1 1.23 A 2 1 4.5 3 1024 1 C° 2 21 2
aIT
(b) (i) P 5 VI 5 1 120 V 2 1 1.35 A 2 5 162 W
(ii) P 5 1 120 V 2 1 1.23 A 2 5 148 W
rL
1
1
1
1
5
1
1 .
so each piece has resistance R / 3. For resistors in parallel,
A
Req
R1
R2
R3
1
1
9
53
5 . Req 5 R / 9. This can also be obtained as decreasing L by 13 and increasing A by 3, for an
Solve:
Req
R
R/3
overall decrease of 91 .
1 2
19.76. Set Up: R 5
19.77. Set Up: The circuit is sketched in Figure 19.77. rtotal is the combined internal resistance of both batteries.
The power delivered to the bulb is I 2R. Energy 5 Pt.
1.5 V
+
1.5 V
+
rtotal
R 5 17 V
Figure 19.77
Solve: (a) rtotal 5 0. The loop rule gives
1.5 V 1 1.5 V 2 I 1 17 V 2 5 0. I 5 0.1765 A. P 5 I 2R 5 1 0.1765 A 2 2 1 17 V 2 5 0.530 W.
This is also 1 3.0 V 2 1 0.1765 A 2 .
(b) Energy 5 1 0.530 W 2 1 5.0 h 2 1 3600 s / h 2 5 9540 J
0.530 W
0.265 W
P
5 0.265 W. P 5 I 2R so I 5
5
5 0.125 A.
2
Å R Å 17 V
3.0 V 2 1 0.125 A 2 1 17 V 2
The loop rule gives 1.5 V 1 1.5 V 2 IR 2 Irtotal . rtotal 5
5 7.0 V.
0.125 A
Reflect: When the power to the bulb has decreased to half its initial value, the total internal resistance of the two batteries is nearly half the resistance of the bulb. Compared to a single battery, using two identical batteries in series
doubles the emf but also doubles the total internal resistance.
(c) P 5
19-20
Chapter 19
19.78. Set Up: The power dissipated in a resistor is I 2R. The power for an emf is EI. When the current passes
through an emf from the 2 to the 1 terminal, chemical energy is converted to electrical energy. When the current
passes through the emf from the 1 to the 2 terminal, electrical energy is converted to chemical energy.
Solve: (a) Assume the current in the circuit is counterclockwise. The loop rule gives:
12.0 V 2 8.0 V 2 I 1 1.0 V 1 8.0 V 1 1.0 V 2 5 0. I 5 0.40 A.
(b) P 5 I 2R 1 I 2r1 1 I 2r2 5 I 2 1 R 1 r1 1 r2 2 5 1 0.40 A 2 2 1 8.0 V 1 1.0 V 1 1.0 V 2 5 1.6 W
(c) In E1 chemical energy is being converted to electrical energy, at the rate 1 12.0 V 2 1 0.40 A 2 5 4.8 W.
19.79. Set Up: Electrical energy is deposited in his body at the rate P 5 I 2R. Heat energy Q produces a
temperature change DT according to Q 5 mcDT, where c 5 4190 J / kg # C°.
Solve: (a) P 5 I 2R 5 1 25,000 A 2 2 1 1.0 kV 2 5 6.25 3 1011 W. The energy deposited is
Pt 5 1 6.15 3 1011 W 2 1 40 3 1026 s 2 5 2.5 3 107 J.
Find DT when Q 5 2.5 3 107 J.
DT 5
Q
2.5 3 107 J
5
5 80 C°
mc
1 75 kg 2 1 4190 J / kg # C° 2
(b) An increase of only 63 C° brings the water in the body to the boiling point; part of the person’s body will be
vaporized.
Reflect: Even this approximate calculation shows that being hit by lightning is very dangerous.
rL
.
A
Solve: (a) The tube of seawater has resistance
19.80. Set Up: V 5 IR. R 5
R5
1 0.13 V # m 2 1 0.20 m 2
rL
3.0 V
V
5
5 260 V. I 5 5
5 11.5 mA
24
2
A
R
260 V
1.0 3 10 m
(b) The object and the remaining tube of seawater are in series. The 10 cm tube of seawater has resistance 130 V.
The object is 10 cm long, has the same cross-sectional area and half the resistance of seawater, so its resistance is
65 V. The total equivalent resistance of the tube is 130 V 1 65 V 5 195 V. The current now is
V
3.0 V
5
5 15.4 mA.
R
195 V
The increase in current due to the proximity of the object is 15.4 mA 2 11.5 mA 5 3.9 mA.
I5
19.81. Set Up: The network is sketched in Figure 19.81. If the current in R1 is I, it is I / 2 for R2 and R3 . P 5 I 2R.
Since the current is greatest in R1 , that resistor dissipates the most power.
R2
/
I2
I
R1
R3
/
I2
Figure 19.81
Solve: P1 5 32.0 W so I 5
32.0 W
P
5
5 4.00 A. The total power dissipated therefore is
Å R Å 2.00 V
1 4.00 A 2 2 1 2.00 V 2 1 1 2.00 A 2 2 1 2.00 V 2 1 1 2.00 A 2 2 1 2.00 V 2 5 48.0 W.
Reflect: R1 is dissipating its maximum of 32.0 W but R2 and R3 are dissipating only 8.0 W each.
Current, Resistance, and Direct-Current Circuits
19-21
19.82. Set Up: Use a network of resistors that has Req 5 1000 V. Each resistor can dissipate less than 2.00 W but
the total power is the sum of the powers for each resistor so it can be 2.00 W.
Solve: (a) and (b) One possible network is two in parallel in series with two in parallel, as shown in Figure 19.82.
Each parallel combination has an equivalent resistance of 500 V so these two units in series have an equivalent resistance of 1000 V. The current is the same in each resistor so each resistor dissipates the same power. If each resistor
dissipates 0.50 W, then the network dissipates 2.00 W. Other combinations also give the desired resistance and power
rating, such as two in series in parallel with two in series.
1000 V
1000 V
1000 V
1000 V
Figure 19.82
19.83. Set Up: Since they are in parallel, the voltage across each device is 120 V. P 5 VI. The total current drawn
from the outlet is the sum of the individual currents.
75 W
P
1800 W
1400 W
5 15.0 A. Frypan: I 5
5 11.7 A. Lamp: I 5
5 0.625 A.
Solve: (a) Toaster: I 5 5
V
120 V
120 V
120 V
(b) I 5 15.0 A 1 11.7 A 1 0.625 A 5 27.3 A. This is greater than 20 A and the circuit breaker will blow.
19.84. Set Up: The resistance of each half of a wire is 0.500 V. The combination can be represented by the resistor
network shown in Figure 19.84.
0.500 V
0.500 V
0.500 V
0.500 V
Figure 19.84
Solve: The two resistors in parallel have an equivalent resistance of 0.250 V. Then the three in series have an equivalent resistance of 0.500 V 1 0.250 V 1 0.500 V 5 1.25 V.
V2
.
Req
V2
Solve: When the resistors are in series, Req 5 3R and Ps 5
. When the resistors are in parallel, Req 5 R / 3.
3R
19.85. Set Up: Let R be the resistance of each resistor. Ptot 5
V2
V2
5 9Ps 5 9 1 27 W 2 5 243 W.
53
R
R/3
Reflect: In parallel, the voltage across each resistor is the full applied voltage V. In series, the voltage across each
resistor is V / 3 and each resistor dissipates less power.
Pp 5
19.86. Set Up: For resistors in series, the currents are the same and the voltages add. Req 5 R1 1 R2 1 c For
resistors in parallel, the currents add and the voltages are the same.
1
1
1
5
1
1 c
Req
R1
R2
19-22
Chapter 19
Solve: (a) Replacing series and parallel combinations of resistors by their equivalents gives the equivalent networks
as shown in Figure 19.86. The equivalent resistance of the network is 8.0 V.
2.4 A 8.0 V
20.0 V
16.0 V
16.0 V
x
y
9.0 V
6.0 V
a
18.0 V
(a)
4.0 V
20.0 V
24.0 V
8.0 V
y
x
x
y
x
y
a
6.0 V
12.0 V
6.0 V
(b)
(c)
(d)
Figure 19.86
(b) The voltage across the 8.0 V resistor in Figure 19.86a is 1 2.4 A 2 1 8.0 V 2 5 19.2 V. This is the voltage across
the 4.0 V resistor in Figure 19.86b. The current through the 4.0 V resistor is 4.8 A. This is also the current through
the 20.0 V resistor and the voltage across that resistor is 96.0 V. Therefore, the voltage between points x and y is
19.2 V 1 96.0 V 5 115.2 V. This voltage is divided equally between the 6.0 V resistors in Figure 19.86b, so the
voltage between points x and y is 1 115.2 V 2 / 2 5 57.6 V. The voltmeter will read 57.6 V.
rL
. Table 19.1 gives the resistivities of copper and aluminum to be rc 5 1.72 3 1028 V # m
A
and ra 5 2.63 3 1028 V # m. For the cables in series (end-to-end), Req 5 Rc 1 Ra . For the cables in parallel the
1
1
1
5
1 . Note that in the two configurations the copper and aluminum
equivalent resistance Req is given by
Req
Rc
Ra
sections have different lengths. And, for the parallel cables the cross sectional area of each cable is half what it is for
the end to end configuration.
Solve: end-to-end: L 5 0.50 3 103 m for each cable.
19.87. Set Up: R 5
Rc 5
Ra 5
1 1.72 3 1028 V # m 2 1 0.50 3 103 m 2
r cL
5
5 0.172 V
A
0.500 3 1024 m2
1 2.63 3 1028 V # m 2 1 0.50 3 103 m 2
r aL
5
5 0.263 V
A
0.500 3 1024 m2
Req 5 0.172 V 1 0.263 V 5 0.435 V
parallel: Now L 5 1.00 3 103 m for each cable. L is doubled and A is halved compared to the other configuration, so
1
1
1
1
1
Rc 5 4 1 0.172 V 2 5 0.688 V and Ra 5 4 1 0.263 V 2 5 1.052 V.
5
1
5
1
and
Req
Rc
Ra
0.688 V
1.052 V
Req 5 0.416 V. The least resistance is for the cables in parallel.
Current, Resistance, and Direct-Current Circuits
19-23
Reflect: The parallel combination has less equivalent resistance even though both cables contain the same volume of
each metal.
19.88. Set Up: Because of the polarity of each emf, the current in the 7.00 V resistor must be in the direction
shown in Figure 19.88. Let I be the current in the 24.0 V battery.
3.80 A
2.00 A
I
24.0 V
+
1.8 A
+
+
+
24.0 V
E
122
E
7.00 V
3.00 V
7.00 V
3.00 V
2.00 V
1.80 A
2.00 V
112
(b)
(a)
Figure 19.88
Solve: The loop rule applied to loop (1) gives: 124.0 V 2 1 1.80 A 2 1 7.00 V 2 2 I 1 3.00 V 2 5 0. I 5 3.80 A. The
junction rule then says that the current in the middle branch is 2.00 A, as shown in Figure 19.88b. The loop rule
applied to loop (2) gives: 1E 2 1 1.80 A 2 1 7.00 V 2 1 1 2.00 A 2 1 2.00 V 2 5 0 and E 5 8.6 V.
19.89. Set Up: There is no current in the middle branch because there is not a complete conducting path for that
branch. There is only a single current in the circuit, as shown in Figure 19.89. To find Vab 5 Va 2 Vb , start at point b
and travel to point a.
I
1.00 V
12.0 V
+
2.00 V
1.00 V
1.00 V
I
a
10.0 V
+
3.00 V
I
b
2.00 V
2.00 V
8.0 V
+
1.00 V
I
Figure 19.89
Solve: 12.0 V 2 8.0 V 2 I 1 2.00 V 1 2.00 V 1 1.00 V 1 2.00 V 1 1.00 V 1 1.00 V 2 5 0.
I 5 0.444 A. Going from a to b through the 12.0 V battery gives
Va 1 I 1 2.00 V 2 1 I 1 1.00 V 2 2 12.0 V 1 I 1 1.00 V 2 1 10.0 V 5 Vb .
Va 2 Vb 5 12.0 V 2 10.0 V 2 1 0.444 A 2 1 4.00 V 2 5 10.22 V.
Or, going from a to b through the 8.0 V battery gives
Va 2 I 1 2.00 V 2 2 I 1 1.00 V 2 2 8.0 V 2 I 1 2.00 V 2 1 10.0 V 5 Vb and
Va 2 Vb 5 22.0 V 1 1 0.444 A 2 1 5.00 V 2 5 10.22 V.
19-24
Chapter 19
The voltmeter reads 0.22 V. Va 2 Vb is positive, so point a is at higher potential.
Reflect: Since there is no current in the middle branch, the two resistances in that branch could be removed without
affecting the potential difference Vab . The 10.0 V emf doesn’t affect the current but does affect Vab
q
. The charge is reduced to 1 / e of its maximum value when t 5 t 5 RC.
C
Solve: (a) At t 5 0, q 5 iRC 5 1 0.250 A 2 1 4600 V 2 1 0.800 3 1029 F 2 5 9.20 3 1027 C 5 0.920 mC.
(b) t 5 t 5 1 4600 V 2 1 0.800 3 1029 F 2 5 3.68 3 1026 s 5 3.68 ms.
19.90. Set Up: iR 5
q
. At t 5 0, q 5 0. t 5 RC.
C
E
400.0 V
5 5.00 3 105 V.
Solve: (a) At t 5 0, E 5 iR and R 5 5
i
0.800 3 1023 A
t
6.00 s
5 1.20 3 1025 F.
(b) t 5 RC so C 5 5
R
5.00 3 105 V
19.91. Set Up: E 5 iR 1
19.92. Set Up: Zero current through the galvanometer means the current I1 through N is also the current through M
and the current I2 through P is the same as the current through X . And it means that points b and c are at the same
potential, so I1N 5 I2P.
E
E
. Using these expressions in
Solve: (a) The voltage between points a and d is E, so I1 5
and I2 5
N1M
P1X
E
E
I1N 5 I2P gives
N5
P. N 1 P 1 X 2 5 P 1 N 1 M 2 . NX 5 PM and X 5 MP / N.
N1M
P1X
1 850.0 V 2 1 33.48 V 2
MP
5
5 1897 V
(b) X 5
N
15.00 V