28

CHAPTER OUTLINE

28.1 Electromotive Force28.2 Resistors in Series and Parallel28.3 Kirchhoff’s Rules28.4 RC28.5 Electrical Meters28.6 Household Wiring and Electrical Safety

Circuits

Direct Current Circuits

ANSWERS TO QUESTIONS

Q28.1 The load resistance in a circuit is the effective resistance of all ofthe circuit elements excluding the emf source. In energy terms,it can be used to determine the energy delivered to the load byelectrical transmission and there appearing as internal energyto raise the temperature of the resistor. The internal resistanceof a battery represents the limitation on the efficiency of thechemical reaction that takes place in the battery to supplycurrent to the load. The emf of the battery represents itsconversion of chemical energy into energy which it puts out byelectric transmission; the battery also creates internal energywithin itself, in an amount that can be computed from itsinternal resistance. We model the internal resistance asconstant for a given battery, but it may increase greatly as thebattery ages. It may increase somewhat with increasing currentdemand by the load. For a load described by Ohm’s law, theload resistance is a precisely fixed value.

Q28.2 The potential difference between the terminals of a battery will equal the emf of the battery whenthere is no current in the battery. At this time, the current though, and hence the potential dropacross the internal resistance is zero. This only happens when there is no load placed on thebattery—that includes measuring the potential difference with a voltmeter! The terminal voltagewill exceed the emf of the battery when current is driven backward through the battery, in at itspositive terminal and out at its negative terminal.

Q28.3 No. If there is one battery in a circuit, the current inside it will be from its negative terminal to itspositive terminal. Whenever a battery is delivering energy to a circuit, it will carry current in thisdirection. On the other hand, when another source of emf is charging the battery in question, it willhave a current pushed through it from its positive terminal to its negative terminal.

Q28.4 Connect the resistors in series. Resistors of 5.0 kΩ, 7.5 kΩ and 2.2 kΩ connected in series presentequivalent resistance 14.7 kΩ.

Q28.5 Connect the resistors in parallel. Resistors of 5.0 kΩ, 7.5 kΩ and 2.2 kΩ connected in parallel presentequivalent resistance 1.3 kΩ.

129

130 Direct Current Circuits

Q28.6

Q28.7 In series, the current is the same through each resistor. Without knowing individual resistances,nothing can be determined about potential differences or power.

Q28.8 In parallel, the potential difference is the same across each resistor. Without knowing individualresistances, nothing can be determined about current or power.

Q28.9 In this configuration, the power delivered to one individual resistor is significantly less than if onlyone equivalent resistor were used. This decreases the possibility of component failure, and possibleelectrical disaster to some more expensive circuit component than a resistor.

Q28.10 Each of the two conductors in the extension cord itself has a small resistance. The longer theextension cord, the larger the resistance. Taken into account in the circuit, the extension cord willreduce the current from the power supply, and also will absorb energy itself in the form of internalenergy, leaving less power available to the light bulb.

Q28.11 The whole wire is very nearly at one uniform potential. There is essentially zero difference inpotential between the bird’s feet. Then negligible current goes through the bird. The resistancethrough the bird’s body between its feet is much larger than the resistance through the wirebetween the same two points.

Q28.12 The potential difference across a resistor is positive when it is measured against the direction of thecurrent in the resistor.

Q28.13 The bulb will light up for a while immediately after the switch is closed. As the capacitor charges, thebulb gets progressively dimmer. When the capacitor is fully charged the current in the circuit is zeroand the bulb does not glow at all. If the value of RC is small, this whole process might occupy a veryshort time interval.

Q28.14 An ideal ammeter has zero resistance. An ideal voltmeter has infinite resistance. Real meters cannotattain these values, but do approach these values to the degree that they do not alter the current orpotential difference that is being measured within the accuracy of the meter. Hooray forexperimental uncertainty!

Q28.15 A short circuit can develop when the last bit of insulation frays away between the two conductors ina lamp cord. Then the two conductors touch each other, opening a low-resistance branch in parallelwith the lamp. The lamp will immediately go out, carrying no current and presenting no danger. Avery large current exists in the power supply, the house wiring, and the rest of the lamp cord up tothe contact point. Before it blows the fuse or pops the circuit breaker, the large current can quicklyraise the temperature in the short-circuit path.

Chapter 28 131

Q28.16 A wire or cable in a transmission line is thick and made of material with very low resistivity. Onlywhen its length is very large does its resistance become significant. To transmit power over a longdistance it is most efficient to use low current at high voltage, minimizing the I R2 power loss in thetransmission line. Alternating current, as opposed to the direct current we study first, can be steppedup in voltage and then down again, with high-efficiency transformers at both ends of the powerline.

Q28.17 Car headlights are in parallel. If they were in series, both would go out when the filament of onefailed. An important safety factor would be lost.

Q28.18 Kirchhoff’s junction rule expresses conservation of electric charge. If the total current into a pointwere different from the total current out, then charge would be continuously created or annihilatedat that point.

Kirchhoff’s loop rule expresses conservation of energy. For a single-loop circle with tworesistors, the loop rule reads + − − =ε IR IR1 2 0 . This is algebraically equivalent to q qIR qIRε = +1 2 ,where q I t= ∆ is the charge passing a point in the loop in the time interval ∆t . The equivalentequation states that the power supply injects energy into the circuit equal in amount to that whichthe resistors degrade into internal energy.

Q28.19 At their normal operating temperatures, from P =∆V

R

2

, the bulbs present resistances

RV

= = =∆

Ω2 2120

60240

P V W

a f

, and 12075

1902 V

W

a f= Ω , and

120200

722 V

W

a f= Ω . The nominal 60 W lamp

has greatest resistance. When they are connected in series, they all carry the same small current.Here the highest-resistance bulb glows most brightly and the one with lowest resistance is faintest.This is just the reverse of their order of intensity if they were connected in parallel, as they aredesigned to be.

Q28.20 Answer their question with a challenge. If the student is just looking at a diagram, provide thematerials to build the circuit. If you are looking at a circuit where the second bulb really is fainter, getthe student to unscrew them both and interchange them. But check that the student’sunderstanding of potential has not been impaired: if you patch past the first bulb to short it out, thesecond gets brighter.

Q28.21 Series, because the circuit breaker trips and opens the circuit when the current in that circuit loopexceeds a certain preset value. The circuit breaker must be in series to sense the appropriate current(see Fig. 28.30).

Q28.22 The hospital maintenance worker is right. A hospital room is full of electrical grounds, including thebed frame. If your grandmother touched the faulty knob and the bed frame at the same time, shecould receive quite a jolt, as there would be a potential difference of 120 V across her. If the 120 V isDC, the shock could send her into ventricular fibrillation, and the hospital staff could use thedefibrillator you read about in Section 26.4. If the 120 V is AC, which is most likely, the current couldproduce external and internal burns along the path of conduction. Likely no one got a shock fromthe radio back at home because her bedroom contained no electrical grounds—no conductorsconnected to zero volts. Just like the bird in Question 28.11, granny could touch the “hot” knobwithout getting a shock so long as there was no path to ground to supply a potential differenceacross her. A new appliance in the bedroom or a flood could make the radio lethal. Repair it ordiscard it. Enjoy the news from Lake Wobegon on the new plastic radio.

132 Direct Current Circuits

Q28.23 So long as you only grab one wire, and you do not touch anything that is grounded, you are safe(see Question 28.11). If the wire breaks, let go! If you continue to hold on to the wire, there will be alarge—and rather lethal—potential difference between the wire and your feet when you hit theground. Since your body can have a resistance of about 10 kΩ, the current in you would be sufficientto ruin your day.

Q28.24 Both 120-V and 240-V lines can deliver injurious or lethal shocks, but there is a somewhat bettersafety factor with the lower voltage. To say it a different way, the insulation on a 120-V line can bethinner. On the other hand, a 240-V device carries less current to operate a device with the samepower, so the conductor itself can be thinner. Finally, as we will see in Chapter 33, the last step-down transformer can also be somewhat smaller if it has to go down only to 240 volts from the highvoltage of the main power line.

Q28.25 As Luigi Galvani showed with his experiment with frogs’ legs, muscles contract when electriccurrent exists in them. If an electrician contacts a “live” wire, the muscles in his hands and fingerswill contract, making his hand clench. If he touches the wire with the front of his hand, his hand willclench around the wire, and he may not be able to let go. Also, the back of his hand may be drierthan his palm, so an actual shock may be much weaker.

Q28.26 Grab an insulator, like a stick or baseball bat, and bat for a home run. Hit the wire away from theperson or hit them away from the wire. If you grab the person, you will learn very quickly aboutelectrical circuits by becoming part of one.

Q28.27 A high voltage can lead to a high current when placed in a circuit. A device cannot supply a highcurrent—or any current—unless connected to a load. A more accurate sign saying potentially highcurrent would just confuse the poor physics student who already has problems distinguishingbetween electrical potential and current.

Q28.28 The two greatest factors are the potential difference between the wire and your feet, and theconductivity of the kite string. This is why Ben Franklin’s experiment with lightning and flying akite was so dangerous. Several scientists died trying to reproduce Franklin’s results.

Q28.29 Suppose ε = 12 V and each lamp has R = 2 Ω. Before the switch is closed the current is 12

2 V

6 A

Ω= .

The potential difference across each lamp is 2 2 4 A Va fa fΩ = . The power of each lamp is2 4 8 A V Wa fa f = , totaling 24 W for the circuit. Closing the switch makes the switch and the wires

connected to it a zero-resistance branch. All of the current through A and B will go through theswitch and (b) lamp C goes out, with zero voltage across it. With less total resistance, the (c) current

in the battery 12

3 V

4 A

Ω= becomes larger than before and (a) lamps A and B get brighter. (d) The

voltage across each of A and B is 3 2 6 A Va fa fΩ = , larger than before. Each converts power3 6 18 A V Wa fa f = , totaling 36 W, which is (e) an increase.

Q28.30 The starter motor draws a significant amount of current from the battery while it is starting the car.This, coupled with the internal resistance of the battery, decreases the output voltage of the batterybelow its the nominal 12 V emf. Then the current in the headlights decreases.

Chapter 28 133

Q28.31 Two runs in series: . Three runs in parallel: . Junction of one lift and

two runs: .

Gustav Robert Kirchhoff, Professor of Physics at Heidelberg and Berlin, was master of theobvious. A junction rule: The number of skiers coming into any junction must be equal to thenumber of skiers leaving. A loop rule: the total change in altitude must be zero for any skiercompleting a closed path.

SOLUTIONS TO PROBLEMS

Section 28.1 Electromotive Force

P28.1 (a) P =∆V

Ra f2

becomes 20 011 6 2

..

W V

=a f

R

so R = 6 73. Ω .

(b) ∆V IR=

so 11 6 6 73. . V = I Ωa fand I = 1 72. A

ε = +IR Ir

so 15 0 11 6 1 72. . . V V A= + a frr = 1 97. Ω .

FIG. P28.1

P28.2 (a) ∆V IRterm =

becomes 10 0 5 60. . V = I Ωa fso I = 1 79. A .

(b) ∆V Irterm = −ε

becomes 10 0 1 79 0 200. . . V A = −ε a fa fΩso ε = 10 4. V .

P28.3 The total resistance is R = =3 00

5 00.

. V

0.600 A Ω .

(a) R R rlamp batteries = − = − =5 00 0 408 4 59. . .Ω Ω Ω

(b)PPbatteries

total

= = =

0 408

5 000 081 6 8 16%

2

2

.

.. .

Ω

Ω

a fa f

I

I FIG. P28.3

134 Direct Current Circuits

P28.4 (a) Here ε = +I R ra f , so IR r

=+

=+

=ε 12 6

0 080 02 48

..

. V

5.00 A

Ω Ωb g .

Then, ∆ ΩV IR= = =2 48 5 00 12 4. . . A Va fa f .

(b) Let I1 and I2 be the currents flowing through the battery and theheadlights, respectively.Then, I I1 2 35 0= + . A , and ε − − =I r I r1 2 0

so ε = + + =I I2 235 0 0 080 0 5 00 12 6. . . . A Vb gb g a fΩ Ω

giving I2 1 93= . A .

Thus, ∆ ΩV2 1 93 5 00 9 65= =. . . A Va fa f .

FIG. P28.4

Section 28.2 Resistors in Series and Parallel

P28.5 ∆V I R R= =1 1 12 00. Aa f and ∆ ΩV I R R R= + = +2 1 2 11 60 3 00b g a fb g. . A

Therefore, 2 00 1 60 3 001 1. . . A A a f a fb gR R= + Ω or R1 12 0= . Ω .

P28.6 (a) Rp = +=

11 7 00 1 10 0

4 12. .

.

Ω Ω

Ωb g b gR R R Rs = + + = + + =1 2 3 4 00 4 12 9 00 17 1. . . . Ω

(b) ∆V IR=34 0 17 1. . V = I Ωa fI = 1 99. A for 4 00. Ω , 9 00. Ω resistors.

Applying ∆V IR= , 1 99 4 12 8 18. . . A Va fa fΩ =

8 18 7 00. . V = I Ωa fso I = 1 17. A for 7 00. Ω resistor

8 18 10 0. . V = I Ωa fso I = 0 818. A for 10 0. Ω resistor.

FIG. P28.6

P28.7 For the bulb in use as intended,

IV

= = =P∆

75 00 625

..

W120 V

A

and RVI

= = =∆

Ω120

192 V

0.625 A .

Now, presuming the bulb resistance is unchanged,

I = =120

0 620 V

193.6 A

Ω. .

Across the bulb is ∆ ΩV IR= = =192 119 0.620 A Va fso its power is P = = =I V∆ 0 620 73 8. . A 119 V Wa f .

FIG. P28.7

Chapter 28 135

P28.8 1201 2 3 4

V eq= = + + +FHG

IKJIR I

A A A Aρ ρ ρ ρ

, or IA A A A

ρ =+ + +

1201 1 1 1

1 2 3 4

Va fe j

∆VIA A A A A A

22 2

1 1 1 1

12029 5

1 2 3 4

= =+ + +

=ρ V

Va f

e j.

P28.9 If we turn the given diagram on its side, we find that it is the same as figure(a). The 20 0. Ω and 5 00. Ω resistors are in series, so the first reduction isshown in (b). In addition, since the 10 0. Ω , 5 00. Ω , and 25 0. Ω resistors arethen in parallel, we can solve for their equivalent resistance as:

Req

=+ +

=1

2 941

10 01

5 001

25 0. . .

.Ω Ω Ω

Ωc h .

This is shown in figure (c), which in turn reduces to the circuit shown infigure (d).

Next, we work backwards through the diagrams applying IV

R=∆

and

∆V IR= alternately to every resistor, real and equivalent. The 12 94. Ωresistor is connected across 25.0 V, so the current through the battery inevery diagram is

IV

R= = =∆

Ω25 0

1 93.

. V

12.94 A .

In figure (c), this 1.93 A goes through the 2 94. Ω equivalent resistor to give apotential difference of:

∆ ΩV IR= = =1 93 2 94 5 68. . . A Va fa f .

From figure (b), we see that this potential difference is the same across ∆Vab ,the 10 Ω resistor, and the 5 00. Ω resistor.

(b) Therefore, ∆Vab = 5 68. V .

(a) Since the current through the 20 0. Ω resistor is also the currentthrough the 25 0. Ω line ab,

IV

Rab

ab= = = =∆

Ω5 68

0 227 227.

. V

25.0 A mA . FIG. P28.9

*P28.10 We assume that the metal wand makes low-resistance contact with the person’s hand and that theresistance through the person’s body is negligible compared to the resistance Rshoes of the shoe soles.The equivalent resistance seen by the power supply is 1 00. M shoesΩ+ R . The current through both

resistors is 50.0 V

1.00 M shoesΩ+ R. The voltmeter displays

∆ ΩΩ

Ω∆V I

RV= =

+=1 00

50 01 00

..

. M

V 1.00 M M shoes

a f a f.

(a) We solve to obtain 50 0 1 00. . V 1.00 M M shoesΩ ∆ Ω ∆a f a f b g= +V V R

RV

Vshoes M 50.0

=−1 00. Ω ∆

∆a f

.

(b) With Rshoes → 0 , the current through the person’s body is50 0

50 0.

. V

1.00 M A

Ω= µ The current will never exceed 50 Aµ .

136 Direct Current Circuits

P28.11 (a) Since all the current in the circuit must pass through the series100 Ω resistor, P = I R2

Pmax max= RI 2

so IRmax

..= = =

P 25 00 500

W100

AΩ

R

V R I

eq

eq

= + +FHG

IKJ =

= =

−

1001

1001

100150

75 0

1

V

Ω Ω Ω

∆ max max .

(b) P = = =I V∆ 0 500 75 0 37 5. . . A V Wa fa f total power

P

P P

1

2 32 2

25 0

100 0 250 6 25

=

= = =

.

. .

W

A WRI Ωa fa f

FIG. P28.11

P28.12 Using 2.00-Ω, 3.00-Ω, 4.00-Ω resistors, there are 7 series, 4 parallel, and 6 mixed combinations:Series2.00 Ω 6.00 Ω3.00 Ω 7.00 Ω4.00 Ω 9.00 Ω5.00 Ω

Parallel Mixed0.923 Ω 1.56 Ω1.20 Ω 2.00 Ω1.33 Ω 2.22 Ω1.71 Ω 3.71 Ω

4.33 Ω5.20 Ω

The resistors may be arranged in patterns:

P28.13 The potential difference is the same across either combination.

∆V IR IR

= =+

31

1 1500c h so R

R1 1

5003+FHGIKJ =

1500

3+ =R

and R = =1 000 1 00 kΩ Ω. .

FIG. P28.13

*P28.14 When S is open, R1 , R2 , R3 are in series with the battery. Thus:

R R R1 2 3 36

106+ + = =−

V A

kΩ . (1)

When S is closed in position 1, the parallel combination of the two R2 ’s is in series with R1 , R3 , and thebattery. Thus:

R R R1 2 3 312

61 2 10

5+ + =×

=− V

A k

.Ω . (2)

When S is closed in position 2, R1 and R2 are in series with the battery. R3 is shorted. Thus:

R R1 2 36

2 103+ =

×=−

V A

kΩ . (3)

From (1) and (3): R3 3= kΩ .

Subtract (2) from (1): R2 2= kΩ .

From (3): R1 1= kΩ .

Answers: R R R1 2 31 00 2 00 3 00= = =. . . k , k , kΩ Ω Ω .

Chapter 28 137

P28.15 R

R

IV

R

p

s

s

= +FHG

IKJ =

= + + =

= = =

−13 00

11 00

0 750

2 00 0 750 4 00 6 7518 0

2 67

1

. ..

. . . ..

.

V

6.75 Abattery

Ω

Ω Ω∆

Ω

a f

P = I R2 : P222 67 2 00= . . A a f a fΩ

P2 14 2= . W in 2.00 Ω

P

P

P

42

2

4

2 4 3 1

33

2

3

2

11

1

2

2 67 4 00 28 4

2 67 2 00 5 33

2 67 4 00 10 67

18 0 2 00

2 003 00

1 33

2 001 00

4 00

= =

= =

= =

= − − = = =

= = =

= = =

. . .

. . .

. . .

. .

..

.

..

.

A A W in 4.00

A V,

A V

V V

V

W in 3.00

V

W in 1.00

a f a fa fa fa fa f

b gb g a f

b g a f

Ω

∆ Ω

∆ Ω

∆ ∆ ∆ ∆ ∆

∆

ΩΩ

∆

ΩΩ

V

V

V V V V V

VR

VR

p

FIG. P28.15

P28.16 Denoting the two resistors as x and y,

x y+ = 690, and 1

1501 1

= +x y

1150

1 1690

690690

690 103 500 0

690 690 414 000

2470 220

2

2

= +−

=− +−

− + =

=± −

= =

x xx x

x x

x x

x

x y

a fa f

a f

Ω Ω

*P28.17 A certain quantity of energy ∆Eint time= P a f is required to raise the temperature of the water to

100°C. For the power delivered to the heaters we have P = =I VVR

∆∆a f2

where ∆Va f is a constant.

Thus comparing coils 1 and 2, we have for the energy ∆ ∆ ∆ ∆V t

RV t

Ra f a f2

1

2

2

2= . Then R R2 12= .

(a) When connected in parallel, the coils present equivalent resistance

RR R R R

Rp = +

=+

=1

1 11

1 1 22

31 2 1 1

1 . Now ∆ ∆ ∆ ∆V t

R

V t

Rpa f a f2

1

2

12 3= ∆

∆t

tp =

23

.

(b) For the series connection, R R R R R Rs = + = + =1 2 1 1 12 3 and ∆ ∆ ∆ ∆V t

RV t

Rsa f a f2

1

2

13=

∆ ∆t ts = 3 .

138 Direct Current Circuits

P28.18 (a) ∆V IR= : 33 0 11 01. . V = I Ωa f 33 0 22 02. . V = I Ωa fI3 3 00= . A I2 1 50= . A

P = I R2 : P123 00 11 0= . . A a f a fΩ P2

21 50 22 0= . . A a f a fΩP1 99 0= . W P2 49 5= . W

The 11.0- resistor uses more power.Ω

(b) P P1 2 148+ = W P = = =I V∆a f a fa f4 50 33 0 148. . W

FIG. P28.18(a)

(c) R R Rs = + = + =1 2 11 0 22 0 33 0. . . Ω Ω Ω

∆V IR= : 33 0 33 0. . V = I Ωa f , so I = 1 00. A

P = I R2 : P121 00 11 0= . . A a f a fΩ P2

21 00 22 0= . . A a f a fΩP1 11 0= . W P2 22 0= . W

The 22.0- resistor uses more power.Ω

FIG. P28.18(c)

(d) P P1 22

1 221 00 33 0 33 0+ = + = =I R Rb g a f a f. . . A WΩ

P = = =I V∆a f a fa f1 00 33 0 33 0. . . A V W

(e) The parallel configuration uses more power.

*P28.19 (a) The resistors 2, 3, and 4 can be combined to a single 2R resistor. This is in series withresistor 1, with resistance R, so the equivalent resistance of the whole circuit is 3R. In series,

potential difference is shared in proportion to the resistance, so resistor 1 gets 13

of the

battery voltage and the 2-3-4 parallel combination get 23

of the battery voltage. This is the

potential difference across resistor 4, but resistors 2 and 3 must share this voltage. 13

goes to

2 and 23

to 3. The ranking by potential difference is ∆ ∆ ∆ ∆V V V V4 3 1 2> > > .

(b) Based on the reasoning above the potential differences are

∆ ∆ ∆ ∆V V V V1 2 3 4329

49

23

= = = =ε ε ε ε

, , , .

(c) All the current goes through resistor 1, so it gets the most. The current then splits at theparallel combination. Resistor 4 gets more than half, because the resistance in that branch isless than in the other branch. Resistors 2 and 3 have equal currents because they are inseries. The ranking by current is I I I I1 4 2 3> > = .

(d) Resistor 1 has a current of I. Because the resistance of 2 and 3 in series is twice that ofresistor 4, twice as much current goes through 4 as through 2 and 3. The current through

the resistors are I I I II

II

1 2 3 4323

= = = =, , .

continued on next page

Chapter 28 139

(e) Increasing resistor 3 increases the equivalent resistance of the entire circuit. The current inthe circuit, which is the current through resistor 1, decreases. This decreases the potentialdifference across resistor 1, increasing the potential difference across the parallelcombination. With a larger potential difference the current through resistor 4 is increased.With more current through 4, and less in the circuit to start with, the current throughresistors 2 and 3 must decrease. To summarize, I I I I4 1 2 3 increases and and decrease, , .

(f) If resistor 3 has an infinite resistance it blocks any current from passing through that branch,and the circuit effectively is just resistor 1 and resistor 4 in series with the battery. The circuit

now has an equivalent resistance of 4R. The current in the circuit drops to 34

of the original

current because the resistance has increased by 43

. All this current passes through resistors 1

and 4, and none passes through 2 or 3. Therefore II

I I II

1 2 3 434

034

= = = =, , .

Section 28.3 Kirchhoff’s Rules

P28.20 + − − =15 0 7 00 2 00 5 00 01. . . .a f a fa fI

5 00 7 00 1. .= I so I1 0 714= . A

I I I3 1 2 2 00= + = . A

0 714 2 002. .+ =I so I2 1 29= . A

+ − − =ε 2 00 1 29 5 00 2 00 0. . . .a f a f ε = 12 6. VFIG. P28.20

P28.21 We name currents I1 , I2 , and I3 as shown.

From Kirchhoff’s current rule, I I I3 1 2= + .

Applying Kirchhoff’s voltage rule to the loop containing I2 and I3 ,

12 0 4 00 6 00 4 00 0

8 00 4 00 6 003 2

3 2

. . . .

. . .

V V− − − =

= +

a f a fa f a f

I I

I I

Applying Kirchhoff’s voltage rule to the loop containing I1 and I2 ,

− − + =6 00 4 00 8 00 02 1. . .a f a fI I V 8 00 4 00 6 001 2. . .a f a fI I= + .FIG. P28.21

Solving the above linear system, we proceed to the pair of simultaneous equations:

8 4 4 68 4 6

1 2 2

1 2

= + += +

RSTI I I

I Ior

8 4 101 33 0 667

1 2

2 1

= += −

RSTI I

I I. .

and to the single equation 8 4 13 3 6 671 1= + −I I. .

I114 7

0 846= =.

. V

17.3 A

Ω. Then I2 1 33 0 846 0 667= −. . . Aa f

and I I I3 1 2= + give I I I1 2 3846 462 1 31= = = mA, mA, A. .

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

140 Direct Current Circuits

P28.22 The solution figure is shown to the right.

FIG. P28.22

P28.23 We use the results of Problem 28.21.

(a) By the 4.00-V battery: ∆ ∆ ∆U V I t= = − = −a f a fa f4 00 0 462 120 222. . V A s J .

By the 12.0-V battery: 12 0 1 31 120 1 88. . . V A s kJa fa f = .

(b) By the 8.00-Ω resistor: I R t2 20 846 8 00 120 687∆ Ω= =. . A s Ja f a f .

By the 5.00-Ω resistor: 0 462 5 00 120 1282. . A s Ja f a fΩ = .

By the 1.00-Ω resistor: 0 462 1 00 120 25 62. . . A s Ja f a fΩ = .

By the 3.00-Ω resistor: 1 31 3 00 120 6162. . A s Ja f a fΩ = .

By the 1.00-Ω resistor: 1 31 1 00 120 2052. . A s Ja f a fΩ = .

(c) − + =222 1 88 1 66 J kJ kJ. . from chemical to electrical.

687 128 25 6 616 205 1 66 J J J J J kJ+ + + + =. . from electrical to internal.

P28.24 We name the currents I1 , I2 , and I3 as shown.

[1] 70 0 60 0 3 00 2 00 02 1. . . .− − − =I I k kΩ Ωa f a f[2] 80 0 4 00 60 0 3 00 03 2. . . .− − − =I I k kΩ Ωa f a f[3] I I I2 1 3= +

(a) Substituting for I2 and solving the resulting simultaneousequations yields

I R

I R

I R

1 1

3 3

2 2

0 385

2 69

3 08

=

=

=

.

.

.

mA through

mA through

mA through

b gb gb g

(b) ∆ ΩVcf = − − = −60 0 3 08 3 00 69 2. . . . V mA k Va fa fPoint is at higher potential.c

FIG. P28.24

Chapter 28 141

P28.25 Label the currents in the branches as shown in the first figure.Reduce the circuit by combining the two parallel resistors as shownin the second figure.

Apply Kirchhoff’s loop rule to both loops in Figure (b) to obtain:

2 71 1 71 2501 2. .R I R Ia f a f+ =

and 1 71 3 71 5001 2. .R I R Ia f a f+ = .

With R = 1 000 Ω , simultaneous solution of these equations yields:

I1 10 0= . mA

and I2 130 0= . mA.

From Figure (b), V V I I Rc a− = + =1 2 1 71 240b ga f. V .

Thus, from Figure (a), IV V

Rc a

4 4240

60 0=−

= = V

4 000 mA

Ω. .

Finally, applying Kirchhoff’s point rule at point a in Figure (a)gives:

I I I= − = − = +4 1 60 0 10 0 50 0. . . mA mA mA,

or I a e= 50 0. mA from point to point .

(a)

(b)

FIG. P28.25

P28.26 Name the currents as shown in the figure to the right. Then w x z y+ + = . Loopequations are

− − + =− + + − =+ − − + =

200 40 0 80 0 080 0 40 0 360 20 0 0360 20 0 70 0 80 0 0

w xx y

y z

. .. . .

. . .

FIG. P28.26

Eliminate y by substitution.x w

x w zw x z

= +− − − =− − − =

RS|T|

2 50 0 500400 100 20 0 20 0 0440 20 0 20 0 90 0 0

. .. .

. . .

Eliminate x .350 270 20 0 0430 70 0 90 0 0

− − =− − =

RSTw zw z

.. .

Eliminate z w= −17 5 13 5. . to obtain 430 70 0 1 575 1 215 0− − + =. w w

w = =70 070 0

1 00..

. A upward in 200 Ω .

Now z = 4 00. A upward in 70.0 Ω

x = 3 00. A upward in 80.0 Ω

y = 8 00. A downward in 20.0 Ω

and for the 200 Ω, ∆ ΩV IR= = =1 00 200 200. A Va fa f .

142 Direct Current Circuits

P28.27 Using Kirchhoff’s rules,

12 0 0 010 0 0 060 0 0

10 0 1 00 0 060 01 3

2 3

. . .

. . .

− − =

+ − =

b g b ga f a f

I I

I I 0

and I I I1 2 3= +

12 0 0 010 0 070 0

10 0 1 00 0 060 0 02 3

2 3

. . .

. . .

− − =

+ − =

0 0a f a fa f b g

I I

I I FIG. P28.27

Solving simultaneously,

I2 0 283= . A downward in the dead battery

and I3 171= A downward in the starter.

The currents are forward in the live battery and in the starter, relative to normal starting operation.The current is backward in the dead battery, tending to charge it up.

P28.28 ∆

∆

∆

V I I I

V I I I I I

V I I I I I

ab

ab

ab

= + −

= + + − +

= − + − +

1 00 1 00

1 00 1 00 5 00

3 00 5 00

1 1 2

1 2 1 2

1 1 2

. .

. . .

. .

a f a fb ga f a f a fb ga fb g a fb g

Let I = 1 00. A , I x1 = , and I y2 = .

Then, the three equations become:

∆V x yab = −2 00. , or y x Vab= −2 00. ∆

∆V x yab = − + +4 00 6 00 5 00. . .

and ∆V x yab = − +8 00 8 00 5 00. . . .

Substituting the first into the last two gives:

7 00 8 00 5 00. . .∆V xab = + and 6 00 2 00 8 00. . .∆V xab = + .

Solving these simultaneously yields ∆Vab =2717

V .

Then, RVIab

ab= =∆ 27

17 V1.00 A

or Rab =2717

Ω .

FIG. P28.28

P28.29 We name the currents I1 , I2 , and I3 as shown.

(a) I I I1 2 3= +

Counterclockwise around the top loop,

12 0 2 00 4 00 03 1. . . V − − =Ω Ωa f a fI I .

Traversing the bottom loop,

8 00 6 00 2 00 02 3. . . V − + =Ω Ωa f a fI I

I I1 33 0012

= −. , I I2 343

13

= + , and I3 909= mA .

(b) V Va b− =0 909 2 00. . A a fa fΩV Vb a− = −1 82. V

FIG. P28.29

Chapter 28 143

P28.30 We apply Kirchhoff’s rules to the second diagram.

50 0 2 00 2 00 01 2. . .− − =I I (1)

20 0 2 00 2 00 03 2. . .− + =I I (2)

I I I1 2 3= + (3)

Substitute (3) into (1), and solve for I1 , I2 , and I3

I1 20 0= . A ; I2 5 00= . A ; I3 15 0= . A.

Then apply P = I R2 to each resistor:

2 00 1. Ωa f : P = = =I12 22 00 20 0 2 00 800. . . A WΩ Ωa f a f a f

4 00. Ωa f : P = FHGIKJ =

5 002

4 00 25 02.

. . A WΩa f(Half of I2 goes through each)

2 00 3. Ωa f : P = = =I32 22 00 15 0 2 00 450. . . A WΩ Ωa f a f a f .

FIG. P28.30

Section 28.4 RC Circuits

P28.31 (a) RC = × × =−1 00 10 5 00 10 5 006 6. . . F sΩe je j

(b) Q C= = × =−ε µ5 00 10 30 0 1506. . C V Ce ja f

(c) I tR

e t RCa fe je j

= =×

FHG

IKJ

−

× ×

L

NMM

O

QPP =

−−

εµ

30 0 10 0

1 00 10 5 00 104 06

6 6

.exp

.

. ..

1.00 10 A6

FIG. P28.31

P28.32 (a) I t I e t RCa f = − −0

IQ

RC

I t

0

6

9

6

9

5 10 10

2 00 101 96

1 969 00 10

2 00 1061 6

= =×

×=

= −− ×

×

L

NMM

O

QPP = −

−

−

−

−

.

..

. exp.

..

C

1 300 F A

A s

1 300 F mA

Ω

Ω

b ge j

a f a f b ge j

(b) q t Qe t RCa f b g a fe j= =

− ×

×

L

NMM

O

QPP =

−−

−5 10

8 00 10

2 00 100 235

6

9. exp

.

.. C

s

1 300 F Cµ µ

Ω

(c) The magnitude of the maximum current is I0 1 96= . A .

P28.33 U C V=12

2∆a f and ∆VQC

= .

Therefore, UQ

C=

2

2 and when the charge decreases to half its original value, the stored energy is one-

quarter its original value: U Uf =14 0 .

144 Direct Current Circuits

P28.34 q t Q e t RCa f = − −1 soq tQ

e t RCa f= − −1

0 600 1 0 900. .= − −e RC or e RC− = − =0 900 1 0 600 0 400. . .

−=

0 9000 400

.ln .

RCa f thus RC =

−=

0 9000 400

0 982.

ln ..a f s .

*P28.35 We are to calculate

e dtRC

edt

RCRC

eRC

e eRC RCt RC t RC t RC−

∞−

∞− ∞ −∞z z= − −FHG

IKJ = − = − − = − − = +2

0

2

0

20

0

22

2 2 20 1

2.

P28.36 (a) τ = = × × =−RC 1 50 10 10 0 10 1 505 6. . . F sΩe je j

(b) τ = × × =−1 00 10 10 0 10 1 005 6. . . F sΩe je j

(c) The battery carries current10 0

200. V

50.0 10 A3×

=Ω

µ .

The 100 kΩ carries current of magnitude I I e et RC t= =×

FHG

IKJ

− −0

1.0010 0. V100 10 3

s

Ω.

So the switch carries downward current 200 100 1.00 A A sµ µ+ −b ge t .

P28.37 (a) Call the potential at the left junction VL and at the right VR . After a“long” time, the capacitor is fully charged.

VL = 8 00. V because of voltage divider:

I

V

L

L

= =

= − =

10 02 00

10 0 2 00 1 00 8 00

..

. . . .

V5.00

A

V A VΩ

Ωa fa fLikewise, VR =

+FHG

IKJ =

2 008 00

10 0 2 00.

.. .

2.00

V VΩ

Ω Ωa f

or IR = =10 0

1 00.

. V

10.0 A

Ω

VR = − =10 0 8 00 1 00 2 00. . . . V A Va f a fa fΩ .

Therefore, ∆V V VL R= − = − =8 00 2 00 6 00. . . V .

FIG. P28.37(a)

(b) Redraw the circuit R =+

=1

1 9 00 1 6 003 60

. ..

Ω ΩΩb g b g

RC = × −3 60 10 6. s

and e t RC− =1

10

so t RC= =ln .10 8 29 sµ . FIG. P28.37(b)

Chapter 28 145

*P28.38 (a) We model the person’s body and street shoesas shown. For the discharge to reach 100 V,

q t Qe C V t C V et RC t RCa f a f= = =− −∆ ∆ 0

∆∆

VV

e t RC

0= − ∆

∆VV

e t RC0 = + tRC

VV

= FHGIKJln

∆∆

0

150 pF 80 pF 5 000 MΩ

3 000 V

FIG. P28.38(a)

t RCVV

= FHGIKJ = × × F

HGIKJ =

−ln ln .∆∆

Ω0 6 125 000 10 230 103 000100

3 91 F se j

(b) t = × × =−1 10 230 10 30 7826 12 V A C V se j ln µ

P28.39 (a) τ = = × × =−RC 4 00 10 3 00 10 12 06 6. . . F sΩe je j

(b) IR

e et RC t= =×

− −ε 12 04 00 106

12 0..

. s

q C e e

q e I e

t RC t

t t

= − = × −

= − =

− − −

− −

ε

µ µ

1 3 00 10 12 0 1

36 0 1 3 00

6 12 0

12 0 12 0

. .

. .

.

. .

a f C A

FIG. P28.39

P28.40 ∆VQC0 =

Then, if q t Q e t RC( ) = − ∆ ∆V t V e t RC( ) = −0b g

and ∆∆V tV

e t RC( )

0b g =− .

When ∆ ∆V t V( ) =12 0b g, then e t RC− =

12

− = FHGIKJ = −

tRC

ln ln12

2.

Thus, Rt

C=

ln 2a f .

Section 28.5 Electrical Meters

P28.41 ∆V I r I I Rg g g p= = −e j , or RI r

I I

I

I Ip

g g

g

g

g

=−

=−e ja fe j

60 0. Ω

Therefore, to have I = =0 100 100. A mA when I g = 0 500. mA :

Rp = =0 500 60 0

99 50 302

. ..

. mA

mA

a fa fΩΩ .

FIG. P28.41

146 Direct Current Circuits

P28.42 Applying Kirchhoff’s loop rule, − + − =I I I Rg g p75 0 0. Ωa f e j .

Therefore, if I = 1 00. A when I g = 1 50. mA ,

RI

I Ip

g

g

=−

=×

− ×=

−

−

75 0 1 50 10 75 0

1 00 1 50 100 113

3

3

. . .

. ..

A

A A

Ω ΩΩ

a fe j

e ja f.

FIG. P28.42

P28.43 Series Resistor → Voltmeter

∆V IR= : 25 0 1 50 10 75 03. . .= × +− Rsb gSolving, Rs = 16 6. kΩ .

FIG. P28.43

P28.44 (a) In Figure (a), theemf sees anequivalentresistance of200 00. Ω .

I =

=

6 000 0200 000 030 000

..

.

V

AΩ

6.0000 V

20.000 Ω

180.00 Ω

180.00 Ω 180.00 Ω

AV AV

(a) (b) (c)

20.000 Ω 20.000 Ω

FIG. P28.44

The terminal potential difference is ∆ ΩV IR= = =0 030 000 180 00 5 400 0. . . A Vb ga f .

(b) In Figure (b), Req = +FHG

IKJ =−

1180 00

120 000

178 391

..

Ω ΩΩ .

The equivalent resistance across the emf is 178 39 0 500 00 20 000 198 89. . . . Ω Ω Ω Ω+ + = .

The ammeter reads IR

= = =ε 6 000 0

0 030 167.

. V

198.89 A

Ω

and the voltmeter reads ∆ ΩV IR= = =0 030 167 178 39 5 381 6. . . A Vb ga f .

(c) In Figure (c),1

180 501

20 000178 89

1

..

Ω ΩΩ+

FHG

IKJ =−

.

Therefore, the emf sends current through Rtot = + =178 89 20 000 198 89. . .Ω Ω Ω .

The current through the battery is I = =6 000 0

0 030 168.

. V

198.89 A

Ω

but not all of this goes through the ammeter.

The voltmeter reads ∆ ΩV IR= = =0 030 168 178 89 5 396 6. . . A Vb ga f .

The ammeter measures current IV

R= = =∆

Ω5 396 6

0 029 898.

. V

180.50 A .

The connection shown in Figure (c) is better than that shown in Figure (b) for accuratereadings.

Chapter 28 147

P28.45 Consider the circuit diagram shown, realizing thatI g = 1 00. mA . For the 25.0 mA scale:

24 0 1 00 25 01 2 3. . . mA mA a fb g a fa fR R R+ + = Ω

FIG. P28.45

or R R R1 2 325 024 0

+ + = FHGIKJ

.

. Ω . (1)

For the 50.0 mA scale: 49 0 1 00 25 01 2 3. . . mA mA a fb g a fb gR R R+ = +Ω

or 49 0 25 01 2 3. .R R R+ = +b g Ω . (2)

For the 100 mA scale: 99 0 1 00 25 01 2 3. . . mA mA a f a fb gR R R= + +Ω

or 99 0 25 01 2 3. .R R R= + + Ω . (3)

Solving (1), (2), and (3) simultaneously yields

R R R1 2 30 260 0 261 0 521= = =. . . , , Ω Ω Ω .

P28.46 ∆V IR=

(a) 20 0 1 00 10 60 031. . . V A = × +−e jb gR Ω

R141 994 10 19 94= × =. . kΩ Ω

FIG. P28.46

(b) 50 0 1 00 10 60 032 1. . . V A = × + +−e jb gR R Ω R2 30 0= . kΩ

(c) 100 1 00 10 60 033 1 V A = × + +−. .e jb gR R Ω R3 50 0= . kΩ

P28.47 Ammeter: I r Ig g= −0 500 0 220. . A e ja fΩor I rg + =0 220 0 110. . VΩa f (1)

Voltmeter: 2 00 2 500. V = +I rg Ωb g (2)

Solve (1) and (2) simultaneously to find:

I g = 0 756. mA and r = 145 Ω .

FIG. P28.47

Section 28.6 Household Wiring and Electrical Safety

P28.48 (a) P = = FHGIKJ =

× ⋅

×=

−

−I R I

A2 2

2 8

3 2

1 00 1 70 10 16 0 0 304 8

0 512 100 101

ρ

π

. . . .

..

A m ft m ft

m W

a f e ja fb ge j

Ω

(b) P = = =I R2 100 0 101 10 1. . WΩa f

148 Direct Current Circuits

P28.49 (a) P = I V∆ : So for the Heater, IV

= = =P∆

1 50012 5

W120 V

A. .

For the Toaster, I = =750

6 25 W

120 V A. .

And for the Grill, I = =1 000

8 33 W

120 V A. .

(b) 12 5 6 25 8 33 27 1. . . .+ + = A

The current draw is greater than 25.0 amps, so this circuit breaker would not be sufficient.

P28.50 I R I RAl2

Al Cu2

Cu= so IRR

I IAlCu

AlCu

Cu

AlCu A= = = = =

ρρ

1 702 82

20 0 0 776 20 0 15 5..

. . . .a f a f

P28.51 (a) Suppose that the insulation between either of your fingers and the conductor adjacent is achunk of rubber with contact area 4 mm2 and thickness 1 mm. Its resistance is

RA

= ≈⋅

×≈ ×

−

−ρ 10 10

4 102 10

13 3

615

m m

m 2

ΩΩ

e je j.

The current will be driven by 120 V through total resistance (series)

2 10 10 2 10 5 1015 4 15 15× + + × ≈ × Ω Ω Ω Ω .

It is: IV

R=

×−∆

Ω~ ~

12010 14 V

5 10 A15 .

(b) The resistors form a voltage divider, with the center of your hand at potential Vh

2, where Vh

is the potential of the “hot” wire. The potential difference between your finger and thumb is∆ ΩV IR= − −~ ~10 10 1014 4 10 A Ve je j . So the points where the rubber meets your fingers are

at potentials of

~Vh

210 10+ − V and ~

Vh

210 10− − V .

Additional Problems

P28.52 The set of four batteries boosts the electric potential of each bit of charge that goes through them by4 1 50 6 00× =. . V V . The chemical energy they store is

∆ ∆U q V= = =240 6 00 1 440 C J C Ja fb g. .

The radio draws current IV

R= = =∆

Ω6 00

0 030 0.

. V

200 A .

So, its power is P = = = =∆V Ia f a fb g6 00 0 030 0 0 180 0 180. . . . V A W J s.

Then for the time the energy lasts, we have P =E

t∆: ∆ t

E= = = ×P

1 4408 00 103 J

0.180 J s s. .

We could also compute this from IQ

t=∆

: ∆ tQI

= = = × =240

8 00 10 2 223 C0.030 0 A

s h. . .

Chapter 28 149

P28.53 IR r

=+ε

, so P = =+

I RR

R r2

2

2εa f or R r R+ =

FHGIKJa f2

2εP

.

Let x ≡ε 2

P, then R r xR+ =a f2 or R r x R r2 22 0+ − − =a f .

With r = 1 20. Ω , this becomes R x R2 2 40 1 44 0+ − − =. .a f ,

which has solutions of Rx x

=− − ± − −2 40 2 40 5 76

2

2. . .a f a f.

(a) With ε = 9 20. V and P = 12 8. W, x = 6 61. :

R =+ ± −

=4 21 4 21 5 76

23 84

2. . ..

a f Ω or 0 375. Ω .

(b) For ε = 9 20. V and P = 21 2. W , x ≡ =ε 2

3 99P

.

R =+ ± −

=± −1 59 1 59 5 76

21 59 3 22

2

2. . . . .a f.

The equation for the load resistance yields a complex number, so there is no resistance

that will extract 21.2 W from this battery. The maximum power output occurs when

R r= = 1 20. Ω , and that maximum is: Pmax .= =ε 2

417 6

r W .

P28.54 Using Kirchhoff’s loop rule for the closed loop, + − − =12 0 2 00 4 00 0. . .I I , so I = 2 00. A

V Vb a− = + − − = −4 00 2 00 4 00 0 10 0 4 00. . . . . V A Va fa f a fa fΩ Ω .

Thus, ∆Vab = 4 00. V and point is at the higher potentiala .

P28.55 (a) R Req = 3 IR

=ε

3Pseries = =ε

εI

R

2

3

(b) RR R R

Req =

+ +=

11 1 1 3b g b g b g I

R=

3εPparallel = =ε

εI

R3 2

(c) Nine times more power is converted in the parallel connection.

*P28.56 (a) We model the generator as a constant-voltage power supply.Connect two light bulbs across it in series. Each bulb is designed to

carry current IV

= = =P∆

1000 833

W120 V

A. . Each has resistance

RVI

= = =∆

Ω120

144 V

0.833 A . In the 240-V circuit the equivalent

resistance is 144 144 288 Ω Ω Ω+ = . The current is

IV

R= = =∆

Ω240

0 833 V

288 A. and the generator delivers power

P = = =I V∆ 0 833 240 200. A V Wa f .

FIG. P28.56(a)

continued on next page

150 Direct Current Circuits

(b) The hot pot is designed to carrycurrent

IV

= = =P∆

5004 17

W120 V

A. .

It has resistance

RVI

= = =∆

Ω120

28 8 V

4.17 A . .

28.8 Ω

240 V 144 Ω

FIG. P28.56(b)

In terms of current, since 4 17

5. A

0.833 A= , we can place five light bulbs in parallel and the hot

pot in series with their combination. The current in the generator is then 4 17. A and it

delivers power P = = =I V∆ 4 17 240 1 000. A V Wa f .

P28.57 The current in the simple loop circuit will be IR r

=+ε

.

(a) ∆V IrR

R rter = − =+

εε

and ∆Vter →ε as R→∞ .

(b) IR r

=+ε

and Ir

→ε

as R→ 0 .

(c) P = =+

I RR

R r2 2

2ε a fddR

R

R r R r

P=

−

++

+=

20

2

3

2

2ε εa f a f

Then 2R R r= + and R r= .

FIG. P28.57

P28.58 The potential difference across the capacitor ∆ ∆V t V e t RCa f e j= − −max 1 .

Using 1 Farad = 1 s Ω , 4 00 10 0 13 00 10 0 10 6

. .. .

V V s s

= −LNM

OQP

− × −a f a f e jeR Ω

.

Therefore, 0 400 1 003 00 105

. ..

= −− ×

eR Ωe j .

Or eR− ×=

3 00 105

0 600.

. Ωe j .

Taking the natural logarithm of both sides, −×

=3 00 10

0 6005.

ln . Ω

Ra f

and R = −×

= + × =3 00 10

0 6005 87 10 587

55.

ln ..

k

ΩΩ Ωa f .

Chapter 28 151

P28.59 Let the two resistances be x and y.

Then, R x yIs

s= + = = =P

2 2225

9 00 W

5.00 A a f . Ω y x= −9 00. Ω

and Rxy

x y Ipp

=+

= = =P

2 250 0

2 00.

. W

5.00 A a f Ω

sox x

x x9 009 00

2 00..

.

ΩΩ

Ω−

+ −=

a fa f x x2 9 00 18 0 0− + =. . .

Factoring the second equation, x x− − =6 00 3 00 0. .a fa fso x = 6 00. Ω or x = 3 00. Ω .

Then, y x= −9 00. Ω gives y = 3 00. Ω or y = 6 00. Ω .

The two resistances are found to be 6 00. Ω and 3 00. Ω .

x y

x

y

FIG. P28.59

P28.60 Let the two resistances be x and y.

Then, R x yIs

s= + =P

2 and Rxy

x y Ipp

=+

=P

2 .

From the first equation, yI

xs= −P

2 , and the second

becomes x I x

x I x Is

s

pP

P

P2

2 2

−

+ −=

e je j

or xI

xI

s s p22 4 0− FHGIKJ + =

P P P.

Using the quadratic formula, xI

s s s p=

± −P P P P2

2

4

2.

Then, yI

xs= −P

2 gives yI

s s s p=

−P P P P∓ 2

2

4

2.

The two resistances are P P P Ps s s p

I

+ −2

2

4

2 and

P P P Ps s s p

I

− −2

2

4

2.

x y

x

y

FIG. P28.60

P28.61 (a) ε ε ε− − + =∑I Rc h b g1 2 0

40 0 4 00 2 00 0 300 0 300 6 00 6 00 0. . . . . . . V A V− + + + − + =a f a f a fR Ω ; so R = 4 40. Ω

(b) Inside the supply, P = = =I R2 24 00 2 00 32 0. . . A Wa f a fΩ .

Inside both batteries together, P = = =I R2 24 00 0 600 9 60. . . A Wa f a fΩ .

For the limiting resistor, P = =4 00 4 40 70 42. . . A Wa f a fΩ .

(c) P = + = + =I ε ε1 2 4 00 6 00 6 00 48 0b g a f a f. . . . A V W

152 Direct Current Circuits

*P28.62 (a) ∆ ∆V V I R I R

I I I II RR

IR R

R

IIR

R R

II RR

IRR R

I

1 2 1 1 2 2

1 2 11 1

21

2 1

2

12

1 2

21 1

2

1

1 22

= =

= + = + =+

=+

= =+

=

IR1

R2

I1

I2

FIG. P28.62(a)

(b) The power delivered to the pair is P = + = + −I R I R I R I I R12

1 22

2 12

1 12

2b g . For minimum power

we want to find I1 such that ddIP

10= .

ddI

I R I I RP

11 1 1 22 2 1 0= + − − =b ga f I R IR I R1 1 2 1 2 0− + =

IIR

R R12

1 2=

+

This is the same condition as that found in part (a).

P28.63 Let Rm = measured value, R = actual value,

IR = current through the resistor R

I = current measured by the ammeter.

(a) When using circuit (a), I R V I IR R= = −∆ 20 000b g or RI

IR= −

LNMOQP

20 000 1 .

But since IV

Rm=∆

and IV

RR =∆

, we haveI

IR

RR m=

(a)

(b)

FIG. P28.63

and RR R

Rm

m=

−20 000b g

. (1)

When R Rm> , we requireR R

Rm−

≤b g

0 050 0. .

Therefore, R Rm ≥ −1 0 050 0.b g and from (1) we find R ≤ 1 050 Ω .

(b) When using circuit (b), I R V IR R= −∆ Ω0 5. a f .

But since IV

RRm

=∆

, R Rm = +0 500.a f . (2)

When R Rm > , we requireR R

Rm −

≤b g

0 050 0. .

From (2) we find R ≥ 10 0. Ω .

Chapter 28 153

P28.64 The battery supplies energy at a changing ratedEdt

IR

e RC= = = FHGIKJ

−P ε εε 1 .

Then the total energy put out by the battery is dER

tRC

dtt

z z= −FHGIKJ

=

∞ ε 2

0

exp

dER

RCt

RCdtRC

Ct

RCC Cz z= − −FHG

IKJ −FHGIKJ = − −FHG

IKJ = − − =

∞ ∞ε

ε ε ε2

0

2

0

2 20 1a f exp exp .

The power delivered to the resistor isdEdt

V I I R RR

tRCR= = = = −FHGIKJP ∆ 2

2

22ε

exp .

So the total internal energy appearing in the resistor is dER

tRC

dtz z= −FHGIKJ

∞ ε 2

0

2exp

dER

RC tRC

dtRC

C tRC

C Cz z= −FHGIKJ −FHG

IKJ −FHGIKJ = − −FHG

IKJ = − − =

∞ ∞ε ε ε ε2

0

2

0

2 2

22 2

22

20 1

2exp exp .

The energy finally stored in the capacitor is U C V C= =12

12

2 2∆a f ε . Thus, energy of the circuit is

conserved ε ε ε2 2 212

12

C C C= + and resistor and capacitor share equally in the energy from the

battery.

P28.65 (a) q C V e t RC= − −∆ 1e j

q e= × −LNM

OQP =

− − × × −

1 00 10 10 0 1 9 936 10 0 2 00 10 1.00 106 6

. . .. .

F V Ce ja f e je j µ

(b) Idqdt

VR

e t RC= = FHGIKJ

−∆

I e=×

FHG

IKJ = × =− −10 0

2 00 103 37 10 33 76

5 00 8..

. .. V

A nAΩ

(c)dUdt

ddt

qC

qC

dqdt

qC

I=FHGIKJ =FHGIKJ = FHG

IKJ

12

2

dUdt

=×

×

FHG

IKJ × = × =

−

−− −9 93 10

3 37 10 3 34 10 3346

68 7.

. . C

1.00 10 C V A W nWe j

(d) Pbattery A V W nW= = × = × =− −Iε 3 37 10 10 0 3 37 10 3378 7. . .e ja f

154 Direct Current Circuits

P28.66 Start at the point when the voltage has just reached 23∆V

and the switch has just closed. The voltage is 23∆V and is

decaying towards 0 V with a time constant R C2

∆ ∆V t V eCt R Ca f = LNMOQP

−23

2 .

We want to know when ∆V tC a f will reach 13∆V .

Therefore,13

23

2∆ ∆V V e t R C= LNMOQP

−

or e t R C− =212

or t R C1 2 2= ln .

V

R1

R2

∆V

+

C ∆Vc

Voltagecontrolledswitch

FIG. P28.66

After the switch opens, the voltage is 13∆V , increasing toward ∆V with time constant R R C1 2+b g :

∆ ∆ ∆V t V V eCt R R Ca f b g= − LNMOQP

− +23

1 2 .

When ∆ ∆V t VC a f = 23

23

23

1 2∆ ∆ ∆V V Ve t R R C= − − +b g or e t R R C− + =1 212

b g .

So t R R C2 1 2 2= +b g ln and T t t R R C= + = +1 2 1 22 2b g ln .

P28.67 (a) First determine the resistance of each light bulb: P =∆V

Ra f2

RV

= = =∆

Ωa f a f2 2120

60 0240

P V W

.

.

We obtain the equivalent resistance Req of the network of light

bulbs by identifying series and parallel equivalent resistances: FIG. P28.67

R RR Req = +

+= + =1

2 3

11 1

240 120 360b g b g Ω Ω Ω .

The total power dissipated in the 360 Ω is P = = =∆

ΩV

Ra f a f2 2120

36040 0

eq

V

W. .

(b) The current through the network is given by P = I R2eq : I

R= = =

P

eq

W360

A40 0 1

3.Ω

.

The potential difference across R1 is ∆ ΩV IR1 113

240 80 0= = FHGIKJ = A Va f . .

The potential difference ∆V23 across the parallel combination of R2 and R3 is

∆Ω Ω

V IR23 2313

11 240 1 240

40 0= = FHGIKJ +

FHG

IKJ = A

Vb g b g . .

Chapter 28 155

*P28.68 (a) With the switch closed, current exists in a simple seriescircuit as shown. The capacitors carry no current. For R2

we have

P = I R22 I

R= =

⋅=

P

2

2 4018 5

..

V A7 000 V A

mA .

The potential difference across R1 and C1 is

∆V IR= = × =−1

21 85 10 4 000 74 1. . A V A Ve jb g .

The charge on C1

FIG. P28.68(a)

Q C V= = × =−1

63 00 10 74 1 222∆ . . C V V Ce ja f µ .

The potential difference across R2 and C2 is

∆ ΩV IR= = × =−2

21 85 10 7 000 130. A Ve jb g .

The charge on C2

Q C V= = × =−2

66 00 10 130 778∆ . C V V Ce ja f µ .

The battery emf is

IR I R Req = + = × + =−1 2

21 85 10 7 000 204b g b g. A 4 000 V A V .

(b) In equilibrium after the switch has been opened, no currentexists. The potential difference across each resistor is zero. Thefull 204 V appears across both capacitors. The new charge C2

Q C V= = × =−2

66 00 10 204 1 222∆ . C V V Ce ja f µ

for a change of 1 222 778 444 C C Cµ µ µ− = . FIG. P28.68(b)

*P28.69 The battery current is

150 45 14 4 213+ + + =a f mA mA.

(a) The resistor with highest resistance is thatcarrying 4 mA. Doubling its resistance willreduce the current it carries to 2 mA. Thenthe total current is

FIG. P28.69

150 45 14 2 211+ + + =a f mA mA, nearly the same as before. The ratio is 211213

0 991= . .

(b) The resistor with least resistance carries 150 mA. Doubling its resistance changes this currentto 75 mA and changes the total to

75 45 14 4 138+ + + =a f mA mA. The ratio is 138213

0 648= . , representing a much larger

reduction (35.2% instead of 0.9%).

(c) This problem is precisely analogous. As a battery maintained a potential difference in parts(a) and (b), a furnace maintains a temperature difference here. Energy flow by heat isanalogous to current and takes place through thermal resistances in parallel. Each resistancecan have its “R-value” increased by adding insulation. Doubling the thermal resistance ofthe attic door will produce only a negligible (0.9%) saving in fuel. Doubling the thermalresistance of the ceiling will produce a much larger saving. The ceiling originally has the

smallest thermal resistance.

156 Direct Current Circuits

*P28.70 From the hint, the equivalent resistance of .

That is, RR R

RTL eq

eq++

=1

1 1

RR R

R RR

R R R R R R R R R

R R R R R

RR R R R

TL eq

L eqeq

T L T eq L eq L eq eq

eq T eq T L

eqT T T L

++

=

+ + = +

− − =

=± − −

2

2

2

0

4 1

2 1

a fb ga f

Only the + sign is physical:

R R R R Req T L T T= + +FH IK12

4 2 .

For example, if RT = 1 Ω .

And RL = 20 Ω , Req = 5 Ω .

P28.71 (a) After steady-state conditions have been reached, there is no DC current through thecapacitor.

Thus, for R3: IR30= steady-stateb g .

For the other two resistors, the steady-state current is simply determined by the 9.00-V emfacross the 12-kΩ and 15-kΩ resistors in series:

For R1 and R2 : IR RR R1 2

1 2

9 0015 0

333+ =+

=+

=b g a f b gεµ

..

V12.0 k k

A steady-stateΩ Ω

.

(b) After the transient currents have ceased, the potentialdifference across C is the same as the potentialdifference across R IR2 2=b g because there is no voltagedrop across R3 . Therefore, the charge Q on C is

Q C V C IRR= = =

=

∆ Ωa f b g b gb ga f2 2 10 0 333 15 0

50 0

. .

. .

F A k

C

µ µ

µFIG. P28.71(b)

continued on next page

Chapter 28 157

(c) When the switch is opened, the branch containing R1

is no longer part of the circuit. The capacitor dischargesthrough R R2 3+b g with a time constant ofR R C2 3 15 0 3 00 10 0 0 180+ = + =b g a fb g. . . . k k F sΩ Ω µ . The

initial current Ii in this discharge circuit is determinedby the initial potential difference across the capacitorapplied to R R2 3+b g in series:

IV

R RIR

R RiC=

+=

+=

+=

∆ ΩΩ Ω

a fb g b g

b ga fa f2 3

2

2 3

333 15 015 0 3 00

278 A k k k

Aµ

µ.

. ..

FIG. P28.71(c)

Thus, when the switch is opened, the current through R2 changes instantaneously from333 Aµ (downward) to 278 Aµ (downward) as shown in the graph. Thereafter, it decaysaccording to

I I e e tR it R R C t

22 3 278 00 180= = >− + −b g a fb g a f A for sµ . .

(d) The charge q on the capacitor decays from Qi to Qi

5 according to

q Q eQ

Q e

et

t

it R R C

ii

t

t

=

=

=

=

= =

− +

−

2 3

55

5180

0 180 5 290

0 180

0 180

b g

b g

a fa f

.

.

ln

. ln

s

s

ms s ms

*P28.72 (a) First let us flatten the circuit on a 2-D planeas shown; then reorganize it to a formateasier to read. Notice that the five resistorson the top are in the same connection asthose in Example 28.5; the same argumenttells us that the middle resistor can beremoved without affecting the circuit. Theremaining resistors over the three parallelbranches have equivalent resistance

Req = + +FHG

IKJ =−1

20120

110

5 001

. Ω .

(b) So the current through the battery is

∆Ω

VReq

= =12 0

2 40.

. V

5.00 A .

FIG. P28.72(a)

158 Direct Current Circuits

P28.73 ∆V e t RC= −ε

so lnε∆V RC

tFHGIKJ =FHGIKJ

1.

A plot of lnε∆VFHGIKJ versus t should

be a straight line with slope equal

to 1

RC.

Using the given data values:

FIG. P28.73

(a) A least-square fit to this data yields the graph above.

xi∑ = 282 , xi2 41 86 10∑ = ×. ,

x yi i∑ = 244, yi∑ = 4 03. , N = 8

Slope =−

−=

∑ ∑ ∑∑ ∑

N x y x y

N x x

i i i i

i i

c h c hc he j c h2 2 0 011 8.

Intercept =−

−=

∑ ∑ ∑ ∑∑ ∑

x y x x y

N x x

i i i i i

i i

2

2 2 0 088 2e jc h c hc h

e j c h.

t s V V Va f a f b g∆ ∆ln ε04.87

11.119.430.846.667.3

102.2

6.195.554.934.343.723.092.471.83

00.1090.2280.3550.5090.6950.9191.219

The equation of the best fit line is: ln . .ε∆V

tFHGIKJ = +0 011 8 0 088 2b g .

(b) Thus, the time constant is τ = = = =RC1 1

0 011 884 7

slope s

..

and the capacitance is CR

= =×

=τ

µ84 7

10 0 108 476

..

. s

F

Ω.

P28.74 (a) For the first measurement, the equivalent circuit is asshown in Figure 1.

R R R R Rab y y y= = + =1 2

so R Ry =12 1. (1)

For the second measurement, the equivalent circuit isshown in Figure 2.

Thus, R R R Rac y x= = +212

. (2)

Substitute (1) into (2) to obtain:

R R Rx2 112

12

= FHGIKJ + , or R R Rx = −2 1

14

.

(b) If R1 13 0= . Ω and R2 6 00= . Ω , then Rx = 2 75. Ω .

a bcRyRy Rx

R1

Figure 1

a c

RxRyRy

R2

Figure 2

FIG. P28.74

The antenna is inadequately grounded since this exceeds the limit of 2 00. Ω .

Chapter 28 159

P28.75 The total resistance between points b and c is:

R =+

=2 00 3 002 00 3 00

1 20. .. .

. k k k k

kΩ ΩΩ Ω

Ωa fa f

.

The total capacitance between points d and e is:C = + =2 00 3 00 5 00. . . F F Fµ µ µ .

The potential difference between point d and e in this series RCcircuit at any time is:

∆V e et RC t= − = −− −ε 1 120 0 1 1 000 6. Va f .

Therefore, the charge on each capacitor between points d and e is:

+ -a

b c d e

f

S120 V

2.00 kΩ

3.00 kΩ

C1 = 2.00 µF

C2 = 3.00 µF

FIG. P28.75

q C V e et t1 1

1 000 6 1 000 62 00 120 0 1 240 1= = − = −− −∆ . . F V Cµ µb ga f b gand q C V e et t

2 21 000 6 1 000 63 00 120 0 1 360 1= = − = −− −∆a f b ga f b g. . F V Cµ µ .

*P28.76 (a) Let i represent the current in the battery and ic the current charging the capacitor. Then i ic− is

the current in the voltmeter. The loop rule applied to the inner loop is + − − =ε iRqC

0 . The loop

rule for the outer perimeter is ε − − − =iR i i rcb g 0 . With idqdtc = , this becomes ε − − + =iR ir

dqdt

r 0 .

Between the two loop equations we eliminate iR

qRC

= −ε

by substitution to obtain

εε

ε ε

ε

− + −FHGIKJ + =

−+

++

+ =

−+

+ ++

=

R rR

qRC

dqdt

r

R rR

R rRC

qdqdt

r

rR r

qC

RrR r

dqdt

a f 0

0

0

This is the differential equation required.

(b) To solve we follow the same steps as on page 875.

dqdt R

R rRrC

qR rRrC

qrC

R r

dqq rC R r

R rRrC

dt qrc

R rR rRrC

t

q rc R rrc R r

R rRrC

t qrc

R rrc

R re

qr

r RC e R

RrR r

q t q t

R r RrC t

t R C

= −+

= −+

−+

FHG

IKJ

− += −

+−

+FHG

IKJ = −

+

− +− +

FHG

IKJ = −

+−

+= −

+

=+

− =+

z z− +

−

ε ε

εε

εε

ε ε

ε

a fa fa fe j

a f0 0 0 0

1

ln

ln

eq where eq

The voltage across the capacitor is VqC

rr R

eCt R C= =

+− −ε 1 eqe j .

(c) As t →∞ the capacitor voltage approaches r

r Rr

r R+− =

+ε

ε1 0a f . If the switch is then opened,

the capacitor discharges through the voltmeter. Its voltage decays exponentially according

to r

r Re t rCε

+− .

160 Direct Current Circuits

ANSWERS TO EVEN PROBLEMS

P28.2 (a) 1 79. A; (b) 10 4. V P28.42 0 113. Ω

P28.44 (a) 30 000. mA , 5 400 0. V ;P28.4 (a) 12 4. V ; (b) 9 65. V(b) 30 167. mA, 5 381 6. V ;

P28.6 (a) 17 1. Ω ; (b) 1 99. A in 4 Ω and 9 Ω;1 17. A in 7 Ω; 0 818. A in 10 Ω

(c) 29 898. mA ; 5 396 6. V

P28.46 see the solutionP28.8 29 5. V

P28.48 (a) 0.101 W; (b) 10.1 WP28.10 (a) see the solution; (b) no

P28.50 15.5 AP28.12 see the solution

P28.52 2 22. hP28.14 R1 1 00= . kΩ; R2 2 00= . kΩ ; R3 3 00= . kΩ

P28.54 a is 4 00. V higherP28.16 470 Ω and 220 Ω

P28.56 (a) see the solution; 833 mA; 200 W;P28.18 (a) 11 0. Ω ; (b) and (d) see the solution; (b) see the solution; 4.17 A; 1.00 kW

(c) 220 Ω; (e) ParallelP28.58 587 kΩ

P28.20 I1 714= mA ; I2 1 29= . A ; ε = 12 6. V

P28.60P P P Ps s s p

I

+ −2

2

4

2 and

P P P Ps s s p

I

− −2

2

4

2P28.22 see the solution

P28.24 (a) 0 385. mA in R1 ; 2 69. mA in R3;3 08. mA in R2 ; (b) c higher by 69 2. V P28.62 (a) I

IRR R1

2

1 2=

+b g ; IIR

R R21

1 2=

+;

(b) see the solutionP28.26 1 00. A up in 200 Ω ; 4 00. A up in 70 Ω ;3 00. A up in 80 Ω ; 8 00. A down in 20 Ω ;200 V P28.64 see the solution

P28.66 R R C1 22 2+b g lnP28.28 see the solution

P28.30 800 W to the left-hand resistor; 25 0. W toeach 4 Ω; 450 W to the right-hand resistor

P28.68 (a) 222 Cµ ; (b) increase by 444 Cµ

P28.70 see the solutionP28.32 (a) −61 6. mA ; (b) 0 235. Cµ ; (c) 1 96. A

P28.72 (a) 5 00. Ω; (b) 2.40 AP28.34 0 982. s

P28.74 (a) R RR

x = −21

4; (b) no; Rx = 2 75. ΩP28.36 (a) 1 50. s ; (b) 1 00. s ;

(c) 200 100 1.00 A A sµ µ+ −b ge t

P28.76 (a) and (b) see the solution; (c) r

r Re t rCε

+−

P28.38 (a) 3.91 s; (b) 0.782 ms

P28.40t

C ln 2