CHAPTER 28—ELECTRIC CIRCUITS

ActivPhysics can help with these problems: Section 12, “DC Circuits”

Section 28-1:—Circuits and Symbols

Problem1. Sketch a circuit diagram for a circuit that includes a resistor R1 connected to the positive terminal of a battery, a pair

of parallel resistors R R2 3 and connected to the lower-voltage end of R1 , then returned to the battery’s negative terminal, and a capacitor across R2 .

SolutionA literal reading of the circuit specifications results in connections like those in sketch (a). Because the connecting wires are assumed to have no resistance (a real wire is represented by a separate resistor), a topologically equivalent circuit diagram is shown in sketch (b).

Problem 1 Solution (a).

Problem 1 Solution (b).

Problem2. A circuit consists of two batteries, a resistor, and a capacitor, all in series. Sketch this circuit. Does the description

allow any flexibility in how you draw the circuit?

SolutionIn a series circuit, the same current must flow through all elements. One possibility is shown. The order of elements and the polarity of the battery connections are not specified.

Problem 2 Solution.

976 CHAPTER 28

Problem3. Resistors R1 and R2 are connected in series, and this series combination is in parallel with R3. This parallel

combination is connected across a battery whose internal resistance is Rint . Draw a diagram representing this circuit.

SolutionThe circuit has three parallel branches: one with R1 and R2 in series; one with just R3; and one with the battery (an ideal emf in series with the internal resistance).

Problem 3 Solution.

Section 28-2:—Electromotive Force

Problem4. What is the emf of a battery that delivers 27 J of energy as it moves 3.0 C between its terminals?

SolutionFrom the definition of emf (as work per unit charge), E W q= =27 3 9 J C V.

Problem5. A 1.5-V battery stores 4.5 kJ of energy. How long can it light a flashlight bulb that draws 0.60 A?

SolutionThe average power, supplied by the battery to the bulb, multiplied by the time equals the energy capacity of the battery. For an ideal battery, P E I, therefore EIt 4 5. , kJ or t 4 5 1 5 0 60 5 10 1393. ( . )( . ) . kJ V A s h.=

Problem6. If you accidentally leave your car headlights (current drain 5 A) on for an hour, how much of the 12-V battery’s

chemical energy is used up?

SolutionThe power delivered by an emf is EI, so (if the voltage and current remain constant) the energy converted is EIt ( )( )( ) .12 5 3600 216 V A s kJ

Problem7. A battery stores 50 W h of chemical energy. If it uses up this energy moving 3 0 104. C through a circuit, what is

its voltage?

SolutionThe emf is the energy (work done going through the source from the negative to the positive terminal) per unit charge: E ( )( ) ( ) .50 3600 3 10 64 W h s/h C V= (This is the average emf; the actual emf may vary with time.)

Section 28-3:—Simple Circuits: Series and Parallel Resistors

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Problem8. A 47-k resistor and a 39-k resistor are in parallel, and the pair is in series with a 22-k resistor. What is the

resistance of the combination?

SolutionFrom Equations 28-1 and 3c, R 22 47 39 47 39 43 3 k k k ( ) ( ) . .=

Problem9. What resistance should be placed in parallel with a 56-k resistor to make an equivalent resistance of 45 k?

SolutionThe solution for R2 in Equation 28-3a is R R R R R2 1 1 56 45 56 45 229 parallel parallel k k .= =( ) ( )( ) ( )

Problem10. In Fig. 28-49 all resistors have the same value, R. What will be the resistance measured (a) between A and B or

(b) between A and C?

FIGURE 28-49 Problems 10 and 11.

Solution(a) The resistance between A and B is equivalent to two resistors of value R in series with the parallel combination of resistors of values R and 2R. Thus, R R R R R R R RAB ( ) ( ) .2 2 8 3= = (b) RAC is equivalent to just one resistor of value R in series with the parallel combination of R and 2R (since the resistor at point B carries no current, i.e., its branch is an open circuit). Thus R R R R R RAC ( ) .2 3 5 3= =

Problem11. In Fig. 28-49, take all resistors to be 1 0. . If a 6.0-V battery is connected between points A and B, what will be the

current in the vertical resistor?

SolutionThe circuit in Fig. 28-49, with a battery connected across points A and B, is similar to the circuit analysed in Example 28-4. In this case, R|| ( )( ) ( ) ( ) , 1 2 1 2 2

3 = and Rtot 1 1 23

83 . The total current (that through the

battery) is I Rtot tot V A E= =6 83

94( ) ( ) . The voltage across the parallel combination is I Rtot A V|| ( )( ) , 9

423

32

which is the voltage across the vertical 1 resistor. The current through this resistor is then ( ) ( ) . .32 1 1 5V A=

Problem12. A defective starter motor in a car draws 300 A from the car’s 12-V battery, dropping the battery terminal voltage to

only 6 V. A good starter motor should draw only 100 A. What will the battery terminal voltage be with a good starter?

SolutionThe starter circuit contains all the resistances in series, as in Fig. 28-10. (We assume RL includes the resistance of the cables, connections, etc., as well as that of the motor.) With the defective starter, V IRT E int 6 12 V V ( ) ,int300 A R so Rint . .0 02 With a good starter, VT 12 100 0 02 10 V A V( )( . ) .

978 CHAPTER 28

Problem13. What is the internal resistance of the battery in the preceding problem?

SolutionThe solution of the preceding problem (or the reasoning of Example 28-2) gives Rint .0 02 (i.e., ( ) ).12 6 300 V V A =

Problem14. Three 1.5-V batteries, with internal resistances of 0 01. , 0.1 , and 1 , each have 1- resistors connected across

their terminals. To three significant figures, what is the voltage across each resistor?

SolutionThe circuit diagram is like Fig. 28-10, and the voltage across the load (from Kirchhoff’s voltage law) is V IRL E int . Since I R R V R R RL L L L E E= =( ), ( )int int (as for a voltage divider). With the given numerical values, V R RL ( . )( ) ( ) .49 .int int1 5 1 1 1 0 01 V V, 1.36 V, and 0.750 V, for , 0.1 , and 1 = respectively.

Problem15. When a 9-V battery is temporarily short-circuited, a 200-mA current flows. What is the internal resistance of the

battery?

SolutionFrom the equation for a battery short-circuited, in the subsection “Real Batteries,” R Iint . E= =9 0 2 45 V A .

Problem16. What possible resistance combinations can you form using three resistors whose values are 1 0. , 2.0 , and 3.0 ?

(Use all three resistors.)

SolutionThere are eight combinations using all three resistors. (a) One in series with two in parallel: 3 2 3 11 3 4 11 4 = = = =, , 2 3 and .1 6 5 11 5 = = (b) One in parallel with two in series: 3 3 3 3 3 2 2 4 2 4 4 3 1 5 1 5 = = = =( ) , ( ) , ( ) and 5 6= . (c) Three in series: 1 2 3 6 . (d) Three in

parallel: ( )1 2 3 6 111 1 1 1 = .

Problem17. A partially discharged car battery can be modeled as a 9-V emf in series with an internal resistance of 0 08. . Jumper

cables are used to connect this battery to a fully charged battery, modeled as a 12-V emf in series with a 0 02. - internal resistance. How much current flows through the discharged battery?

SolutionTerminals of like polarity are connected with jumpers of negligible resistance. Kirchhoff’s voltage law gives E E1 2 IR IR I R R1 2 1 2 1 20 12 9 0 02 0 08 30 , ( ) ( ) ( ) ( . . ) or V A.E E = =

Problem 17 Solution.

Problem18. You have a number of 50- resistors, each capable of dissipating 0.50 W without overheating. How many resistors

would you need, and how would you connect them, so as to make a 50- combination that could be connected safely

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across a 12-V battery?

SolutionThe 50 combination must be capable of dissipating P E 2 212 50 2 8= =R ( ) . V W of power (when connected across an ideal 12 V battery), so combinations with at least six resistors (capable of dissipating 0 5. W each) must be considered. In order to get the same total resistance as each individual resistor, n parallel branches of n resistors in series are needed (or n branches in series of n resistors in parallel), making a total of n2 resistors. The smallest n2 greater than 6 is for n 3. (Alternatively, one could argue that the total current is 12 50 0 24 V A,= . while the maximum current in

each resistor element is 0 5 50 0 1. . W A,= so at least three equal branches in the circuit are needed.)

Problem19. What is the equivalent resistance between A and B in each of the circuits shown in Fig. 28-50? Hint: In (c), think about

symmetry and the current that would flow through R2 .

Solution(a) There are two parallel pairs ( )1

2 1R in series, so R R R RAB 12 1

12 1 1. (b) Here, there are two series pairs ( )2 1R in

parallel, so R R R R R RAB ( )( ) ( ) .2 2 2 21 1 1 1 1= (c) Symmetry requires that the current divides equally on the right and left sides, so points C and D are at the same potential. Thus, no current flows through R2 , and the circuit is equivalent to (b). (Note that the reasoning in parts (a) and (b) is easily generalized to resistances of different values; the generalization in

part (c) requires the equality of ratios of resistances which are mirror images in the plane of symmetry.)

FIGURE 28-50 Problem 19 Solution.Problem20. A 6.0-V battery has an internal resistance of 2 5. . If the battery is short circuited, what is the rate of energy

dissipation in its internal resistance?

SolutionFor a short-circuited battery, I RE= int , so the dissipated power is P E I R R2 2 26 2 5 14int int ( ) . .4= = V W.

Problem21. How many 100-W, 120-V light bulbs can be connected in parallel before they below a 20-A circuit breaker?

SolutionThe circuit breaker is activated if I R 120 20 V A,= min or if Rmin 6 . The resistance of each light bulb is

R V 2 2120 100= =P ( ) V W 144 , and n bulbs in parallel have resistance R R n|| . = Therefore R R|| min implies n 144 6 24= , so more than 24 bulbs would blow the circuit.

Problem22. What is the current through the 3- resistor in the circuit of Fig. 28-51? Hint: This is trivial. Can you see why?

980 CHAPTER 28

FIGURE 28-51 Problem 22.

SolutionThe current is I V R3 3 3 6 3 2 = = V A, from Ohm’s law. The answer is trivial because the potential difference across the 3 resistor is evident from the circuit diagram. (However, if the 6 V battery had internal resistance, an argument like that in Example 28-5 must be used.)

Problem23. Take E 12 2701 V and R in the voltage divider of Fig. 28-5. (a) What should be the value of R2 in order that

4.5 V appear across R2? (b) What will be the power dissipation in R2?

Solution(a) For this voltage divider, Equation 28-2b gives V R R R R R V V2 2 1 2 2 1 2 2 270 4 5 12 4 5 E E= = =( ), ( ) ( )( . ) ( . ) or 162 . (b) The power dissipated (Equation 27-9b) is P2 2

22

24 5 162 125 V R= =( . ) V mW.

Problem24. A voltage divider consists of two 1 0. -k resistors connected in series across a 160-V emf. If a 10-k resistor is

connected across one of the 1 0. -k resistors, what will be the voltage across it?

SolutionResistors of 1 k and 10 k connected in parallel have a combined resistance of ( )10 11= k (Equation 28-3c), so the altered circuit is a voltage divider (Fig. 28-5) with R1 1 k and R2 10 11 ( ) .= k Equation 28-2b gives V R R R2 2 1 2 160 10 11 1 10 11 76 2 E = = = =( ) ( )( ) ( ) . . V V

Problem25. In the circuit of Fig. 28-52, R1 is a variable resistor, and the other two resistors have equal resistances R. (a) Find an

expression for the voltage across R1, and (b) sketch a graph of this quantity as a function of R1 as R1 varies from 0 to 10R. (c) What is the limiting value as R1 ?

FIGURE 28-52 Problem 25.

Solution(a) The resistors in parallel have an equivalent resistance of R RR R R|| ( ). 1 1= The other R, and R|| , is a voltage divider in series with E, so Equation 28-2 gives V R R R R R R|| || ||( ) ( ). E E= =1 12 (b) and (c) If R1 0 (the second resistor

shorted out), V|| ,0 while if R1 (open circuit), V|| 12 E (the value when R1 is removed). If R R V1 10 10 21 , ( )|| = E

(as in Problem 24).

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Problem 25 Solution.

Problem26. In the circuit of Fig. 28-53 find (a) the current supplied by the battery and (b) the current through the 6- resistor.

Solution(a) With reference to the solution of the next problem, the resistance of the three parallel resistors in ( ) ,12 11= so the current supplied by the battery is I R R E= = =( ) ( ) ( ) . .||1 6 23 11 2 87 V A (b) The voltage drop across the resistors in parallel is V IR IR|| || , E 1 and the current through the 6 resistor is I V6 6 || .= Thus, I6 2 87 2 11 A ( . )( )= 522 mA.

FIGURE 28-53 Problem 26 Solution, and Problem 27.

Problem27. In the circuit of Fig. 28-53, how much power is being dissipated in the 4- resistor?SolutionThe three resistors in parallel have an effective resistance of 1 1 2 1 4 1 6 1= = = =R|| ( ) , or R|| ( ) . 12 11= Equation 28-2

gives the voltage across them as V|| ( )( ) ( ) ( ) . 6 12 11 1 12 11 72 23 V V= = = = Thus, P42

42 272 23 4 V R|| ( )= = = V

2.45 W.

Problem28. A 50- resistor is connected across a battery, and a 26-mA current flows. When the 50- resistor is replaced with a

22- resistor, a 43-mA current flows. What are the battery’s voltage and internal resistance?

SolutionThe circuit diagram is like Fig. 28-10, and Kirchhoff’s voltage law is E IR IRLint .0 For the two cases given, this may be written as E ( ) ( )( ),int26 26 50 mA mA R and E ( ) ( )( ).int43 43 22 mA mA R Solving for E and Rint , we find

Rint . , ( )( . ) . .

26 50 43 22

43 2620 8 26 50 20 8 184 and mA VE

Section 28-4:—Kirchhoff’s Laws and Multiloop Circuits

Problem29. In the circuit of Fig. 28-54 it makes no difference whether the switch is open or closed. What is E3 in terms of the

other quantities shown?

982 CHAPTER 28

SolutionIf the switch is irrelevant, then there is no current through its branch of the circuit. Thus, points A and B must be at the same potential, and the same current flows through R1 and R2 . Kirchhoff’s voltage law applied to the outer loop, and to the left-hand loop, gives E E1 1 2 2 0 IR IR , and E E1 1 3 0 IR , respectively. Therefore,

E EE E

EE E

3 1 11 2

1 21 1

2 1 1 2

1 2

FHG

IKJ

IRR R

RR R

R R.

FIGURE 28-54 Problem 29 Solution.

Problem30. What is the current through the ammeter in Fig. 28-55?

SolutionIf the ammeter has zero resistance, the potential difference across it is zero, or nodes C and D are at equal potentials. If I is the current through the battery, 1

2 I must go through each of the 2 -resistors connected at node A (because

V V I V VA C A D 12 2( ) ). At node B, the 2 -resistor inputs twice the current of the 4 -resistor, or 2

3 I and 13 I

respectively (because V V I I V VC B D B 23

132 4( ) ( ) ). Therefore 1

6 I must go through the ammeter from D to C, as required by Kirchhoff’s current law. To find the value of I, note that the upper pair of resistors are effectively in parallel ( )V VC D as is the lower pair. The effective resistance between A and B is Reff 2 2 2 2 2 4 =( ) ( ) ( ) ( ) .2 4 1 4

373 Thus I V R = eff , and the ammeter current is 1

616

73

376 0 V A AI ( ) ( ) ( ) .429 .=

FIGURE 28-55 Problem 30 Solution.

Problem31. In Fig. 28-56, what is the equivalent resistance measured between points A and B?

SolutionThe effective resistance is determined by the current which would flow through a pure emf if it were connected between A and B: R IAB E=. Since I is but one of six branch currents, the direct solution of Kirchhoff’s circuit laws is tedious (6 6 determinants). (The method of loop currents, not mentioned in the text, involves more tractable 3 3 determinants.) However, because of the special values of the resistors in Fig. 28-56, a symmetry argument greatly simplifies the calculation.

The equality of the resistors on opposite sides of the square implies that the potential difference between A and C equals that between D and B, i.e., V V V VA C D B . Equivalently, V V V VA D C B . Since V V I R V VA C A D 1 , I R2 2( ), etc., the symmetry argument requires that both R-resistors on the perimeter carry the same current, I1, and both 2R-resistors carry current I2 . Then Kirchhoff’s current law implies that the current through E is I I1 2 , and the current through the central resistor is I I1 2 (as added to Fig. 28-56). Now there are only two independent

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branch currents, which can be found from Kirchhoff’s voltage law, applied, for example, to loops ACBA, E I R I R1 2 2 0( ) , and ACDA, I R I I R I R1 1 2 2 2 0( ) ( ) . These equations may be rewritten as I I R1 22 E= and 2 3 01 2I I , with solution I R1 3 7 E= and I R2 2 7 E= . Therefore, I I I R 1 2 5 7E= , and R I RAB E= =7 5. (The configuration of resistors in Fig. 28-56 is called a Wheatstone bridge.)

FIGURE 28-56 Problem 31 Solution.

Problem32. Find all three currents in the circuit of Fig. 28-18, but now with E2 1 0 . . V

SolutionThe currents are (see solution to the next problem):

I R R R

I R R R

I R R

R R R R R R

1 2 3 1 3 21

2 3 1 1 3 21

3 2 1 1 21

1 2 2 3 3 12

4 1 6 1 1 14 2 07

1 6 2 1 1 14 0 214

4 6 2 1 14 186

2 4 4 1 1 2 14

[( ) ] [( )( ) ( )( )] .

[ ( ) ] [( )( ) ( )( )] .

( ) [( )( ) ( )( )] .

( )( ) ( )( ) ( )( ) .

E E

E E

E E

A A

A A and

A A, where

=

=

=

Problem33. Find all three currents in the circuit of Fig. 28-18 with the values given, but with battery E2 reversed.

SolutionThe general solution of the two loop equations and one node equation given in Example 28-5 can be found using determinants (or I1 and I2 can be found in terms of I3 , as in Example 28-5). The equations and the solution are:

I R I R

I R I R

I I I

1 1 3 3 1

2 2 3 3 2

1 2 3

0 1

0 2

0

E

E

( ),

( ),

( );

loop

loop

node A

R R

R R R R R R R R

I

R

R RR R R

I

R R

RR R R

I

R

RR R

1 3

2 3 1 2 2 3 3 1

1

1 3

2 2 31 2 3 2 3

2

1 1 3

2 31 3 2 1 3

3

1 1

2 22 1 1 2

0

0

1 1 1

10

0 1 1

10

1 0 1

10

0

1 1 0

,

( ),

( ),

.

E

EE E

E

EE E

E

EE E

984 CHAPTER 28

With the particular values of emf’s and resistors in this problem, we find currents of I1 4 1 6 1 9 14 2 79 [( ) ( )] . A A,= I2 1 6 2 1 9 14 2 36 [ ( )( )] . A A,= and I3 4 6 2 9 14 0 [ ( )] .429 . A A= Or, one could retrace the reasoning of Example 28-5, with E2 9 V replacing the original value in loop 2. Then, everything is the same until the equation 9 2 6 3 03 3( ) ,I I or I I3

37 2

12

376 3 33 14 ( ) ( ) ( ) A, A A,= and I I I1 2 3 39 14 ( )= A.

Problem34. In Fig. 28-57, take E E E1 2 3 1 2 3 46 0 1 5 4 5 270 150 560 820 . , . , . , , , , . V V V and R R R R Find

the current in R3 , and give its direction.SolutionThe general expressions for the branch currents can be found from the solution to the next problem. Here, we only need

Ib

( . )( ) ( . . )( )

( )( ) ( )( ). .

6 1 5 820 4 5 1 5 420

560 820 420 13804 77

V V V V

mA

A negative current is downward through E2 in Fig. 28-57.

FIGURE 28-57 Problem 34 Solution, and Problems 35 and 36.

Problem35. With all the values except E2 in Fig. 28-57 as given in the preceding problem, find the condition on E2 that will make

the current in R3 flow upward.

SolutionLet us choose the positive sense for each of the three branch currents in Fig. 28-57 as upward through their respective emf’s (at least one must be negative, of course), and consider the two smaller loops shown. Kirchhoff’s circuit laws give:

I I I

R R I R I

R I R I

a b c

a b

b c

0

1 2 3 1 2

3 4 3 2

top node

left loop

(right loop

( )

( ) ( )

).

E E

E E

Solve for Ia and Ic from the loop equations and substitute into the node equation:

( ) ( ).

E E E E1 2 3

1 2

3 2 3

4

0

R I

R RI

R I

Rb

bb

Then

IR R R

R R R R R Rb

4 2 1 1 2 2 3

3 4 1 2 3 4

( ) ( )( )

( )( ),

E E E E

with similar expressions for Ia and Ic . One can see that Ib is positive if R R R4 2 1 1 2 2 3 0( ) ( )( ) ,E E E E or

EE E

24 1 1 2 3

1 2 4

820 6 420 4 5

12405

R R R

R R R

( ) ( )( ) ( )( . )

( ).49 .

V V

V

Problem36. Suppose that all resistors in Fig. 28-57 have the same value R and that E E E1 3 and E E2 2 . Find expressions for

the currents in the four resistors, and give their directions.

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SolutionWhen all the resistors are equal to R and E E E E1 3

12 2 , the solution to the previous problem gives I Rb 3 5E= , and

IR I

R R R R

IR I

R R R

ab

cb

( )

( ).

E E E E E

E E E E E

1 2 3

1 2

3 2 3

4

3 5

5

3 5 2

5

=

=

2

(Positive currents are upward through their respective emf’s in Fig. 28-57.)

Problem37. Figure 28-58 shows a portion of a circuit used to model the electrical behavior of long, cylindrical biological cells

such as muscle cells or the axons of neurons. Find the current through the emf E3 , given that all resistors have the same value R 1 5. M and that E E1 275 45 mV mV,, and E3 20 mV. Be sure to specify the direction of the current.

SolutionThe circuits in Figs. 28-57 and 58 each have three branches (with emf’s and total resistances E E Ea a b b c cR R R, , , , , and ) connected in parallel between two nodes. (The emf’s and currents in each branch are taken as positive upward in the figures.) The expression for Ib in the solution to Problem 35 can be used for any permutation of indices a, b, and c (any order of parallel branches is equivalent). For example, in Fig. 28-58, take b to be the third branch etc., so that E E E Eb b a a c cR R R R R R 3 20 2 75 2 45 mV mV mV and , , , , , , with R 1 5. . M Then

IR R

R R R R R R

R R

R R RRb

b a c b c a

a b b c c a

b a b cb c a

( ) ( ) ( ) ( )( ) ( )

.

E E E E E E E EE E E

2

4 2 23 2 8 60 90 75 12

8 75

2 2 2 = = mV M

nA.

FIGURE 28-58 Problems 37, 38.

Problem38. An electrochemical impulse traveling along the cell modeled in Fig. 28-58 changes the value of emf E3 so that now it

supplies an upward current of 40 nA. Assuming the rest of the circuit remains as described in the preceding problem, what is the new value of E3?

SolutionThe relation between Ib and the circuit emf’s and resistances, given in the solution to the previous problem, can be solved for E Eb 3 in Fig. 28-58, resulting in E E Eb b c aRI 1

3 8 2( ). For Ib 40 nA and the rest of the circuit elements the

same, E313 8 1 5 40 90 75 215 ( . ) . M nA mV mV mV

Section 28-5:—Electrical Measuring Instruments

Problem39. A voltmeter with 200-k resistance is used to measure the voltage across the 10-k resistor in Fig. 28-59. By what percentage is the measurement in error because of the finite meter resistance?

986 CHAPTER 28

FIGURE 28-59 Problems 39 and 40.

SolutionThe voltage across the 10 k resistor in Fig. 28-59 is ( )( ) ( )150 10 10 5 100 V V= (the circuit is just a voltage divider as described by Equations 28-2a and b), as would be measured by an ideal voltmeter with infinite resistance. With the real voltmeter connected in parallel across the 10 k resistor, its effective resistance is changed to R| | ( )( ) 10 200 k k ( ) . ,210 9 52 k k and the voltage reading is only ( )( . ) ( . ) .4 ,150 9 52 9 52 5 98 V V= or about 1.64% lower.

Problem40. An ammeter with 100- resistance is inserted in the circuit of Fig. 28-59. By what percentage is the measurement in

error because of the nonzero meter resistance?

SolutionThe current in the circuit of Fig. 28-59 is I ( ) ( ) .150 5 10 10 V k mA= With the ammeter inserted, the resistance is increased and the current drops to ( ) ( . ) .150 5 10 0 1 9 93 V k mA,= about 0.662% lower.

Problem41. A neophyte mechanic foolishly connects an ammeter with 0 1. - resistance directly across a 12-V car battery whose

internal resistance is 0 01. . What is the power dissipation in the meter? No wonder it gets destroyed!

SolutionThe current through the misconnected ammeter is I R Rm E=( ),int so the power dissipated in it is P I Rm

2

E2 2 212 0 11 0 1 119R R Rm m= =( ) ( . ) ( . ) .int V kW (comparable to a small toaster-oven).

Problem42. The voltage across the 30-k resistor in Fig. 28-60 is measured with (a) a 50-k voltmeter, (b) a 250-k voltmeter,

and (c) a digital meter with 10-M resistance. To two significant figures, what does each read?

FIGURE 28-60 Problem 42 Solution.SolutionWith a meter of resistance Rm connected as indicated, the circuit reduces to two pairs of parallel resistors in series. The total resistance is R R Rm mtot k k k ( ) ( ) .30 30 40 2 = = The voltage reading is V R I R Im m m m ( )30 k tot ( ),30 k Rm where I Rtot tot V ( )100 = (the expression for Vm follows from Equation 28-2, with R1 and R2 as the above pairs, or from Im as a fraction of Itot , as in the solution to Problem 65). For the three voltmeters specified, I tot mA, 2 58. 2 14. mA, and 2.00 mA, while Vm 48 57 3.4 , . , V V and 59.9 V, respectively. (After checking the calculations, round off to two figures. Of course, 60 V is the ideal voltmeter reading.)

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Problem43. In Fig. 28-61 what are the meter readings when (a) an ideal voltmeter or (b) an ideal ammeter is connected between

points A and B?

FIGURE 28-61 Problem 43.

Solution(a) An ideal voltmeter has infinite resistance, so AB is still an open circuit (as shown on Fig. 28-61) when such a voltmeter is connected. The meter reads the voltage across the 20 k resistor (part of a voltage divider), or ( ) ( )30 20 20 10 V = 20 V (see Equation 28-2a or b). (b) An ideal ammeter has zero resistance, and thus measures the current through the points A and B when short-circuited (i.e., no current flows through the 20 k resistor). In Fig. 28-61, this would be I AB 30 10 3 V mA.= (Such a connection does not measure the current in the original circuit, since an ammeter should be connected in series with the current to be measured.)

Problem44. A resistor draws 1.00 A from an ideal 12.0-V battery. (a) If an ammeter with 0 10. - resistance is inserted in the

circuit, what will it read? (b) If this current is used to calculate the resistance, how will the calculated value compare with the actual value?

Solution(a) The internal resistance of an ideal battery is zero, so the resistor has a value of R 12 1 12 V A = . With the ammeter in place, I V R Rm = =( ) . . .12 12 1 0 992 V A (b) If the resistance of the ammeter were neglected in the calculation, one would obtain R 12 0 992 12 1 V A =. . (off by 0.83%), but subtraction of Rm is a correction easily included.

Section 28-6:—Circuits with Capacitors

Problem45. Show that the quantity RC has the units of time (seconds).

SolutionThe SI units for the time constant, RC, are ( )( ) ( )( ) ( )( ) , F V A C V s C C s = = = as stated.

Problem46. If capacitance is given in F, what will be the units of the RC time constant when resistance is given in (a) , (b) k, (c) M? Your answers eliminate the need for tedious power-of-10 conversions.

Solution(a) ( )( ) F s (see previous problem), (b) ( )( ) ,k F s ms 10 103 6 (c) (M F s)( ) .

Problem47. Show that a capacitor is charged to approximately 99% of the applied voltage in five time constants.

SolutionAfter five time constants, Equation 28-6 gives a voltage of V eC=E ' 1 1 6 74 10 99 3%5 3. . of the applied

988 CHAPTER 28

voltage.

Problem48. An uncharged 10- F capacitor and a 470-k resistor are connected in series, and 250 V applied across the

combination. How long does it take the capacitor voltage to reach 200 V?

SolutionFor the RC circuit described, Equation 28-6 gives the voltage across the capacitor, as a function of time. Thus, VC E( )1 e t RC= or t RC VC ln[ ( )] ( )( ) ln( ) . .E E= =10 470 250 50 7 56 F k s

Problem49. Figure 28-62 shows the voltage across a capacitor that is charging through a 4700- resistor in the circuit of

Fig. 28-29. Use the graph to determine (a) the battery voltage, (b) the time constant, and (c) the capacitance.

FIGURE 28-62 Problem 49 Solution.

Solution(a) For the circuit considered, the voltage across the capacitor asymptotically approaches the battery voltage after a long time (compared to the time constant). In Fig. 28-62, this is about 9 V. (b) The time constant is the time it takes the capacitor voltage to reach 1 63 2%1 e . of its asymptotic value, or 5.69 V in this case. From the graph, ' 1 5. . ms (c) The time constant is RC, so C 1 5 4700 0 319. . . ms F=

Problem50. The voltage across a charging capacitor in an RC circuit rises to 1 1 =e of the battery voltage in 5.0 ms. (a) How long

will it take to reach 1 1 3 =e of the battery voltage? (b) If the capacitor is charging through a 22-k resistor, what is its capacitance?

Solution(a) Equation 28-6 and the given circuit characteristics imply that the time constant is RC 5 0. ms. Therefore, in three time constants, or 15 ms, the capacitor is charged to 1 3 e of the battery voltage. (b) Evidently, C R = =5 22 ms k 0 227. . F

Problem51. A 1 0. - F capacitor is charged to 10.0 V. It is then connected across a 500-k resistor. How long does it take (a) for

the capacitor voltage to reach 5.0 V and (b) for the energy stored in the capacitor to decrease to half its initial value?

SolutionA capacitor discharging through a resistor is described by exponential decay, with time constant RC (see Equation 28-8),

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and, of course, U t CV t CV e U eCt RC

Ct RC( ) ( ) ( ) 1

22 1

2 02 2 20= = is the energy stored (see Equation 26-8b).

(a) V t V( ) ( )= =0 1 2 implies t RC ln ( )( )( . ) .2 500 1 0 693 347 k F ms (b) U t Uc c( ) ( )= =0 1 2 implies t RC 1

2 2 173ln . ms

Problem52. A capacitor used to provide steady voltages in the power supply of a stereo amplifier charges rapidly to 35 V every

1 60= of a second. It must then hold that voltage to within 1.0 V for the next 1 60= s while it discharges through the amplifier circuit. If the circuit draws 1.2 A from the 35-V supply (a) what is its effective resistance and (b) what value of capacitance is needed?

Solution(a) The effective resistance of a circuit that draws 1.2 A from a constant 35 V supply is 35 1 2 29 2 V A =. . . (b) To keep the voltage within the prescribed range for the discharging capacitor (Equation 28-8), the time constant must satisfy V V e t RC= ==

0 34 35 , or RC t = =ln( ).35 34 For t 1 60= s and R 29 2. , one finds C 19 7. . mF

Problem53. A capacitor is charged until it holds 5.0 J of energy. It is then connected across a 10-k resistor. In 8.6 ms, the resistor

dissipates 2.0 J. What is the capacitance?

SolutionEquation 28-8 gives a voltage V V e t RC

0= for a capacitor discharging through a resistor. If 2 J is dissipated in time t, the

energy stored in the capacitor drops from U0 5 J to U 3 J (assuming there are no losses due to radiation, etc.). Since U CV U U V V e t RC 1

22

0 02 2, ( ) = = = and we may solve for C C t R U U: ln( ) ( . ) ( ) ln( ) . . ms k F 2 2 8 6 10 5 3 3 370= = = =

Problem54. A 2 0. - F capacitor is charged to 150 V. It is then connected to an uncharged 1 0. - F capacitor through a 2 2. -k

resistor, by closing switch S in Fig. 28-63. Find the total energy dissipated in the resistor as the circuit comes to equilibrium. Hint: Think about charge conservation.

FIGURE 28-63 Problem 54.

SolutionWhen current stops flowing (at t ), the potential difference across the capacitors is equal, but the total charge is just the initial charge. Thus, V V V1 2( ) ( ) ( ), and Q Q Q2 1 20( ) ( ) ( ). Since Q CV C V C V C V , ( ) ( ) ( ) 1 1 2 2 2 2 0

or V V C C C V( ) ( ) ( ) ( ). 2 2 1 223 20 0= The energy stored in the capacitors is initially U C V( ) ( )0 01

2 2 22

12

22 150 22 5( )( ) . , F V mJ and finally U C C V( ) ( ) ( ) ( )( ) . 12 1 2

2 12

23 100 15 0 F V mJ. The difference,

U 7 50. mJ, is dissipated in the resistor (except for a negligible amount of radiated energy).

Problem55. For the circuit of Example 28-9, take E 100 4 0 6 01 2 V k and k, . , .R R , and assume the capacitor is initially

uncharged. What are the currents in both resistors and the voltage across the capacitor (a) just after the switch is closed and (b) a long time after the switch is closed? Long after the switch is closed it is again opened. What are I1, I2, and VC (c) just after this switch opening and (d) a long time later?

990 CHAPTER 28

SolutionIn addition to the explanation in Example 28-9, we note that when the switch is in the closed position, Kirchhoff’s voltage law applied to the loop containing both resistors yields E I R I R1 1 2 2 , and to the loop containing just R2 and C, V I RC 2 2

. (a) If the switch is closed at t 0 , Example 28-9 shows that V I I RC ( ) , ( ) , ( )0 0 0 0 0 100 42 1 1 and V kE= = 25 mA. (b) After a long time, t , Example 28-9 also shows that I I R R1 2 1 2 100 10( ) ( ) ( ) E= = V k 10 mA, and V I RC ( ) ( ) ( )( ) . 2 2 10 6 60 mA k V (c) Under the conditions stated, the fully charged capacitor ( )VC 60 V simply discharges through R2. (R1 is in an open-circuit branch, so I1 0 for the entire discharging process.) The initial discharging current is I V RC2 2 60 6 10 = = V k mA . (d) I2 and VC decay exponentially to zero.

Problem56. In the circuit of Fig. 28-64 the switch is initially open and both capacitors initially uncharged. All resistors have the

same value R. Find expressions for the current in R2 (a) just after the switch is closed and (b) a long time after the switch is closed. (c) Describe qualitatively how you except the current in R3 to behave after the switch is closed.

FIGURE 28-64 Problem 56 Solution.

Solution(a) An uncharged capacitor acts instantaneously like a short circuit (see Example 28-9), so initially ( )t 0 all of the current from the battery goes through R1 and C1, and none goes through R2 and R3. Thus, I R1 0( ) ,E= and I I2 30 0 0( ) ( ) . (b) A fully charged capacitor acts like an open circuit (when responding to a constant applied emf ), so after a long time ( )t , all of the current goes through R1 and R2 in series, and none goes through R3. Thus I I R1 2 2( ) ( ) , E= and I3 0( ) . (c) One can easily guess that I1 and I2 respectively decrease and increase monotonically from their initial to their final values, and that I3 first increases from, and then decreases to zero. (One can use the loop and node equations to solve for the currents. They turn out to be linear combinations of two decaying exponentials with different time constants.)

Problem57. In the circuit for Fig. 28-65 the switch is initially open and the capacitor is uncharged. Find expressions for the current

I supplied by the battery (a) just after the switch is closed and (b) a long time after the switch is closed.

FIGURE 28-65 Problem 57.

Solution(a) Just after the switch is closed, the uncharged capacitor acts instantaneously like a short circuit and the resistors act like two parallel pairs in series. The effective resistance of the combination is 2 2 2 4 3 ( )( ) ( ) ,R R R R R= = and the current supplied by the battery is I R( ) .0 3 4 E= (b) A long time after the switch is closed, the capacitor is fully charged and acts

like an open circuit. Then the resistors act like two series pairs in parallel, with an effective resistance of ( ) ( )12 2 R R

3 2R=. The battery current is I R( ) . 2 3E=

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Problem58. Obtain an expression for the rate ( )dV dt= at which the voltage across a charging capacitor increases. Evaluate your

result at time t 0, and show that if the capacitor continued charging steadily at this rate it would be fully charged in exactly one time constant.

SolutionThe loop law for a battery charging a capacitor through a resistor is E IR VC 0. Differentiate this and use Equation 28-4 to obtain ( ) ( ) ( ) ( ) .dV dt d IR dt R dI dt R I RC I CC= = = = = E ( I t( ) is given in Equation 28-5.) For an initially uncharged capacitor, I R0 E= (an uncharged capacitor acts like a short circuit), so the initial rate of increase of the capacitor’s voltage is ( ) .dV dt RCC= = =0 E E If the capacitor were to charge steadily at this rate (i.e., if the voltage were a linear function of time, ( ) ),dV dt t tC= =0 E the voltage would reach its final value ( ( ) )VC E in just one time constant (i.e., V tC ( ) E= for t ).

Paired Problems

Problem59. A 3 3. -k resistor and a 4 7. -k resistor are connected in parallel, and the pair is in series with a 1 5. -k resistor. What

is the resistance of the combination?

SolutionEquation 28-3c for the resistance of the parallel pair, combined with Equation 28-1 for the resistance of this in series with the third resistor gives an effective resistance of Reff k k k 1 5 3 3 4 7 3 3 4 7 3. ( . )( . ) ( . . ) .44 . =

Problem60. Find the value of R in Fig. 28-66 that will make the resistance between points A and B equal to R.

FIGURE 28-66 Problem 60.

SolutionAs in the previous problem, R R Reff 1 3 3 ( ) ( ),= so R Reff implies ( )( ) ( ) .R R R 1 3 3 Solving

this quadratic equation for the positive value of the resistance, we find R 12 1 13 2 30( ) . .

Problem61. A battery’s voltage is measured with a voltmeter whose resistance is 1000 ; the result is 4.36 V. When the

measurement is repeated with a 1500- meter the result is 4.41 V. What are (a) the battery voltage and (b) its internal resistance?

SolutionThe internal resistance of the battery ( )Ri and the resistance of the voltmeter ( )Rm are in series with the battery’s emf, so the current is I R Ri m E=( ). The potential drop across the meter (its reading) is V IR R R Rm m m i m E =( ). From the given data, 4 36 1 1 4 1 5 1 5. ( ) ( ) .41 ( . ) ( . ) V k k and V k k E E = =R Ri i , which can be solved simultaneously for E and Ri. One obtains Ri 1 1 4 36 k k V E( . )= and Ri 1 5 1 5 4. ( . .41 ), k k V E = or

E FH IK

( . ).

.41 ..1 5 1

1 5

4

1

4 364 51

1

k k k

V

k

V V

992 CHAPTER 28

and Ri ( . )( . ) . .4 51 1 4 36 1 35 2 V k V k =

Problem62. An ammeter with a resistance of 1.4 is connected momentarily across a battery (not the way to treat an ammeter!)

and it reads 9.78 A. When the measurement is repeated with a meter whose resistance is 2 1. the reading is 7.46 A. What are (a) the battery voltage and (b) its internal resistance?

SolutionThe internal resistance of the battery ( )Ri and the ammeter’s resistance ( )Rm are still in series with the battery (as in the previous problem), but the ammeter reading is the total current I R Ri m E=( ). Thus, the given data require E ( .4 )( . ) ( . )( .46 )R Ri i 1 9 78 2 1 7 A A , which have as solutions

Ri

( . )( .46) ( .4 )( . )

( . .46). ( . .4)( . ) . .

2 1 7 1 9 78

9 78 70 851 0 851 1 9 78 22 0

, and V V

E

Problem63. In Fig. 28-67, take E E E1 2 3 1 2 312 6 0 3 0 1 0 2 0 4 0 V V V and , . , . , . , . , . .R R R Find the current

in R2 and give its direction.

SolutionAn obvious reconfiguration of the circuit in Fig. 28-67 results in a circuit like that in Fig. 28-57, with R1 replacing ( ),R R R1 2 2 for R R3 3, for R4 , and E3 for E3. Thus, the solution to Problem 35, properly altered, gives

IR R

R R R R R Rb

( ) ( ) ( )( ) ( )( )

( )( ) ( )( ).

E E E E2 1 3 2 3 1

1 2 1 3 2 3

6 12 4 6 3 1

1 2 4 2 41 07A A.

(A negative current is opposite to the direction of the emf E2 .)

FIGURE 28-67 Problem 63.

Problem64. (a) With all values except E2 as given in the preceding problem, find E2 such that there is no current in this battery.

(b) What are the currents in R1 and R2 under these conditions?

Solution(a) Ib in the previous problem is zero if ( ) ( )E E E E3 2 1 1 2 3 R R , or E E E2 1 3 3 1 1 3 ( ) ( )R R R R= ( ) ( ) .12 4 3 1 1 4 9 V V= (b) With no current in the branch containing E2 and R2, the current in the outer loop (use the loop law) is just I R R ( ) ( ) ( ) ( ) .E E1 3 1 3 12 3 1 4 3= = V A

Problem65. In Fig. 28-68 what are the meter readings when (a) an ideal voltmeter or (b) an ideal ammeter is connected between

points A and B?

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Solution(a) An ideal voltmeter has Rm (AB open circuited), so VAB is just the voltage across the 5 6. k resistor. This is part of a voltage divider (in series with the 4 7. k resistor), so Equation 28-2 gives VAB ( )( . ) ( . . ) . .24 5 6 5 6 4 7 13 0 V V= (b) An ideal ammeter has Rm 0 (AB short circuited), so IAB is just the current through the 3 3. k resistor. This is part of parallel combination, R|| ( . )( . ) ( . . ) . , 3 3 5 6 3 3 5 6 2 08 k k = which, in series with the 4 7. k resistor, draws a total current of Itot V k mA 24 2 08 4 7 3 54=( . . ) . . Now, two resistors in parallel form a current divider, each one taking a fraction of the total current, given by I R R I I R R I1 1 2 2 ( ) , ( ) ,|| ||= =tot tot and respectively. (This follows directly from Ohm’s law: V I R I R I R|| || .) tot 1 1 2 2 Therefore, in the case of the ideal ammeter, I R IAB ( . ) ( . . )||= =3 3 2 08 3 3 k tot ( . ) . .3 54 2 23 mA mA

FIGURE 28-68 Problem 65 Solution.

Problem66. In Fig. 28-68 what are the meter readings when (a) a voltmeter with 50-k resistance or (b) an ammeter with

150- resistance is connected between points A and B?

SolutionWith realistic instruments in the circuit of the previous problem, we will need to find the current in the branch containing AB. This has resistance 3 3. , k Rm and is in parallel with 5 6. , k yielding an effective resistance R Rm|| ( . ) 3 3 k ( . ) ( . ).5 6 8 9 k k = Rm The total resistance across the battery terminals is R Rtot k || . ,4 7 so I Rtot tot V ( ) .24 = As in part (b) of the previous solution, I I R RAB m tot k|| ( . ).=3 3 (a) For the voltmeter with R Rm 50 5 07 k k , . ,|| R Itot tot k mA, 9 77 2. , .46 and IAB 0 234. . mA The voltmeter reading is V I RAB AB m ( . )( )0 234 50 mA k 11 7. . V (b) For the ammeter with R R R Im 150 2 13 6 83 3 51 k k mA,tot tot , . , . , .|| and IAB 2 17. mA, which is the ammeter reading.

Problem67. An initially uncharged capacitor in an RC circuit reaches 75% of its full charge in 22.0 ms. What is the time constant?

SolutionFrom Equation 28-6, V eC

t= =E 75% 1 , which implies e t= 4, or t= =ln ln . .4 22 4 15 9 ms ms

Problem68. Find the resistance needed in an RC circuit to bring a 20- F capacitor from zero charge to 45% of its full charge in

140 ms.

SolutionAs in the previous solution, RC t= = = =ln[ ( ln( . )1 1 45%)] 140 1 0 55 234 ms ms, so R 234 20 11 7 ms F k= . .

Supplementary Problems

Problem69. Suppose the currents into and out of a circuit node differed by 1 A . If the node consists of a small metal sphere with

diameter 1 mm, how long would it take for the electric field around the node to reach the breakdown field in air (3 MV/m)?

994 CHAPTER 28

SolutionThe charge on the node (whether positive or negative) accumulates at a rate of 1 1 A C per second, so q t t( ) ( ) 1 A (where we assume that q( ) ).0 0 If the node is treated approximately as an isolated sphere, the electric

field strength at its surface, k q r k t r A= =2 21 ( ) , equals the breakdown field for air, when

t ( / )( . ) ( )( ) . .3 0 5 9 10 1 83 32 9 MV m mm m/F A s=

Problem70. A problem on dairy farms is “stray voltage,” caused by corroded wiring, poor wiring practices, or ground currents

associated with nearby power lines. These conditions can result in several volts of potential difference between metal objects such as watering bowls, feed troughs, or milking equipment, and the ground. Cows feel slight shocks that make them nervous resulting in reduced milk output. As a result, farmers can face serious financial losses. Figure 28-69 shows a circuit model for a stray voltage situation; the 1 5. -k resistor represents the resistance of corroded connections and poor wiring; you can assume the ground has negligible resistance. (a) The resistance from a cow’s mouth to hoof is approximately 350 . How much current will flow through the cow in the situation shown? (It takes only about 1 mA to affect milk production.) In an effort to diagnose the problem, the farmer moves the cow aside and connects a multimeter between the watering bowl and ground. What will it read if it’s set to measure (b) voltage and (c) current?

FIGURE 28-69 Stray voltage can bankrupt a dairy farm. (Problem 70)

Solution(a) The equivalent circuit for the stray voltage contains just an emf in series with resistances representing the corroded connections and the cow, so the current is I 5 0 1500 350 2 70. ( ) . . V mA= Suppose that the farmer’s multimeter behaves like an ideal meter (i.e., Rm is or 0 when set for voltage or current) and replace the cow’s resistance with Rm. Then (b) when set for voltage, it reads 5.0 V (really V IR R Rm m m m ( . ) ( . )5 0 1 5 V k= as Rm ), and (c) when set for current it reads 5 0 1 5 3 33. . . V k mA= (really 5 0 1 5. ( . ) V k= Rm as Rm 0).

Problem71. In Fig. 28-70, what is the current in the 4- resistor when each of the following circuit elements is connected between

points A and B: (a) an ideal ammeter; (b) an ideal voltmeter; (c) another 4.0- resistor; (d) an uncharged capacitor, right after it’s connected; (e) long after the capacitor of part (d) is connected; (f ) an ideal 12-V battery, with its positive terminal at A; (g) a capacitor initially charged to 12 V, right after it’s connected with its positive plate at A; (h) long after the capacitor in part (g) is connected?

Solution(a) An ideal ammeter connected across AB short-circuits the 4 resistor, so I4 0 . (This is a dangerous connection unless the ammeter can handle 3 A.) (b) An ideal voltmeter across AB maintains the open circuit, so I4 6 2 V=( 4) 1 A. (c) Another 4 resistor in parallel across AB divides the total current equally; Itot V 6 2 4 4 4= =[ ( 4)] 1 5. A, and I4 0 75 . . A (d) The instantaneous voltage across an uncharged capacitor is zero, so AB is momentarily short-circuited and I4 0 , as in (a). (e) When the capacitor is fully charged ( )dq dt= 0 it behaves like an open circuit, so I4 1 A, as in (b). (f ) Kirchhoff’s voltage law applied to the loop containing just AB (with the ideal 12 V emf) and the 4 resistor, gives 12 44 V I ( ), or I4 3 A. (g) Instantaneously, a capacitor charged to 12 V appears like the battery

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in (f ) when first connected, so I4 3 A. (h) Charge leaves the capacitor until, after a long time, dq dt IC= 0 and it appears like an open circuit. Then I4 1 A, as in (b) and (e).

FIGURE 28-70 Problem 71 Solution.

Problem72. A resistance R is connected across a battery with internal resistance Rint . Show that the maximum power dissipation in

R occurs when R R int . Note: This is not the way to treat a battery! But it is the basis for matching loads in amplifiers and other devices; for example, a stereo amplifier designed to drive 8- speakers has internal resistance close to 8 .

SolutionThe current for the connection described is I R R E=( ),int so the power dissipated in the resistor is P I R2

E2 2R R R=( ) .int This inherently positive function of R is zero for R 0 and R , so it must have a maximum at some

intermediate value of R. The condition for the maximum is d dRP= 0, which implies ( ) ( ) ,int intR R R R R 2 2 0 or R R int .

Problem73. A parallel-plate capacitor is insulated with a material of dielectric constant and resistivity . Since the resistivity is

finite, the capacitor “leaks” charge and can be modeled as an ideal capacitor in parallel with a resistor. (a) Show that the time constant of the capacitor is independent of its dimensions (provided the spacing is small enough that the usual parallel-plate approximation applies) and is given by 0 . (b) If the insulating material is polystyrene ( . , 2 6 1016 m), how long will it take for the stored energy in the capacitor to decrease by a factor of 2?

Solution(a) In the parallel plate approximation, the electric field between the plates is constant, E V d = , the leakage current density is uniform, J E = , and the capacitance is C A d 0 = . These relations imply that the resistance of the dielectric slab between the plates is R V I V JA V EA d A = = = = =( ) , and the time constant for discharging through the

dielectric is RC d A A d( )( ) .= =0 0 (b) The stored energy in the capacitor, U CVC C 12

2 , decays with half the

voltage time constant, i.e., U C V e CV e U eCt RC t RC t RC 1

2 02 1

2 02 2

02( ) .( )= = = = To decay by 50% takes time t

( ) ( ) .RC= =2 2 2 20ln ln With the values given for polystyrene, t p ( . )( . )( ) ln .2 6 8 85 10 2 7 97 1016 12

4 F/m m s 22 2. h.

Problem74. Of the total energy drawn from a battery in charging an RC circuit, show that only half ends up as stored energy in the

capacitor. Hint: What happens to the rest of it? You will need to integrate.

SolutionThe power supplied by the battery charging a capacitor (initially uncharged) in an RC circuit is P E E I R e t RC( ) ,2= =

where we used I from Equation 28-5. The total energy supplied is U dt R e dt C e et RCbat z z

02

02 0P E E( ) ( )= =

CE2 . This is twice the energy stored in the fully charged capacitor, U CV C UC ( ) ( ) . 12

2 12

2 12E bat The other half

of the energy supplied by the battery is dissipated in the resistor. ( ( ) . )U I R dt R e dt CRt RCz z

02 2

02 1

22 E E= =

Problem75. Write the loop and node laws for the circuit of Fig. 28-71, and show that the time constant for this circuit is

996 CHAPTER 28

R R C R R1 2 1 2=( ).

SolutionConsider the loops and node added to Fig. 28-71. Kirchhoff’s laws are E I R I R V I RC1 1 2 2 2 2, , and I I IC 1 2 . Since V q C I dq dtC C = = and , the equations can be combined to yield

E E E E FHGIKJ

I R I R I I R I R I R

V

RR R I R

q

CR R RC CC

C1 1 2 2 2 1 2 2 12

1 2 12 1 2

0( ) ( )( )

.=

This is exactly in the same form as the first equation, solved in the text, in the section “The RC Circuit: Charging” (with I I R R C CR R RC , ( )), and 1 2 1 2= so the time constant for the circuit is CR R R R1 2 1 2=( ) (the ratio of the coefficients of IC and q).

FIGURE 28-71 Problem 75 Solution.

Problem76. The circuit in Fig. 28-72 extends forever to the right, and all the resistors have the same value R. Show that the

equivalent resistance measured across the two terminals at left is 12 1 5R( ). Hint: You don’t need to sum an infinite

series.

FIGURE 28-72 Problem 76.

SolutionSince the circuit line is infinite, the addition or deletion of one more element leaves the equivalent resistance unchanged. Diagrammatically:

Problem 78 Solution.

The right-hand picture represents R in series with the parallel combination R and Req, therefore R R RR R Req eq eq =( ).

Solving for Req , one finds R RR R R Req eq eq or 2 2 120 1 5 , ( ) (only the positive root is physically meaningful for

a resistance).