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Chapter 18

Direct-Current Circuits

Quick Quizzes

1. (a), (d). Bulb R1 becomes brighter. Connecting a wire from b to c provides a nearly zero resistance path from b to c and decreases the total resistance of the circuit from R1 + R2 to just R1. Ignoring internal resistance, the potential difference maintained by the battery is unchanged while the resistance of the circuit has decreased. The current passing through bulb R1 increases, causing this bulb to glow brighter. Bulb R2 goes out because essentially all of the current now passes through the wire connecting b and c and bypasses the filament of Bulb R2.

2. (b). When the switch is opened, resistors R1 and R2 are in series, so that the total circuit resistance is larger than when the switch was closed. As a result, the current decreases.

3. (a). When the switch is closed, resistors R1 and R2 are in parallel, so that the total circuit resistance is smaller than when the switch was open. As a result, the total current increases.

4. (a) unchanged; (b) unchanged; (c) increase; (d) decrease. Neglecting internal resistance, the terminal potential difference of the battery remains constant (equal to the emf) as bulbs are added. Since each bulb is added in parallel with the battery, the current in each bulb, and hence the brightness of the bulb, is unchanged as new bulbs are added. The total current supplied by the battery increases as bulbs are added in parallel. Thus, the power delivered by the battery increases and its lifetime decreases.

5. ((a) decrease; (b) decrease; (c) decrease; (d) increase. Adding bulbs in series increases the total resistance of the circuit and results in a decrease in the current in the bulbs. Thus, the brightness of individual bulbs deceases as bulbs are added. Neglecting internal resistance, the terminal potential difference of the battery remains constant (equal to the emf) and the power supplied by the battery decreases as bulbs are added. This means that the battery will supply energy at a lower rate, and its lifetime will be increased.

6. (c). After the capacitor is fully charged, current flows only around the outer loop of the circuit. This path has a total resistance of 3 Ω , so the 6-V battery will supply a current of 2 Amperes.

99

100 CHAPTER 18

Answers to Even Numbered Conceptual Questions

2.

4. (a) ii (b) i

6. (a) ii (b) i

8. A short circuit can develop when the last bit of insulation frays away between the two conductors in a lamp cord. Then the two conductors touch each other, opening a low resistance branch in parallel with the lamp. The lamp will immediately go out, carrying no current and presenting no danger. A very large current will be produced in the power source, the house wiring, and the wire in the lamp cord up to and through the short. The circuit breaker will interrupt the circuit quickly but not before considerable heating and sparking is produced in the short-circuit path.

10. A wire or cable in a transmission line is thick and made of material with very low resistivity. Only when its length is very large does its resistance become significant. To transmit power over a long distance it is most efficient to use low current at high voltage, minimizing the power loss in the transmission line. 2I R

12. The bulbs of set A are wired in parallel. The bulbs of set B are wired in series, so removing one bulb produces an open circuit with infinite resistance and zero current.

14. (a) ii. The power delivered may be expressed as P , and while resistors connected in series have the same current in each, they may have different values of resistance. (b) ii. The power delivered may also be expressed as

2I R=

( )2V R= ∆P , and while resistors connected in parallel have the same potential difference across them, they may have different values of resistance.

16. Compare two runs in series to two resistors connected in series. Compare three runs in parallel to three resistors connected in parallel. Compare one chairlift followed by two runs in parallel to a battery followed immediately by two resistors in parallel.

The junction rule for ski resorts says that the number of skiers coming into a junction must be equal to the number of skiers leaving. The loop rule would be stated as the total change in altitude must be zero for any skier completing a closed path.

18. Because water is a good conductor, if you should become part of a short circuit when fumbling with any electrical circuit while in a bathtub, the current would follow a pathway through you, the water, and to ground. Electrocution would be the obvious result.

Direct-Current Circuits 101

20. Even if the fuse were to burn out or the circuit breaker to trip, there would still be a current path through the device, and it would not be protected.

22. When connected in series, all bulbs carry the same current. Thus, the one with the lowest resistance dissipates the least power 2I R = P and glows the dimmest. This is seen to be

the bulb that is labeled “200 W”. If they were connected in parallel, all bulbs would have the same potential difference across them. Then, the one with the lowest resistance will

dissipate the most power ( )2V = ∆P R and glow the brightest.

24. The break is closer to A.

102 CHAPTER 18

Answers to Even Numbered Problems

2. (a) (b) 1.0 A (c) 24 Ω 4 8 122.18 , 6.0 A, 3.0 A, 2.0 AI I IΩ = = =

4. (a) 30 V (b) 2.3 V

6. 15 Ω

8. (a) (b) 4.53 V 5.13 Ω

10. 6.0 , 3.0A BR R= Ω = Ω

I

12. (a) 75.0 V (b) 25.0 W, 6.25 W, 6.25 W; 37.5 W

14. 14.2 W, 28.4 W, 1.33 W, 4.00 W

16. 0.714 A, 1.29 A, 12.6 V

18. 5.4 V with a at a higher potential than b

20. 2 12.0 A (flows from toward ), 1.0 AI b a= − =

22. 0.50 W

24. (a) (b) 0.0816 or 8.16% 4.59 Ω

26. 0.28 A (dead battery), 1. (starter) 27 10 A×

28. (a) 9.0 V with b at a higher potential than a (b) 0.42 A directed from b to a

32. (a) 2.00 ms (b) 180 Cµ (c) 114 Cµ

34. 47 Fµ

36. (a) 10.0 Fµ (b) 415 Cµ

38. (a) 8.0 A (b) 120 V (c) 0.80 A (d) 25.8 10 W×

40. No. Their combined power requirements exceed the 1 800 W available.

42. (a) (b) 114.1 10 J−× 0.56 Aµ

44. 0.865 or 86.5%

46. (a) (b) 56 W (c) 2.0 A 14 Ω

48. (a) 40.0 W (b) 80.0 V, 40.0 V, 40.0 V


50. (a) 0 in the resistor and 3-kΩ 333 Aµ in the others (b) 50.0 Cµ

52. 14 Ω

54. 14 s

56. 20 or 98R R= Ω = Ω

60. ( )( )1 000 6.0 s1 240 C 1 tQ eµ −= − , ( )( )1 000 6.0 s

2 360 C 1 tQ eµ −= −

62. (a)

(b) 1.

(c) 9

1 2 3 1000.099 9 , 50.0 A, 45.0 mAR R RI I I IΩ = = = =

1 2 3 10009 , 4.55 A, 45.5 mAR R RI I I IΩ = = = =

1 2 3 100.99 , 0.450 A, 50.0 mAR R RI I I IΩ = = = =

64. (a) The two bulbs have the same current, power delivered, and hence same brightness. (b) Again, the two bulbs have the same current, power delivered, and brightness. (c) The bulbs in case 2 are brighter than those in case 1 because each bulb carries twice the

current it had in case 1. (d) In case 1, both bulbs go out if either bulb fails because the current in the circuit would

be zero. In case 2, if one bulb fails, the other bulb is unaffected and maintains its original current and brightness.

104 CHAPTER 18

Problem Solutions

18.1 From ( )V I R r+∆ = , the internal resistance is

9.00 V

72.0 4.92 0.117 A

= − Ω = ΩV

r RI∆

= −

18.2 (a) 1 2 3 4.0 8.0 12 24 eq R R= + + = + Ω + Ω + Ω = ΩR R

(b) The same current exists in all resistors in a series combination.

24 V

1.0 A24 eq

VI

R∆

= = =Ω

(c) If the three resistors were connected in parallel,

1 1

1 2 3

1 1 1 1 1 12.18

4.0 8.0 12 eqRR R R

− − = + + = + + = Ω Ω Ω Ω

Resistors in parallel have the same potential difference across them, so

4I4

24 V6.0 A

4.0 V

R∆

= = =Ω

, 8

24 V3.0 A

8.0 I = =

Ω, and 12

24 V2.0 A

12 I = =

Ω

18.3 For the bulb in use as intended,

( ) ( )2 2120 V

192 75.0 Wbulb

VR

∆= = =

PΩ

Now, presuming the bulb resistance is unchanged, the current in the circuit shown is

120 V

0.620 A0.800 192 0.800 eq

VI

R∆

= = =Ω + Ω + Ω

( ) ( )

and the actual power dissipated in the bulb is 22 0.620 A 192 73.8 WbulbI R= = Ω =P


18.4 (a) The current through this series combination is

( ) 12 V

2.0 A6.0

bc

bc

VI

R

∆= = =

Ω

(

Therefore, the terminal potential difference of the power supply is )( )2.0 A 9.0 6.0 30 VeqV I R∆ = = Ω + Ω =

(b) When connected in parallel, the potential difference across either resistor is the voltage setting of the power supply. Thus, ( )( )9 9 0.25 A 9.0 2.3 VV I R∆ = = Ω =

18.5 (a) The equivalent resistance of the two parallel resistors is

11 1

4.12 7.00 10.0 pR

− = + = Ω Ω

Ω

Thus, ( )4 9 4.00 4.12 9.00 17.1 ab pR R R R= + + = + + Ω = Ω

(b) ( ) 34.0 V

1.99 A17.1

abab

ab

V

R

∆=

ΩI = = , so 4 9 1.99 AI I= =

Also, ( ) ( )( )4.12 8.18 VΩ =1.99 Aab pp

V I R∆ = =

(

Then, )

77

8.18 V1.17 A

7.00 p

VI

R

∆= = =

Ω

(

and )

1010

8.18 V0.818 A

10.0 p

VI

R

∆= = =

Ω

106 CHAPTER 18

18.6 The equivalent resistance of the parallel combination of three resistors is

11 1 1

3.0 18 9.0 6.0 pR

− = + + = Ω Ω Ω

Ω

Hence, the equivalent resistance of the circuit connected to the 30 V source is 12 12 3.0 15 eq pR R R= + = Ω + Ω = Ω

18.7 If a potential difference is applied between points a and b, the vertical resistor with a free end is not part of any closed current path. Hence, it has no effect on the circuit and can be ignored. The remaining four resistors between a and b reduce to a single equivalent resistor, 2.5eqR = R , as shown below:

18.8 (a) The rules for combining resistors in series and parallel are used to reduce the circuit to an equivalent resistor in the stages shown below. The result is 5.13 eqR = Ω .

(b) From ( )2eqV R= ∆P , the emf of the power source is

( )( )4.00 W 5.13 4.53 VeqV R∆ = ⋅ = Ω =P


18.9 Turn the circuit given in Figure P18.9 90° counterclockwise to observe that it is equivalent to that shown in Figure 1 below. This reduces, in stages, as shown in the following figures.

From Figure 4,

25.0 V

1.93 A12.9

VI

R∆

= = =Ω

(b) From Figure 3, ( )

( )( )1.93 A 2.94 5.68 V

babaV I R∆ =

= Ω =

(a) From Figures 1 and 2, the current through the 20.0 Ω resistor is

( )

20

5.68 V0.227 A

25.0 ba

bca

VI

R

∆= = =

Ω

108 CHAPTER 18

18.10 First, consider the parallel case. The resistance of resistor B is

( ) 6.0 V

3.0 2.0 A

BB

B

VR

I

∆= = =

( ) ( ) ( ) 6.0B battery A

V V V∆ = ∆ − ∆ =

( )

Ω

In the series combination, the potential difference across B is given by

The current through the series combination is then

V-4.0 V 2.0 V=

2.0 V 2A

3.0 3B

sB

VI

R

∆= = =

Ω

and the resistance of resistor A is ( ) 4.0 V

6.0 2 3 A

AA

s

VR

I

∆= = = Ω

18.11 The equivalent resistance is eq pR R R= + , where pR is the total resistance of the three

parallel branches;

( )( )1 1 30 5.0 1 1 1

120 40 5.0 35 p

RR

R

− − Ω + Ω = + = + = Ω Ω + Ω + Ω

( )

15.0

+ + Ω

( )

130 R RΩ

Thus, ( )2 230 5 65 150

75 35

RR

R RΩ + + + Ω

Ω = +Ω

(

.0 3

R RΩ=

+ Ω

5Ω+

which reduces to )2 2 2 475R R −10 0− Ω Ω = or ( )( )55 45 0R R− Ω + Ω = . Only the positive solution is physically acceptable, so 55 R = Ω


18.12 (a) The total current from a to b is equal to the maximum current allowed in the 100 Ω series resistor adjacent to point a. This current has a value of

maxmax

250.500 A

1I

R= = =

P .0 W00 Ω

The total resistance is

1

100100 abR

− = Ω + Ω Ω

)(

1 1 1

100+

(

00 50.0 150= Ω + Ω = Ω

)

Thus, ( ) maxmax 0 A 150 75.0 VabV I∆ = Ω =0.50R =

(b) The power dissipated in the series resistor is 21 max 25.0 WI R= =P , and the power

dissipated in each of the identical parallel resistors is

( ) ( ) ( )2 22 3 max 2 0.250 A 100 6.25 WI R= = = Ω =P P

( )

The total power delivered is 1 2 3 25.0 6.25 6.25 W 37.5 W= + + = + + =P P P P

110 CHAPTER 18

18.13 The resistors in the circuit can be combined in the stages shown below to yield an equivalent resistance of ( )63 11 ad = ΩR .

From Figure 5, ( )

( )18 V

3.14 A63 11

ad

ad

V

R

∆= = =

Ω

( )

I

Then, from Figure 4, ( )( )3.14 A 30 11 8.57 Vbdbd

V I R∆ = = Ω =

Now, look at Figure 2 and observe that

( )

2

8.57 V1.71 A

3.0 2.0 5.0 bd

VI

∆= = =

Ω + Ω Ω

(

so ) ( )( )2 1.71 A 3.0 5.14bebe

V I R∆ = = Ω = V

Finally, from Figure 1, ( )

1212

5.1412

beV

IR

∆= =

V0.43 A=

Ω


Ω18.14 The resistance of the parallel combination of the

resistors is

3.00 and 1.00 Ω

11 10.750

1.00

− = + = Ω Ω

4.00 6.75pR R+ + Ω =

3.00pR

2.00eq = Ω

Ω

Ω

The equivalent resistance of the circuit connected to the battery is

and the current supplied by the battery is

18.0

eq

VI

R∆

= = V

2.67 A6.75

=Ω

2.00-

The power dissipated in the Ω resistor is

( ) ( )22.67 A 2.00 14.2Ω =

4.00-

22 2I R= =P W

and that dissipated in the Ω resistor is

( ) ( )22.67 A 4.00 28.4Ω =2

4 4I R= =P W

The potential difference across the parallel combination of the resistors is

3.00 and 1.00 Ω Ω

( ) ( )( )2.67 A 0.750 2.00 VppV I R∆ = = Ω =

( )

Thus, the power dissipation in these resistors is given by

( )2 2

33

2.00 V

3.00 p

V

R

∆= = =

ΩP 1.33 W

( )

and ( )2 2

11

2.00 V

1.00 p

V

R

∆= = =

ΩP 4.00 W

112 CHAPTER 18

18.15 (a) Connect two 5 to get 0- resistors in parallelΩ 25 Ω . Then connect that parallel

combination in series with a 20- resistorΩ for a total resistance of . 45 Ω

(b) Connect two 5 to get 25 0- resistors in parallelΩ Ω .

Also, connect two 20- resistors in parallelΩ to get 10 Ω .

Then, connect these two parallel combinations in series to obtain . 35 Ω

18.16 Going counterclockwise around the upper loop, applying Kirchhoff’s loop rule, gives ( ) ( )( )115.0 V 7.00 5.00 2.00 A 0I+ − − =

or 1

15.0 V 10.0 V0.714 A

7.00 I

−= =

Ω

1 2 2.00 AI I

From Kirchhoff’s junction rule, 0+ − = so 2 12.00 A 2.00 A 0.714I I= − = −

(

A 1.29 A=

Going around the lower loop in a clockwise direction gives ) ( )( )22.00 5.00 2.00 AI+ − − =E

(

0

or )( ) ( )( )2.00 1.29 A 5.00 = Ω +E 2.00 A 1Ω 2.6 V=


18.17 We name the currents as shown. Using Kirchhoff’s loop rule on the rightmost loop gives

1 2 3, , and I I I

( )( )

3

25.00 1.00

I

I− +

12.0 V- 1.00+3.00

4.00 V 0

+

− =

(

or ) ( )3 2 4.00 VI I =2.00 3.00+

(

(1)

Applying the loop rule to the leftmost loop yields ) ( )2 14.00 V+ 1.00 0I I

(

+5.00 8.00+ − = or ) ( )1 2 2.00 VI I =

I I

4.00 3.00−

1 2 3+ =I

(2) From Kirchhoff’s junction rule, (3) Solving equations (1), (2) and (3) simultaneously gives 1 2 3=0.846 A, 1.31 AI I I ==0.462 A, and

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

18.18 Observe that the center branch of this circuit, that is the branch containing points a and b, is not a continuous conducting path, so no current can flow in this branch. The only current in the circuit flows counterclockwise around the perimeter of this circuit. Going counterclockwise around the this outer loop and applying Kirchhoff’s loop rule gives ( ) ( ) ( ) ( )8.0 V 2.0 3.0 +12 V 10 5.0 0I I I I− − Ω − Ω − Ω − Ω =

or 12 V 8.0 V

0.20 A20

I−

= =Ω

(

Now, we start at point b and go around the upper panel of the circuit to point a, keeping track of changes in potential as they occur. This gives )( ) ( )( ) ( )( )4.0 V+ 6.0 0 3.0 0.20 A 12 Vab a bV V V∆ = − = − Ω − Ω +

0abV∆ >

10 0.20 A 5.4 V− Ω = + Since , point is 5.4 V higher in potential than point a b

114 CHAPTER 18

18.19 (a) Applying Kirchhoff’s loop rule, as you go clockwise around the loop, gives ( ) ( ) ( )20.0 V 2 000 30.0 V 2 500 25.0 V 500 0I I+ − − − + − I = ,

or 33.00 10 A 3.00 mAI −= × =

(b) Start at the grounded point and move up the left side, recording changes in potential as you go, to obtain ( )( ) ( )( )3 320.0 V 2 000 3.00 10 A 30.0 V 1 000 3.00 10 AAV − −= + − Ω × − − Ω ×

or 19.0 VAV = −

(c) ( ) ( )( )31 500 1 500 3.00 10 A 4.50 VV −∆ = Ω × =

(The upper end is at the higher potential.)

18.20 Following the path of from a to b, and recording changes in potential gives

1I

( )( )24 V 6.0 3.0 A 6.0 Vb aV V− = + − Ω = +

2I

(

Now, following the path of from a to b, and recording changes in potential gives ) 23.0 6.0 Vb aV V I− = − Ω = + , or 2 2.I = −

2

0 A

Thus, is directed from I from toward b a

(

and has magnitude of 2.0 A.

Applying Kirchhoff’s junction rule at point a gives )3 1 2 3.0 2.0 A 1.0 AI I I= + = + − = A


18.21 First simplify the circuit by combining the series resistors. Then, apply Kirchhoff’s junction rule at point a to find Next, we apply Kirchhoff’s loop rule to the rightmost loop to obtain

1 2 2.00 AI I+ =

( ) ( )1 212.0 0I I−

( )

8.00 V 6.00− +

( )

= or ( )1 12.0 2.00I − 1I =8.00 V 6.00− +

(

A 0− This yields

Finally, apply Kirchhoff’s loop rule to the leftmost loop to obtain

1 1.78 AI =

)( ) ( )1 10 A 6.00 I4.00 2.0 8.00 V 0ε+ −

( )(

−

) ( )(

+ =

)

or 1 4.00 2.00ε A 6.00 1.78= + A 8.00 V 10.7 V− =

18.22 From Kirchhoff’s point rule, note that 1I I I2= − in the circuit shown at the right. Going counterclockwise around the upper pane of the circuit, Kirchhoff’s loop rule gives ( ) ( ) ( )( )2 0I I− =2 250 30 90 I I− Ω − Ω + Ω

or 2

170 90

I IΩ = Ω

(1)

Now, going counterclockwise around the outer perimeter of the circuit, Kirchhoff’s loop rule gives

( ) ( ) ( )2 212 V 50 30 20 0I I+ − Ω − Ω − Ω =

( )

I or, using Equation (1)

2

170 50 30 20 12 V

90 I

Ω Ω + Ω + Ω = Ω which gives 2 0.10 AI =

The power delivered to the 50-Ω resistor is ( ) ( ) ( )22

2 50 A 50 0.50 WI Ω =P 0.10= Ω =

116 CHAPTER 18

18.23 (a) We name the currents as shown. Applying Kirchhoff’s loop rule to loop , gives

1 2 3, , and I I I

1 1 0R Iabcfa

1 2 2 2R Iε ε+ − − −

2 12 10.0 mAI I+ =

3 3 3 2R I R I

=

edcfe

2 2 0

or (1) Applying the loop rule to loop yields

3

ε ε+ − − −

2 34 20.0 mAI I+ =

2 1 3I I I= +

= or (2) Finally, applying Kirchhoff’s junction rule at junction c gives (3) Solving equations (1), (2), and (3) simultaneously yields

3

1 20.385 mA, I I= = 3 2.69I3.08 mA, and mA=

(b) Start at point c and go to point f, recording changes in potential to obtain

( )( )3 32 2 2 60.0 V 3.00 10 3.08 10 A 69.2 Vf cV V R Iε −− = − − = − − × Ω × = −

or 69.2 V and point is at the higher potentialc fV c∆ =

18.24 (a) Applying Kirchhoff’s loop rule to the circuit gives ( )( )3.00 V 0.255 0.153 0.600 A 0R+ − Ω + Ω + =

or ( )3.00 V0.255 0.153 4.59

0.600 AR = − Ω + Ω = Ω


(b) The total power input to the circuit is ( ) ( )( )1 2 1.50 V 1.50 V 0.600 A 1.80 Winput Iε ε= + = + =P

( ) ( ) ( )221 2 0.600 A 0.255 0.153 0.147loss I r r= + = Ω + Ω =P

Thus, the fraction of the power input that is dissipated internally is

W

0.147 W0.081 6 or 8.16%

1.80 Wloss

input

= =P

P

18.25 Applying Kirchhoff’s junction rule at junction a gives (1) Using Kirchhoff’s loop rule on the upper loop yields

1 2I I I= +

(

3

) ( )1 3.0 4.0 3.0 0I I+ − =

.0 A

24 V 2+ −

1 32 8I I+ =

(

or (2) and for the lower loop, ) ( )3 2.0 1.0 5.0 0I I− + =

.0 A

12 V 3+ +

2 32 4I I− =

or (3) Solving equations (1), (2), and (3) simultaneously gives 1 2 3, 2.5 A, and 1.0I I I= =3.5 A A=

118 CHAPTER 18

18.26 Using Kirchhoff’s loop rule on the outer perimeter of the circuit gives ( ) ( )1 312 V 0.01 0.06 0I I+ − − =

31 36 1.2 10 AI I+ = ×

or (1) For the rightmost loop, the loop rule gives ( ) ( )2 310 V 1.00 0.06 0I I+ + − =

2 30.06 10 AI I− = −

1 2 3I I I= +

or (2) Applying Kirchhoff’s junction rule at either junction gives (3) Solving equations (1), (2), and (3) simultaneously yields

( ) 22 3 and 1.7 10 A (in starter)I I= = ×0.28 A in dead battery

18.27 Assume currents I I in the directions shown. Then, using Kirchhoff’s junction rule at junction a gives (1) Applying Kirchhoff’s loop rule on the upper loop,

1 2 3, , and I

2I

(

3 1I I= +

) ( )1 2.00 10.0 V 0I I −20.0 V+ −

1 26 I I− =

30.0 5+

2.00 A

= or (2) and for the lower loop, ( ) ( )2 3 0I I10.0 V

.00 A

5.00 20.0+ − − = or (3) Solving equations (1), (2), and (3) simultaneously yields

2 34 2I I+ =

1 2 A, 0.118I I= = 3 A, and 0.471I =0.353 A

!"


18.28 (a) Since there is not a continuous path in the center branch, no current exists in that part of the circuit. Then, applying Kirchhoff’s loop rule to the outer perimeter gives

( )18 V 36 V 1.0 4.0 3.0 8.0 2.0 0I + + − + + + + Ω =

or 54 V

3.0 A18

I = =Ω

(

Now, start at point b and go around the lower loop to point a, recording changes in potential to obtain )( ) ( )( )36 V 4.0 1.0 3.0 A 6.0 a bV V− = − + Ω + Ω + 5.0 0 12 V 9.0 VΩ+ Ω + = − or 9.0 V w

ab∆ = ith point at a higher potential than V b a

(b) Assume currents as shown in the modified circuit. Applying Kirchhoff’s loop rule to the upper loop gives ( ) ( ) ( ) 111 12 V 7.0 13 18 V=0I I I− + − − +

118 13 30 AI I+ =

( )( )

or (1) For the lower loop, the loop rule yields ( ) ( ) =0I+15.0 36 V 7.0 12 V 11I I I− − + + −

123 5 24 AI I− = −

or (2) Solving equations (1) and (2) simultaneously gives I1 2.9 A= , and Thus, the current in the

0.42 AI = −

7.0-Ω resistor is 0.42 A fl to b aowing from .

#$%&'

120 CHAPTER 18

18.29 Applying Kirchhoff’s junction rule at junction a gives (1) Using Kirchhoff’s loop rule on the leftmost loop yields

3 1I I I= +

(

2

) ( )3 14.00 5.00 12.0 V 0I I− +

9.00 A

( ) (

3.00 V− −

1 35 4I I+ =

= or (2) and for the rightmost loop, )3 24.00 3.00 2.00 18.0I I− + +

15.0 A

1 2 3 A, 1.523 A, and 1.846 AI I I= = =

3.00 V− −

2 35 4I I+ =

0.323

(

V 0=

or (3) Solving equations (1), (2), and (3) simultaneously gives Therefore, the potential differences across the resistors are )2 2 2V I∆ = .00 3.05 VΩ = , ( )3 2 3.00 4.5V I∆ = Ω 7 V=

( )4 3 4V I∆ = .00 7.3Ω 8 V= , and ( )5 1 5V I∆ = .00 1.62 VΩ =

18.30 The time constant is RCτ =

( )(

. Considering units, we find

) Volts Coulombs CoulombsOhm

Amperes Volts Amperes

CoulombsSecond

Coulombs Second

RC → = =

= =

RC

s Farads

or τ = has units of time.

18.31 (a) ( )( )6 62.0 10 6.0 10 F 12 sRCτ −= = × Ω × =

(b) ( )( ) 46max 6.0 10 F 20 V 1.2 10 Cε −−= = × = ×Q C


18.32 (a) ( )( )6 3100 20.0 10 F 2.00 10 s 2.00 msRCτ − −= = Ω × = × =

(b) ( )( )6 4max 20.0 10 F 9.00 V 1.80 10 C 180 CQ C µε − −= = × = × =

(c) ( ) ( )max max max

11 1 1 11te Q e Q

eτ τ τ 4 CQ Q µ− − = − = − = − =

18.33 ( )( )6 4max 5.0 10 F 30 V 1.5 10 CQ Cε − −= = × = ×

(

, and

)( )6 61.0 10 5.0 10 F 5RCτ −= = × Ω × =

10 s 2t

.0 s

Thus, at τ= =

( ) ( )( )4 2max 1 1.5 10 C 1tQ Q e eτ− − −= − = × − 41.3 10 C−= ×

18.34 The charge on the capacitor at time t is ( )max 1 teQ Q τ−= − , where

( )Q C V= ∆ and Qmax Cε= . Thus, ( )1 tV e τε −∆ = − or ( )1te Vτ ε− = − ∆

We are given that, 12 Vε = , and at t 1.0 s= , 10 VV∆ =

Therefore, 1.0e− s τ 10 121

12 1210

61.0

−= − = = or 1.0 s 6.0e τ+ =

Taking the natural logarithm of each side of the equation gives

( )6.01.0 s

lnτ

= or ( )

1.0ln

τ s0.56 s

6.0= =

Since the time constant is RCτ = , we have

5 F 47 3

s0.5612 10

CR

4.7 10

Fτ µ−= = = ×

× Ω=

122 CHAPTER 18

18.35 From ( )max 1 teQ Q τ−= − , we have at 0.900 st = ,

0.900 s

max

Qe

Q−= −1 0τ = .600

Thus, 0.900 s 0.400e τ− = , or ( )ln 0.400

( )

0.900τ

s− =

giving the time constant as 0.

982 sln

900 s0.

0.400=τ = −

18.36 (a) max RI

ε= , so the resistance is

4-3

max

48.0 V9.60 10

0.500 10 AR

Iε

= = = ××

RC

Ω

The time constant is τ = , so the capacitance is found to be

54

0.960 s10 F 10.0 F

9.60 10C

R1.00

τ µ−= = × ==

× Ω

(b) ( )( )max 10.0 F 48.0 V 480 CQ C µ µε= = =1.92 st =

, so the charge stored in the capacitor at is

( ) ( ) ( )( )1.92 s

20.960 smax 1 480 C 1 480 C 1 415 Cte e eτQ Q µ µ µ

−− − = − = − = − =

18.37 (a) The current drawn by each appliance is

Heater: 1 300 W

10.8 A120 V

IV

= = =∆P

Toaster: 1 000 W

8.33 A120 V

IV

= = =∆P

Grill: 1 500 W

12.5 A120 V

IV

= = =∆P

(b) If the three appliances are operated simultaneously, they will draw a total current of ( )10.8 8.33 12.5 A 31.7 AtotalI = + + = . Therefore, a 30 ampere circuit breaker is

insufficient to handle the load .


18.38 (a) The equivalent resistance of the parallel combination is

1 1

1 2 3

1 1 1 1 1 115

150 25 50 eqRR R R

− − = + + = + + = Ω Ω Ω Ω

so the total current supplied to the circuit is

120 V

8.0 A15 total

eq

VI

R∆

= = =Ω

(b) Since the appliances are connected in parallel, the voltage across each one is 120 VV∆ = .

(c) 120 V

0.80 A150 lamp

lamp

VR∆

=Ω

I = =

(d) ( ) ( )2 2

2120 V5.8 10 W

25 heaterheater

VR∆

= ×Ω

= =P

18.39 From ( )2V R= ∆

( )

P , the resistance of the element is

( )2 2240 V

19.2 3 000 W

VR

∆= = =

PΩ

When the element is connected to a 120-V source, we find that

(a) 120 V

6.25 A19.2

VR∆

=Ω

I = = , and

(b) ( ) ( )( )120 V 6.25 A 750 WV I = == ∆P

18.40 The maximum power available from this line is ( ) ( )( )max max 120 V 15 A 1 800 WV I= ∆ = =P Thus, the combined power requirements (2 400 W) exceeds the available power, and you

cannot operate the two appliances together .

124 CHAPTER 18

18.41 (a) The area of each surface of this axon membrane is

( ) ( ) ( )6 62 0.10 m 2 10 10 m 2 10 mA rπ π π− − = = × = × 2

and the capacitance is

( )6 2

12 2 20 -8

2 10 m3.0 8.85 10 C N m

1.0 10 mA

Cd

πκ− ×

= ∈ = × ⋅ ×

( ) ( )

81.67 10 F− −= ×

In the resting state, the charge on the outer surface of the membrane is

( )8 3 91.67 10 F 70 10 V 1.17 10 C 1.2 10 i iQ C V − − −= ∆ = × × = × → × 9− C

The number of potassium ions required to produce this charge is

9

9 +-19K

1.17 10 C7.3 10 K ions

1.6 10 CiQ

Ne

+

−×= = = ×

×

and the charge per unit area on this surface is

( )

9 2

-6 2 -19

1.17 10 C 1 10 m2 10 m 1.6 10 C 1 Å

iQA

σπ

− −× = = × ×

0 2

22 4 2

1 1 8.6 10 Å 290 Å

e e = = ×

2

e

This corresponds to a low charge density of one electronic charge per square of side 290 Å, compared to a normal atomic spacing of one atom per several Å .

(b) In the resting state, the net charge on the inner surface of the membrane is , and the net positive charge on this surface in the excited state

is

91.17 10 CiQ −− = − ×

( ) ( )( )8 37 10 F 30 10 V 5.0 10 C− −× + × = + ×

( )

10−1.6f fQ C V= ∆ =

The total positive charge which must pass through the membrane to produce the excited state is therefore

105.0 10 C

f iQ Q Q∆ = −

= + × 9 91.17 10 C 1.67 10 C 1.7− − −− − × = × → × 910 C−

corresponding to

+

910 +

-19 +Na

1.67 10 C1.0 10 Na ions

1.6 10 C Na ionQ

Ne

−∆ ×= = = ×

×


(c) If the sodium ions enter the axon in a time of 2.0 mst∆ = , the average current is

9

73

1.67 10 C8.3 10 A 0.83

2.0 10 sQ

It

Aµ−

−−

∆ ×= = = × =∆ ×

(d) When the membrane becomes permeable to sodium ions, the initial influx of sodium ions neutralizes the capacitor with no required energy input. The energy input required to charge the now neutral capacitor to the potential difference of the excited state is

( ) ( )( )22 8 31 11.67 10 F 30 10 V 7.5 10 J

2 2fW C V − −= ∆ = × × = × 12−

18.42 The capacitance of the 10 cm length of axon was found to be C in the solution of Problem 18.41.

81.67 10 F−= ×

(a) When the membrane becomes permeable to potassium ions, these ions flow out of the axon with no energy input required until the capacitor is neutralized. To maintain this outflow of potassium ions and charge the now neutral capacitor to the resting action potential requires an energy input of

( ) ( )( )22 8 31 11.67 10 F 70 10 V 4.1 10 J

2 2W C V − −= ∆ = × × = × 11−

−

.

(b) As found in the solution of Problem 18.41, the charge on the inner surface of the membrane in the resting state is and the charge on this surface in the

excited state is . Thus, the positive charge which must flow out of the axon as it goes from the excited state to the resting state is , and the average current during the 3.0 ms required to return to the resting state is

91.17 10 C−− ×

9 C 1.67 10= ×

105.0 10 C−+ ×

100 C 1.17 1+ × 95.0 1 0 CQ − −∆ = ×

9

3

67 10 C5.6

3.0 10 s

−

−

∆ ×=

∆ ×70 A 0.56 µ−× =

1.1 A

QI

t= =

18.43 From Figure 18.28, the duration of an action potential pulse is 4.5 ms. From the solution Problem 18.41, the energy input required to reach the excited state is W . The energy input required during the return to the resting state is found in Problem 18.42 to be W . Therefore, the average power input required during an action potential pulse is

121 7.5 10 J−= ×

112 4.1 10 J−= ×

12 118

3

7.5 10 J+4.1 10 J1.1 10 W

4.5 10 s

− −−

−

× ×= = ×

×P 1 2totalW W W

t t+

= =∆ ∆

11 nW=

126 CHAPTER 18

18.44 From ( )max 1 teQ Q τ−= − , the ratio maxQ Q at t 2τ= is found to be

22

max

11 1 0.8

Qe

Q eτ τ−= − = − = 65 , or Q is 86.5% of Q at tmax 2τ=

18.45 The resistive network between a an b reduces, in the stages shown below, to an equivalent resistance of 7.5 eq = ΩR .

18.46 (a) The circuit reduces as shown below to an equivalent resistance of 14 eqR = Ω .

(b) The power dissipated in the circuit is ( ) ( )2 228 V

56 W14 eq

VR∆

= = =Ω

P

(c) The current in the original 5.0-Ω resistor (in Figure 1) is the total current supplied by the battery. From Figure 6, this is

28 V

2.0 A14 eq

VI

R∆

= = =Ω


18.47 (a) The resistors combine to an equivalent resistance of eqR = 15 Ω as shown.

(b) From Figure 5, 1

15 V1.0 A

15 ab

eq

VI

R∆

= = =Ω

(

Then, from Figure 4, )1 6.0 6.0 VI= Ω =ac dbV V∆ = ∆ and ( )1 3.0 3.0 VcdV I∆ = Ω =

From Figure 3, 2 3 6.0 6.cdV

I I3.0 V

0.50 A0

∆= = =

Ω Ω=

From Figure 2, ( )3 3.6edV I∆ = Ω 1.8= V

Then, from Figure 1, 4 6.0V

I1.8 V

6.ed 0.30 A

0 ∆

= =Ω Ω

=

and 5

1.8 V9.0 9.0

fd edV V

I∆ ∆

= = =Ω Ω

0.29.0

=Ω

0 A

(c) From Figure 2, ( )3 2.4 1.2 VceV I∆ = Ω = . All the other needed potential differences

were calculated above in part (b). The results were 6.0 V=ac dbV V∆ = ∆ ; 3.0 VcdV∆ = ; and 1.8 Vfd edV V∆ = ∆ =

128 CHAPTER 18

(d) The power dissipated in each resistor is found from ( )2V R= ∆P with the following results:

( ) ( )2 26.0 V

6.0 W6.0

acac

ac

V

R

∆= = =

ΩP

( ) ( )2 21.2 V0.60 W

2.4 ce

cece

V

R

∆= = =

ΩP

( ) ( )2 21.8 V

0.54 W6.0

eded

ed

V

R

∆= = =

ΩP

( ) ( )2 21.8 V

0.36 W9.0

fdfd

fd

V

R

∆= = =

ΩP

( ) ( )2 23.0 V

1.5 W6.0

cdcd

cd

V

R

∆= = =

ΩP

( ) ( )2 26.0 V6.0 W

6.0 db

dbdb

V

R

∆= = =

ΩP

18.48 (a) From ( )2V R= ∆

( )

P , the resistance of each of the three bulbs is given by

( )2 2120 V

240 60.0 W

VR

∆= = =

P

2 3 and R R 1R

Ω

As connected, the parallel combination of

is in series with . Thus, the equivalent resistance of the circuit is

1 1

2 3

1 1 1 10 36

240 240 R R

− − = + + Ω + + = Ω Ω Ω

( )

1eqR R 24= 0

The total power delivered to the circuit is

( )2 2120 V

W360

= =Ω

40.0=eq

VR∆

P

(b) The current supplied by the source is 120 V 1

A360 3eq

VR

I∆

= = =Ω

. Thus, the potential

difference across is

1R

( ) ( )11

1 A 240 80.0 V

3V I R ∆ = = Ω =

( ) ( ) ( ) ( )

The potential difference across the parallel combination of is then

2 and R 3R

12 3 120 V 80.0 V 40.source

V V V V∆ = ∆ = ∆ − ∆ = − = 0 V


(

)18.49 (a) From ( loadI r Rε = + , the current supplied when the headlights are the entire load is

( )

12.6 V2.48 A

0.080+5.00 load

Ir Rε

= = =+ Ω

( )( )

The potential difference across the headlights is then 2.48 A 5.00 12.4 VloadV I R∆ = = Ω =

(b) The starter motor connects in parallel with the headlights. If is the current supplied to the headlights, the total current delivered by the battery is

The terminal potential difference of the battery is

hlI

I r

35.0 AhlI I= +

V ε∆ = − , so the total current is

( )I V rε= − ∆ while the current to the headlights is 5.00 hlI V= ∆ Ω . Thus, becomes

35.0 AhlI I= +

35.0 A0

V V5.0r

ε − ∆ ∆= +

Ω

( )( )

which yields

( )( )

( ) ( )35

1V

r.0 A 12.6 V 35.0 A 0.080

5.00 1+ 0.080 5.00 r

9.65 Vε − −

=+ Ω Ω Ω

Ω=∆ =

18.50 (a) After steady-state conditions have been reached, there is no current in the branch containing the capacitor.

Thus, for : 3R ( )3

0 steady-stateRI =

For the other two resistors, the steady-state current is simply determined by the 9.00-V emf across the 12.0-kΩ and 15.0-kΩ resistors in series: For and :

1R 2R

( ) ( )1 2

9.00 V333 A steady-state

(12.0 k 15.0 k )I

R Rµ

1 2R R

ε+ = = =

+ Ω + Ω

130 CHAPTER 18

(b) When the steady-state has been reached, the potential difference across C is the same as the potential difference across because there is no change in potential across . Therefore, the charge on the capacitor is

2R

3R

( )

( ) ( )( )( )

2

6 32 10.0 F 333 10 A 0 10 50.0 C

RV

R 15.

Q C

C I µ µ−= = × × Ω =

= ∆

18.51 Applying Kirchhoff’s junction rule at junction a gives (1) Applying Kirchhoff’s loop rule on the leftmost loop yields

2 1I I I= +

(

3

) ( )1 25.0 4.0 V 10 0I I− − =

.0 A

9.0 V+ −

1 22 1I I+ =

(

or (2) For the rightmost loop, ) ( )2 34.0 V 10 14 V 0I I+ − =

A

10+ +

2 3 1.0I I+ =

or (3) Solving equations (1), (2) and (3) simultaneously gives 1 20,I I= = 3 0.50 AI =


18.52 With the switch open, the circuit may be reduced as follows:

With the switch closed, the circuit reduces as shown below:

Since the equivalent resistance with the switch closed is one-half that when the switch is open, we have

( )118 50

2R R+ Ω = + Ω , which yields 14 R = Ω

18.53 When a generator with emf ε and internal resistance r supplies current I, its terminal voltage is V I rε∆ = −

0 V I

. If when , then 11V∆ = 10.0 A= ( )110 V 10.0 A rε= − (1) Given that ∆ = when 106 VV 30.0 AI = , yields ( )106 V 30.0 A rε= − (2) Solving equations (1) and (2) simultaneously gives 112 V and 0.2r 00 ε = = Ω

18.54 At time t, the charge on the capacitor will be ( )1 tmaxQ Q e τ−= − where

( )( )6 62.0 10 3.0 10 F 6RCτ −= = × Ω × =

0.90 maxQ Q=

.0 s

When , this gives 0.90 1 te τ−= −

or 0.10te τ− = Thus, ( )ln 0.10tτ

( ) ( )

− =

giving 6.0 s ln 0.10 14 st = − =

132 CHAPTER 18

18.55 (a) For the first measurement, the equivalent circuit is as shown in Figure 1. From this, 1 2ab y y yR R R R R= = + =

so 1

12yR R= (1)

For the second measurement, the equivalent circuit is shown in Figure 2. This gives

2

12ac y xR R R R= = + (2)

Substitute (1) into (2) to obtain

2 1

1 12 2 xR R R = +

, or 2 1

14xR R R= −

(b) If R , then 1 213 and 6.0 R= Ω = Ω 2.8 xR = Ω

Since this exceeds the limit of 2.0 Ω , the antenna is inadequately grounded .

18.56 Assume a set of currents as shown in the circuit diagram at the right. Applying Kirchhoff’s loop rule to the leftmost loop gives ( ) ( )( )175 5.0 30 0I I I+ − − − =

17 6 15I I− =

(

or (1) For the rightmost loop, the loop rule gives

) ( )( )1 140 30 0R I I I− + + − = , or 1

73 30

RI = +

I (2)

Substituting equation (2) into (1) and simplifying gives (3)

Also, it is known that , so

( )1 1310 7 450I I R+ =

21 20 WR I R= =P 1

1

20 WI R

I= (4)


Substitution of equation (4) into (3) yields

11

140310 450I

I+ = or 2

1 1310 450 140 0I I− + =

Solving this quadratic equation gives two possible values for the current . These are

. Then, from

1I

1 A and 0.452I I= 11.0 A= 21

20 W=R

I, we find two possible values for the

resistance R. These are 20 or 98R R= Ω = Ω

18.57 When connected in series, the equivalent resistance is 1 2eq nR R R R nR= + + ⋅ ⋅ ⋅ + = . Thus,

the current is ( ) ( )s eqI V R V n= ∆ = ∆

( )( )

R , and the power consumed by the series

configuration is

( ) ( )2 2

2eq

V VI R nR

nRnR

∆ ∆= = =2

s sP

( )

For the parallel connection, the power consumed by each individual resistor is

2VR

( )

1

∆=P , and the total power consumption is

2

1p

n Vn

R∆

= =P P

Therefore, ( )

( )

2

2 2

1s

p

V RnR nn V

∆= ⋅ =

∆P

P or 2

1s pn=P P

134 CHAPTER 18

18.58 Consider a battery of emf ε connected between points a and b as shown. Applying Kirchhoff’s loop rule to loop gives

acbea

( ) ( )( )1 1 31.0 0I1.0I I ε− − − +

1 32 I I

= or ε− =

adbea

(

(1) Applying the loop rule to loop gives ) ( )( )2 2 33.0 0I I I5.0 ε− − + +

2 38 5I I

= or ε+

adca

(

= (2) For loop , the loop rule yields ) ( ) ( )2 3 13.0 1.0 0I I− + +1.0I = or 1 3 3I I I2+ = (3) Solving equations (1), (2) and (3) simultaneously gives

1 2

4, ,

27I I 3

127

13 and

27Iε ε ε=−= =

Then, applying Kirchhoff’s junction rule at junction a gives

1 2

4 1727 27

I I I1327ε ε= + + = ε= . Therefore,

( )27

17 27 17IabR

ε εε= = = Ω

18.59 (a) and (b) - With R the value of the load resistor, the current in a series circuit composed of a 12.0 V battery, an internal resistance of 10.0 Ω , and a load resistor is

12.0 V

10.0 I

R=

+ Ω

and the power delivered to the load resistor is

( )

( )

2

2

V

10.0

R

+ Ω2

144L I R

R= =P


Some typical data values for the graph are

)

R ( ) Ω LP (W)

1.00 1.19 5.00 3.20 10.0 3.60 15.0 3.46 20.0 3.20 25.0 2.94 30.0 2.70

The curve peaks at P at a load resistance of R3.60 WL= 10.0 = Ω .

18.60 The total resistance in the circuit is

11

1 2

1 1 1 11.2 k

2.0 k 3.0 kR

R R

−−

= + = + = Ω Ω

1 2 2.0 F+3.0 F=5.C C C

Ω

and the total capacitance is 0 Fµ µ µ= + =

(

Thus, )( )max 5.0 F 120 V 600 CQ C µ µε= = =

( )( )

and 3 6 3 6.0 s5.0 10 F 6.0 10 s

1 000− −Ω × = × =1.2 10 RCτ = = ×

The total stored charge at any time t is then ( )1 2 max 1Q Q Q Q te τ−− ( )= + = or ( )1 000 6.0 ste−1 2Q Q

( )

600 C 1µ+ = − (1)

Since the capacitors are in parallel with each other, the same potential difference exists across both at any time.

Therefore, 1 2

1 2C

Q QV

C C∆ = = , or 2

2 1 11

1.5C

QC

= =

( )

Q Q (2)

Solving equations (1) and (2) simultaneously gives

( )1 000 6.0 s1 240 C 1 tQ eµ −= −

( )

and ( )1 000 6.0 s2 360 C 1 tQ eµ −= −

136 CHAPTER 18

18.61 (a) Using the rules for combining resistors in series and parallel, the circuit reduces as shown below:

From the figure of Step 3, observe that

25.0 V

1.93 A12.94 A

I = = ( ) ( )( ) and 2.94 1.93 A 2.94 5.68 VabV I∆ = Ω = Ω =

(b) From the figure of Step 1, observe that 1

5.68 V0.227 A

25.0 25.0 abV

I∆

= = =Ω Ω

18.62 (a) When the power supply is connected to points A and B, the circuit reduces as shown below to an equivalent resistance of 0.099 9 eqR = Ω .

From the center figure above, observe that 1 1

5.00 V50.0 A

0.100 RI I= = =Ω

and 2 3 100

5.00 V0.045 0 A 45.0 mA

111 R RI I I= = = = =Ω


(b) When the power supply is connected to points A and C, the circuit reduces as shown below to an equivalent resistance of 1.09 eqR = Ω .

From the center figure above, observe that 1 2 1

5.00 V4.55 A

1.10 R RI I I= = = =Ω

and 3 100

5.00 V0.045 5 A 45.5 mA

110 RI I= = = =Ω

(c) When the power supply is connected to points A and D, the circuit reduces as shown below to an equivalent resistance of 9.99 eqR = Ω .

From the center figure above, observe that 1 2 3 1

5.00 V0.450 A

11.1 R R RI I I I= = = = =Ω

and 100

5.00 V0.050 0 A 50.0 mA

100 I = = =

Ω

18.63 In the circuit diagram at the right, note that all points labeled a are at the same potential and equivalent to each other. Also, all points labeled c are equivalent. To determine the voltmeter reading, go from point e to point d along the path ecd, keeping track of all changes in potential to find: 4.50 V 6.00 V 1.50 Ved d eV V V∆ = − = − + = +

138 CHAPTER 18

Apply Kirchhoff’s loop rule around loop abcfa to find ( ) ( ) 36.00 6.00 0I I− Ω + Ω = or I3 I= (1) Apply Kirchhoff’s loop rule around loop abcda to find ( ) ( ) 2 0I IΩ =6.00 6.00 V 10.0− Ω + − or 2 0.600 A 0.600I I= − (2) Apply Kirchhoff’s loop rule around loop abcea to find ( ) ( ) 1 0I IΩ =6.00 4.50 V 5.00− Ω + − or 1 0.900 A 1.20I I= − (3) Finally, apply Kirchhoff’s junction rule at either point a or point c to obtain (4) Substitute equations (1), (2), and (3) into equation (4) to obtain the current through the ammeter. This gives or and

3 1 2I I I I+ = +

0.900 A 1.20 0.600I I I+ = − +

3.80 1.50 AI =

A 0− .600I

1.50 A 3.80 0.395 AI = =

18.64 In the figure given below, note that all bulbs have the same resistance, .

R

*

*

(a) In the series situation, Case 1, the same current I flows through both bulbs. Thus,

the same power, , is supplied to each bulb. Since the brightness of a bulb is proportional to the power supplied to it, they will have the same brightness. We conclude that the

12

1 1I R=P

bulbs have the same current ower supplied, and brightness, p .

(b) In the parallel case, Case 2, the same potential difference V∆ is maintained across each of the bulbs. Thus, the same current 2I V R= ∆ will flow in each branch of

this parallel circuit. This means that, again, the same power is supplied

to each bulb, and the two bulbs will have

22 2I R=P

equal brightness .


(c) The total resistance of the single branch of the series circuit (Case 1) is 2R. Thus, the current in this case is 1 2I V R= ∆ . Note that this is one half of the current that flows through each bulb in the parallel circuit (Case 2). Since the power supplied is proportional to the square of the current, the power supplied to each bulb in Case 2 is four times that supplied to each bulb in Case 1. Thus, the bulbs in

2I

Case 2 are

much brighter than those in Case 1.

(d) If either bulb goes out in Case 1, the only conducting path of the circuit is broken and all current ceases. Thus, in the series case, the other bulb must also go out . If

one bulb goes out in Case 2, there is still a continuous conducting path through the other bulb. Neglecting any internal resistance of the battery, the battery continues to maintain the same potential difference V∆ across this bulb as was present when both bulbs were lit. Thus, in the parallel case, the second bulb remains lit with

unchanged current and brightness when one bulb fails.

140 CHAPTER 18

chapter 18 direct-current circuits - university of · pdf filechapter 18 direct-current...

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