Download Chapter 18 ELECTRIC CURRENT AND CIRCUITS

Chapter 18
ELECTRIC CURRENT AND CIRCUITS
Conceptual Questions
1.
2. Voltmeters have very large internal resistances, so there is usually no danger of a large current. Ammeters on the
other hand have very small internal resistances, so they may draw very large currents in a circuit if they are not
connected in series with any other significant resistance.
3. The resistance of a light-bulb filament increases significantly as it heats up. The largest current therefore flows
through the bulb when it is cold. Thus, it is most likely to burn out just after being switched on.
4. If he connects three 300  resistors in parallel, the equivalent resistance will be the desired 100 .
5. This statement is not exactly true. The current flowing through a branch in a circuit is inversely proportional to the
resistance of the branch. Thus, more current follows the path of least resistance than follows any other path, but
every path has some current.
6. An ideal ammeter has zero resistance so as to have no effect on the current it is supposed to measure when
connected in series in a circuit. An ideal voltmeter has an infinite resistance so that it does not perturb the voltage
it is supposed to measure when connected in parallel in a circuit.
7. As the temperature increases, the atoms in the metal begin to vibrate with greater amplitudes. The chance of an
electron colliding with one of the atoms is therefore increased. This effectively reduces the mean free path of the
electrons and increases the resistance of the metal.
8. Some of the energy is dissipated as heat by the resistors and some of it gets stored in the electric fields of the
capacitors as they are charged up.
9. Electric stoves and clothes dryers require relatively large amounts of power to operate. Supplying them with
240 V instead of 120 V decreases the magnitude of required current to supply the power. This reduces the rate
(P = I2R) at which energy is dissipated in the wiring.
10. Elements connected in series in a circuit have the identical current flowing through them, while elements in
parallel have identical potential differences. Therefore, ammeters are connected in series and voltmeters in
parallel. It is true that if the element under consideration is a resistor, a measurement of the current allows one to
calculate the voltage via Ohm’s law and vice versa. So, for a resistor, an ammeter could be used to measure the
voltage in a sense. This doesn’t work for other elements though. A fully charged capacitor for example has no
current flowing through it and an ammeter connected in series would tell us the current but give us no information
about the voltage. Similarly, a voltmeter connected in parallel to a capacitor would allow us to measure the
voltage but give us no information about the current flowing through it.
11. Most of the electrical resistance of the body is due to the skin. The resistance of wet skin is much lower than the
resistance of dry skin. Thus, it is more dangerous to touch a “live” electrical wire with wet hands because more
current will flow through the body.
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12. An electric field is required to start a current flowing within a conductor. The electric field inside a conductor is
therefore not equal to zero if a current is flowing through it.
13. Batteries convert chemical energy into electrical energy. As a battery is used, its supply of chemical energy is
depleted. Recharging the battery does not actually put additional charges back into the battery, but instead
converts electrical energy into chemical energy.
14. Current flows from B to C through the clock. Current flows from D to A through the battery. Terminal A of the
battery is at the higher potential. Side B of the clock is at the higher potential. Current can be made to flow across
a circuit element from a lower to a higher potential if work is done by an external agent such as a battery.
15. The total resistance of two resistors connected in series is equal to the sum of their resistances. If resistance is
proportional to length, then the above statement tells us that the total resistance of two resistors of length L/2
connected in series is proportional to the length of the combined resistor L. If resistance had any other relationship
to length, the total resistance would not agree with that given by the first statement.
16. The inverse of the total resistance of two resistors connected in parallel is equal to the sum of the inverse
resistances of each resistor. If the resistance of a wire is inversely proportional to its cross-sectional area, then the
above statement tells us that the total resistance of two resistors of cross-sectional area A connected in parallel is
proportional to 0.5A. If resistance had any other relationship to cross-sectional area, the total resistance would not
agree with that given by the first statement.
17. An electrician working on live wiring wears insulated shoes to avoid being grounded and therefore to reduce the
chance of starting a flow of current through the body if a live wire is touched. Similarly, an electrician using two
hands would risk completing a circuit so that current flows from one hand to the other through the body and near
the heart.
18. A 20-A circuit breaker would allow more current into the circuit than it was designed to handle. The additional
current may allow additional appliances to operate, but resistive heating in the wires due to the increased current
may be hazardous.
19. The bird perched on the power line is at the same electric potential as the line but it is isolated from the ground so
that no current flows through its body. When a person standing on the ground touches a power line with a metal
pole, a potential difference exists between the line and the person’s grounded feet—a current will therefore flow
through their body.
20. The total emf produced by several batteries in series is the sum of the emf of each battery—each battery may have
any emf. If batteries connected in parallel have different emfs, significantly larger currents will flow through the
batteries than if they had the same emf. One way to see this is that in parallel the batteries must have the same
potential drop across them, and usually the internal resistance of a battery is very small. For example, consider
two batteries connected in parallel, one with an emf of 9 V and one with 10 V, and each with an internal
resistance of 0.1 . Since the potential drops across the batteries must be equal, there must be a difference of 10 A
of current flowing through the 9-V battery as compared to the current flowing through the 10-V battery, to make
up for the 1-V difference. Even if the batteries are disconnected from the rest of the circuit, there will be
significant currents flowing around the loop composed of just the batteries in parallel. The result in either case is a
significant loss of power to resistive heating in the batteries—a waste of the energy stored in the batteries.
Furthermore, these large currents could result in overheating of the batteries, which could be dangerous. To avoid
this problem, batteries connected in parallel should have the same emf.
21. (a) It increases.
(b) It decreases.
(c) It increases.
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22. (a) Bulbs C and D are equally bright and they are brighter than bulbs A or B. Bulbs A and B are also equally
bright, but less so than bulbs C and D.
(b) The brightness of bulb B increases.
(c) The brightness of bulb C remains the same.
23. (a) Bulbs B and C get brighter.
(b) Bulb A gets dimmer and bulb C gets brighter. With all three resistances in the circuit, the currents are
IA  2V/(3R), IB  IC  V/(3R). When B is removed, the current in both A and C is V/(2R).
(c) Bulb A gets brighter, bulb C stops glowing entirely.
Problems
1. Strategy Use the definition of electric current.
Solution Compute the total charge.
q
I
, so q  I t  (3.0 A)(4.0 h)(3600 s h)  4.3 10 4 C .
t
2. (a) Strategy Use the definition of electric current.
Solution Compute the charge.
q  I t  (0.500 A)(10.0 s)  5.00 C .
(b) Strategy Divide the charge found in part (a) by the magnitude of the charge of an electron.
Solution Compute the number N of electrons.
q
5.00 C
N

 3.12  1019 electrons
C
e 1.602  1019
electron
3. (a) Strategy and Solution The electrons flow from the filament to the anode; since they are negatively charged,
the current flows from the anode to the filament.
(b) Strategy Use the definition of electric current.
Solution Compute the current in the tube.
q
I
 q  f  (1.602  1019 C)(6.0  1012 s 1 )  0.96 A
t
4. Strategy Use the definition of electric current.
Solution Compute the beam current.
q
I
 q  f  [2(1.602  1019 C)](3.0  1013 s 1 )  9.6 A
t
5. Strategy Use the definition of electric current and the elementary charge of an electron.
Solution Find the number of electrons per second that hit the screen.
I
q Ne
N I
320  106 A

, so
 
 2.0  1015 electrons/s .
C
t t
t e 1.602  1019
electron
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6. Strategy Since the negatively charged electrons and positive ions move in opposite directions, they both
contribute to the current in the same direction. Use the definition of electric current.
Solution Compute the current in the tube.
I
q (3.8  1016  1.2  1016 )(1.602  1019 C)

 8.0 mA
t
1.0 s
7. Strategy Since the oppositely charged ions move in opposite directions, they both contribute to the current in the
same direction. Use the definition of electric current.
Solution Compute the current in the solution.
I
q Ne [2(3.8  1016 )  6.2  1016 ](1.602  1019 C)


 22.1 mA
t t
1.0 s
8. Strategy The energy delivered by each battery is equal to the total work done by each battery. Use Eq. (18-2).
Solution Compute the energy delivered by each battery, assuming they are ideal.
Scooter:
W  %q  (12 V)(4.0 kC)  48 kJ
Automobile:
W  (12 V)(30.0 kC)  360 kJ
9. Strategy The total energy stored in a battery is equal to the total work the battery is able to do. Use Eq. (18-2).
Solution Compute the energy stored in the battery.
W  %q  (1.20 V)(675 C)  810 J
10. (a) Strategy A coulomb is an A  s. Convert A  h to A  s to find the amount of charge that can be pumped by
the battery.
Solution Compute the charge.
(180.0 A  h)(3600 s h )  6.480 105 C
(b) Strategy The total energy stored in a battery is equal to the total work the battery is able to do. Use Eq.
(18-2).
Solution Compute the electrical energy that the battery can supply.
W  %q  (12.0 V)(6.480  105 )  7.78 MJ
(c) Strategy Use the definition of electric current.
Solution Find the time required to drain the battery.
t 
q 6.480  105 C  1 h 


  54.5 h
I
3.30 A
 3600 s 
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11. (a) Strategy Use the definition of electric current.
Solution Compute the amount of charge pumped by the battery.
q  I t  (220.0 A)(1.20 s)  264 C
(b) Strategy The electrical energy supplied is equal to the work done by the battery. Use Eq. (18-2).
Solution Compute the amount of electrical energy supplied by the battery.
W  %q  (12.0 V)(264 C)  3.17 kJ
12. (a) Strategy Calculate the work done using Eq. (18-2) and the definition of electric current.
Solution Compute the amount of electrical energy supplied by the solar cell.
W  %q  %I t  (0.45 V)(18.0  103 A)(9.0 h)(3600 s h )  260 J
(b) Strategy The power is equal to the rate at which the solar cell supplies electrical energy.
Solution Find the average power by dividing the energy supplied by the time.
W %I t
P

 %I  (0.45 V)(18.0  103 A)  8.1 mW
t
t
13. Strategy Use Eq. (18-3).
Solution Form a proportion.
I1
I2
1
neA1v1
neA2 v2


  d12v1 , so v   d2 2 v   2 2 v  4v .
1 
2
 2  1  2
ne  14  d 22  v2 d 22 v2
 d1 
ne 14  d12 v1
The relationship between the drift speeds is v1  4v2 .
14. Strategy Use Eq. (18-3).
Solution Find the drift speed of the conduction electrons in the wire.
I
I
2.50 A
I  neAvD , so vD 


 5.86  105 m s .
2
28

3
neA ne( r ) (8.47  10 m )(1.602  1019 C) (0.00100 m) 2
15. Strategy Use Eq. (18-3) and x  vD t.
Solution Find the drift speed of the conduction electrons in the wire.
I
I
I
I  neAvD , so vD 


.
2
neA ne( r )  ner 2
Find the time to travel 1.00 m along the wire.
t 
x
x
 ner 2 x  (8.47  1028 m 3 )(1.602  1019 C)( 0.00100 m 2 ) 2 (1.00 m)  1 min 





I
vD
I
10.0 A
 60 s 
2
 ner
 17.8 min
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16. Strategy Find the time of travel t using x  vD t and Eq. (18-3).
Solution Find the time of travel.
28
3
19 C) 1  (0.0010 m) 2 (0.010 m)
x neAd (5.8  10 m )(1.602  10
 1 min 
4
t 



  8.1 min
vD
I
0.15 A
 60 s 
17. Strategy Let h be the thickness of the strip so that the cross-sectional area is A = hw, where w is the width. Use
Eq. (18-3).
Solution Find the thickness of the strip.
I  neAvD  ne(hw)vD , so h 
I
130  106 A

 81 m .
newvD (8.8  1022 m 3 )(1.602  1019 C)(260  106 m)(0.44 m s)
18. Strategy The cross-sectional area of the wire is 14  d 2 . Use Eq. (18-3).
Solution Find the current in the wire.
I  neAvD  ne


4
 14  d 2  vD  4 ned 2vD
(5.90  1028 m 3 )(1.602  1019 C)(0.00050 m) 2 (6.5 10 6 m s)  12 mA
19. Strategy Use Eq. (18-3) and n  1.3 N A M , the number of electrons per unit volume.
Solution Find the drift speed of the conduction electrons.
I
IM
(2.0 A)(64 g mol)
vD 


6
3
neA 1.3 N A eA 1.3(9.0  10 g m )(6.022  1023 mol1 )(1.602  1019 C)(1.00  106 m 2 )
 0.11 mm s
20. Strategy Find the average time t it takes for an electron to move 12 m along the wire using x  vD t , Eq.
(18-3), and n  3.5  N A M , the number of electrons per unit volume. The cross-sectional area of the wire is
1  d 2.
4
Solution Find t.
x neAx 3.5  N A eAx
t 


vD
I
MI

3.5(2.7  106 g m3 )(6.022  1023 mol 1 )(1.602 1019 C) 14  (0.0026 m)2 (12 m)
(27 g mol)(12 A)(3600 s h )
21. Strategy Use the definition of resistance.
Solution Compute the current through the resistor.
V
V 16 V
R
, so I 

 1.3 A .
I
R 12 
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Chapter 18: Electric Current and Circuits
22. (a) Strategy Use the definition of resistance.
Solution Compute the resistance.
V
4.50 V
R

 54 
I
0.083 A
(b) Strategy and Solution The current flows from right to left through the battery (from low to high potential).
Thus, the current flows from left to right through the resistor.
23. Strategy Use Eq. (18-8). The cross-sectional areas of the wires are given by 14  d 2 .
Solution Form a proportion to find the ratio of diameters.
 Al
RAl
1
RCu
Cu
L
AAl
L
ACu

 Al ACu  Al dCu 2
d
Cu
1.67

, so Cu 

 0.794 .
Cu AAl Cu d Al 2
d Al
Al
2.65
24. Strategy Use the definition of resistance and Eq. (18-8). The cross-sectional area of the wire is 14  d 2 .
Solution Find V , the potential difference between the bird’s feet.
V  IR 
I  L (150 A)(2.65  108   m)(0.020 m)

 0.25 mV
1  (0.020 m) 2
A
4
25. (a) Strategy Use the definition of resistance.
Solution Compute the required potential difference between the electrician’s hands.
V  IR  (50 mA)(1 k)  50 V
(b) Strategy and Solution An electrician working on a “live” circuit keeps one hand behind his or her back
to avoid becoming part of the circuit.
26. Strategy Use the definition of resistance.
Solution
(a) Compute the resistance given the data at point 1.
V
0.30 V
R

 15 
I
0.020 A
(b) Compute the resistance given the data at point 2.
0.40 V
R
 10 
0.040 A
27. Strategy Use Eq. (18-8). The cross-sectional area of the wire is 14  d 2 .
Solution Find the diameter of the nichrome wire.
R
L
L
L
(108  108   m)(46 m)

, so d  2
2
 2.5 mm .
1d2
A
R
 (10.0 )
4
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College Physics
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28. Strategy As found in Example 18.4, R R0  1  T .
Solution Find  .
1   T 

R
1  R
1
 25.0  
, so  
 1 
 1  3.8  103 C1 .


R0
T  R0  85.0 C  15.0 C  19.8  
29. Strategy As found in Example 18.4, R R0  1  T . Find T using this and the definition of resistance.
Solution Estimate the temperature of the tungsten filament.
 2.90 V


R
1 R
1
 0.300 A  1  20.0C  1750C .
1   T 
, so T  
 1  T0 
R0
  R0 
4.50  103 C1  1.10  
30. Strategy The terminal voltage is V  % Ir.
Solution Solve for I, the maximum current.
% V 1.5 V  1.0 V
I

 5A
r
0.10 
31. Strategy The current is equal to the terminal voltage divided by the resistance of the resistor connected across the
battery terminals. Use Eq. (18-10).
Solution Find the terminal voltage, V.
V
%
12.0 V
V  IR and V  % Ir , so V  % r , or V 

 4.0 V .
r
2.0 
R
1  R 1  1.0

Compute the current through the 1.0- resistor.
V 4.0 V
I 
 4.0 A
R 1.0 
32. Strategy For wires of the same length and diameter, R   .
Solution Find the ratios; then compare.
(a)
(b)
RAg
RCu
RAl
RCu


 Ag
Cu

1.59  108   m
1.67  108   m
2.65  108   m
1.67  108   m
 0.952
 1.59
(c) The material with the lowest resistivity is the best conductor. That material is silver.
33. Strategy Use the definition of resistance, the relationship between voltage and uniform electric field, and Eq.
(18-8).
Solution V = IR = EL and R   L A . Find E.
V  EL  IR  I 
L
I
, so E   , where  is the resistivity .
A
A
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34. Strategy As found in Example 18.4, R R0  1  T . Use Eq. (18-8) and the definition of resistance.
Solution LAl  3LCu , rAl  2rCu , Cu  0.6  Al , and  Cu   Al .
(a) Form a proportion.
RAl
RCu

LAl
AAl
L
Cu ACu
Cu
 Al

2
 Al LAl ACu
Al (3LCu ) rCu
3
, so RAl 
R 
(24 )  30  .
2 Cu
Cu LCu AAl
(0.6  Al ) LCu  (2rCu )
0.6(2)2
(b) At I = 10 A, V = 300 V, so R 
V 300 V

 30  .
I
10 A
(c) Find T.
1   T 

R
1 R
1
 30  
, so T  
 1  T0 
 1  20C  80C .


1
R0
  R0 
0.004C  24  
35. Strategy and Solution E does not depend upon T, but  is directly proportional to T, and vD  I . So,
the electric field stays the same, the resistivity decreases, and the drift speed increases.
36. (a) Strategy Sum the individual emfs with those with their left terminal at the higher potential being positive.
Solution Compute the equivalent emf.
%eq  3.0 V  4.5 V  1.5 V  2.0 V  5.0 V  3.0 V
(b) Strategy Use the definition of resistance.
Solution Compute the current through the resistor.
%eq 3.0 V
V %eq
R

, so I 

 0.94 A .
I
I
R
3.2 
37. (a) Strategy Sum the individual emfs with those with their left terminal at the higher potential being positive.
Solution Compute the equivalent emf.
%eq  3.0 V  3.0 V  2.5 V  1.5 V  7.0 V
(b) Strategy Use the definition of resistance.
Solution Find the value of the resistor.
V %eq
7.0 V
R


 18 
I
I
0.40 A
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Chapter 18: Electric Current and Circuits
38. (a) Strategy Ceq  Ci for the capacitors, which are in parallel.
Solution Compute the equivalent capacitance.
Ceq  2.0 F  6.0 F  3.0 F  11.0 F
(b) Strategy Use Eq. (17-14).
Solution Compute the charge on the capacitor.
Q  C V  C % (6.0  106 F)(44.0 V)  260 C .
39. (a) Strategy Ceq  Ci for the capacitors, which are in parallel.
Solution Compute the equivalent capacitance.
Ceq  4.0 F  2.0 F  3.0 F  9.0 F  5.0 F  23.0 F
(b) Strategy Use Eq. (17-14).
Solution Compute the charge on the equivalent capacitor.
Q  C V  Ceq % (23.0  106 F)(16.0 V)  368 C .
(c) Strategy Use Eq. (17-14).
Solution Compute the charge on the capacitor.
Q  C V  C % (3.0  106 F)(16.0 V)  48 C .
40. (a) Strategy Use Eqs. (18-13) and (18-17).
Solution Compute the equivalent resistance between points A and B.
1 
 1
Req  15   


 12  24  
1
 23 
(b) Strategy Label the currents on a diagram. Use Kirchhoff’s rules.
Solution The current through the emf is I  % Req  I1  I 2 , where the
currents labeled 1 and 2 are shown in the diagram. Applying the loop rule,
R
we have I 2 R2  I1R1  0, so I 2  1 I1. Solve for the current through the
R2
12- resistor, I1.

R
I  I1  I 2  I1  1 I1   1 
R2


R1 
 I1 , so I1  1 
R2 

R1 

R2 
719
1

I   1 

R1 

R2 
1
%  12 
 1  
Req  24 
1
276 V
 8.0 A .
23 
College Physics
Chapter 18: Electric Current and Circuits
41. (a) Strategy Use Eqs. (18-13) and (18-17).
Solution Compute the resistance between points A and B.
1
 1

Req  


 2.0  1.0   1.0  
1
 4.0   5.0 
(b) Strategy Label the currents on a diagram. Use Kirchhoff’s rules.
Solution The current through the emf is I  % Req  I1  I 2 , where the
currents labeled 1 and 2 are shown in the diagram. Applying the loop rule,
R
2.0 
I  1.0 I1.
we have I 2 (2 R2 )  I1R1  0, so I 2  1 I1 
2 R2
2(1.0 ) 1
Solve for the current through the 2.0- resistor, I1.
I
%
20 V
I  I1  I 2  I1  1.0 I1  2.0 I1 , so I1 


 2.0 A .
2.0 2.0 Req 2.0(5.0 )
42. Strategy Use Eqs. (18-13) and (18-17). The current through the 4.0- resistor is the same as that through the
emf, I  % Req .
Solution Find Req .
Req 
1
1
2.0 
1
 R 1.0

Find R.
%
I
, so Req 
Req
1
1
2.0 
1
 R 1.0

 4.0 
1
1
2.0 

1
R 1.0 
 4.0  
% 93.5 V

 5.5  .
I
17 A
 1.5 
1
1
1


2.0  R  1.0  1.5 
1
1
1


R  1.0  1.5  2.0 
1
1 
 1
R

  1.0   5.0 
 1.5  2.0  
43. (a) Strategy Use Eqs. (18-15) and (18-18).
Solution Find the equivalent capacitance of the circuit.
1
Ceq 
 1.5 F
1 
1
1
4.0 F

1  1
1.0 F 1.0 F

 2.0 F
μF
(b) Strategy The charge on the 4.0- capacitor is the same charge as on the equivalent capacitor. Use
Eq. (17-14).
Solution Find the charge.
Q  C V  Ceq % (1.54 F)(24 V)  37 C
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College Physics
Chapter 18: Electric Current and Circuits
44. (a) Strategy Ceq  1.63 F. Use Eqs. (18-15) and (18-18).
Solution Find the unknown capacitance.
1
1
1


1
1.63 F 4.0 F
1  1
 2.0 F
1.0 F C

 1
1
 

1.0

F
C


1
 1
1 



1.63

F
4.0
F 


1
 2.0 F
1

1  1
1 
 

  2.0 F
C  1.63 F 4.0 F 



1

1
1.0 F
1
1


1

1 
1 
  1

C 


2.0

F

  3.0 F

1.0 F 

  1.63 F 4.0 F 



μF
(b) Strategy The charge on the 4.0- capacitor is the same charge as on the equivalent capacitor. Use Eq.
(17-14).
Solution Find the charge.
Q  C V  Ceq % (1.63 F)(24 V)  39 C
45. Strategy Use the concept of equivalent resistance. The equivalent resistance of two identical resistances R in
parallel is half or R 2.
Solution
(a) The two 2.0- resistors are in series, so their equivalent resistance is 2.0   2.0   4.0  . These two
resistors are in parallel with the rightmost 4.0- resistor. Because the resistances of each branch of this
parallel circuit are equal, the current is split evenly. Let the current through each branch be called I3 . We
must determine I3 . Now, the equivalent resistance of this parallel circuit is 2.0  , and this is in series with
the rightmost 3.0- resistor and the rightmost 1.0- , so the equivalent series resistance is 6.0  . This
resistance is in parallel with the 6.0- resistor, so the current is again split evenly. Let it be called I 2 ; then,
I 3  I 2 2. The equivalent resistance of this parallel circuit is 3.0-, and this is in series with the middle
1.0- resistor, so the equivalent series resistance is 4.0  . This equivalent resistance is in parallel with the
leftmost 4.0- resistor, so the current is again split evenly. Let it be called I1; then, I 3  I 2 2  I1 4. The
equivalent resistance of this parallel circuit is 2.0- , and this is in series with the leftmost 1.0- and 3.0-
resistors, so the equivalent resistance of the entire circuit is 6.0  . If the current through the emf is I; then,
I 3  I 2 2  I1 4  I 8. The current though the emf is given by I  % Req . Compute the current through one
of the 2.0- resistors.
I
%
24 V
I3  

 0.50 A
8 8 Req 8(6.0 )
(b) The current through the 6.0- resistor is I 2 , which is one-fourth of the current through the emf.
I
%
24 V
I2  

 1.0 A
4 4 Req 4(6.0 )
721
College Physics
Chapter 18: Electric Current and Circuits
(c) The current through the leftmost 4.0- resistor is I1 , which is half of the current through the emf.
I1 
I
%
24 V


 2.0 A
2 2 Req 2(6.0 )
46. (a) Strategy Use Eqs. (8-13) and (8-17).
Solution Compute the resistance between points A and B.
1
Req  1.0  
 3.0 
1
1

1
2.0 1.0 
3.3 
 4.01   8.01  
(b) Strategy The current through the 1.0- resistor connected directly to point A is the same as the current
through the emf.
Solution Find the current.
% 18 V
I

 6.0 A
Req 3.0 
(c) Strategy Use the concept of equivalent resistance and Kirchhoff’s rules. The equivalent resistance of
resistances R and 2R in parallel is 2 R 3.
Solution The 4.0- and 8.0- resistances are in parallel and eight is twice four, so the equivalent resistance
of this parallel circuit is 2.67  . Let the current through the 4.0- resistor (directed to the right) be I 3 and
that through the 8.0- resistor (directed to the right) be I 4 . Let the current entering this parallel circuit from
the left be I 2 ; then, I 2  I 3  I 4 . According to the loop rule, I 3 (4.0 )  I 4 (8.0 )  0, so I 3  2.0 I 4 .
Find I 2 in terms of I 4 .
I 2  I 3  I 4  2.0 I 4  I 4  3.0 I 4
The 3.3- resistor and the 2.67- equivalent resistance are in series, so the equivalent resistance of these
two is 6.0  . This resistance is in parallel with the series resistance of the 2.0- resistor and the rightmost
1.0- resistor with an equivalent resistance of 3.0  . Since six is twice three, the equivalent resistance of
the 6.0- and 3.0- resistances is 2.0  . This resistance is in series with the leftmost 1.0- resistor, so
the equivalent resistance of the entire circuit is 1.0   2.0   3.0  . Let the current through the 2.0-
resistor and the rightmost 1.0- resistor be I1 and the current through the emf be I; then, I  I1  I 2 .
According to the loop rule, I1 (3.0 )  I 2 (6.0 )  0, so I1  2.0 I 2 . Find I 4 in terms of I.
I  I1  I 2  2.0 I 2  I 2  3.0 I 2  3.0(3.0 I 4 )  9.0 I 4 , so I 4 
I
%
I
%
18 V
, so I 4 


 0.67 A .
Req
9.0 9.0 Req 9.0(3.0 )
722
I
. The current through the emf is
9.0
College Physics
Chapter 18: Electric Current and Circuits
47. (a) Strategy The resistors are in parallel; they all begin at A and end at B, or vice versa.
Solution Find the equivalent resistance.
1 1 1 1 1 1 1 1
Req          
R R R R R R R R
1

R
8
(b) Strategy and Solution There is a “short” circuit between points B and C, so R  0 .
(c) Strategy The potential across each resistor is 32 V.
Solution Compute the current in one of the resistors.
V 32 V
I 
 16 A
R 2.0 
48. (a) Strategy Use Eqs. (18-13) and (18-17).
Solution Compute the resistance between points A and B.
1
Req 
 3.0 
1 
1
1
6.0 
2.0 
1

1.0 1.0 
 3.0 
1
 4.01   4.01  
1
(b) Strategy Use the concept of equivalent resistance and Kirchhoff’s rules. The equivalent resistance of two
identical resistances R in parallel is half or R 2.
Solution The 4.0- resistors are in parallel, along with the series combination of the two 1.0- resistors, so
the potential difference across them is the same. Let this potential difference be V. The equivalent resistance
of the two 4.0- resistors is 2.0  . The equivalent resistance of the two 1.0- resistors is 2.0  , as well.
Thus, the equivalent resistance of all four resistors is 1.0  . This resistance is in series with the 2.0- and
3.0- resistors, so their equivalent resistance is 6.0  . This resistance is in parallel with the 6.0-
resistance, so their equivalent resistance is 3.0  , which agrees with the result obtained in part (a). The
current through the emf is I 
% 12 V

 4.0 A. This current is split evenly between the 6.0-
Req 3.0 
equivalent resistance and the actual 6.0- resistor. So, 2.0 A flows through the 2.0- and 3.0- resistors
and 1.0- equivalent resistance (of the 1.0- and 4.0- resistors). According to the loop rule (taking the
outer loop that excludes the 6.0- resistor), 12 V  (2.0 A)(2.0 )  V  (2.0 A)(3.0 )  0, so the potential
difference across the 4.0- resistors is V  12 V  (2.0 A)(2.0 )  (2.0 A)(3.0 )  2.0 V .
(c) Strategy and Solution As found in part (b), the current through the 3.0- resistor is 2.0 A.
723
College Physics
Chapter 18: Electric Current and Circuits
49. (a) Strategy Use Eqs. (18-15) and (18-18).
Solution Find the equivalent capacitance.
 1

1
Ceq  


12

F
12

F

12

F


1
 8.0 F
(b) Strategy Since the capacitor at the left side of the diagram (1) is in series with the parallel combination of
the other two capacitors (2), the charge Q on the capacitor 1 is the same as that on capacitor 2. (Think of the
parallel combination as one capacitor with capacitance C2  12 F  12 F  24 F.) Use the definition of
capacitance.
Solution Find the potential difference across C1. Let this potential difference be V1 and the potential
difference across C2 be V2 . Then, % V1  V2 . Form a proportion.
V2 % V1 %
Q C2 C1
%
25 V

 1 

, so V1 

 17 V .
C
12 F
1
V1
V1
V1
Q C1 C2
1
1
C2
24 F
(c) Strategy The charge on the capacitor at the far right of the circuit (1) is half of the charge on the capacitor at
the left of the circuit (2).
Solution Find the charge on the capacitor.
1
1
Q2  C2V2  2Q1 , so Q1  C2V2  (12  106 F)(17 V)  1.0  104 C .
2
2
50. Strategy Use Eqs. (18-15) and (18-18). Draw a circuit diagram.
Solution The circuit diagram:
Ceq 
1
1
9.0 pF
 9.0 pF1 9.0 pF
 6.0 pF
51. (a) Strategy Use Eqs. (18-13) and (18-17).
Solution Find the equivalent resistance.
1
Req 
1 
1
4.00 
1
2.00 
1 
4.00 
 2.00 
1
1 1
 1
2.00  


 4.00  4.00  
(b) Strategy Use the definition of resistance and the equivalent resistance found in part (a).
Solution Compute the current that flows through the emf.
% 6.00 V
I

 3.00 A
Req 2.00 
724
College Physics
Chapter 18: Electric Current and Circuits
(c) Strategy Use Eqs. (18-13) and (18-17) and the definition of resistance. Redraw the circuit.
Solution Let R  2.00 ; then Rc  4.00   2 R is the 4.00- resistor at the bottom. The emf has been
left out for convenience.
 1
1 
From part (a), Req  


4.00

R

L
the circuit. Find RL .
1
 2.00 , where RL is the equivalent resistance of the left branch of
1
1
1
1
1 
 1


, so RL  

  4.00 .
2.00  4.00  RL
 2.00  4.00  
So, the resistances are the same for the left and right branches. Since 3.00 A flows through the emf, 1.50 A
flows through each branch. This current is equal to I1  I 2. Find I1/I 2 considering that V is the same
across each branch.
1

I1
2R
1.50 A
1  
 1

 1, or I1  I 2 
 0.750 A.
V  I1  R  

   I 2 (2 R ), so
I2 R  R
2
 2 R 2 R  

This current is split evenly again since the parallel circuit containing Rc has equal resistances.
So, Rc 
0.750 A
 0.375 A .
2
52. Strategy Use Kirchhoff’s rules. Let I1 be the top branch, I 2 be the middle branch, and I 3 be the bottom branch.
Assume that each current flows right to left.
Solution Find the current in each branch of the circuit.
(1) I1   I 2  I 3 (2) 0  5.00 V  (56 ) I 2  (22 ) I1 (3) 0  1.00 V  (56 )I 2  (75 )I 3
Substitute (1) into (2).
0  5.00 V  (56 )I 2  (22 )(  I 2  I 3 )  5.00 V  (78 )I 2  (22 )I 3 (4)
Solve (3) for I 2 and substitute into (4).
(56 )I 2  (75 )I 3  1.00 V, so I 2 
75
1
I 
A.
56 3 56
725
College Physics
Chapter 18: Electric Current and Circuits
Solve for I3 .
1
5
39
 75

0  5.00 V  (78 )  I 3 
A   (22 ) I 3 
A
(75 I3  1 A)  I 3  0.164 A  5.75 I 3 , so
56
56
22
616


I 3  0.0285 A.
Calculate I 2 .
75
1
I 2  (0.0285 A) 
A  0.0560 A
56
56
Calculate I1.
I1   I 2  I 3  0.0560 A  0.0285 A  0.0845 A
To two significant figures, the currents are:
Branch
I (mA)
Direction
Top
85
right to left
Middle
56
left to right
Bottom
29
left to right
53. Strategy Use Kirchhoff’s rules. Let I1 be the top branch, I 2 be the middle branch, and I 3 be the bottom branch.
Assume that each current flows right to left.
Solution Find the current in each branch of the circuit.
(1) I1   I 2  I 3 (2) 0  25.00 V  (5.6 )I 2  (122 )I1
(3) 0  25.00 V  5.00 V  (75 )I 3  (122 ) I1
Substitute (1) into (2).
(5) 0  25.00 V  (122   5.6 )I 2  (122 ) I 3
Multiply (4) by 5 and subtract from (5).
(4) 0  5.00 V  (75 )I 3  (5.6 ) I 2
0  [122   5.6   5(5.6 )]I 2  [122   5(75 )]I3 , so I 2 
Substitute the result above into (4).
0  5.00 V  (75 )I3  (5.6 )(1.626I3 ), so I 3 
5(75 )  122 
I  1.6 I3 (1.626 I3 ).
122   5.6   5(5.6 ) 3
5.00 V
  0.076 A.
1.626(5.6 )  75 
So, I 2  1.626(0.076 A)  0.12 A and I1  (0.12 A)  (0.076 A)  0.20 A.
Branch
I (A)
Direction
AB
0.20
right to left
FC
0.12
left to right
ED
0.076
left to right
726
College Physics
Chapter 18: Electric Current and Circuits
54. Strategy Use Kirchhoff’s rules. Let I be the current flowing up through the 5.00- resistor and I1 be the current
flowing to the left through the 4.00- resistor.
Solution Find the unknown emf and the unknown currents.
(1) I  I1  0.0500 A
0  % (1.00 )I1  1.20 V  (4.00 ) I1  (2.00 )(0.0500 A)
(2) 0  % 1.10 V  (5.00 )I1
0  % (5.00 )I  1.00 V  (2.00 )(0.0500 A)
(3) 0  % 1.10 V  (5.00 )I
Subtract (3) from (2).
0  2.20 V  (5.00 )( I  I1 ), so I  I1 
Set this result equal to (1).
0.440 A  I1  I1  0.0500 A, so I1 
2.20 V
, or I  0.440 A  I1.
5.00 
0.490 A
 0.245 A.
2
Find I.
I  I1  0.0500 A  0.245 A  0.0500 A  0.195 A
Solve for %.
0  % 1.10 V  (5.00 )I , so % 1.10 V  (5.00 )(0.195 A)  0.13 V.
The current through the 5.00- resistor flows upward and is 0.195 A. The current through the 4.00- resistor
flows right to left and is 0.245 A. The emf is 0.13 V.
55. Strategy Use Kirchhoff’s rules. Let the current on the left be I, the one in the middle be I1 , and the one on the
right be I 2 . I1 flows downward.
Solution Find the unknown emf and the unknown resistor.
I1  I  I 2  1.00 A  10.00 A  11.00 A
Loop ABCFA:
0   % (6.00 )(1.00 A)  (4.00 )(11.00 A)  125 V, so % 75 V .
Loop ABCDEFA:
0   % (6.00 )(1.00 A)  (10.00 A)R  75 V  6.00 V  (10.00 A)R, so R 
81 V
 8.1  .
10.00 A
56. Strategy Use Kirchhoff’s rules and Eq. (18-10). Draw a diagram showing the currents.
Solution Find the current in each branch of the circuit.
(1)  I 2  I  I1
0  14.0 V  12.0 V  I1 (0.0150 )  I (0.0850 )
(2) 0  2.0 V  I1 (0.0150 )  I (0.0850 )
(3) 0  12.0 V  I 2 (0.850 )  I1 (0.0150 )
Add (2) to (3).
0  14.0 V  I 2 (0.850 )  (0.0850 )I
14.0 V
0.850 
I

I
0.0850  0.0850  2
2800
(4) I 
A  10.0 I 2
17
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College Physics
Chapter 18: Electric Current and Circuits
Solve (3) for I1.
I1 (0.0150 )  12.0 V  I 2 (0.850 )
170
(5)
I1  800 A 
I
3 2
Substitute (4) and (5) into (1) and solve for I 2 .
2800
170
A  10.0 I 2  800 A 
I
17
3 2
170 
2800

 1 I 2  800 A 
A
10.0 
3
17


800 A  2800
A
17
I2 
 14.3 A (  14.26 A)
11.0  170
3
I2 
Calculate I and I1.
I
2800
A  10.0(14.26 A)  22.1 A and I1   I 2  I  14.26 A  22.14 A  7.9 A.
17
Branch
I (A)
Direction
alternator
22.1
up
battery
7.9
down
R
14.3
down
Find the terminal voltage.
V  % I1r  12.0 V  (7.88 A)(0.0150 )  12.1 V
Since V  % (the current flows from high potential to low potential), the battery is charging.
57. Strategy Use Eq. (18-20).
Solution Compute the power dissipated by the resistor.
P  %I  (2.00 V)(2.0 A)  4.0 W
58. Strategy Use Eq. (18-21a).
Solution Compute the power dissipated by the resistor.
P  I 2 R  (2.0 A)2 (5.00 )  20 W
59. Strategy Use Eq. (18-20).
Solution Compute the current in the bulb.
P 60.0 W
I 
 0.50 A
% 120 V
60. Strategy Use Eq. (18-21b).
Solution Compute the resistance of the lightbulb.
R
V 2 (120 V) 2

 360 
P
40.0 W
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College Physics
Chapter 18: Electric Current and Circuits
61. Strategy and Solution Yes; the power rating can be determined by P  IV  (5.0 A)(120 V)  600 W .
62. Strategy % 1.50 V  1.50 V  1.50 V  4.50 V since the batteries are in series. Use Eq. (18-20).
Solution Find the maximum power consumed by the CD player.
P  I % (0.2500 A)(4.50 V)  1.13 W
63. Strategy % 2.00 V  2.00 V  4.00 V since the emfs are in series. Use Eq. (18-21b) and the fact that power is
the rate of change of work, P  W t .
Solution Find the work done by the batteries in every 10.0-s time interval.
W  Pt 
%2
(4.00 V) 2 (10.0 s)
t 
 80.0 J
R
2.00 
64. Strategy Amperes are coulombs per second; amperes times ohms is volts; volts are joules per coulomb; and watts
are joules per second.
Solution Show that A 2    W.
A 2    A(A  )  A  V 
C J J
  W
s C s
65. (a) Strategy Use Eqs. (8-13) and (8-17) to find the equivalent resistance. Then draw the diagram.
Solution Compute the equivalent resistance.
1
1
1


Req  20.0   50.0   

  106.0 
 70.0   20.0  40.0   20.0  
The simplest equivalent circuit contains the emf and one 106.0- resistor.
(b) Strategy Use the definition of resistance.
Solution Compute the current that flows from the battery.
%
120 V
I

 1.1 A
Req 106.0 
(c) Strategy Compute the resistance between A and B. Then use the definition of resistance to find the potential
difference.
Solution Compute the resistance. It is in series with the two resistors not between A and B.
106.0   20.0   50.0   36.0 
Compute the potential difference.
V  IR  (1.13 A)(36.0 )  41 V
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College Physics
Chapter 18: Electric Current and Circuits
(d) Strategy The current that flows through the battery is shared by each branch between points A and B. Draw a
diagram with equivalent resistances. Use Kirchhoff’s rules.
Solution The diagram is shown.
60.0
I  I1  I 2 and I1 (60.0 )  I 2 (90.0 )  0, so I 2 
I and
90.0 1
60.0
 60.0 
I  I1 
I1  1 
 I1.
90.0
 90.0 
1
1
 60.0 
 60.0 
So, the current through the upper branch is I1   1 
 I  1 
 (1.13 A)  0.68 A , and the
 90.0 
 90.0 
60.0
60.0
current through the lower branch is I 2 
I 
(0.68 A)  0.45 A .
90.0 1 90.0
(e) Strategy Use P  I 2 R.
Solution Determine the power dissipated in the resistors.
P50  (1.13 A)2 (50.0 )  64 W , P70  (0.45 A)2 (70.0 )  14 W , and
P40  (0.68 A)2 (40.0 )  18 W .
66. (a) Strategy Use Eqs. (18-13) and (18-17).
Solution Compute the equivalent resistance.
1
 1

Req  15.0   


 15.0  20.0   30.0  
1
 10.0   36.5 
(b) Strategy The current that flows through R1 is the same current that flows through the emf. Use the
definition of resistance.
Solution Compute the current that flows through R1.
I
%
24.0 V

 0.657 A
Req 36.54 
(c) Strategy R2 is in parallel with the two resistors to its right, so the voltage drop across R2 is the voltage drop
across the equivalent resistance of these three resistors. Use Kirchhoff’s rules.
Solution Compute the equivalent resistance, R3 .
1
 1

R3  


 15.0  20.0   30.0  
Find the potential difference, V3 .
1
 11.5 
% I (15.0 )  V3  IR1  0, so
V3  % I (15.0 )  IR1  24.0 V  (0.657 A)(15.0 )  (0.657 A)(10.0 )  7.58 V .
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College Physics
Chapter 18: Electric Current and Circuits
(d) Strategy Use the definition of resistance.
Solution Compute the current through R2 , I 2 .
V
7.58 V
I2  3 
 0.505 A
R2 15.0 
(e) Strategy Use Eq. (18-19).
Solution Determine the power dissipated in R2 .
P  I V  I 2V3  (0.505 A)(7.58 V)  3.83 W
67. (a) Strategy Use Eq. (18-21b).
Solution Find the resistance of the heater when it is turned on.
P
V2
V 2 (120 V)2
, so R 

 6.5  .
R
P
2200 W
(b) Strategy Use Eq. (18-19).
Solution Find the current in the heater.
P 2200 W
P  IV , so I  
 18 A .
V
120 V
(c) Strategy Use Eqs. (18-8) and (18-9). The cross-sectional area of the wire is 14  d 2 .
Solution Find the diameter of the wire when it is hot.
L
L
R    0 (1   T )
, so
1d2
A
4
d
4 0 (1   T ) L
R

4(108  108   m)[1  (0.00040 K 1 )(400 K)](3.0 m)
 0.86 mm .
 (6.545 )
(d) Strategy From Example 18.4, R R0  1   T . Use the definition of resistance.
Solution Find the resistance of the wire when the heater is first turned on.
R
R  R0 (1   T )  R0 [1  (0.00040 K 1 )(400 K)]  1.16 R0 , so R0 
.
1.16
Compute the current.
V 1.16V 1.16V 1.16 P 1.16(2200 W)
I




 21 A
R0
R
V
120 V
V2 P
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College Physics
Chapter 18: Electric Current and Circuits
68. Strategy Use Kirchhoff’s rules and Eq. (18-21a).
Solution The currents are defined in the table.
Branch
AB
Currents
I; direction of flow
I1; left
CD
I 2 ; left
EF
I 3 ; right
Find the currents.
(1) I1  I 2  I 3
0  9.00 V  2.00 V  I 2 (5.00 )  I1 (4.00 )
(2) 0  7.00 V  I 2 (5.00 )  I1 (4.00 )
(3) 0  9.00 V  I 3 (8.00 )  I1 (4.00 )
Subtract (2) from (3).
0  2.00 V  I3 (8.00 )  I 2 (5.00 )
(4) I 2  0.400 A  1.60 I3
Substitute (1) into (2).
0  7.00 V  I 2 (5.00 )  ( I3  I 2 )(4.00 )
(5) 0  7.00 V  I 2 (9.00 )  I3 (4.00 )
Substitute (4) into (5).
0  7.00 V  (0.400 A  1.60 I 3 )(9.00 )  I3 (4.00 )
0  7.00 V  3.60 V  I 3 (18.4 )
I 3  0.576 A
I 2  0.400 A  1.60(0.576 A)  0.522 A
I1  I 3  I 2  0.576 A  (0.522 A)  1.098 A
Calculate the rate at which electrical energy is converted to internal energy in the chosen resistors; this is the same
as the power dissipated.
P4  I12 (4.00 )  (1.098 A)2 (4.00 )  4.82 W
P5  I 22 (5.00 )  (0.522 A)2 (5.00 )  1.36 W
69. (a) Strategy Use Eq. (18-10).
Solution Compute the voltage across the terminals of the battery.
V  % Ir  6.00 V  (1.20 A)(0.600 )  5.28 V
(b) Strategy Use Eq. (18-19).
Solution Compute the power supplied by the battery.
P  IV  (1.20 A)(5.28 V)  6.34 W
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College Physics
Chapter 18: Electric Current and Circuits
70. (a) Strategy Use Eq. (18-21b). Form a proportion and assume R doesn’t change.
Solution Find the power consumed by the lightbulb.
P108
P120

V1202
2
2
V 
 108 
, so P108   108  P120  
 (100.0 W)  81 W .
 120 
R
 V120 
V1082 R
(b) Strategy and Solution If the filament is at a lower temperature, its resistance will be lower, so its power
output will be higher for a particular voltage. Thus, the power drop will be less than that found in part (a).
71. (a) Strategy and Solution According to Eq. (18-10), Vterminal  % Ir because of the voltage drop due to the
internal resistance.
(b) Strategy Use Eqs. (18-10) and P  I V .
Solution Derive Eq. (18-22).
P  I V  I ( % Ir ), so P  %I  I 2 r .
%I is the rate that an ideal emf supplies power, and I 2 r is the rate of energy dissipation by the internal
resistance.
(c) Strategy and Solution According to Eq. (18-19), the rate at which electrical energy is converted to chemical
energy in the recharging battery is P  I V  %I .
(d) Strategy and Solution The power supplied must be sufficient to charge the battery while overcoming its
internal resistance, so P  %I  I 2r .
72. Strategy Ammeters are connected in series.
Solution Redraw the circuit to include the ammeters.
(a)
(b)
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College Physics
Chapter 18: Electric Current and Circuits
73. Strategy Voltmeters are connected in parallel.
Solution Redraw the circuit to include the voltmeters.
(a)
(b)
74. Strategy Ammeters are connected in series.
Solution
(a) Redraw the circuit to include the ammeter.
(b) The ammeter is connected in series with the 1.40-k resistor.
Find the new current through the resistor, assuming the ammeter
to be ideal (R  0).
(1) I1  I 2  I 3
(2) 0  9.00 V  I1(35 )  I 2 (1.40  103   240 )
0  I 2 (1.40  103   240 )  I3 (99.0  103 )
I3 
1.40  103   240 
I2
99.0  103 
(3) I3  0.0166 I 2
Substitute (3) into (1).
I1  I 2  0.0166 I 2
(4) I1  1.0166 I 2
Substitute (4) into (2).
0  9.00 V  1.0166 I 2 (35 )  I 2 (1.40  103   240 ), so
9.00 V
I2 
 5.37 mA .
1.0166(35 )  1.40  103   240 
734
College Physics
Chapter 18: Electric Current and Circuits
(c) The ammeter is connected in series with the 1.40-k resistor.
Find the new current through the resistor.
(1) I1  I 2  I 3
(2) 0  9.00 V  I1(35 )  I 2 (1.40  103   120 )
0  I 2 (1.40  103   120 )  I3 (99.0  103 )
I3 
1.40  103   120 
I2
99.0  103 
(3) I3  0.01535I 2
Substitute (3) into (1).
I1  I 2  0.01535I 2
(4) I1  1.01535I 2
Substitute (4) into (2).
0  9.00 V  1.01535I 2 (35 )  I 2 (1.40  103   120 ), so
9.00 V
I2 
 5.79 mA .
1.01535(35 )  1.40  103   120 
75. Strategy Ammeters are connected in series. Voltmeters are connected in parallel.
Solution
(a) Redraw the circuit to include the voltmeter.
(b) Find the reading of the voltmeter. Assume the voltmeter to be
ideal (R  0). Use Kirchhoff’s rules.
(1) I1  I 2  I 3
(2) 0  9.00 V  I1(35 )  I 2 (1.40  103 )
0  I 2 (1.40  103 )  I3 (99.0  103 )
I3 
1.40  103 
I2
99.0  103 
1.40
(3) I3 
I2
99.0
Substitute (3) into (1).
1.40
I1  I 2 
I2
99.0
 1.40 
(4) I1   1 
 I2
 99.0 
Substitute (4) into (2).
9.00 V
 1.40 
0  9.00 V  1 
I 2 (35 )  I 2 (1.40  103 ), so I 2 
 6.27 mA.

1.40
 99.0 
1  99.0 (35 )  1.40  103 

735

College Physics
Chapter 18: Electric Current and Circuits
Find I3 using (3).
μA
1.40
1.40
I3 
I2 
(6.27 mA)  88.7
99.0
99.0
The reading of the voltmeter is V  IR  (88.7
μA)(83.0
)
 
7.36 V .
(c) Find the reading of the voltmeter. Use Kirchhoff’s rules.
(1) I1  I 2  I 3
(2) 0  9.00 V  I1(35 )  I 2 (1.40  103 )
0  I 2 (1.40  103 )
1

1
1

 

 I 3 16.0  103   



 83.0  103  1.00  106   
1.40  103 
I3 
I
1 2
3
1
1
16.0  10  

3
6
(3) I 3  0.01511I 2
Substitute (3) into (1).
I1  I 2  0.01511I 2
(4) I1  1.01511I 2
Substitute (4) into (2).

83.010 
1.0010 

0  9.00 V  1.01511I 2 (35 )  I 2 (1.40  103 ), so I 2 
9.00 V
1.01511(35 )  1.40  103 
 6.27 mA.
Find I3 using (3).
μA
I3  0.01511I 2  0.01511(6.27 mA)  94.7 μA)(83.0 ) 7.86 mV .
 
The reading of the voltmeter is V  IR  (94.7
76. Strategy The resistances are in parallel, so the voltages across the galvanometer and shunt resistor are the same.
Solution Find the required resistance of the shunt resistor.
V  (10.0 A  0.250  103 A)RS  (0.250  10 3 A)(50.0 ), so RS 
(0.250  103 A)(50.0 )
10.0 A  0.250  103 A
77. Strategy The resistances are in series, so V  IReq .
Solution Find the required resistance of the series resistor.
100.0 V
(0.120  103 A)(RS  34.0 )  100.0 V, so RS 
 34.0   833 k .
0.120  103 A
78. Strategy The resistances are in series, so V  IReq .
Solution Find the required resistance of the series resistor.
100.0 V
(2.0  103 A)(RS  75 )  100.0 V, so RS 
 75   50 k .
2.0  103 A
736
 1.25 m .
College Physics
Chapter 18: Electric Current and Circuits
79. Strategy The resistances are in series, so V  IReq.
Solution Find the required resistances of the series resistors.
(a) (2.0  103 A)(RS  75 )  50.0 V, so RS 
(b) RS 
500.0 V
2.0  103 A
50.0 V
2.0  103 A
 75   25 k .
 75   250 k
80. (a) Strategy To measure the current in a circuit, an ammeter must be in series in the circuit. When 12.0 A pass
through the ammeter, the meter should deflect full scale. Therefore, 10.0 A should pass through the ammeter
and 2.0 A should pass through the resistor.
Solution Since the current must be split between the ammeter and the resistor, the resistor must be placed
in parallel with the ammeter. The voltages across the ammeter and the resistor are the same. Find the size of
the resistor, R.
I
R
(10.0 A)(24 )
V  I Ammeter RAmmeter  IR, so R  Ammeter Ammeter 
 120  .
I
2.0 A
(b) Strategy and Solution When 12.0 A is measured, 10.0 A is the reading on the meter. 12.0 10.0  1.20, so
the meter readings should be multiplied by 1.20 to get the correct current values .
81. Strategy The resistances are in series, so V  IReq .
Solution The two equations to solve simultaneously are
I (9850   RG )  25.0 V and I (3850   RG )  10.0 V.
Find RG .
I (9850   RG ) 25.0 V

 2.50
I (3850   RG ) 10.0 V
9850   RG  2.50(3850   RG )
2.50(3850 )  9850 
RG 
 150 
1.50
Find I.
25.0 V
I (9850   150 )  25.0 V, so I 
 2.50 mA .
10.0  103 
82. Strategy When fully charged, the voltage across the capacitor is equal to that of the emf. According to
Kirchhoff’s loop rule, when the capacitor is discharging, the voltage across the capacitor is equal to the voltage
across the resistor. Use Eqs. (18-24) and (18-26).
Solution Compute the voltage across the resistor.
VR  VC  %e t
( RC )
 (90.0 V)e 8.410
3
s [(30.0103  )(0.10106 F)]
737
 5.5 V
College Physics
Chapter 18: Electric Current and Circuits
83. Strategy Solve Eq. (18-23) for t.
Solution Find the time it takes for the voltage across the capacitor to be 15.0 V.
VC (t )  %(1  e t  )
 V (t ) 
ln et ( RC )  ln 1  C 
% 

 VC (t ) 
 15.0 
t   RC ln 1 
   (1.00 M)(2.00 F) ln 1 
  2.77 s
% 
 20.0 

84. Strategy Solve for R using VC  %(1  e t /  ) where   RC.
Solution Find the required resistance.
VC  %(1  e t /  )
V
1  C  et / 
%
t
 VC 
ln 1 
   RC
%


t
1.80 s
R

 8.04 k
6
VC
V
(125  10 F) ln 1  10.0
C ln 1  %
12.0 V




85. (a) Strategy The energy dissipated by the resistor is equal to the energy initially stored in the capacitor. Use
Eqs. (18-24), (18-25), and (17-18b), and the definition of resistance.
Solution Find the time constant.
I (t   )  I 0e 1  0.368I 0  0.368(100.0 mA)  36.8 mA, so   12.8 ms  RC.
V0
9.0 V
 0.0128 s

 90  ; C  
 140 F ;

3
I 0 100.0  10 A
R
90 
1
1
U  CV02  (142  106 F)(9.0 V)2  5.8 mJ
2
2
R
(b) Strategy The energy is directly proportional to the voltage across the capacitor squared. Solve Eq. (18-26)
for t.
Solution The energy is half its initial value when
2
1
V
V 
VC2  V02   0  or VC  0 .
2
2
2


V
t
1
1
1
VC  V0e t /   0 , so   ln
or t   ln 2  (12.8 ms)(0.693)  4.4 ms .

2
2
2
2
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College Physics
Chapter 18: Electric Current and Circuits
(c) Strategy Use Eq. (18-26).
Solution Substitute numerical values and graph the voltage across the capacitor.
VC (t )  V0e t /   (9.0 V)e t /(12.8 ms)
86. (a) Strategy Use Eq. (17-18b).
Solution Find the initial energy stored in the capacitor.
1
1
U 0  CV02  (50.0  106 F)(6.0  103 V)2  900 J
2
2
(b) Strategy Use the definition of resistance.
Solution Compute the initial current.
I0 
V0 6.0  103 V

 25 A
R
240 
(c) Strategy The voltage across the capacitor at t  1.0 ms is given by VC  V0e t /  . The energy stored in the
capacitor at this time is U  12 CVC2. The energy dissipated in the patient is the difference U 0  U .
Solution Find the energy dissipated.
6
1
U 0  U  U 0  CV02e 2t /   U 0 1  e 2t /( RC )   (900 J) 1  e 2(0.0010 s) /(240 )/(50.010 F)   140 J





2
(d) Strategy Use the definition of average power. Form a proportion.
Solution
U0
U 0tpatient (900)(1.0  103 )
Psource
t
 source


 0.0033
Ppatient t U
U tsource
(138)(2.0)
patient
The average power supplied by the power source is 0.0033 times that delivered to the patient.
(e) Strategy and Solution
The capacitor can deliver a much larger burst of current (and a much higher average power) to the patient
than the power source could.
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College Physics
Chapter 18: Electric Current and Circuits
87. (a) Strategy Use Eq. (17-18b).
Solution Find the required initial potential difference.
1
2U
2(20.0 J)
C (V ) 2 , so V 

 632 V .
2
C
100.0  106 F
(b) Strategy Use the definition of capacitance.
U
Solution Find the initial charge.
Q  C (V )  (100.0  106 F)(632 V)  63.2 mC
(c) Strategy Solve for R using I  I 0e t /  where   RC.
Solution Find the resistance of the lamp.
0.050 I 0  I 0et / 
t
ln 0.050  
RC
t
0.0020 s
R

 6.7 
C ln 0.050
(100.0  106 F) ln 0.050
88. (a) Strategy Find the equivalent resistance using Eq. (18-13) and the equivalent capacitance using Eqs. (18-15)
and (18-18).
Solution Find the equivalent resistance.
Req  25   33   58 
Find the equivalent capacitance.

1
1 
Ceq  


12 F  23 F 46 F 
Draw the equivalent circuit.
1
 20 F
(b) Strategy When the capacitors are fully charged, their chargers are no longer changing, so the current
‘through’ them is zero.
Solution Because it is in series with a combination of fully charged capacitors, the current in the resistor is
0 .
To find the charge on C1, first the charge on the equivalent capacitor. The entire battery voltage is dropped
across Ceq, so Q  Ceq V  (20  106 F)(6.0 V)  1.2  10 4 C
This is the charge on C3, so the voltage drop across C3 is V3 = Q3/C3 = (1.2 × 10–4 C)/(46 × 10–6 F) = 2.61 V.
The voltage drop across the parallel combination of C1 and C2 is then 6.0 V – 2.61 V = 3.39 V. Then the
charge on C1 is Q1 = C1 V1 = (12 × 10–6 F)(3.39 V) = 4.1 × 10–5 C. [Check: find the charge on C2 and check
that Q1 + Q2 = Q3.]
(c) Strategy Use Eq. (18-24).
Solution Find the time constant of the circuit.
  RC  Req Ceq  (58 )(20 F)  1.2 ms
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College Physics
Chapter 18: Electric Current and Circuits
(d) Strategy Solve Eq. (18-23) for t, setting the voltage across the capacitors equal to 50% of the emf.
Solution Find the time it takes to charge the capacitors to 50% of their final value.
V   (1  e t  )
0.50  1  et 
e
t ( ReqCeq )
 0.50
t   Req Ceq ln 0.50  (58 )(20  106 F) ln 0.50  8.0  104 s
89. Strategy Use Eqs. (18-24) and (18-25) and the definition of resistance.
Solution
(a) Initially (t  0), the capacitor has nearly zero resistance. Find the currents and the voltages.
V
12 V
I1  I 2  
 0.30 mA and V1  V2  12 V .
R 40.0  103 
(b) Calculate the time constant.
  (40.0  103 )(5.0  108 F)  2.0 ms
Find the currents.
I1  I 2  I 0et /   (3.0  104 A)e(1.0 ms)/(2.0 ms)  0.18 mA
Find V1 and V2.
V1  12 V and V2  I 2 R  (1.82  104 A)(40.0  103 )  7.3 V .
(c) Find the currents and the voltages.
I1  I 2  (3.0  104 A)e(5.0 ms)/(2.0 ms)  25 A , V1  12 V , and
V2  I 2 R  (2.463  105 A)(40.0  103 )  0.99 V .
90. Strategy Use Eqs. (18-24), (18-26), and (17-18b).
Solution
(a) The time constant is   RC  (0.40  103 )(0.50 F)  0.20  103 s.
Find V0.
U0 
1
CV02 , so V0 
2
2U 0

C
2(25 J)
 10 V.
0.50 F
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College Physics
Chapter 18: Electric Current and Circuits
3
The voltage across the resistor is given by VR (t )  V0e t /   (10 V)e t /(0.2010 s) .
(b) Solve for t.
1
1
U  CV 2  CV02e 2t /   U 0e 2t / 
2
2
U0
2t / 
e

U
U
2t
 ln 0

U
 U 0 0.20  103 s
25 J
t  ln

ln
 5.0 min
2 U
2
1.25 J
91. (a) Strategy Use the definitions of resistance and capacitance and Eqs. (18-24) and (18-25).
Solution Find the time constant.
  RC  (5  103 )(20  106 F)  0.1 s
Find I 0.
I0 
V0 Q0 Q0 200  106 C



 2 mA
R RC

0.1 s
The current through the resistor is given by I  I 0e t /   (2 mA)e t /(0.1 s) .
(b) Strategy Use Eq. (18-21a).
Solution Compute the initial power dissipated in the resistor.
P0  I 02 R  (0.002 A)2 (5  103 )  20 mW
(c) Strategy Use Eq. (17-18c).
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College Physics
Chapter 18: Electric Current and Circuits
Solution Compute the total energy dissipated.
Q02 (200  106 C) 2

 1 mJ
2C
2(20  106 F)
U
92. Strategy Use the definition of capacitance and Eqs. (18-23) and (18-25).
Solution Find the number of time constants that have elapsed in each situation.
(a) Solve for t.
Q(t )  CVC (t )  C %(1  e t /  )  Q0 (1  e t /  )
Q
 1  et / 
Q0
et /   1  0.990
t
  ln 0.010

t   ln 0.010  4.6
(b) t   ln(1  0.9990)  6.9
(c) Solve for t.
% t / 
e
R
0.010 I 0  I 0e t / 
1
et /  
0.010
 1 
t   ln 
  4.6
 0.010 
I
93. (a) Strategy According to the figure, I 0  0.070 A. Use Eq. (18-25).
Solution Compute the current at t   .
I (t   )  I 0e 1  (0.070 A)e 1  0.026 A
So, according to the figure,   0.060 s. The final charge is Q  I 0t  I 0  (0.070 A)(0.060 s)  4.2 mC .
(b) Strategy Use the definition of capacitance.
Solution Find the capacitance.
Q 0.0042 C
C 
 470 F
V
9.0 V
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College Physics
Chapter 18: Electric Current and Circuits
(c) Strategy Use Eq. (18-24).
Solution Find the total resistance in the circuit.

0.060 s
R 
 130 
C 470  106 F
(d) Strategy Use Eqs. (17-18b) and (18-23) and the fact that U  U 0 2.
Solution Solve for t.
1
1
U  CV 2  CV02 (1  e t /  )2  U 0 (1  e t /  )2
2
2
U

 1  et / 
U0
et /   1 

U
U0

U 
 ln 1 



U 0 


U
t   ln 1 

U0

t


1
   (0.060 s) ln 1 
  32 ms or 74 ms
2 


t cannot be negative, so the answer is 74 ms .
94. (a) Strategy According to the figure, I 0  95 mA and   2.5 ms (since e   0.368). Use the definitions of
resistance and capacitance and Eqs. (18-24), (18-25), and (17-18c).
Solution Find the current at t   .
I
0.095 A
I 0 
 0.035 A. Find Q0.
e
2.718
Q0  CV0  C ( I 0 R)  I 0 ( RC )  I 0  (0.095 A)(0.0025 s)  2.4  104 C
Find C.
Q02
Q 2 I 2 2 (0.095 A)2 (0.0025 s) 2
, so C  0  0 
 140 F .
2C
2U
2U
2(2.0  104 J)
Find R.

0.0025 s
R 
 18 
C 140  106 F
U
(b) Strategy Use Eqs. (17-18b) and (18-26).
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College Physics
Chapter 18: Electric Current and Circuits
Solution Find t.
1
1
U  C (V ) 2  CV02e 2t /   U 0e 2t / 
2
2
U0
2t / 
e

U
U
2t
 ln 0

U
 U 0 0.0025 s 2.0 104 J
t  ln

ln
 1.7 ms
2 U
2
5.0  105 J
95. (a) Strategy Use the definition of resistance.
Solution Compute the current that passes through Oscar.
V
100.0 V
I 
 50 mA
R 2.0  103 
(b) Strategy I1 passes through the 15  resistor, and I 2 passes through Oscar. They are in parallel, so the
voltage across each is the same, 100.0 V. Use Kirchhoff’s rules.
Solution
(1) 1.00 A  I1  I 2
V  I1R1  I 2 R2 , so
(2) (15 )I1  (2.0  103 )I 2
Solve (2) for I1 and substitute it into (1).
(15 )I1  (2.0  103 )I 2 , so I1 
Substitute.
1.00 A 
2.0  103
I 2.
15
2.0  103
1.00 A
I 2  I 2 , so I 2 
 7.4 mA .
2.0
103  1
15
15
96. Strategy Use Eq. (18-19).
Solution The maximum current that can be supplied by the batteries is
P
5.0 W
I max  max 
 0.050 A.
V
100.0 V
V
100.0 V
I 
 0.10 A  I max , so the current that passes through her is 50 mA .
R 1.0  103 
97. (a) Strategy and Solution The circuit breaker should be placed at D . This will best protect the household
against a short circuit in case any or all of the appliances overload the circuit or short out.
(b) Strategy Use Eq. (18-19).
Solution Compute the current.
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College Physics
Chapter 18: Electric Current and Circuits
I
P 1500 W  300 W  1200 W

 25 A
V
120 V
The devices cannot all be operated at the same time since the total current would be 25 A,
which is greater than the rated 20.0 A.
98. Strategy and Solution
(a) If the person receives a shock, the microwave is not grounded.
(b) If the cord begins to smoke, the wires are too small to handle the current and thus overheat.
(c) If a fuse blows out, too much current is drawn, and the appliance has a short circuit .
(d) An electrical fire breaking out inside the kitchen wall is likely the result of poor household wiring .
99. Strategy Use the definition of average power and Eq. (18-19).
Solution Find the amount of chemical energy that is converted to electrical energy.
E
P
 IV , so E  IV t  (0.30 A)(1.5 V)(4.0 h)(3600 s h)  6.5 kJ .
t
100. Strategy Add the potential differences.
Solution Find the potential at point X.
VX  4 V  (12 V)  8 V
101. Strategy Use the definition of resistance and Eqs. (18-13) and (18-17).
Solution
(a) The current through A1 is the same as that through the emf.
V
10.0 V
I

 2.00 A
1
Req 2.00  
1
 1  1
 2.00 
 2.00 
1 
 1
(b) Since 


 3.00  6.00  
2.00 A
I
 1.00 A .
2
1
3.00 
6.00 

 2.00 , the current is split evenly at the first junction to the right of A1. So,
102. Strategy Use the definition of resistance, Eqs. (18-13) and (18-17), and Kirchhoff’s junction rule.
Solution
(a) The current through A1 is the same as that through the emf.
V
10.0 V
I

 1.91 A
1
Req 2.00   0.200   1
 2.00 
1
2.00 

1 
 1
0.200  


 3.00  6.00  
746
1
College Physics
Chapter 18: Electric Current and Circuits
(b) The resistance due to A 2 is in series with the parallel combination of the 3.00- and 6.00- resistors. The
combination of these resistors ( RR ) is in parallel with the rightmost 2.00- resistor ( RL ).
1
1 
 1
RR  0.200   

  2.20 
 3.00  6.00  
Since these resistances are in parallel, the voltage is the same across each, so I L RL  I R RR , where I R is the
current through A 2. Also, I  I L  I R .
RL
2.00 
I L RL  ( I  I R ) RL  I R RR , so I R 
I
(1.9056 A)  0.907 A .
RL  RR
2.00   2.20 
103. Strategy The average discharge current is the charge on the capacitor divided by the time to discharge. Use the
definition of capacitance.
Solution Find the average discharge current.
I av 
Q CV (25  106 F)(1.0 V)


 31 A
t
t
0.80 s
104. Strategy Use Eq. (18-19) and the graph shown with Problem 26.
Solution Compute the power dissipated at points 1 and 2.
P1  I1V1  (0.020 A)(0.30 V)  6.0 mW and P2  (0.040 A)(0.40 V)  16 mW .
105. Strategy Use Eq. (18-19).
Solution Compute the current drawn by the motor.
P 1.5 hp
I 
(745.7 W hp )  9.3 A
V 120 V
106. Strategy Use Eqs. (18-17), (18-21a), and (18-21b).
Solution
(a) P  V 2 Req , so we need to design the circuit such that V is maximized and Req is minimized. If the
batteries are placed in series, V  2%. If the light bulbs are connected in parallel,
1
R
1 1
Req      ,
R
R
2


which is the smallest possible value. The potential across each is V  2%, which is the largest possible value.
(b) The power dissipated by each bulb is the same.
P
V 2 (2%)2
4%2


R
R
R
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College Physics
Chapter 18: Electric Current and Circuits
(c) The power through each bulb is given by P  I 2 R. The circuit must be designed so that the current through
the brighter bulb is larger than that through the dimmer bulb. In the circuit below, the maximum current
passes through the bulb on the right, whereas only a fraction of that current passes through the bulb on the
left, so the bulb on the right is brighter.
107. Strategy Use Eq. (18-21b).
Solution Since the resistances are equal and in series, the voltage is dropped by half by each resistor in Circuit 1.
V 2 (% 2)2 %2


 5.0 W.
R
R
4R
Since the bulbs are connected in parallel in Circuit 2, the voltage across each is % and, thus, the power dissipated
So, the power is P1 
by each is P2 
V 2 %2

 4P1  4(5.0 W)  20 W .
R
R
108. Strategy Use Eq. (18-21b).
Solution The equivalent emf for identical emfs in parallel is the same as one emf, so the total power dissipated in
the circuit is P 
V2
%2

.
Req
2R
109. Strategy Use Eqs. (18-21a) and (18-21b).
Solution
(a) P  V 2 R , so R  V 2 P . Compute the resistances.
R60 
(120 V) 2
(120 V)2
 240  and R100 
 140  .
60.0 W
100.0 W
(b) Since P  I 2 R, and I is the same through each bulb (connected in series), the bulb with the larger resistance
dissipates more power and, thus, shines brighter. So, the 60.0-W bulb shines brighter (240   140 ).
(c) When the bulbs are connected in parallel, the voltage across each is the same; and since P  V 2 R , the bulb
with the smaller resistance dissipates the most power, therefore, the 100.0-W bulb shines brighter
(140   240 ).
748
College Physics
Chapter 18: Electric Current and Circuits
110. Strategy Use Eq. (18-21b).
Solution
(a) Form a proportion.
2
2
Pf Vf 2 R  Vf 
 110 
 2
  
 0.84

Pi Vi R  Vi 
 120 
1  0.84  0.16, so the heat output decreases by 16% .
(b) Form a proportion.
2
Pf Vf 2 Rf  Vf  Ri
R
 2
 
 0.84 i
Pi Vi Ri  Vi  Rf
Rf
At lower temperature (at the lower voltage), the resistance is lower, so Rf  Ri , or Ri Rf  1. Therefore,
0.84( Ri Rf )  0.84 and the actual drop in heat output is smaller than that calculated in part (a).
111. (a) Strategy Use the definition of resistance.
Solution If no current flows through the galvanometer, the potential difference between points A and B is
zero. This means that the potential drop across the 45- resistor is the same as that across the 67- resistor,
and the drop across the 234- resistor is the same as that across Rx . The first drop is equal to
I1(45 )  I 2 (67 ). The second is equal to I1(234 )  I 2 Rx . Eliminate I1 and I 2.
Rx
I1 67 
67


, so Rx 
(234 )  350  .
I 2 45  234 
45
(b) Strategy and Solution Since no current flows through the galvanometer, its resistance does not affect the
measurement. The answer is no.
112. Strategy Draw a circuit diagram showing the currents. Use Kirchhoff’s rules.
Solution Let R1  R2  R3  R4  R.
(1) I1  I 2  I 3
(2) 0  % I1R  I 2 R
(3) 0  % I1R  I 3 (2 R)
Subtract (2) from (3).
0  2 RI3  I 2 R
(4) I 2  2I3
Substitute (4) into (1).
(5) I1  2 I3  I3  3I 3
Substitute (5) into (3) and solve for I3.
0  % 3I3R  2I 3R  % 5I 3R, so I 3 
% 150 V

 3.0 A .
5R 5(10 )
113. (a) Strategy Use Eq. (18-21b).
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College Physics
Chapter 18: Electric Current and Circuits
Solution Find the resistance of the hair dryer.
P
V2
V 2 (120 V)2
, so R 

 9.6  .
R
P
1500 W
(b) Strategy Use the definition of resistance.
Solution Find the current through the hair dryer.
V 120 V
I 
 13 A
R 9.6 
(c) Strategy The total energy used by the hair dryer is equal to its power times the time of usage.
Solution Find the cost to run the hair dryer for five minutes.
1h
10 cents
Cost  E  rate  Pt  rate  (1.5 kW)(5.00 min)

 1.3 cents
60 min kW  h
(d) Strategy Use Eq. (18-21b).
Solution Find the power used.
P
V 2 (240 V) 2

 6.0 kW
R
9.6 
(e) Strategy Use the definition of resistance.
Solution Find the current through the hair dryer.
V 240 V
I 
 25 A
R 9.6 
114. (a) Strategy The voltage across each bulb is 120 V. Use Eq. (18-21b).
Solution Find the resistance of each light bulb.
P
V2
V 2 (120 V) 2
, so R 

 1600  .
R
P
9.0 W
(b) Strategy Use the definition of resistance.
Solution Find the current through each bulb.
V 120 V
I 
 0.075 A
R 1600 
(c) Strategy The bulbs are connected in parallel, so the total current is the sum of individual currents.
Solution The total current is 25(0.075 A)  1.9 A .
(d) Strategy Let the number of 10.4-W bulbs be n. Then, the number of 9.0-W bulbs is 25  n. Use the
definition of resistance and Eq. (18-19).
Solution The current through a single bulb is given by I  P V . The total current for each kind of bulb is
(25  n) P1 V and nP2 V , where P1  9.0 W and P2  10.4 W. The total current must be less than or equal to
2.0 A. Find n.
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College Physics
Chapter 18: Electric Current and Circuits
(25  n) P1
nP
 2  2.0 A
V
V
25 P1  nP1  nP2  (2.0 A)V
n( P2  P1 )  (2.0 A)V  25P1
(2.0 A)V  25 P1 (2.0 A)(120 V)  25(9.0 W)
n

 10.7
P2  P1
10.4 W  9.0 W
Thus, up to 10 9.0-W bulbs can be replaced with 10.4-W bulbs without blowing a fuse.
115. (a) Strategy Use Eq. (17-18b).
Solution Find the total energy stored in the two capacitors.
1
1
U 2  CV 2  (2.00  106 )(5.00 V)2  2.50  105 J and
2
2
1
U 3  (3.00  106 F)(10.0 V)2  1.50  104 J, so U total  U 2  U 3  1.75  104 J  175 J .
2
(b) Strategy Once the capacitors are connected together, the voltage across each must be the same, so charge
flows from one capacitor to the other to equalize the voltage. The total amount of charge is conserved. Use
the definition of capacitance and Eq. (17-18b).
Solution Find the original total charge Q.
Q2i  C2V2i  (2.00  106 F)(5.00 V)  10.0 C and Q3i  C3V3i  (3.00  106 F)(10.0 V)  30.0 C, so
Q  40.0 C. The final charges are Q2f  C2V and Q3f  C3V . We also know that Q2f  Q3f  Q.
Q
Q
Eliminating V gives 2f  3f . Find Q2f .
C2
C3
Q2f

C2
 1
1 
Q2f 


 C2 C3 
Q2f 
Q3f Q  Q2f

C3
C3
Q
C3

1
C2
Q

 C1 C3
3

Q
C3
C2
1

40.0 C
 16.0 C
3.00  1
2.00
So, Q3f  Q  Q2f  40.0 C  16.0 C  24.0 C .
Find V.
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College Physics
Chapter 18: Electric Current and Circuits
Q2f 16.0  106 C

 8.00 V
C2
2.00  106 F
Find the total energy.
1
1
1
1
U total  C2V 2  C3V 2  V 2 (C2  C3 )  (8.00 V)2 (2.00  106 F  3.00  106 F)  160 J
2
2
2
2
V
(c) Strategy and Solution The “missing” energy is due to heat loss in the wires connecting the capacitors
during the time that current flows.
116. (a) Strategy and Solution From the diagram, it is obvious that bulbs 2 and 3 are connected in parallel to the
high potential side of the battery. This eliminates (b). Bulbs 1 and 3 are connected in parallel to the low
potential side of the battery. This eliminates (c) and (d). Bulbs 1 and 2 are connected in series.
(a), (e), and (f) are equivalent, and they all represent the circuit.
(b) Strategy Use Eq. (18-21b).
Solution Since bulb 3 is in parallel with the series combination of bulbs 1 and 2, the voltage drop across bulb
3 is the same as that of the series combination. P  V 2 R , and the light bulbs are identical
( R1  R2  R3  R). If the potential across the battery is %, the power dissipated in bulb 3 is P3  %2 R . The
voltage is dropped by half across each bulb in the series combination, since they are identical. So,
P1  P2  (% 2)2 R  %2 (4R)  P3 4. Thus, bulb 3 is the brightest, and bulb 1 and bulb 2 are the same .
(c) Strategy Use the definition of resistance.
Solution Through bulb 3:
% 6.0 V
I3  
 0.25 A
R 24.0 
Through bulbs 1 and 2:
%2
%
6.0 V
I1  I 2 


 0.13 A
R
2R 2(24.0 )
117. Strategy and Solution
(a) Olivia should use one of the 1.5 V batteries to oppose the 6.0 V battery; 6.0 V  (1.5 V)  4.5 V.
(b) The current will flow in the wrong direction through the 1.5 V-battery. This current may be too large for the
battery to handle, since the 1.5-V battery is not meant to be recharged .
118. Strategy For the axon: r  radius; t  thickness of the membrane; L  length. Use Eq. (18-8).
Solution
(a) R  
L
L
0.010 m
  2  (2.0   m)
 250 M
A
r
 (5.0  106 m)2
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College Physics
(b) R  
Chapter 18: Electric Current and Circuits
t
t
8.0  109 m

 (2.5  107   m)
 640 k
A
2 rL
2 (5.0  106 m)(0.010 m)
(c) Set Ra  Rb and solve for L.
Ra  a
L
t
 Rb   b
, so L 
2
2 rL
r
 btr
(2.5  107   m)(8.0  109 m)(5.0  106 m)

 0.50 mm .
2 a
2(2.0   m)
119. Strategy Use Eq. (18-8).
Solution Initially, R0   L0 A0 , and finally, R   L A . The volume is constant, so V  A0 L0  AL, or
A0  AL L0 . Divide R by R0.
2
2
L
 AL  L 
 L 
 3L0 
AL
R
 LA , so R  0 R0  

 R0    R0  
 R0  9R0 .
R0  0
AL0
L0  AL0 
L0 
L0 



A
0
120. Strategy Redraw the circuit diagram with the currents and voltmeter included. Use Kirchhoff’s rules.
Solution The new circuit diagram:
The voltmeter reads V4  I 4 (83.0 k), so I 4 must be determined. From Kirchhoff’s junction rule,
(1) I1  I 2  I 3 and (2) I3  I 4  I 5.
Find I1.
I1 
%

Req 35  
9.00 V
1

1.40103 
 6.367  103 A
1
1

16.0103   
1

1


1

 83.0103  670103  
The 1.40-k resistor is in parallel with the equivalent resistance of the combination of the 16.0-k, 83.0-k,
and 670-k resistances, so the voltage drop across these two is the same.
1

1
1

 
V  (1.40 k)I 2  I 3 16.0  103   

 

 83.0  103  670  103   
This gives I 2  64.18I3. Combining this with (1) results in I 3  0.01534 I1  9.767  105 A. Now, using
Kirchhoff’s loop rule for the rightmost loop, we have 0  I 4 (83.0 k)  I5 (670 k), so I 5  0.1239 I 4.
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College Physics
Chapter 18: Electric Current and Circuits
Combining this result with (2) gives I 4  0.8898 I 3  8.691 105 A. Thus,
V4  (8.691 105 A)(83.0  103 )  7.21 V .
121. Strategy The current is related to the drift speed by I  neAvD . Form a proportion.
Solution Compare the drift speeds.
(3nAu )vAl 3vAl
I Al
n eAv
n v
 Al Al  Al Al 

 1, so vAu  3vAl .
I Au nAu eAvAu nAu vAu
nAu vAu
vAu
122. (a) Strategy Use Eq. (18-21a).
Solution Compute the power loss.
P  I 2 R  (50 A) 2 (0.15  km)(500 km)  1.9  105 W for both materials.
(b) Strategy Use Eq. (18-8).
Solution Find the necessary cable diameters.
Copper:
L
L

1.67  108   m

,
so
d

2

2
 1.2 cm .
3
1 d2
A
 RL

(0.15

10

m)
4
Aluminum:
R
d 2
2.65  108   m
 (0.15  103  m)
 1.5 cm
(c) Strategy In this calculation,  is the mass density, not the resistivity.
Solution Find the mass per unit length for a cable made of each material.
m
m
m
m
  r 2 .
 

, so
L
V
AL  r 2 L
Copper:
m
  (0.0060 m)2 (8920 kg m3 )  1.0 kg m
L
Aluminum:
m
  (0.0075 m)2 (2702 kg m3 )  0.48 kg m
L
123. (a) Strategy Initially, the resistance of the capacitors is zero, so the entire voltage is dropped across the resistor.
Use the definition of resistance.
Solution Compute the initial charging current.
V
3.0 V
I 
 30 A
R 0.10  106 
(b) Strategy Use the definition of capacitance.
Solution When the current stops, the voltage drop across the resistor is zero, so points B and C are at the
same potential. There is no potential drop between the battery and point A, so the voltage of point A is 3.0 V
Since B and C are at the same potential, the magnitude of the charge on the plates of both capacitors must be
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College Physics
Chapter 18: Electric Current and Circuits
C
the same. So, Q  C1V1  C2V2 , or V1  2 V2. The voltage drop across both capacitors must be
C1
C

C
V  3.0 V, so V  V1  V2  2 V2  V2   2  1 V2. Calculate V2.
C1
 C1 
V
3.0 V
V2 

 0.86 V
C2
1  5.0
1
2.0
C1
Since the voltage across C2 is 0.86 V and C2 is connected to ground, point C must be at 0.86 V, as must
point B.
124. (a) Strategy Use the definition of capacitance and Eq. (17-15).
Solution Find the charge on the upper plate.
Q  CV 
e0 A
d
V
e0 L2
d
V
[8.854  1012 C 2 (N  m 2 )](0.10 m) 2 (10.0 V)
89  106 m
 9.9 nC
(b) Strategy Use the definition of capacitance and Eqs. (17-15), (18-24), and (18-25).
Solution Find the time constant.
  RC  R
e0 A

(0.100  106 )[8.854  1012 C 2 (N  m 2 )](0.10 m) 2
d
Compute the initial current.
%
10.0 V

 100 A
R 0.100  106 
89  106 m
 9.9  105 s
5
The current is given by I  (100 A)et (9.910 s).
(c) Strategy The energy dissipated in R is that stored in the capacitor, U. Use Eqs. (17-15) and (17-18b).
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College Physics
Chapter 18: Electric Current and Circuits
Solution Compute the energy dissipated over the whole discharging process.
U
1
1e A
[8.854  1012 C2 (N  m 2 )](0.10 m) 2 (10.0 V) 2
CV 2   0  %2 
 50 nJ
2
2 d 
2(89  106 m)
125. (a) Strategy It is okay to treat the Earth-ionosphere system as a parallel plate capacitor, since
d
5.0  104 m

 102 ; locally, the Earth is flat when compared with the distance between the “plates”.
R 6.371 m
Use Eq. (17-15).
Solution Find the capacitance.
C
e0 A
d

[8.854  1012 C 2 (N  m 2 )]4 (6.371 106 m) 2
5.0  104 m
 0.090 F
(b) Strategy Use Eqs. (17-10) and (17-18b).
Solution Find the energy stored in the capacitor.
1
1
1
U  CV 2  C ( Ed )2  (0.090 F)(150 V m)2 (5.0  104 m)2  2.5 TJ
2
2
2
(c) Strategy Use the definition of resistance and Eqs. (18-8) and (17-10).
Solution Compute the resistance.
L
5.0  104 m
 (3.0  1014   m)
 29 k
A
4 (6.371 106 m) 2
Compute the current.
R
I


V (5.0  104 m)
150 m
V Ed


 260 A .
R
R
29,408 
(d) Strategy The system can be modeled by an RC circuit. The voltage across the capacitor while it is
discharging is given by V  V0e t /( RC ) . Since Q = CV, Q  Q0et /( RC ) assuming C doesn’t change.
Solution Solve for t.
Q  Q0e t /( RC )
Q
et /( RC )  0
Q
Q
t
 ln 0
RC
Q
Q
1   1 min 

t  RC ln 0  (29  103 )(0.090 F) ln
 200 min
Q 
0.01   60 s 
126. (a) Strategy and Solution If the electric field is directed downward, the positive ions move down and the
negative ions move up.
(b) Strategy and Solution The current due to the positive ions is down and that due to the negative ions is
down, so the total current is down.
(c) Strategy Use Eqs. (17-10), (18-3), and (18-8), and the definition of resistance.
Solution I  neAvD , so vD  I (neA). I is given by I  V R , and V  EL for a uniform electric field, so
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College Physics
Chapter 18: Electric Current and Circuits
EL R
EL

.
neA
neAR
The resistivity is given, so R can be replaced with R   L A .
vD 
vD 
EL
 
neA 
L
A

E
100.0 V m

ne  (600.0 cm 3  500.0 cm 3 )(100 cm m)3 (1.602 10 19 C)(4.0  1013   m)
 0.014 m s
(d) Strategy Use the result of part (c), the surface area of a sphere, and Eq. (18-3).
Solution Compute the total current due to the movement of ions in the air.
 E  AE 4 RE2 E 4 (6.371 106 m) 2 (100.0 V m)
I  neAvD  neA 


 1.3 kA



4.0  1013   m
 ne  
127. Strategy Use the definition of resistance and Eq. (18-21b).
Solution
(a) Since the diode and the resistor are in parallel, the voltage drop across each is % 1.0 V  VD . According to
the figure, for VD  1.0 V, I D  2 mA .
V
1.0 V
 500 . The current through the battery is
(b) The resistance of the diode is RD  D 
I D 0.002 A
I
 1
%
1
1
 1

 %
   (1.0 V) 

  3 mA .
3
Req
 500  1.0  10  
 RD R 
(c) Compute the total power dissipated in the diode and resistor.
Ptotal 
%2
1
 1

 (1.0 V) 2 

  3 mW
3
Req
 500  1.0  10  
(d) The slope of the I D vs. VD graph is increasing. Since I D  (1 RD )VD , the slope of the graph is the inverse of
the resistance. Therefore, RD decreases as VD increases. Form a proportion.
R
PR
V2 R
R
PR .
 2
 D , or PD 
RD
PD V RD
R
Since R is fixed and RD decreases, the ratio R RD increases. Thus, doubling the power dissipated in the
resistor will more than double, or increase by a factor greater than two , that dissipated in the diode.
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College Physics
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128. Strategy Use Poiseuille’s law, Eq. (9-15), and the concepts of series and parallel resistance.
Solution
(a) Show that Poiseuille’s law can be written in the form P  IR.
If I 
V  r 4P
8 L
8 L

, then P  I 4  IR, where R  4 is the fluid flow resistance.
t
8 L
r
r
(b) From (a), R 
8 L
 r4
.
(c) Since the volume flow rate is the same in all of the pipes, the total pressure drop is
Ptot  P1  P2    IR1  IR2    I ( R1  R2  )  IReq by definition.
So, the answer is yes , Req  R1  R2 .
(d) Since the pressure drops across all of the pipes are the same, the total volume flow rate is
 1
 P
P P
1
I tot 

   P  
  
by definition.
R1 R2
R
R
2
 1
 Req
1
1
1
So, the answer is yes ,


.
Req R1 R2
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