858 CHAPTE R 28 • Direct Current Circuits

Chapter 28

Direct Current Circuits

CHAPTE R OUTL I N E

28.1 Electromotive Force

28.2 Resistors in Series andParallel

28.3 Kirchhoff’s Rules

28.4 RC Circuits

28.5 Electrical Meters

28.6 Household Wiring andElectrical Safety

! An assortment of batteries that can be used to provide energy for various devices.Batteries provide a voltage with a fixed polarity, resulting in a direct current in a circuit, thatis, a current for which the drift velocity of the charges is always in the same direction.(George Semple)

858

This chapter is concerned with the analysis of simple electric circuits that containbatteries, resistors, and capacitors in various combinations. We will see some circuits inwhich resistors can be combined using simple rules. The analysis of more complicatedcircuits is simplified using two rules known as Kirchhoff’s rules, which follow from thelaws of conservation of energy and conservation of electric charge for isolated systems.Most of the circuits analyzed are assumed to be in steady state, which means thatcurrents in the circuit are constant in magnitude and direction. A current that isconstant in direction is called a direct current (DC). We will study alternating current(AC), in which the current changes direction periodically, in Chapter 33. Finally, wedescribe electrical meters for measuring current and potential difference, and discusselectrical circuits in the home.

28.1 Electromotive Force

In Section 27.6 we discussed a closed circuit in which a battery produces a potentialdifference and causes charges to move. We will generally use a battery in our discus-sion and in our circuit diagrams as a source of energy for the circuit. Because thepotential difference at the battery terminals is constant in a particular circuit, thecurrent in the circuit is constant in magnitude and direction and is called directcurrent. A battery is called either a source of electromotive force or, more commonly, asource of emf. (The phrase electromotive force is an unfortunate historical term, describ-ing not a force but rather a potential difference in volts.) The emf of a batteryis the maximum possible voltage that the battery can provide between itsterminals. You can think of a source of emf as a “charge pump.” When an electricpotential difference exists between two points, the source moves charges “uphill”from the lower potential to the higher.

Consider the circuit shown in Figure 28.1, consisting of a battery connected toa resistor. We shall generally assume that the connecting wires have no resistance.

!

859

+

Resistor

Battery–

Figure 28.1 A circuit consisting of a resistorconnected to the terminals of a battery.

The positive terminal of the battery is at a higher potential than the negative termi-nal. Because a real battery is made of matter, there is resistance to the flow of chargewithin the battery. This resistance is called internal resistance r. For an idealized bat-tery with zero internal resistance, the potential difference across the battery (called itsterminal voltage) equals its emf. However, for a real battery, the terminal voltage is notequal to the emf for a battery in a circuit in which there is a current. To understandwhy this is so, consider the circuit diagram in Figure 28.2a, where the battery ofFigure 28.1 is represented by the dashed rectangle containing an ideal, resistance-freeemf in series with an internal resistance r. Now imagine moving through the batteryfrom a to b and measuring the electric potential at various locations. As we pass fromthe negative terminal to the positive terminal, the potential increases by an amount .However, as we move through the resistance r, the potential decreases by an amountIr, where I is the current in the circuit. Thus, the terminal voltage of the battery"V # Vb $ Va is1

"V # $ Ir (28.1)

From this expression, note that is equivalent to the open-circuit voltage—that is,the terminal voltage when the current is zero. The emf is the voltage labeled on a battery—for example, the emf of a D cell is 1.5 V. The actual potential difference between theterminals of the battery depends on the current in the battery, as described by Equa-tion 28.1.

Figure 28.2b is a graphical representation of the changes in electric potential as thecircuit is traversed in the clockwise direction. By inspecting Figure 28.2a, we see thatthe terminal voltage "V must equal the potential difference across the external resis-tance R, often called the load resistance. The load resistor might be a simple resistivecircuit element, as in Figure 28.1, or it could be the resistance of some electrical device(such as a toaster, an electric heater, or a lightbulb) connected to the battery (or, inthe case of household devices, to the wall outlet). The resistor represents a load on thebattery because the battery must supply energy to operate the device. The potential dif-ference across the load resistance is "V # IR. Combining this expression with Equation28.1, we see that

# IR % Ir (28.2)

Solving for the current gives

(28.3)

This equation shows that the current in this simple circuit depends on both the loadresistance R external to the battery and the internal resistance r. If R is much greaterthan r, as it is in many real-world circuits, we can neglect r.

If we multiply Equation 28.2 by the current I, we obtain

I # I 2R % I 2r (28.4)

This equation indicates that, because power ! # I "V (see Eq. 27.22), the total poweroutput I of the battery is delivered to the external load resistance in the amount I 2Rand to the internal resistance in the amount I 2r.

!

!

& #!

R % r

!

!!

!

!

1 The terminal voltage in this case is less than the emf by an amount Ir. In some situations, theterminal voltage may exceed the emf by an amount Ir. This happens when the direction of the current isopposite that of the emf, as in the case of charging a battery with another source of emf.

860 CHAPTE R 28 • Direct Current Circuits

a c

(b)

Rr

db

V

IRIr

ε

ε

εa

d R

I

br– +

c

(a)

I

Active Figure 28.2 (a) Circuitdiagram of a source of emf (inthis case, a battery), of internalresistance r, connected to anexternal resistor of resistance R.(b) Graphical representationshowing how the electric potentialchanges as the circuit in part (a) istraversed clockwise.

At the Active Figures linkat http://www.pse6.com, youcan adjust the emf andresistances r and R to see theeffect on the current and on thegraph in part (b).

!

! PITFALL PREVENTION28.1 What Is Constant in a

Battery?It is a common misconceptionthat a battery is a source ofconstant current. Equation 28.3clearly shows that this is not true.The current in the circuitdepends on the resistanceconnected to the battery. It is alsonot true that a battery is a sourceof constant terminal voltage,as shown by Equation 28.1. Abattery is a source of constantemf.

Quick Quiz 28.1 In order to maximize the percentage of the power that isdelivered from a battery to a device, the internal resistance of the battery should be (a) as low as possible (b) as high as possible (c) The percentage does not depend onthe internal resistance.

S ECT I O N 28 . 1 • Electromotive Force 861

Example 28.1 Terminal Voltage of a Battery

A battery has an emf of 12.0 V and an internal resistance of0.05 '. Its terminals are connected to a load resistance of3.00 '.

(A) Find the current in the circuit and the terminal voltageof the battery.

Solution Equation 28.3 gives us the current:

and from Equation 28.1, we find the terminal voltage:

"V # $ Ir # 12.0 V $ (3.93 A)(0.05 ') #

To check this result, we can calculate the voltage across theload resistance R :

"V # IR # (3.93 A)(3.00 ') # 11.8 V

(B) Calculate the power delivered to the load resistor, thepower delivered to the internal resistance of the battery, andthe power delivered by the battery.

Solution The power delivered to the load resistor is

!R # I 2R # (3.93 A)2 (3.00 ') #

The power delivered to the internal resistance is

!r # I 2r # (3.93 A)2 (0.05 ') # 0.772 W

46.3 W

11.8 V!

3.93 AI #!

R % r#

12.0 V3.05 '

#

Hence, the power delivered by the battery is the sumof these quantities, or 47.1 W. You should check this result,using the expression ! # I .

What If? As a battery ages, its internal resistanceincreases. Suppose the internal resistance of this batteryrises to 2.00 ! toward the end of its useful life. How doesthis alter the ability of the battery to deliver energy?

Answer Let us connect the same 3.00-' load resistor to thebattery. The current in the battery now is

and the terminal voltage is

"V # $ Ir # 12.0 V $ (2.40 A) (2.00 ') # 7.2 V

Notice that the terminal voltage is only 60% of the emf. Thepowers delivered to the load resistor and internal resistanceare

!R # I 2R # (2.40 A)2 (3.00 ') #

!r # I 2r # (2.40 A)2 (2.00 ') # 11.5 W

Notice that 40% of the power from the battery is deliveredto the internal resistance. In part (B), this percentage is1.6%. Consequently, even though the emf remains fixed,the increasing internal resistance significantly reduces theability of the battery to deliver energy.

17.3 W

!

I #!

R % r#

12.0 V(3.00 ' % 2.00 ')

# 2.40 A

!

Interactive

Example 28.2 Matching the Load

Show that the maximum power delivered to the load resis-tance R in Figure 28.2a occurs when the load resistancematches the internal resistance—that is, when R # r.

Solution The power delivered to the load resistance isequal to I 2R, where I is given by Equation 28.3:

When ! is plotted versus R as in Figure 28.3, we find that! reaches a maximum value of 2/4r at R # r. When R islarge, there is very little current, so that the power I 2Rdelivered to the load resistor is small. When R is small,the current is large and there is significant loss of powerI 2r as energy is delivered to the internal resistance. WhenR # r, these effects balance to give a maximum transfer ofpower.

We can also prove that the power maximizes at R # r bydifferentiating ! with respect to R, setting the result equal

!

! # I 2R #!2R

(R % r)2

to zero, and solving for R. The details are left as a problemfor you to solve (Problem 57).

At the Interactive Worked Example link at http://www.pse6.com, you can vary the load resistance and internal resistance,observing the power delivered to each.

r 2r 3rR

!max

!

Figure 28.3 (Example 28.2) Graph of the power ! deliveredby a battery to a load resistor of resistance R as a function of R.The power delivered to the resistor is a maximum when theload resistance equals the internal resistance of the battery.

2 The term voltage drop is synonymous with a decrease in electric potential across a resistor and isused often by individuals working with electric circuits.

28.2 Resistors in Series and Parallel

Suppose that you and your friends are at a crowded basketball game in a sports arena anddecide to leave early. You have two choices: (1) your group can exit through a single doorand push your way down a long hallway containing several concession stands, each sur-rounded by a large crowd of people waiting to buy food or souvenirs; or (2) each memberof your group can exit through a separate door in the main hall of the arena, where eachwill have to push his or her way through a single group of people standing by the door. Inwhich scenario will less time be required for your group to leave the arena?

It should be clear that your group will be able to leave faster through the separatedoors than down the hallway where each of you has to push through several groups ofpeople. We could describe the groups of people in the hallway as being in series, becauseeach of you must push your way through all of the groups. The groups of peoplearound the doors in the arena can be described as being in parallel. Each member ofyour group must push through only one group of people, and each member pushesthrough a different group of people. This simple analogy will help us understand thebehavior of currents in electric circuits containing more than one resistor.

When two or more resistors are connected together as are the lightbulbs in Figure28.4a, they are said to be in series. Figure 28.4b is the circuit diagram for the lightbulbs,which are shown as resistors, and the battery. In a series connection, if an amount ofcharge Q exits resistor R1, charge Q must also enter the second resistor R2. (This isanalogous to all members of your group pushing through each crowd in the singlehallway of the sports arena.) Otherwise, charge will accumulate on the wire betweenthe resistors. Thus, the same amount of charge passes through both resistors in a giventime interval. Hence,

The potential difference applied across the series combination of resistors will dividebetween the resistors. In Figure 28.4b, because the voltage drop2 from a to b equals IR1and the voltage drop from b to c equals IR2, the voltage drop from a to c is

"V # IR1 % IR2 # I(R1 % R2)

for a series combination of two resistors, the currents are the same in both resis-tors because the amount of charge that passes through R1 must also pass throughR2 in the same time interval.

862 CHAPTE R 28 • Direct Current Circuits

+ –

(a)

Battery

R1 R2

∆V

(c)

I

+ –

a c

(b)

I

R1 R2

I

+ –

a b c

I1 = I2 = I Req = R1 + R2

∆V

Active Figure 28.4 (a) A series connection of two lightbulbs with resistances R1 andR2. (b) Circuit diagram for the two-resistor circuit. The current in R1 is the same as thatin R2. (c) The resistors replaced with a single resistor having an equivalent resistanceReq # R1 % R2.

At the Active Figures linkat http://www.pse6.com, youcan adjust the battery voltageand resistances R1 and R2 tosee the effect on the currentsand voltages in the individualresistors.

The potential difference across the battery is also applied to the equivalent resistanceReq in Figure 28.4c:

"V # IReq

where we have indicated that the equivalent resistance has the same effect on thecircuit because it results in the same current in the battery as the combination of resis-tors. Combining these equations, we see that we can replace the two resistors in serieswith a single equivalent resistance whose value is the sum of the individual resistances:

"V # IReq # I(R1 % R2) 9: R eq # R1 % R2 (28.5)

The resistance Req is equivalent to the series combination R1 % R2 in the sense thatthe circuit current is unchanged when Req replaces R1 % R2.

The equivalent resistance of three or more resistors connected in series is

(28.6)

This relationship indicates that the equivalent resistance of a series connection ofresistors is the numerical sum of the individual resistances and is alwaysgreater than any individual resistance.

Looking back at Equation 28.3, the denominator is the simple algebraic sum of theexternal and internal resistances. This is consistent with the fact that internal andexternal resistances are in series in Figure 28.2a.

Note that if the filament of one lightbulb in Figure 28.4 were to fail, the circuitwould no longer be complete (resulting in an open-circuit condition) and the secondbulb would also go out. This is a general feature of a series circuit—if one device in theseries creates an open circuit, all devices are inoperative.

R eq # R1 % R2 % R3 % ( ( (

S ECT I O N 28 . 2 • Resistors in Series and Parallel 863

The equivalent resistance ofseveral resistors in series

! PITFALL PREVENTION28.2 Lightbulbs Don’t

BurnWe will describe the end of thelife of a lightbulb by saying thatthe filament fails, rather than bysaying that the lightbulb “burnsout.” The word burn suggests acombustion process, which is notwhat occurs in a lightbulb.

Quick Quiz 28.2 In Figure 28.4, imagine positive charges pass first throughR1 and then through R2. Compared to the current in R1, the current in R2 is(a) smaller, (b) larger, or (c) the same.

Quick Quiz 28.3 If a piece of wire is used to connect points b and c in Figure28.4b, does the brightness of bulb R1 (a) increase, (b) decrease, or (c) remain the same?

Quick Quiz 28.4 With the switch in the circuit of Figure 28.5 closed (left),there is no current in R2, because the current has an alternate zero-resistance paththrough the switch. There is current in R1 and this current is measured with the amme-ter (a device for measuring current) at the right side of the circuit. If the switch isopened (Fig. 28.5, right), there is current in R2 . What happens to the reading on theammeter when the switch is opened? (a) the reading goes up; (b) the reading goesdown; (c) the reading does not change.

A

R1

Switch closed

R2 A

R1

Switch open

R2

Figure 28.5 (Quick Quiz 28.4) What happens when the switch is opened?

(c)

I

∆V+ –

b

(b)

I1

R1

R2

∆V+ –

a

II2

+ –

(a)

R1

R2

Battery

∆V1 = ∆V2 = ∆V

Req R1 R2

1 1 1= +

Active Figure 28.6 (a) A parallel connection of two lightbulbs with resistances R1 andR2. (b) Circuit diagram for the two-resistor circuit. The potential difference across R1 isthe same as that across R2. (c) The resistors replaced with a single resistor having anequivalent resistance given by Equation 28.7.

Now consider two resistors connected in parallel, as shown in Figure 28.6. Whencharges reach point a in Figure 28.6b, called a junction, they split into two parts, with somegoing through R1 and the rest going through R2. A junction is any point in a circuitwhere a current can split (just as your group might split up and leave the sports arenathrough several doors, as described earlier.) This split results in less current in each indi-vidual resistor than the current leaving the battery. Because electric charge is conserved,the current I that enters point a must equal the total current leaving that point:

I # I1 % I2

where I1 is the current in R1 and I2 is the current in R2.As can be seen from Figure 28.6, both resistors are connected directly across the

terminals of the battery. Therefore,

when resistors are connected in parallel, the potential differences across the resis-tors is the same.

864 CHAPTE R 28 • Direct Current Circuits

! PITFALL PREVENTION28.3 Local and Global

ChangesA local change in one part of acircuit may result in a globalchange throughout the circuit.For example, if a single resistanceis changed in a circuit containingseveral resistors and batteries, thecurrents in all resistors and batter-ies, the terminal voltages of all bat-teries, and the voltages across allresistors may change as a result.

! PITFALL PREVENTION28.4 Current Does Not

Take the Path ofLeast Resistance

You may have heard a phrase like“current takes the path of leastresistance” in reference to a par-allel combination of currentpaths, such that there are two ormore paths for the current totake. The phrase is incorrect.The current takes all paths.Those paths with lower resistancewill have large currents, but evenvery high-resistance paths willcarry some of the current.

Because the potential differences across the resistors are the same, the expression"V # IR gives

where Req is an equivalent single resistance which will have the same effect on thecircuit as the two resistors in parallel; that is, it will draw the same current from thebattery (Fig. 28.6c). From this result, we see that the equivalent resistance of two resis-tors in parallel is given by

(28.7)

or

R eq #1

1R1

%1

R 2

#R1R 2

R1 % R2

1R eq

#1

R1%

1R2

I # I 1 % I 2 #"VR1

%"VR 2

# "V ! 1R1

%1

R2" #

"VR eq

At the Active Figures linkat http://www.pse6.com, youcan adjust the battery voltageand resistances R1 and R2 tosee the effect on the currentsand voltages in the individualresistors.

An extension of this analysis to three or more resistors in parallel gives

(28.8)

We can see from this expression that the inverse of the equivalent resistance of twoor more resistors connected in parallel is equal to the sum of the inverses of theindividual resistances. Furthermore, the equivalent resistance is always lessthan the smallest resistance in the group.

Household circuits are always wired such that the appliances are connected in par-allel. Each device operates independently of the others so that if one is switched off,the others remain on. In addition, in this type of connection, all of the devices operateon the same voltage.

1R eq

#1

R1%

1R2

%1

R3% ( ( (

S ECT I O N 28 . 2 • Resistors in Series and Parallel 865

The equivalent resistance ofseveral resistors in parallel

Quick Quiz 28.5 In Figure 28.4, imagine that we add a third resistor in serieswith the first two. Does the current in the battery (a) increase, (b) decrease, or (c) remain the same? Does the terminal voltage of the battery (d) increase, (e) decrease, or (f) remain the same?

Quick Quiz 28.6 In Figure 28.6, imagine that we add a third resistor inparallel with the first two. Does the current in the battery (a) increase, (b) decrease,or (c) remain the same? Does the terminal voltage of the battery (d) increase, (e) decrease, or (f) remain the same?

Quick Quiz 28.7 With the switch in the circuit of Figure 28.7 open (left),there is no current in R2. There is current in R1 and this current is measured with theammeter at the right side of the circuit. If the switch is closed (Fig. 28.7, right), there iscurrent in R2. What happens to the reading on the ammeter when the switch is closed?(a) the reading goes up; (b) the reading goes down; (c) the reading does not change.

Switch open

R1

R2

Switch closed

A

R1

R2

A

Figure 28.7 (Quick Quiz 28.7) What happens when the switch is closed?

Conceptual Example 28.3 Landscape Lights

A homeowner wishes to install 12-volt landscape lightingin his back yard. To save money, he purchases inexpensive18-gauge cable, which has a relatively high resistance perunit length. This cable consists of two side-by-side wiresseparated by insulation, like the cord on an appliance.He runs a 200-foot length of this cable from the powersupply to the farthest point at which he plans to position alight fixture. He attaches light fixtures across the two wires

on the cable at 10-foot intervals, so the light fixtures are inparallel. Because of the cable’s resistance, the brightnessof the bulbs in the light fixtures is not as desired. Whichproblem does the homeowner have? (a) All of the bulbsglow equally less brightly than they would if lower-resistance cable had been used. (b) The brightness of thebulbs decreases as you move farther from the powersupply.

866 CHAPTE R 28 • Direct Current Circuits

Solution A circuit diagram for the system appears inFigure 28.8. The horizontal resistors (such as RA and RB)represent the resistance of the wires in the cable betweenthe light fixtures while the vertical resistors (such as RC)represent the resistance of the light fixtures themselves.Part of the terminal voltage of the power supply is droppedacross resistors RA and RB. Thus, the voltage across light

fixture RC is less than the terminal voltage. There is afurther voltage drop across resistors RD and RE. Conse-quently, the voltage across light fixture RF is smaller thanthat across RC . This continues on down the line of lightfixtures, so the correct choice is (b). Each successive lightfixture has a smaller voltage across it and glows less brightlythan the one before.

RA RD

RC RF

RB RE

Resistance oflight fixtures

Resistance inwires of cable

Powersupply

Figure 28.8 (Conceptual Example 28.3) The circuit diagram for a set of landscapelight fixtures connected in parallel across the two wires of a two-wire cable. Thehorizontal resistors represent resistance in the wires of the cable. The vertical resistorsrepresent the light fixtures.

Example 28.4 Find the Equivalent Resistance

Four resistors are connected as shown in Figure 28.9a.

(A) Find the equivalent resistance between points a and c.

Solution The combination of resistors can be reduced insteps, as shown in Figure 28.9. The 8.0-' and 4.0-' resistorsare in series; thus, the equivalent resistance between a and bis 12.0 ' (see Eq. 28.5). The 6.0-' and 3.0-' resistors are inparallel, so from Equation 28.7 we find that the equivalentresistance from b to c is 2.0 '. Hence, the equivalent resis-tance from a to c is 14.0 '.

(B) What is the current in each resistor if a potential differ-ence of 42 V is maintained between a and c?

Solution The currents in the 8.0-' and 4.0-' resistors arethe same because they are in series. In addition, this is thesame as the current that would exist in the 14.0-' equivalentresistor subject to the 42-V potential difference. Therefore,using Equation 27.8 (R # "V/I ) and the result from part(A), we obtain

This is the current in the 8.0-' and 4.0-' resistors. Whenthis 3.0-A current enters the junction at b, however, it splits,with part passing through the 6.0-' resistor (I1) and partthrough the 3.0-' resistor (I2). Because the potential differ-ence is "Vbc across each of these parallel resistors, we seethat (6.0 ')I1 # (3.0 ')I2, or I2 # 2I1. Using this result andthe fact that I1 % I2 # 3.0 A, we find that I1 # 1.0 A and

I #"Vac

R eq#

42 V14.0 '

# 3.0 A

I2 # 2.0 A. We could have guessed this at the start by notingthat the current in the 3.0-' resistor has to be twice that inthe 6.0-' resistor, in view of their relative resistances and thefact that the same voltage is applied to each of them.

As a final check of our results, note that "Vbc #(6.0 ')I1 # (3.0 ')I2 # 6.0 V and "Vab # (12.0 ')I # 36 V;therefore, "Vac # "Vab % "Vbc # 42 V, as it must.

6.0 Ω

3.0 Ω

cb

I1

I2

4.0 Ω8.0 Ω

a

c

2.0 Ω12.0 Ω

ba

14.0 Ω

ca

(a)

(b)

(c)

I

Figure 28.9 (Example 28.4) The original network of resistorsis reduced to a single equivalent resistance.

Example 28.5 Finding Req by Symmetry Arguments

Consider five resistors connected as shown in Figure 28.10a.Find the equivalent resistance between points a and b.

Solution If we inspect this system of resistors, we realize thatwe cannot reduce it by using our rules for series and parallel

connections. We can, however, assume a current enteringjunction a and then apply symmetry arguments. Because ofthe symmetry in the circuit (all 1-' resistors in the outsideloop), the currents in branches ac and ad must be equal;hence, the electric potentials at points c and d must be equal.

Example 28.6 Three Resistors in Parallel

Three resistors are connected in parallel as shown in Figure28.11a. A potential difference of 18.0 V is maintainedbetween points a and b.

(A) Find the current in each resistor.

Solution The resistors are in parallel, and so the potentialdifference across each must be 18.0 V. Applying the relation-ship "V # IR to each resistor gives

2.00 AI3 #"VR3

#18.0 V9.00 '

#

3.00 AI2 #"VR2

#18.0 V6.00 '

#

6.00 AI1 #"VR1

#18.0 V3.00 '

#

(B) Calculate the power delivered to each resistor and thetotal power delivered to the combination of resistors.

Solution We apply the relationship ! # I 2R to each resis-tor and obtain

This shows that the smallest resistor receives the mostpower. Summing the three quantities gives a total power of198 W.

36.0 W9.00-': !3 # I3

2R3 # (2.00 A)2(9.00 ') #

54.0 W6.00-': !2 # I2

2R2 # (3.00 A)2(6.00 ') #

108 W3.00-': !1 # I1

2R1 # (6.00 A)2(3.00 ') #

S ECT I O N 28 . 2 • Resistors in Series and Parallel 867

Figure 28.11 (Example 28.6) (a) Threeresistors connected in parallel. Thevoltage across each resistor is 18.0 V.(b) Another circuit with three resistorsand a battery. Is this equivalent to thecircuit in part (a) of the figure?

Interactive

I1 I2 I3

Ia

b

18.0 V 3.00 Ω 6.00 Ω 9.00 Ω

I1 I2 I3

a

b

3.00 Ω 6.00 Ω 9.00 Ω18.0 V

I

(a)

(b)

This means that "Vcd # 0 and there is no current betweenpoints c and d. As a result, points c and d may be connectedtogether without affecting the circuit, as in Figure 28.10b.Thus, the 5-' resistor may be removed from the circuit and

the remaining circuit then reduced as in Figures 28.10c andd. From this reduction we see that the equivalent resistanceof the combination is 1 '. Note that the result is 1 ' regard-less of the value of the resistor connected between c and d.

(c)

0.5 Ω

ba c,d

0.5 Ω

(a)

1 Ω1 Ω

1 Ω1 Ω

5 Ωba

c

d

1 Ω

5 Ω

1 Ωba c,d

1 Ω

1 Ω

(b) (d)

1 Ω

ba

Figure 28.10 (Example 28.5) Because of the symmetry in this circuit, the 5-' resistordoes not contribute to the resistance between points a and b and therefore can bedisregarded when we calculate the equivalent resistance.

868 CHAPTE R 28 • Direct Current Circuits

Conceptual Example 28.7 Operation of a Three-Way Lightbulb

Figure 28.12 illustrates how a three-way lightbulb is con-structed to provide three levels of light intensity. The socketof the lamp is equipped with a three-way switch for selectingdifferent light intensities. The bulb contains two filaments.When the lamp is connected to a 120-V source, one filamentreceives 100 W of power, and the other receives 75 W.Explain how the two filaments are used to provide threedifferent light intensities.

Solution The three light intensities are made possible byapplying the 120 V to one filament alone, to the other fila-ment alone, or to the two filaments in parallel. When switchS1 is closed and switch S2 is opened, current exists only inthe 75-W filament. When switch S1 is open and switch S2 isclosed, current exists only in the 100-W filament. When bothswitches are closed, current exists in both filaments, and thetotal power is 175 W.

If the filaments were connected in series and one ofthem were to break, no charges could pass through thebulb, and the bulb would give no illumination, regardless ofthe switch position. However, with the filaments connected

in parallel, if one of them (for example, the 75-W filament)breaks, the bulb will still operate in two of the switch posi-tions as current exists in the other (100-W) filament.

120 V

100-W filament

75-W filament

S1

S2

Figure 28.12 (Conceptual Example 28.7) A three-waylightbulb.

Application Strings of Lights

Strings of lights are used for many ornamental purposes,such as decorating Christmas trees.3 Over the years, bothparallel and series connections have been used for stringsof lights powered by 120 V. Series-wired bulbs are safer thanparallel-wired bulbs for indoor Christmas-tree use becauseseries-wired bulbs operate with less energy per bulb and at alower temperature. However, if the filament of a single bulbfails (or if the bulb is removed from its socket), all thelights on the string go out. The popularity of series-wiredlight strings diminished because troubleshooting a failedbulb was a tedious, time-consuming chore that involvedtrial-and-error substitution of a good bulb in each socketalong the string until the defective bulb was found.

In a parallel-wired string, each bulb operates at 120 V.By design, the bulbs are brighter and hotter than those on

a series-wired string. As a result, these bulbs are inherentlymore dangerous (more likely to start a fire, for instance),but if one bulb in a parallel-wired string fails or is removed,the rest of the bulbs continue to glow. (A 25-bulb string of4-W bulbs results in a power of 100 W; the total powerbecomes substantial when several strings are used.)

A new design was developed for so-called “miniature”lights wired in series, to prevent the failure of one bulb fromcausing the entire string to go out. This design creates aconnection (called a jumper) across the filament after itfails. When the filament breaks in one of these miniaturelightbulbs, the break in the filament represents the largestresistance in the series, much larger than that of the intactfilaments. As a result, most of the applied 120 V appearsacross the bulb with the broken filament. Inside the

3 These and other household devices, such as the three-way lightbulb in Conceptual Example 28.7and the kitchen appliances discussed in Section 28.6, actually operate on alternating current (AC), tobe introduced in Chapter 33.

(C) Calculate the equivalent resistance of the circuit.

Solution We can use Equation 28.8 to find Req:

1.64 'R eq #18.0 '11.0

#

1R eq

#1

3.00 '%

16.00 '

%1

9.00 '

What If? What if the circuit is as shown in Figure 28.11binstead of as in Figure 28.11a? How does this affect thecalculation?

Answer There is no effect on the calculation. The physicalplacement of the battery is not important. In Figure 28.11b,the battery still applies a potential difference of 18.0 Vbetween points a and b, so the two circuits in Figure 28.11are electrically identical.

At the Interactive Worked Example link at http://www.pse6.com, you can explore different configurations of the batteryand resistors.

28.3 Kirchhoff’s Rules

As we saw in the preceding section, simple circuits can be analyzed using the expres-sion "V # IR and the rules for series and parallel combinations of resistors. Veryoften, however, it is not possible to reduce a circuit to a single loop. The procedurefor analyzing more complex circuits is greatly simplified if we use two principles calledKirchhoff ’s rules:

S ECT I O N 28 . 3 • Kirchhoff’s Rules 869

lightbulb, a small jumper loop covered by an insulatingmaterial is wrapped around the filament leads. When thefilament fails and 120 V appears across the bulb, an arcburns the insulation on the jumper and connects thefilament leads. This connection now completes the circuitthrough the bulb even though its filament is no longeractive (Fig. 28.13).

Suppose that all the bulbs in a 50-bulb miniature-lightstring are operating. A 2.40-V potential drop occurs acrosseach bulb because the bulbs are in series. A typical powerinput to this style of bulb is 0.340 W. The filament resis-tance of each bulb at the operating temperature is(2.40 V)2/(0.340 W) # 16.9 '. The current in each bulb is2.40 V/16.9 ' # 0.142 A. When a bulb fails, the resistanceacross its terminals is reduced to zero because of the alternate

jumper connection mentioned in the preceding paragraph.All the other bulbs not only stay on but glow more brightlybecause the total resistance of the string is reduced and con-sequently the current in each bulb increases.

Let us assume that the resistance of a bulb remains at16.9 ' even though its temperature rises as a result of theincreased current. If one bulb fails, the potential differenceacross each of the remaining bulbs increases to 120 V/49 #2.45 V, the current increases from 0.142 A to 0.145 A, and thepower increases to 0.355 W. As more bulbs fail, the currentkeeps rising, the filament of each bulb operates at a highertemperature, and the lifetime of the bulb is reduced. For thisreason, you should check for failed (nonglowing) bulbs insuch a series-wired string and replace them as soon as possible,in order to maximize the lifetimes of all the bulbs.

Figure 28.13 (a) Schematic diagram of a modern “miniature” holiday lightbulb, with ajumper connection to provide a current path if the filament breaks. When the filamentis intact, charges flow in the filament. (b) A holiday lightbulb with a broken filament.In this case, charges flow in the jumper connection. (c) A Christmas-tree lightbulb.

Geor

ge S

empl

e

Filament

Jumper

Glass insulator

(b)(a)

I I

I

1. Junction rule. The sum of the currents entering any junction in a circuit mustequal the sum of the currents leaving that junction:

(28.9)

2. Loop rule. The sum of the potential differences across all elements around anyclosed circuit loop must be zero:

(28.10)#closedloop

"V # 0

# I in # # I out

(c)

Kirchhoff’s first rule is a statement of conservation of electric charge. All chargesthat enter a given point in a circuit must leave that point because charge cannot buildup at a point. If we apply this rule to the junction shown in Figure 28.14a, we obtain

I1 # I2 % I3

Figure 28.14b represents a mechanical analog of this situation, in which water flowsthrough a branched pipe having no leaks. Because water does not build up anywherein the pipe, the flow rate into the pipe equals the total flow rate out of the twobranches on the right.

Kirchhoff’s second rule follows from the law of conservation of energy. Let usimagine moving a charge around a closed loop of a circuit. When the chargereturns to the starting point, the charge–circuit system must have the same totalenergy as it had before the charge was moved. The sum of the increases in energy asthe charge passes through some circuit elements must equal the sum of thedecreases in energy as it passes through other elements. The potential energydecreases whenever the charge moves through a potential drop $ IR across a resis-tor or whenever it moves in the reverse direction through a source of emf. Thepotential energy increases whenever the charge passes through a battery from thenegative terminal to the positive terminal.

When applying Kirchhoff’s second rule in practice, we imagine traveling around theloop and consider changes in electric potential, rather than the changes in potential energydescribed in the preceding paragraph. You should note the following sign conventionswhen using the second rule:

• Because charges move from the high-potential end of a resistor toward the low-potential end, if a resistor is traversed in the direction of the current, the poten-tial difference "V across the resistor is $ IR (Fig. 28.15a).

• If a resistor is traversed in the direction opposite the current, the potential differ-ence "V across the resistor is % IR (Fig. 28.15b).

• If a source of emf (assumed to have zero internal resistance) is traversed in thedirection of the emf (from $ to %), the potential difference "V is % (Fig.28.15c). The emf of the battery increases the electric potential as we movethrough it in this direction.

• If a source of emf (assumed to have zero internal resistance) is traversed in thedirection opposite the emf (from % to $ ), the potential difference "V is $(Fig. 28.15d). In this case the emf of the battery reduces the electric potential aswe move through it.

Limitations exist on the numbers of times you can usefully apply Kirchhoff’s rulesin analyzing a circuit. You can use the junction rule as often as you need, so long aseach time you write an equation you include in it a current that has not been used ina preceding junction-rule equation. In general, the number of times you can use thejunction rule is one fewer than the number of junction points in the circuit. You canapply the loop rule as often as needed, as long as a new circuit element (resistor orbattery) or a new current appears in each new equation. In general, in order tosolve a particular circuit problem, the number of independent equations youneed to obtain from the two rules equals the number of unknown currents.

Complex networks containing many loops and junctions generate great numbersof independent linear equations and a correspondingly great number of unknowns.Such situations can be handled formally through the use of matrix algebra. Computersoftware can also be used to solve for the unknowns.

The following examples illustrate how to use Kirchhoff’s rules. In all cases, it isassumed that the circuits have reached steady-state conditions—that is, the currentsin the various branches are constant. Any capacitor acts as an open branch in acircuit; that is, the current in the branch containing the capacitor is zero understeady-state conditions.

!

!

870 CHAPTE R 28 • Direct Current Circuits

(a)

I

a b∆V = –IR

(b)

I

a b∆V = +IR

(c)

εa b

∆V = +ε– +

(d)a b

∆V = –ε–+

ε

ε

ε

Figure 28.15 Rules fordetermining the potentialdifferences across a resistor and abattery. (The battery is assumed tohave no internal resistance.) Eachcircuit element is traversed fromleft to right.

(a)

I1

I2

I3

(b)

Flow inFlow out

Figure 28.14 (a) Kirchhoff’sjunction rule. Conservation ofcharge requires that all chargesentering a junction must leave thatjunction. Therefore, I1 # I2 % I3.(b) A mechanical analog of thejunction rule: the amount of waterflowing out of the branches on theright must equal the amountflowing into the single branch onthe left.

S ECT I O N 28 . 3 • Kirchhoff’s Rules 871

Gustav KirchhoffGerman Physicist (1824–1887)

Kirchhoff, a professor atHeidelberg, and Robert Bunseninvented the spectroscope andfounded the science ofspectroscopy, which we shallstudy in Chapter 42. Theydiscovered the elements cesiumand rubidium and inventedastronomical spectroscopy. (AIPESVA/W.F. Meggers Collection)

P R O B L E M - S O LV I N G H I N T S

Kirchhoff’s Rules• Draw a circuit diagram, and label all the known and unknown quantities. You

must assign a direction to the current in each branch of the circuit. Althoughthe assignment of current directions is arbitrary, you must adhere rigorously tothe assigned directions when applying Kirchhoff’s rules.

• Apply the junction rule to any junctions in the circuit that provide newrelationships among the various currents.

• Apply the loop rule to as many loops in the circuit as are needed to solve forthe unknowns. To apply this rule, you must correctly identify the potentialdifference as you imagine crossing each element while traversing the closedloop (either clockwise or counterclockwise). Watch out for errors in sign!

• Solve the equations simultaneously for the unknown quantities. Do not bealarmed if a current turns out to be negative; its magnitude will be correct and thedirection is opposite to that which you assigned.

Quick Quiz 28.8 In using Kirchhoff’s rules, you generally assign a separateunknown current to (a) each resistor in the circuit (b) each loop in the circuit (c) eachbranch in the circuit (d) each battery in the circuit.

Example 28.8 A Single-Loop Circuit

A single-loop circuit contains two resistors and two batteries,as shown in Figure 28.16. (Neglect the internal resistancesof the batteries.)

(A) Find the current in the circuit.

Solution We do not need Kirchhoff’s rules to analyze thissimple circuit, but let us use them anyway just to see howthey are applied. There are no junctions in this single-loopcircuit; thus, the current is the same in all elements. Let usassume that the current is clockwise, as shown in Figure28.16. Traversing the circuit in the clockwise direction,starting at a, we see that a : b represents a potential differ-ence of % 1, b : c represents a potential difference of$ IR1, c : d represents a potential difference of $ 2, and!

!

d : a represents a potential difference of $ IR2. ApplyingKirchhoff’s loop rule gives

Solving for I and using the values given in Figure 28.16, weobtain

The negative sign for I indicates that the direction of thecurrent is opposite the assumed direction. Notice that theemfs in the numerator subtract because the batteries haveopposite polarities in Figure 28.16. In the denominator, theresistances add because the two resistors are in series.

(B) What power is delivered to each resistor? What power isdelivered by the 12-V battery?

Solution Using Equation 27.23,

Hence, the total power delivered to the resistors is!1 % !2 # 2.0 W.

The 12-V battery delivers power I 2 # 4.0 W. Half of thispower is delivered to the two resistors, as we just calculated.The other half is delivered to the 6-V battery, which is being

!

1.1 W!2 # I 2R2 # (0.33 A)2(10 ') #

0.87 W!1 # I 2R1 # (0.33 A)2(8.0 ') #

$0.33 A(1) I #!1 $ !2

R1 % R2#

6.0 V $ 12 V8.0 ' % 10 '

#

!1 $ IR 1 $ !2 $ IR 2 # 0

# "V # 0

a bI

cd

1 = 6.0 V

+–

R 1 = 8.0 ΩR 2 = 10 Ω

2 = 12 V

+–ε

εFigure 28.16 (Example 28.8) A series circuit containing twobatteries and two resistors, where the polarities of the batteriesare in opposition.

872 CHAPTE R 28 • Direct Current Circuits

charged by the 12-V battery. If we had included the internalresistances of the batteries in our analysis, some of the powerwould appear as internal energy in the batteries; as a result,we would have found that less power was being delivered tothe 6-V battery.

What If? What if the polarity of the 12.0-V battery werereversed? How would this affect the circuit?

Answer While we could repeat the Kirchhoff’s rulescalculation, let us examine Equation (1) and modify itaccordingly. Because the polarities of the two batteries are

now in the same direction, the signs of 1 and 2 are thesame and Equation (1) becomes

The new powers delivered to the resistors are

!1 # I 2R1 # (1.0 A)2(8.0 ') # 8.0 W

!2 # I 2R2 # (1.0 A)2(10 ') # 10 W

This totals 18 W, nine times as much as in the original circuit,in which the batteries were opposing each other.

I #!1 % !2

R1 % R2#

6.0 V % 12 V8.0 ' % 10 '

# 1.0 A

!!

Example 28.9 Applying Kirchhoff’s Rules

Find the currents I1, I2, and I3 in the circuit shown in Figure28.17.

Solution Conceptualize by noting that we cannot simplifythe circuit by the rules of adding resistances in series andin parallel. (If the 10.0-V battery were taken away, we couldreduce the remaining circuit with series and parallel com-binations.) Thus, we categorize this problem as one inwhich we must use Kirchhoff’s rules. To analyze the circuit,we arbitrarily choose the directions of the currents as la-beled in Figure 28.17. Applying Kirchhoff’s junction ruleto junction c gives

(1) I1 % I2 # I3

We now have one equation with three unknowns—I1, I2, andI3. There are three loops in the circuit—abcda, befcb, andaefda. We therefore need only two loop equations to deter-mine the unknown currents. (The third loop equationwould give no new information.) Applying Kirchhoff’s looprule to loops abcda and befcb and traversing these loopsclockwise, we obtain the expressions

(2) abcda 10.0 V $ (6.0 ')I1 $ (2.0 ')I3 # 0

(3) befcb $14.0 V % (6.0 ')I1 $ 10.0 V $ (4.0 ') I2 # 0

Note that in loop befcb we obtain a positive value whentraversing the 6.0-' resistor because our direction of travelis opposite the assumed direction of I1. Expressions (1), (2),and (3) represent three independent equations with threeunknowns. Substituting Equation (1) into Equation (2)gives

10.0 V $ (6.0 ')I1 $ (2.0 ') (I1 % I2) # 0

(4) 10.0 V # (8.0 ')I1 % (2.0 ')I2

Dividing each term in Equation (3) by 2 and rearranginggives

(5) $ 12.0 V # $ (3.0 ')I1 % (2.0 ')I2

Subtracting Equation (5) from Equation (4) eliminates I2,giving

22.0 V # (11.0 ')I1

I1 #

Using this value of I1 in Equation (5) gives a value for I2:

(2.0 ')I2 # (3.0 ')I1 $ 12.0 V

# (3.0 ')(2.0 A) $ 12.0 V # $ 6.0 V

I2 #

Finally,

I3 # I1 % I2 #

To finalize the problem, note that I2 and I3 are both nega-tive. This indicates only that the currents are opposite thedirection we chose for them. However, the numerical valuesare correct. What would have happened had we left thecurrent directions as labeled in Figure 28.17 but traversedthe loops in the opposite direction?

$1.0 A

$3.0 A

2.0 A

14.0 Ve

b

4.0 Ω

– +

10.0 V 6.0 Ω

–+ f

I2

c

I3

I1

2.0 Ωda

Figure 28.17 (Example 28.9) A circuit containing differentbranches.

Interactive

Practice applying Kirchhoff’s rules at the Interactive Worked Example link at http://www.pse6.com.

28.4 RC Circuits

So far we have analyzed direct current circuits in which the current is constant. In DCcircuits containing capacitors, the current is always in the same direction but may varyin time. A circuit containing a series combination of a resistor and a capacitor is calledan RC circuit.

Charging a Capacitor

Figure 28.19 shows a simple series RC circuit. Let us assume that the capacitor in thiscircuit is initially uncharged. There is no current while switch S is open (Fig. 28.19b).If the switch is closed at t # 0, however, charge begins to flow, setting up a current inthe circuit, and the capacitor begins to charge.4 Note that during charging, charges do

4 In previous discussions of capacitors, we assumed a steady-state situation, in which no current waspresent in any branch of the circuit containing a capacitor. Now we are considering the case before thesteady-state condition is realized; in this situation, charges are moving and a current exists in the wiresconnected to the capacitor.

S ECT I O N 28 . 4 • RC Circuits 873

Example 28.10 A Multiloop Circuit

(A) Under steady-state conditions, find the unknown currentsI1, I2, and I3 in the multiloop circuit shown in Figure 28.18.

Solution First note that because the capacitor representsan open circuit, there is no current between g and b alongpath ghab under steady-state conditions. Therefore, whenthe charges associated with I1 reach point g, they all gotoward point b through the 8.00-V battery; hence, Igb # I1.Labeling the currents as shown in Figure 28.18 and applyingEquation 28.9 to junction c, we obtain

(1) I1 % I2 # I3

Equation 28.10 applied to loops defcd and cfgbc, traversedclockwise, gives

(2) defcd 4.00 V $ (3.00 ')I2 $ (5.00 ')I3 # 0

(3) cfgbc (3.00 ')I2 $ (5.00 ')I1 % 8.00 V # 0

From Equation (1) we see that I1 # I3 $ I2, which, whensubstituted into Equation (3), gives

(4) (8.00 ')I2 $ (5.00 ')I3 % 8.00 V # 0

Subtracting Equation (4) from Equation (2), we eliminate I3and find that

Because our value for I2 is negative, we conclude that the di-rection of I2 is from c to f in the 3.00-' resistor. Despite thisinterpretation of the direction, however, we must continueto use this negative value for I2 in subsequent calculationsbecause our equations were established with our originalchoice of direction.

Using I2 # $ 0.364 A in Equations (3) and (1) gives

(B) What is the charge on the capacitor?

Solution We can apply Kirchhoff’s loop rule to loop bghab(or any other loop that contains the capacitor) to find thepotential difference "Vcap across the capacitor. We use thispotential difference in the loop equation without referenceto a sign convention because the charge on the capacitordepends only on the magnitude of the potential difference.Moving clockwise around this loop, we obtain

$ 8.00 V % "Vcap $ 3.00 V # 0

"Vcap # 11.0 V

Because Q # C "Vcap (see Eq. 26.1), the charge on thecapacitor is

Q # (6.00 )F)(11.0 V) #

Why is the left side of the capacitor positively charged?

66.0 )C

1.02 AI3 #1.38 AI1 #

$0.364 AI2 # $4.00 V11.0 '

#

4.00 V

d

c

5.00 Ω

–+8.00 V

3.00 Ω

– + e

I3

f

I1

I25.00 Ω

ha

g

– +

3.00 V

–+

6.00 F

I = 0

b

I3

I1

µFigure 28.18 (Example 28.10) A multiloop circuit. Kirchhoff’sloop rule can be applied to any closed loop, including the onecontaining the capacitor.

not jump across the capacitor plates because the gap between the plates represents anopen circuit. Instead, charge is transferred between each plate and its connecting wiresdue to the electric field established in the wires by the battery, until the capacitor isfully charged. As the plates are being charged, the potential difference across thecapacitor increases. The value of the maximum charge on the plates depends on thevoltage of the battery. Once the maximum charge is reached, the current in the circuitis zero because the potential difference across the capacitor matches that supplied bythe battery.

To analyze this circuit quantitatively, let us apply Kirchhoff’s loop rule to the circuitafter the switch is closed. Traversing the loop in Fig. 28.19c clockwise gives

(28.11)

where q/C is the potential difference across the capacitor and IR is the potential differ-ence across the resistor. We have used the sign conventions discussed earlier for thesigns on and IR. For the capacitor, notice that we are traveling in the direction fromthe positive plate to the negative plate; this represents a decrease in potential. Thus, weuse a negative sign for this potential difference in Equation 28.11. Note that q and I areinstantaneous values that depend on time (as opposed to steady-state values) as thecapacitor is being charged.

We can use Equation 28.11 to find the initial current in the circuit and the maxi-mum charge on the capacitor. At the instant the switch is closed (t # 0), the charge onthe capacitor is zero, and from Equation 28.11 we find that the initial current I0 in thecircuit is a maximum and is equal to

(28.12)

At this time, the potential difference from the battery terminals appears entirely acrossthe resistor. Later, when the capacitor is charged to its maximum value Q , chargescease to flow, the current in the circuit is zero, and the potential difference from thebattery terminals appears entirely across the capacitor. Substituting I # 0 into Equation28.11 gives the charge on the capacitor at this time:

(maximum charge) (28.13)

To determine analytical expressions for the time dependence of the charge andcurrent, we must solve Equation 28.11—a single equation containing two variables, qand I. The current in all parts of the series circuit must be the same. Thus, the currentin the resistance R must be the same as the current between the capacitor plates and the

Q # C !

I0 #!R (current at t # 0)

!

! $qC

$ IR # 0

874 CHAPTE R 28 • Direct Current Circuits

+ –

Resistor

Battery

Capacitor

Switch

(a)

ε(b)

S

t < 0

R

C

(c) t > 0

ε

R

S

Iq–

+ q

Active Figure 28.19 (a) A capacitor in series with a resistor, switch, and battery.(b) Circuit diagram representing this system at time t * 0, before the switch is closed.(c) Circuit diagram at time t + 0, after the switch has been closed.

At the Active Figures linkat http://www.pse6.com, youcan adjust the values of R andC to see the effect on thecharging of the capacitor.

wires. This current is equal to the time rate of change of the charge on the capacitorplates. Thus, we substitute I # dq/dt into Equation 28.11 and rearrange the equation:

To find an expression for q, we solve this separable differential equation. We firstcombine the terms on the right-hand side:

Now we multiply by dt and divide by q $ C to obtain

Integrating this expression, using the fact that q # 0 at t # 0, we obtain

From the definition of the natural logarithm, we can write this expression as

(28.14)

where e is the base of the natural logarithm and we have made the substitution fromEquation 28.13.

We can find an expression for the charging current by differentiating Equation28.14 with respect to time. Using I # dq/dt, we find that

(28.15)

Plots of capacitor charge and circuit current versus time are shown in Figure 28.20.Note that the charge is zero at t # 0 and approaches the maximum value C as t : ,.The current has its maximum value I0 # /R at t # 0 and decays exponentially to zeroas t : ,. The quantity RC, which appears in the exponents of Equations 28.14 and28.15, is called the time constant - of the circuit. It represents the time intervalduring which the current decreases to 1/e of its initial value; that is, in a time interval-, I # e$1I0 # 0.368I 0. In a time interval 2-, I # e$ 2I 0 # 0.135I0, and so forth.Likewise, in a time interval -, the charge increases from zero to C [1 $ e$1] # 0.632C .!!

!!

I(t) #!R

e$t/RC

q(t ) # C!(1 $ e$t/RC) # Q(1 $ e$t/RC)

ln ! q $ C!$C! " # $

tRC

$q

0

dq(q $ C!)

# $1

RC $t

0 dt

dqq $ C ! # $

1RC

dt

!

dqdt

#C!RC

$q

RC# $

q $ C!RC

dqdt

#!R

$q

RC

S ECT I O N 28 . 4 • RC Circuits 875

Charge as a function of time fora capacitor being charged

Current as a function of time fora capacitor being charged

q

τ t

C

0.632C

(a)

I

τ t

0.368I0

(b)

I0 I0 = R

ε

εε

=RCτ

Figure 28.20 (a) Plot of capacitor charge versus time for the circuit shown in Figure28.19. After a time interval equal to one time constant - has passed, the charge is 63.2%of the maximum value C . The charge approaches its maximum value as t approachesinfinity. (b) Plot of current versus time for the circuit shown in Figure 28.19. Thecurrent has its maximum value I0 # /R at t # 0 and decays to zero exponentially as tapproaches infinity. After a time interval equal to one time constant - has passed, thecurrent is 36.8% of its initial value.

!

!

The following dimensional analysis shows that - has the units of time:

Because - # RC has units of time, the combination -/RC is dimensionless, as it must bein order to be an exponent of e in Equations 28.14 and 28.15.

The energy output of the battery as the capacitor is fully charged is Q # C 2.After the capacitor is fully charged, the energy stored in the capacitor is Q # C 2,which is just half the energy output of the battery. It is left as a problem (Problem 64)to show that the remaining half of the energy supplied by the battery appears as inter-nal energy in the resistor.

Discharging a Capacitor

Now consider the circuit shown in Figure 28.21, which consists of a capacitor carryingan initial charge Q , a resistor, and a switch. When the switch is open, a potential differ-ence Q /C exists across the capacitor and there is zero potential difference across theresistor because I # 0. If the switch is closed at t # 0, the capacitor begins to dischargethrough the resistor. At some time t during the discharge, the current in the circuit is Iand the charge on the capacitor is q (Fig. 28.21b). The circuit in Figure 28.21 is thesame as the circuit in Figure 28.19 except for the absence of the battery. Thus, we elim-inate the emf from Equation 28.11 to obtain the appropriate loop equation for thecircuit in Figure 28.21:

(28.16)

When we substitute I # dq/dt into this expression, it becomes

Integrating this expression, using the fact that q # Q at t # 0 gives

(28.17)

Differentiating this expression with respect to time gives the instantaneous current as afunction of time:

(28.18)

where Q /RC # I0 is the initial current. The negative sign indicates that as the capaci-tor discharges, the current direction is opposite its direction when the capacitor wasbeing charged. (Compare the current directions in Figs. 28.19c and 28.21b.) We seethat both the charge on the capacitor and the current decay exponentially at a ratecharacterized by the time constant - # RC.

I(t ) #dqdt

#ddt

(Qe$t/RC ) # $Q

RC e$t/RC

q(t ) # Qe$t/RC

ln ! qQ " # $

tRC

$q

Q dqq

# $1

RC $t

0 dt

dqq

# $1

RC dt

$R dqdt

#qC

$qC

$ IR # 0

!

!12!1

2

!!

[-] # [RC ] # % "VI

.Q

"V & # % QQ / "t & # ["t] # T

876 CHAPTE R 28 • Direct Current Circuits

(a)

S

RC

t < 0

–Q

+Q

R

S

I–q

+qC

(b)t > 0

Active Figure 28.21 (a) Acharged capacitor connected to aresistor and a switch, which is openfor t * 0. (b) After the switch isclosed at t # 0, a current thatdecreases in magnitude with time isset up in the direction shown, andthe charge on the capacitordecreases exponentially with time.

At the Active Figures linkat http://www.pse6.com, youcan adjust the values of R andC to see the effect on thedischarging of the capacitor.

Charge as a function of time fora discharging capacitor

Current as a function of time fora discharging capacitor

Quick Quiz 28.9 Consider the circuit in Figure 28.19 and assume that thebattery has no internal resistance. Just after the switch is closed, the potential differ-ence across which of the following is equal to the emf of the battery? (a) C (b) R(c) neither C nor R. After a very long time, the potential difference across which of thefollowing is equal to the emf of the battery? (d) C (e) R (f) neither C nor R.

S ECT I O N 28 . 4 • RC Circuits 877

Quick Quiz 28.10 Consider the circuit in Figure 28.22 and assume that thebattery has no internal resistance. Just after the switch is closed, the current in the bat-tery is (a) zero (b) /2R (c) 2 /R (d) /R (e) impossible to determine. After a verylong time, the current in the battery is (f) zero (g) /2R (h) 2 /R (i) /R (j) impos-sible to determine.

!!!!!!

ε

C

R R

Figure 28.22 (Quick Quiz 28.10) How does the currentvary after the switch is closed?

Conceptual Example 28.11 Intermittent Windshield Wipers

Many automobiles are equipped with windshield wipers thatcan operate intermittently during a light rainfall. How doesthe operation of such wipers depend on the charging anddischarging of a capacitor?

Solution The wipers are part of an RC circuit whose timeconstant can be varied by selecting different values of R

through a multiposition switch. As it increases with time, thevoltage across the capacitor reaches a point at which ittriggers the wipers and discharges, ready to begin anothercharging cycle. The time interval between the individualsweeps of the wipers is determined by the value of the timeconstant.

Example 28.12 Charging a Capacitor in an RC Circuit

An uncharged capacitor and a resistor are connected inseries to a battery, as shown in Figure 28.23. If # 12.0 V,C # 5.00 )F, and R # 8.00 . 105 ', find the time constantof the circuit, the maximum charge on the capacitor, themaximum current in the circuit, and the charge and currentas functions of time.

Solution The time constant of the circuit is - # RC #(8.00 . 105 ')(5.00 . 10$6 F) # 4.00 s. The maximumcharge on the capacitor is Q # C # (5.00 )F)(12.0 V) #60.0 )C. The maximum current in the circuit is I 0 #

/R # (12.0 V)/(8.00 . 105 ') # 15.0 )A. Using thesevalues and Equations 28.14 and 28.15, we find that

Graphs of these functions are provided in Figure 28.24.

(15.0 )A)e$t/4.00 sI(t ) #

(60.0 )C)(1 $ e$t/4.00 s)q(t ) #

!

!

!

Interactive

R

εC

+ – S

0 1 2 3 4 5 6 70

10

20

30

40

50

60q(µC)

t(s)

0 1 2 3 4 5 6 70

5

10

15I(µA)

t(s)

(a)

(b)

µ

µ

Q = 60.0 µCµ

I 0 = 15.0 µAµt = τ

At the Interactive Worked Example link at http://www.pse6.com, you can vary R, C, and ! and observe the charge andcurrent as functions of time while charging or discharging the capacitor.

Figure 28.23 (Example 28.12) The switch in this series RCcircuit, open for times t * 0, is closed at t # 0.

Figure 28.24 (Example 28.12) Plots of (a) charge versus timeand (b) current versus time for the RC circuit shown in Figure28.23, with # 12.0 V, R # 8.00 . 105 ', and C # 5.00 )F.!

878 CHAPTE R 28 • Direct Current Circuits

Example 28.13 Discharging a Capacitor in an RC Circuit

Consider a capacitor of capacitance C that is being dis-charged through a resistor of resistance R , as shown inFigure 28.21.

(A) After how many time constants is the charge on thecapacitor one-fourth its initial value?

Solution The charge on the capacitor varies with timeaccording to Equation 28.17, q(t) # Qe$t/RC. To find the timeinterval during which q drops to one-fourth its initial value,we substitute q(t) # Q /4 into this expression and solve for t :

Taking logarithms of both sides, we find

(B) The energy stored in the capacitor decreases with timeas the capacitor discharges. After how many time constantsis this stored energy one-fourth its initial value?

Solution Using Equations 26.11 (U # Q2/2C) and 28.17, wecan express the energy stored in the capacitor at any time t as

U #q 2

2C#

Q 2

2C e$2t/RC # U0e$2t/RC

1.39-t # RC (ln 4) # 1.39RC #

$ln 4 # $t

RC

14 # e$t/RC

Q4

# Qe$t/RC

where U0 # Q2/2C is the initial energy stored in the capaci-tor. As in part (A), we now set U # U0/4 and solve for t :

Again, taking logarithms of both sides and solving for t gives

What If? What if we wanted to describe the circuit in termsof the time interval required for the charge to fall to one-halfits original value, rather than by the time constant "? Thiswould give a parameter for the circuit called its half-life t1/2.How is the half-life related to the time constant?

Answer After one half-life, the charge has fallen from Q toQ /2. Thus, from Equation 28.17,

leading to

t1/2 # 0.693-

The concept of half-life will be important to us when westudy nuclear decay in Chapter 44. The radioactive decay ofan unstable sample behaves in a mathematically similarmanner to a discharging capacitor in an RC circuit.

12 # e$t1/2/RC

Q2

# Qe$t 1/2/RC

0.693-t # 12 RC ln 4 # 0.693RC #

14 # e$2t/RC

U0

4 # U0e$2t/RC

Example 28.14 Energy Delivered to a Resistor

A 5.00-)F capacitor is charged to a potential difference of800 V and then discharged through a 25.0-kV resistor. Howmuch energy is delivered to the resistor in the time intervalrequired to fully discharge the capacitor?

Solution We shall solve this problem in two ways. The firstway is to note that the initial energy in the circuit equals theenergy stored in the capacitor, C 2/2 (see Eq. 26.11). Oncethe capacitor is fully discharged, the energy stored in it iszero. Because energy in an isolated system is conserved, theinitial energy stored in the capacitor is transformed into in-ternal energy in the resistor. Using the given values of C and

, we find

The second way, which is more difficult but perhaps moreinstructive, is to note that as the capacitor dischargesthrough the resistor, the rate at which energy is delivered tothe resistor is given by I 2R, where I is the instantaneouscurrent given by Equation 28.18. Because power is definedas the rate at which energy is transferred, we conclude that

1.60 JEnergy # 12C!2 # 1

2(5.00 . 10$6 F)(800 V)2 #

!

!

the energy delivered to the resistor must equal the time inte-gral of I 2R dt :

To evaluate this integral, we note that the initial current I0 isequal to /R and that all parameters except t are constant.Thus, we find

This integral has a value of RC/2 (see Problem 35); hence,we find

which agrees with the result we obtained using the simplerapproach, as it must. Note that we can use this secondapproach to find the total energy delivered to the resistor atany time after the switch is closed by simply replacing theupper limit in the integral with that specific value of t.

Energy # 12C!2

(1) Energy #!2

R $,

0 e$2t/RC dt

!

Energy # $,

0 I 2R dt # $,

0 ($I0e$t/RC )2 R dt

28.5 Electrical Meters

The Galvanometer

The galvanometer is the main component in analog meters for measuring current andvoltage. (Many analog meters are still in use although digital meters, which operate on adifferent principle, are currently in wide use.) Figure 28.25 illustrates the essentialfeatures of a common type called the D’Arsonval galvanometer. It consists of a coil of wiremounted so that it is free to rotate on a pivot in a magnetic field provided by a perma-nent magnet. The basic operation of the galvanometer uses the fact that a torque actson a current loop in the presence of a magnetic field (Chapter 29). The torque experi-enced by the coil is proportional to the current in it: the larger the current, the greaterthe torque and the more the coil rotates before the spring tightens enough to stop therotation. Hence, the deflection of a needle attached to the coil is proportional to thecurrent. Once the instrument is properly calibrated, it can be used in conjunction withother circuit elements to measure either currents or potential differences.

The Ammeter

A device that measures current is called an ammeter. The charges constituting thecurrent to be measured must pass directly through the ammeter, so the ammetermust be connected in series with other elements in the circuit, as shown in Figure28.26. When using an ammeter to measure direct currents, you must connect it sothat charges enter the instrument at the positive terminal and exit at the negativeterminal.

Ideally, an ammeter should have zero resistance so that the current beingmeasured is not altered. In the circuit shown in Figure 28.26, this condition requiresthat the resistance of the ammeter be much less than R1 % R2. Because any ammeteralways has some internal resistance, the presence of the ammeter in the circuit slightlyreduces the current from the value it would have in the meter’s absence.

A typical off-the-shelf galvanometer is often not suitable for use as an ammeter,primarily because it has a resistance of about 60 '. An ammeter resistance this greatconsiderably alters the current in a circuit. You can understand this by consideringthe following example. The current in a simple series circuit containing a 3-Vbattery and a 3-' resistor is 1 A. If you insert a 60-' galvanometer in this circuit tomeasure the current, the total resistance becomes 63 ' and the current is reducedto 0.048 A!

A second factor that limits the use of a galvanometer as an ammeter is the fact thata typical galvanometer gives a full-scale deflection for currents on the order of 1 mA orless. Consequently, such a galvanometer cannot be used directly to measure currentsgreater than this value. However, it can be converted to a useful ammeter by placing ashunt resistor Rp in parallel with the galvanometer, as shown in Figure 28.27. The valueof Rp must be much less than the galvanometer resistance so that most of the currentto be measured is directed to the shunt resistor.

The Voltmeter

A device that measures potential difference is called a voltmeter. The potential differ-ence between any two points in a circuit can be measured by attaching the terminals ofthe voltmeter between these points without breaking the circuit, as shown in Figure28.28. The potential difference across resistor R2 is measured by connecting the volt-meter in parallel with R2. Again, it is necessary to observe the polarity of the instru-ment. The positive terminal of the voltmeter must be connected to the end of theresistor that is at the higher potential, and the negative terminal to the end of the resis-tor at the lower potential.

S ECT I O N 28 . 5 • Electrical Meters 879

Spring

S

Coil

Scale

N

Figure 28.25 The principal compo-nents of a D’Arsonval galvanometer.When the coil situated in a magneticfield carries a current, the magnetictorque causes the coil to twist. Theangle through which the coil rotatesis proportional to the current in thecoil because of the counteractingtorque of the spring.

R1

ε

–

+

R2

A

Figure 28.26 Current can be mea-sured with an ammeter connected inseries with the elements in which themeasurement of a current is desired.An ideal ammeter has zero resistance.

Rp

Galvanometer

60 Ω

Active Figure 28.27 A galva-nometer is represented here by itsinternal resistance of 60 '. When agalvanometer is to be used as anammeter, a shunt resistor Rp isconnected in parallel with thegalvanometer.

At the Active Figures linkat http://www.pse6.com, you canpredict the value of Rp neededto cause full-scale deflection inthe circuit of Figure 28.26, andtest your result.

An ideal voltmeter has infinite resistance so that no current exists in it. InFigure 28.28, this condition requires that the voltmeter have a resistance much greaterthan R2. In practice, if this condition is not met, corrections should be made for theknown resistance of the voltmeter.

A galvanometer can also be used as a voltmeter by adding an external resistor Rs inseries with it, as shown in Figure 28.29. In this case, the external resistor must have avalue much greater than the resistance of the galvanometer to ensure that the gal-vanometer does not significantly alter the voltage being measured.

28.6 Household Wiring and Electrical Safety

Household circuits represent a practical application of some of the ideas presented inthis chapter. In our world of electrical appliances, it is useful to understand the powerrequirements and limitations of conventional electrical systems and the safetymeasures that prevent accidents.

In a conventional installation, the utility company distributes electric power to indi-vidual homes by means of a pair of wires, with each home connected in parallel tothese wires. One wire is called the live wire,5 as illustrated in Figure 28.30, and the otheris called the neutral wire. The neutral wire is grounded; that is, its electric potential istaken to be zero. The potential difference between the live and neutral wires is about120 V. This voltage alternates in time, and the potential of the live wire oscillates rela-tive to ground. Much of what we have learned so far for the constant-emf situation(direct current) can also be applied to the alternating current that power companiessupply to businesses and households. (Alternating voltage and current are discussed inChapter 33.)

A meter is connected in series with the live wire entering the house to record thehousehold’s energy consumption. After the meter, the wire splits so that there areseveral separate circuits in parallel distributed throughout the house. Each circuitcontains a circuit breaker (or, in older installations, a fuse). The wire and circuitbreaker for each circuit are carefully selected to meet the current demands for thatcircuit. If a circuit is to carry currents as large as 30 A, a heavy wire and an appropriatecircuit breaker must be selected to handle this current. A circuit used to power onlylamps and small appliances often requires only 20 A. Each circuit has its own circuitbreaker to provide protection for that part of the entire electrical system of the house.

880 CHAPTE R 28 • Direct Current Circuits

Galvanometer

Rs

60 Ω

Active Figure 28.29 When the galvanometeris used as a voltmeter, a resistor Rs isconnected in series with the galvanometer.

At the Active Figures link athttp://www.pse6.com, you can predictthe value of Rs needed to cause full-scale deflection in the circuit of Figure28.28, and test your result.

R1

ε

V

R2

Figure 28.28 The potential difference acrossa resistor can be measured with a voltmeterconnected in parallel with the resistor. Anideal voltmeter has infinite resistance.

5 Live wire is a common expression for a conductor whose electric potential is above or below groundpotential.

R1

Live120 V

Neutral

0 V

R2

Circuitbreaker

Meter

R3

Figure 28.30 Wiring diagram for ahousehold circuit. The resistancesrepresent appliances or otherelectrical devices that operate withan applied voltage of 120 V.

As an example, consider a circuit in which a toaster oven, a microwave oven, and acoffee maker are connected (corresponding to R1, R2, and R3 in Fig. 28.30). We cancalculate the current in each appliance by using the expression ! # I "V. The toasteroven, rated at 1 000 W, draws a current of 1 000 W/120 V # 8.33 A. The microwaveoven, rated at 1 300 W, draws 10.8 A, and the coffee maker, rated at 800 W, draws6.67 A. If the three appliances are operated simultaneously, they draw a total current of25.8 A. Therefore, the circuit should be wired to handle at least this much current. Ifthe rating of the circuit breaker protecting the circuit is too small—say, 20 A—thebreaker will be tripped when the third appliance is turned on, preventing all threeappliances from operating. To avoid this situation, the toaster oven and coffee makercan be operated on one 20-A circuit and the microwave oven on a separate 20-A circuit.

Many heavy-duty appliances, such as electric ranges and clothes dryers, require 240 Vfor their operation. The power company supplies this voltage by providing a third wirethat is 120 V below ground potential (Fig. 28.31). The potential difference betweenthis live wire and the other live wire (which is 120 V above ground potential) is 240 V.An appliance that operates from a 240-V line requires half as much current compared tooperating it at 120 V; therefore, smaller wires can be used in the higher-voltage circuitwithout overheating.

Electrical Safety

When the live wire of an electrical outlet is connected directly to ground, the circuit iscompleted and a short-circuit condition exists. A short circuit occurs when almost zeroresistance exists between two points at different potentials; this results in a very largecurrent. When this happens accidentally, a properly operating circuit breaker opensthe circuit and no damage is done. However, a person in contact with ground can beelectrocuted by touching the live wire of a frayed cord or other exposed conductor. Anexceptionally effective (and dangerous!) ground contact is made when the personeither touches a water pipe (normally at ground potential) or stands on the groundwith wet feet. The latter situation represents effective ground contact because normal,nondistilled water is a conductor due to the large number of ions associated withimpurities. This situation should be avoided at all cost.

Electric shock can result in fatal burns, or it can cause the muscles of vital organs,such as the heart, to malfunction. The degree of damage to the body depends on themagnitude of the current, the length of time it acts, the part of the body touched bythe live wire, and the part of the body in which the current exists. Currents of 5 mA orless cause a sensation of shock but ordinarily do little or no damage. If the current islarger than about 10 mA, the muscles contract and the person may be unable torelease the live wire. If a current of about 100 mA passes through the body for only afew seconds, the result can be fatal. Such a large current paralyzes the respiratory

S ECT I O N 28 . 6 • Household Wiring and Electrical Safety 881

(a)

+120 V –120 V

(b)

Figure 28.31 (a) An outlet for connection to a 240-V supply. (b) The connections foreach of the openings in a 240-V outlet.

Geor

ge S

empl

e

882 CHAPTE R 28 • Direct Current Circuits

(a)

Motor

“Live”

Walloutlet

Circuitbreaker

120 V

“Neutral”

Ground

“Ouch!”

I

II

(b)

Motor

“Live”

3-wireoutlet

Circuitbreaker

120 V

“Neutral”

GroundI

I

“Ground”

I

I

Figure 28.32 (a) A diagram of the circuit for an electric drill with only two connectingwires. The normal current path is from the live wire through the motor connectionsand back to ground through the neutral wire. In the situation shown, the live wire hascome into contact with the drill case. As a result, the person holding the drill acts as acurrent path to ground and receives an electric shock. (b) This shock can be avoidedby connecting the drill case to ground through a third ground wire. In this situation,the drill case remains at ground potential and no current exists in the person.

muscles and prevents breathing. In some cases, currents of about 1 A can produceserious (and sometimes fatal) burns. In practice, no contact with live wires is regardedas safe whenever the voltage is greater than 24 V.

Many 120-V outlets are designed to accept a three-pronged power cord. (Thisfeature is required in all new electrical installations.) One of these prongs is the livewire at a nominal potential of 120 V. The second is the neutral wire, nominally at 0 V,and carries current to ground. The third, round prong is a safety ground wire thatnormally carries no current but is both grounded and connected directly to the casingof the appliance (see Figure 28.32). If the live wire is accidentally shorted to the casing(which can occur if the wire insulation wears off), most of the current takes the low-resistance path through the appliance to ground. In contrast, if the casing of the appli-ance is not properly grounded and a short occurs, anyone in contact with theappliance experiences an electric shock because the body provides a low-resistancepath to ground.

Special power outlets called ground-fault interrupters (GFIs) are now used inkitchens, bathrooms, basements, exterior outlets, and other hazardous areas of newhomes. These devices are designed to protect persons from electric shock by sensingsmall currents (' 5 mA) leaking to ground. (The principle of their operation isdescribed in Chapter 31.) When an excessive leakage current is detected, the current isshut off in less than 1 ms.

Summary 883

The emf of a battery is equal to the voltage across its terminals when the current iszero. That is, the emf is equivalent to the open-circuit voltage of the battery.

The equivalent resistance of a set of resistors connected in series is

Req # R1 % R2 % R3 % … (28.6)

The equivalent resistance of a set of resistors connected in parallel is foundfrom the relationship

(28.8)

If it is possible to combine resistors into series or parallel equivalents, the precedingtwo equations make it easy to determine how the resistors influence the rest of thecircuit.

Circuits involving more than one loop are conveniently analyzed with the use ofKirchhoff’s rules:

1. Junction rule. The sum of the currents entering any junction in an electric circuitmust equal the sum of the currents leaving that junction:

(28.9)

2. Loop rule. The sum of the potential differences across all elements around anycircuit loop must be zero:

(28.10)

The first rule is a statement of conservation of charge; the second is equivalent to astatement of conservation of energy.

When a resistor is traversed in the direction of the current, the potential difference"V across the resistor is $ IR . When a resistor is traversed in the direction opposite thecurrent, "V # % IR. When a source of emf is traversed in the direction of the emf(negative terminal to positive terminal), the potential difference is % . When asource of emf is traversed opposite the emf (positive to negative), the potential differ-ence is $ . The use of these rules together with Equations 28.9 and 28.10 allows youto analyze electric circuits.

If a capacitor is charged with a battery through a resistor of resistance R , thecharge on the capacitor and the current in the circuit vary in time according to theexpressions

(28.14)

(28.15)

where Q # C is the maximum charge on the capacitor. The product RC is called thetime constant - of the circuit. If a charged capacitor is discharged through a resistorof resistance R , the charge and current decrease exponentially in time according tothe expressions

(28.17)

(28.18)

where Q is the initial charge on the capacitor and I0 # Q /RC is the initial current inthe circuit.

I(t ) # $I0e$t/RC

q(t ) # Qe$t/RC

!

I(t ) #!R

e$t/RC

q(t ) # Q(1 $ e$t/RC )

!

!

#closedloop

"V # 0

# I in # # Iout

1Req

#1

R1%

1R2

%1R3

% ( ( (

S U M M A RY

Take a practice test forthis chapter by clicking onthe Practice Test link athttp://www.pse6.com.

884 CHAPTE R 28 • Direct Current Circuits

SwitchBattery+ –

C

Figure Q28.13

Figure Q 28.19

Henr

y Lea

p an

d Ji

m Le

hman

1. Explain the difference between load resistance in a circuitand internal resistance in a battery.

2. Under what condition does the potential difference acrossthe terminals of a battery equal its emf? Can the terminalvoltage ever exceed the emf? Explain.

Is the direction of current through a battery always fromthe negative terminal to the positive terminal? Explain.

4. How would you connect resistors so that the equivalentresistance is larger than the greatest individual resistance?Give an example involving three resistors.

5. How would you connect resistors so that the equivalentresistance is smaller than the least individual resistance?Give an example involving three resistors.

6. Given three lightbulbs and a battery, sketch as many differ-ent electric circuits as you can.

7. When resistors are connected in series, which of the follow-ing would be the same for each resistor: potential differ-ence, current, power?

8. When resistors are connected in parallel, which of the fol-lowing would be the same for each resistor: potential dif-ference, current, power?

9. What advantage might there be in using two identicalresistors in parallel connected in series with anotheridentical parallel pair, rather than just using a singleresistor?

10. An incandescent lamp connected to a 120-V source with ashort extension cord provides more illumination than thesame lamp connected to the same source with a very longextension cord. Explain.

Why is it possible for a bird to sit on a high-voltage wirewithout being electrocuted?

12. When can the potential difference across a resistor bepositive?

Referring to Figure Q28.13, describe what happens to thelightbulb after the switch is closed. Assume that the capaci-tor has a large capacitance and is initially uncharged, andassume that the light illuminates when connected directlyacross the battery terminals.

13.

11.

3.

14. What is the internal resistance of an ideal ammeter? Ofan ideal voltmeter? Do real meters ever attain theseideals?

15. A “short circuit” is a path of very low resistance in a circuitin parallel with some other part of the circuit. Discuss theeffect of the short circuit on the portion of the circuit itparallels. Use a lamp with a frayed cord as an example.

16. If electric power is transmitted over long distances, theresistance of the wires becomes significant. Why? Whichmethod of transmission would result in less energywasted—high current and low voltage or low current andhigh voltage? Explain your answer.

17. Are the two headlights of a car wired in series or in paral-lel? How can you tell?

18. Embodied in Kirchhoff’s rules are two conservation laws.What are they?

19. Figure Q28.19 shows a series combination of threelightbulbs, all rated at 120 V with power ratings of 60 W,75 W, and 200 W. Why is the 60-W lamp the brightest andthe 200-W lamp the dimmest? Which bulb has the greatestresistance? How would their intensities differ if they wereconnected in parallel?

Q U E S T I O N S

20. A student claims that the second lightbulb in series is lessbright than the first, because the first bulb uses up some ofthe current. How would you respond to this statement?

21. Is a circuit breaker wired in series or in parallel with thedevice it is protecting?

22. So that your grandmother can listen to A Prairie Home Com-panion, you take her bedside radio to the hospital whereshe is staying. You are required to have a maintenanceworker test it for electrical safety. Finding that it develops120 V on one of its knobs, he does not let you take it up toyour grandmother’s room. She complains that she has hadthe radio for many years and nobody has ever gottena shock from it. You end up having to buy a new plasticradio. Is this fair? Will the old radio be safe back in herbedroom?

Problems 885

A

S

B C

ε

Figure Q28.29

Section 28.1 Electromotive Force

A battery has an emf of 15.0 V. The terminal voltageof the battery is 11.6 V when it is delivering 20.0 W ofpower to an external load resistor R. (a) What is the valueof R? (b) What is the internal resistance of the battery?

2. (a) What is the current in a 5.60-' resistor connected to abattery that has a 0.200-' internal resistance if the termi-nal voltage of the battery is 10.0 V? (b) What is the emf ofthe battery?

3. Two 1.50-V batteries — with their positive terminals in thesame direction — are inserted in series into the barrel of aflashlight. One battery has an internal resistance of 0.255 ',the other an internal resistance of 0.153 '. When theswitch is closed, a current of 600 mA occurs in the lamp.(a) What is the lamp’s resistance? (b) What fraction of thechemical energy transformed appears as internal energy inthe batteries?

4. An automobile battery has an emf of 12.6 V and an inter-nal resistance of 0.080 0 '. The headlights togetherpresent equivalent resistance 5.00 ' (assumed constant).What is the potential difference across the headlight bulbs(a) when they are the only load on the battery and

1.

(b) when the starter motor is operated, taking an addi-tional 35.0 A from the battery?

Section 28.2 Resistors in Series and Parallel5. The current in a loop circuit that has a resistance of R1 is

2.00 A. The current is reduced to 1.60 A when an addi-tional resistor R 2 # 3.00 ' is added in series with R1.What is the value of R1?

6. (a) Find the equivalent resistance between points a and bin Figure P28.6. (b) A potential difference of 34.0 V isapplied between points a and b. Calculate the current ineach resistor.

1, 2, 3 = straightforward, intermediate, challenging = full solution available in the Student Solutions Manual and Study Guide

= coached solution with hints available at http://www.pse.com = computer useful in solving problem

= paired numerical and symbolic problems

P R O B L E M S

9.00 Ω4.00 Ω

10.0 Ω

7.00 Ω

baFigure P28.6

23. Suppose you fall from a building and on the way downgrab a high-voltage wire. If the wire supports you as youhang from it, will you be electrocuted? If the wire thenbreaks, should you continue to hold onto an end of thewire as you fall?

What advantage does 120-V operation offer over 240 V?What disadvantages?

25. When electricians work with potentially live wires, theyoften use the backs of their hands or fingers to move wires.Why do you suppose they use this technique?

26. What procedure would you use to try to save a person whois “frozen” to a live high-voltage wire without endangeringyour own life?

27. If it is the current through the body that determineshow serious a shock will be, why do we see warnings ofhigh voltage rather than high current near electricalequipment?

28. Suppose you are flying a kite when it strikes a high-voltagewire. What factors determine how great a shock youreceive?

29. A series circuit consists of three identical lamps connectedto a battery as shown in Figure Q28.29. When the switch Sis closed, what happens (a) to the intensities of lamps Aand B; (b) to the intensity of lamp C; (c) to the current inthe circuit; and (d) to the voltage across the three lamps?(e) Does the power delivered to the circuit increase,decrease, or remain the same?

24.

30. If your car’s headlights are on when you start the ignition,why do they dim while the car is starting?

31. A ski resort consists of a few chair lifts and several inter-connected downhill runs on the side of a mountain, with alodge at the bottom. The lifts are analogous to batteries,and the runs are analogous to resistors. Describe how tworuns can be in series. Describe how three runs can be inparallel. Sketch a junction of one lift and two runs. StateKirchhoff’s junction rule for ski resorts. One of the skiershappens to be carrying a skydiver’s altimeter. She nevertakes the same set of lifts and runs twice, but keeps passingyou at the fixed location where you are working. StateKirchhoff’s loop rule for ski resorts.

7. A lightbulb marked “75 W [at] 120 V” is screwed into asocket at one end of a long extension cord, in which eachof the two conductors has resistance 0.800 '. The otherend of the extension cord is plugged into a 120-V outlet.Draw a circuit diagram and find the actual power deliv-ered to the bulb in this circuit.

8. Four copper wires of equal length are connected inseries. Their cross-sectional areas are 1.00 cm2, 2.00 cm2,3.00 cm2, and 5.00 cm2. A potential difference of 120 V isapplied across the combination. Determine the voltageacross the 2.00-cm2 wire.

Consider the circuit shown in Figure P28.9. Find(a) the current in the 20.0-' resistor and (b) the potentialdifference between points a and b.

9.

10. For the purpose of measuring the electric resistance ofshoes through the body of the wearer to a metal groundplate, the American National Standards Institute (ANSI)specifies the circuit shown in Figure P28.10. The potentialdifference "V across the 1.00-M' resistor is measured witha high-resistance voltmeter. (a) Show that the resistance ofthe footwear is given by

(b) In a medical test, a current through the human bodyshould not exceed 150 )A. Can the current delivered bythe ANSI-specified circuit exceed 150 )A? To decide,consider a person standing barefoot on the ground plate.

R shoes # 1.00 M' ! 50.0 V $ "V"V "

12. Using only three resistors—2.00 ', 3.00 ', and 4.00 '—find 17 resistance values that may be obtained by variouscombinations of one or more resistors. Tabulate the com-binations in order of increasing resistance.

13. The current in a circuit is tripled by connecting a 500-'resistor in parallel with the resistance of the circuit. Deter-mine the resistance of the circuit in the absence of the500-' resistor.

14. A 6.00-V battery supplies current to the circuit shown inFigure P28.14. When the double-throw switch S is open, asshown in the figure, the current in the battery is 1.00 mA.When the switch is closed in position 1, the current in thebattery is 1.20 mA. When the switch is closed in position 2,the current in the battery is 2.00 mA. Find the resistancesR1, R2, and R3.

886 CHAPTE R 28 • Direct Current Circuits

V

1.00 MΩ

50.0 V

Figure P28.10

a

100 Ω

100 Ω

100 Ω

b

Figure P28.11

2.00 Ω

18.0 V3.00 Ω

4.00 Ω

1.00 Ω

Figure P28.15

R1 R2

R3

6.00 V S

R2

1

2

Figure P28.14

11. Three 100-' resistors are connected as shown in FigureP28.11. The maximum power that can safely be deliv-ered to any one resistor is 25.0 W. (a) What is the maxi-mum voltage that can be applied to the terminals a andb? For the voltage determined in part (a), what is thepower delivered to each resistor? What is the total powerdelivered?

20.0 Ω

a 10.0 Ω

10.0 Ω 25.0 V

5.00 Ω

b

5.00 Ω

Figure P28.9

Calculate the power delivered to each resistor in thecircuit shown in Figure P28.15.

15.

Problems 887

16. Two resistors connected in series have an equivalent resis-tance of 690 '. When they are connected in parallel, theirequivalent resistance is 150 '. Find the resistance of eachresistor.

17. An electric teakettle has a multiposition switch and twoheating coils. When only one of the coils is switched on, thewell-insulated kettle brings a full pot of water to a boil overthe time interval "t. When only the other coil is switchedon, it requires a time interval of 2"t to boil the sameamount of water. Find the time interval required to boil thesame amount of water if both coils are switched on (a) in aparallel connection and (b) in a series connection.

18. In Figures 28.4 and 28.6, let R1 # 11.0 ', R2 # 22.0 ',and let the battery have a terminal voltage of 33.0 V. (a) Inthe parallel circuit shown in Figure 28.6, to which resistoris more power delivered? (b) Verify that the sum of thepower (I 2R) delivered to each resistor equals the powersupplied by the battery (! # I"V ). (c) In the seriescircuit, which resistor uses more power? (d) Verify that thesum of the power (I 2R) used by each resistor equals thepower supplied by the battery (! # I"V ). (e) Whichcircuit configuration uses more power?

19. Four resistors are connected to a battery as shown inFigure P28.19. The current in the battery is I, the batteryemf is , and the resistor values are R1 # R , R2 # 2R ,R3 # 4R, R4 # 3R. (a) Rank the resistors according to thepotential difference across them, from largest to smallest.Note any cases of equal potential differences. (b) Deter-mine the potential difference across each resistor in termsof . (c) Rank the resistors according to the current inthem, from largest to smallest. Note any cases of equalcurrents. (d) Determine the current in each resistor interms of I. (e) What If? If R3 is increased, what happens tothe current in each of the resistors? (f) In the limit thatR3 : ,, what are the new values of the current in eachresistor in terms of I, the original current in the battery?

!

!

Determine the current in each branch of the circuitshown in Figure P28.21.

21.

22. In Figure P28.21, show how to add just enough ammetersto measure every different current. Show how to add justenough voltmeters to measure the potential differenceacross each resistor and across each battery.

23. The circuit considered in Problem 21 and shown in FigureP28.21 is connected for 2.00 min. (a) Find the energydelivered by each battery. (b) Find the energy delivered toeach resistor. (c) Identify the types of energy transformationsthat occur in the operation of the circuit and the totalamount of energy involved in each type of transformation.

24. Using Kirchhoff’s rules, (a) find the current in each resistorin Figure P28.24. (b) Find the potential difference betweenpoints c and f. Which point is at the higher potential?

7.00 Ω 15.0 V

5.00 Ω

2.00 Ω εI2

I1

A

Figure P28.20

60.0 V70.0 V 80.0 V

R2

a f e

R3

3.00 kΩ

2.00 kΩ

4.00 kΩcb d

ε1ε ε2ε ε3ε

R1

Figure P28.24

3.00 Ω

1.00 Ω

5.00 Ω

1.00 Ω

4.00 V+

8.00 Ω

12.0 V+

$

$

Figure P28.21 Problems 21, 22, and 23.

R2 = 2R

R3 = 4R

R1 = R

ε

R4 = 3R

I

Figure P28.19

Section 28.3 Kirchhoff’s Rules

20. The ammeter shown in Figure P28.20 reads 2.00 A. FindI1, I2, and .!

Note: The currents are not necessarily in the directionshown for some circuits.

888 CHAPTE R 28 • Direct Current Circuits

25. Taking R # 1.00 k' and # 250 V in Figure P28.25,determine the direction and magnitude of the current inthe horizontal wire between a and e.

!

26. In the circuit of Figure P28.26, determine the current ineach resistor and the voltage across the 200-' resistor.

27. A dead battery is charged by connecting it to the live batteryof another car with jumper cables (Fig. P28.27). Determinethe current in the starter and in the dead battery.

28. For the network shown in Figure P28.28, show that theresistance Rab # (27/17) '.

30. Calculate the power delivered to each resistor shown inFigure P28.30.

Section 28.4 RC CircuitsConsider a series RC circuit (see Fig. 28.19) for which

R # 1.00 M', C # 5.00 )F, and # 30.0 V. Find (a) thetime constant of the circuit and (b) the maximum chargeon the capacitor after the switch is closed. (c) Find thecurrent in the resistor 10.0 s after the switch is closed.

32. A 2.00-nF capacitor with an initial charge of 5.10 )C isdischarged through a 1.30-k' resistor. (a) Calculate thecurrent in the resistor 9.00 )s after the resistor is con-nected across the terminals of the capacitor. (b) Whatcharge remains on the capacitor after 8.00 )s? (c) What isthe maximum current in the resistor?

33. A fully charged capacitor stores energy U0. How muchenergy remains when its charge has decreased to half itsoriginal value?

34. A capacitor in an RC circuit is charged to 60.0% of itsmaximum value in 0.900 s. What is the time constant ofthe circuit?

35. Show that the integral in Equation (1) of Example 28.14has the value RC/2.

36. In the circuit of Figure P28.36, the switch S has been openfor a long time. It is then suddenly closed. Determine thetime constant (a) before the switch is closed and (b) afterthe switch is closed. (c) Let the switch be closed at t # 0.Determine the current in the switch as a function of time.

!31.

1.0 Ω

1.0 Ω 1.0 Ω

5.0 Ω3.0 Ω

a b

Figure P28.28

4.00 Ω

b

a

2.00 Ω

6.00 Ω8.00 V

12.0 V

Figure P28.29

50.0 kΩ

100 kΩ

10.0 VS

10.0 Fµ

Figure P28.36

2.0 Ω

20 V50 V

2.0 Ω

4.0 Ω 4.0 Ω

Figure P28.30

0.01 Ω

Livebattery

+–

+–

1.00 Ω0.06 ΩStarter

Deadbattery

12 V 10 V

Figure P28.27

ε

R

a

b2R

3R4R

c d

e

+–

+– ε2

Figure P28.25

80 Ω200 Ω 20 Ω 70 Ω

40 V 360 V 80 V

Figure P28.26

For the circuit shown in Figure P28.29, calculate (a) thecurrent in the 2.00-' resistor and (b) the potential differ-ence between points a and b.

29.

Problems 889

The circuit in Figure P28.37 has been connected for a longtime. (a) What is the voltage across the capacitor? (b) If thebattery is disconnected, how long does it take the capacitorto discharge to one tenth of its initial voltage?

37. Section 28.5 Electrical MetersAssume that a galvanometer has an internal resistance of60.0 ' and requires a current of 0.500 mA to produce full-scale deflection. What resistance must be connected inparallel with the galvanometer if the combination is toserve as an ammeter that has a full-scale deflection fora current of 0.100 A?

42. A typical galvanometer, which requires a current of1.50 mA for full-scale deflection and has a resistance of75.0 ', may be used to measure currents of much greatervalues. To enable an operator to measure large currentswithout damage to the galvanometer, a relatively smallshunt resistor is wired in parallel with the galvanometer,as suggested in Figure 28.27. Most of the current thengoes through the shunt resistor. Calculate the value of theshunt resistor that allows the galvanometer to be used tomeasure a current of 1.00 A at full-scale deflection.(Suggestion: use Kirchhoff’s rules.)

The same galvanometer described in the previous problemmay be used to measure voltages. In this case a large resis-tor is wired in series with the galvanometer, as suggested inFigure 28.29. The effect is to limit the current in thegalvanometer when large voltages are applied. Most of thepotential drop occurs across the resistor placed in series.Calculate the value of the resistor that allows thegalvanometer to measure an applied voltage of 25.0 V atfull-scale deflection.

44. Meter loading. Work this problem to five-digit precision.Refer to Figure P28.44. (a) When a 180.00-' resistor isconnected across a battery of emf 6.000 0 V and internalresistance 20.000 ', what is the current in the resistor?What is the potential difference across it? (b) Supposenow an ammeter of resistance 0.500 00 ' and a voltmeterof resistance 20 000 ' are added to the circuit as shown inFigure P28.44b. Find the reading of each. (c) What If?Now one terminal of one wire is moved, as shown inFigure P28.44c. Find the new meter readings.

45. Design a multirange ammeter capable of full-scale deflec-tion for 25.0 mA, 50.0 mA, and 100 mA. Assume the metermovement is a galvanometer that has a resistance of25.0 ' and gives a full-scale deflection for 1.00 mA.

46. Design a multirange voltmeter capable of full-scaledeflection for 20.0 V, 50.0 V, and 100 V. Assume themeter movement is a galvanometer that has a resistanceof 60.0 ' and gives a full-scale deflection for a current of1.00 mA.

43.

41.

10.0 V

1.00 Ω 8.00 Ω

2.00 Ω4.00 Ω

1.00 µFµ

Figure P28.37

(a)180.00 Ω

20.000 Ω6.000 0 V

(b)

AV

(c)

AV

Figure P28.44

38. In places such as a hospital operating room and a factoryfor electronic circuit boards, electric sparks must beavoided. A person standing on a grounded floor andtouching nothing else can typically have a body capaci-tance of 150 pF, in parallel with a foot capacitance of80.0 pF produced by the dielectric soles of his or hershoes. The person acquires static electric charge frominteractions with furniture, clothing, equipment, packag-ing materials, and essentially everything else. The staticcharge is conducted to ground through the equivalentresistance of the two shoe soles in parallel with each other.A pair of rubber-soled street shoes can present an equiva-lent resistance of 5 000 M'. A pair of shoes with specialstatic-dissipative soles can have an equivalent resistance of1.00 M'. Consider the person’s body and shoes asforming an RC circuit with the ground. (a) How long doesit take the rubber-soled shoes to reduce a 3 000-V staticcharge to 100 V? (b) How long does it take the static-dissipative shoes to do the same thing?

39. A 4.00-M' resistor and a 3.00-)F capacitor are connectedin series with a 12.0-V power supply. (a) What is the timeconstant for the circuit? (b) Express the current in thecircuit and the charge on the capacitor as functions oftime.

40. Dielectric materials used in the manufacture of capacitorsare characterized by conductivities that are small but notzero. Therefore, a charged capacitor slowly loses its chargeby “leaking” across the dielectric. If a capacitor havingcapacitance C leaks charge such that the potential differ-ence has decreased to half its initial (t # 0) value at a timet, what is the equivalent resistance of the dielectric?

890 CHAPTE R 28 • Direct Current Circuits

47. A particular galvanometer serves as a 2.00-V full-scale volt-meter when a 2 500-' resistor is connected in series withit. It serves as a 0.500-A full-scale ammeter when a 0.220-'resistor is connected in parallel with it. Determine theinternal resistance of the galvanometer and the currentrequired to produce full-scale deflection.

Section 28.6 Household Wiring and Electrical Safety

48. An 8.00-ft extension cord has two 18-gauge copper wires,each having a diameter of 1.024 mm. At what rate isenergy delivered to the resistance in the cord when it iscarrying a current of (a) 1.00 A and (b) 10.0 A?

An electric heater is rated at 1 500 W, a toaster at750 W, and an electric grill at 1 000 W. The threeappliances are connected to a common 120-V householdcircuit. (a) How much current does each draw? (b) Is acircuit with a 25.0-A circuit breaker sufficient in thissituation? Explain your answer.

50. Aluminum wiring has sometimes been used instead ofcopper for economy. According to the National ElectricalCode, the maximum allowable current for 12-gaugecopper wire with rubber insulation is 20 A. What shouldbe the maximum allowable current in a 12-gaugealuminum wire if the power per unit length delivered tothe resistance in the aluminum wire is the same as thatdelivered in the copper wire?

51. Turn on your desk lamp. Pick up the cord, with yourthumb and index finger spanning the width of the cord.(a) Compute an order-of-magnitude estimate for thecurrent in your hand. You may assume that at a typicalinstant the conductor inside the lamp cord next to yourthumb is at potential ( 102 V and that the conductor nextto your index finger is at ground potential (0 V). Theresistance of your hand depends strongly on the thicknessand the moisture content of the outer layers of your skin.Assume that the resistance of your hand between finger-tip and thumb tip is ( 104 '. You may model the cord ashaving rubber insulation. State the other quantities youmeasure or estimate and their values. Explain your rea-soning. (b) Suppose that your body is isolated from anyother charges or currents. In order-of-magnitude termsdescribe the potential of your thumb where it contacts thecord, and the potential of your finger where it touchesthe cord.

Additional Problems52. Four 1.50-V AA batteries in series are used to power a tran-

sistor radio. If the batteries can move a charge of 240 C,how long will they last if the radio has a resistance of200 '?

53. A battery has an emf of 9.20 V and an internal resistanceof 1.20 '. (a) What resistance across the battery willextract from it a power of 12.8 W? (b) a power of 21.2 W?

54. Calculate the potential difference between points a and bin Figure P28.54 and identify which point is at the higherpotential.

49.55. Assume you have a battery of emf and three identical

lightbulbs, each having constant resistance R. What is thetotal power delivered by the battery if the bulbs are con-nected (a) in series? (b) in parallel? (c) For which connec-tion will the bulbs shine the brightest?

56. A group of students on spring break manages to reach adeserted island in their wrecked sailboat. They splashashore with fuel, a European gasoline-powered 240-V gen-erator, a box of North American 100-W 120-V lightbulbs, a500-W 120-V hot pot, lamp sockets, and some insulatedwire. While waiting to be rescued, they decide to use thegenerator to operate some lightbulbs. (a) Draw a diagramof a circuit they can use, containing the minimum numberof lightbulbs with 120 V across each bulb, and no highervoltage. Find the current in the generator and its poweroutput. (b) One student catches a fish and wants to cookit in the hot pot. Draw a diagram of a circuit containingthe hot pot and the minimum number of lightbulbs with120 V across each device, and not more. Find the currentin the generator and its power output.

A battery has an emf and internal resistance r. A variableload resistor R is connected across the terminals of the bat-tery. (a) Determine the value of R such that the potentialdifference across the terminals is a maximum. (b) Deter-mine the value of R so that the current in the circuit is amaximum. (c) Determine the value of R so that the powerdelivered to the load resistor is a maximum. Choosing theload resistance for maximum power transfer is a case ofwhat is called impedance matching in general. Impedancematching is important in shifting gears on a bicycle, inconnecting a loudspeaker to an audio amplifier, in con-necting a battery charger to a bank of solar photoelectriccells, and in many other applications.

58. A 10.0-)F capacitor is charged by a 10.0-V battery througha resistance R. The capacitor reaches a potential differenceof 4.00 V in a time 3.00 s after charging begins. Find R.

59. When two unknown resistors are connected in series witha battery, the battery delivers 225 W and carries a totalcurrent of 5.00 A. For the same total current, 50.0 W isdelivered when the resistors are connected in parallel.Determine the values of the two resistors.

60. When two unknown resistors are connected in series witha battery, the battery delivers total power !s and carries atotal current of I. For the same total current, a total power

!57.

!

2.00 Ω

4.00 Ω

10.0 Ω

4.00 V

12.0 V

a

b

Figure P28.54

Problems 891

!p is delivered when the resistors are connected in paral-lel. Determine the values of the two resistors.

61. A power supply has an open-circuit voltage of 40.0 V andan internal resistance of 2.00 '. It is used to charge twostorage batteries connected in series, each having an emfof 6.00 V and internal resistance of 0.300 '. If thecharging current is to be 4.00 A, (a) what additionalresistance should be added in series? (b) At what rate doesthe internal energy increase in the supply, in the batteries,and in the added series resistance? (c) At what rate doesthe chemical energy increase in the batteries?

62. Two resistors R1 and R2 are in parallel with each other.Together they carry total current I. (a) Determine thecurrent in each resistor. (b) Prove that this division of thetotal current I between the two resistors results in lesspower delivered to the combination than any otherdivision. It is a general principle that current in a directcurrent circuit distributes itself so that the total power delivered tothe circuit is a minimum.

The value of a resistor R is to be determined using theammeter–voltmeter setup shown in Figure P28.63. Theammeter has a resistance of 0.500 ', and the voltmeter hasa resistance of 20 000 '. Within what range of actualvalues of R will the measured values be correct to within5.00% if the measurement is made using the circuit shownin (a) Figure P28.63a and (b) Figure P28.63b?

63.Three 60.0-W, 120-V lightbulbs are connected across a120-V power source, as shown in Figure P28.67. Find(a) the total power delivered to the three bulbs and(b) the voltage across each. Assume that the resistance ofeach bulb is constant (even though in reality the resis-tance might increase markedly with current).

67.

69. Four resistors are connected in parallel across a 9.20-Vbattery. They carry currents of 150 mA, 45.0 mA,14.00 mA, and 4.00 mA. (a) If the resistor with the largestresistance is replaced with one having twice the resis-tance, what is the ratio of the new current in the battery

(a)

V

RA

V

AR

(b)

Figure P28.63

R1

120 V R2 R3

Figure P28.67

C1

R2

R1

C2

S

Figure P28.68

∆V3

2∆V3

Voltage–controlledswitch

(a)

∆V

R1

R2

T

∆Vc(t)

∆V

t(b)

C ∆VcV

Figure P28.66

68. Switch S has been closed for a long time, and the electriccircuit shown in Figure P28.68 carries a constant current.Take C1 # 3.00 )F, C2 # 6.00 )F, R1 # 4.00 k', andR2 # 7.00 k'. The power delivered to R2 is 2.40 W.(a) Find the charge on C1. (b) Now the switch is opened.After many milliseconds, by how much has the charge onC2 changed?

64. A battery is used to charge a capacitor through a resistor,as shown in Figure 28.19. Show that half the energy sup-plied by the battery appears as internal energy in the resis-tor and that half is stored in the capacitor.

The values of the components in a simple series RC circuitcontaining a switch (Fig. 28.19) are C # 1.00 )F, R # 2.00 .106 ', and # 10.0 V. At the instant 10.0 s after the switchis closed, calculate (a) the charge on the capacitor, (b) thecurrent in the resistor, (c) the rate at which energy is beingstored in the capacitor, and (d) the rate at which energy isbeing delivered by the battery.

66. The switch in Figure P28.66a closes when "Vc + 2 "V/3and opens when "Vc * "V/3. The voltmeter reads avoltage as plotted in Figure P28.66b. What is the period Tof the waveform in terms of R1, R2, and C ?

!

65.

892 CHAPTE R 28 • Direct Current Circuits

to the original current? (b) What If? If instead the resis-tor with the smallest resistance is replaced with onehaving twice the resistance, what is the ratio of the newtotal current to the original current? (c) On a Februarynight, energy leaves a house by several heat leaks, includ-ing the following: 1 500 W by conduction through theceiling; 450 W by infiltration (air flow) around thewindows; 140 W by conduction through the basementwall above the foundation sill; and 40.0 W by conductionthrough the plywood door to the attic. To produce thebiggest saving in heating bills, which one of these energytransfers should be reduced first?

70. Figure P28.70 shows a circuit model for the transmissionof an electrical signal, such as cable TV, to a largenumber of subscribers. Each subscriber connects a loadresistance RL between the transmission line and theground. The ground is assumed to be at zero potentialand able to carry any current between any groundconnections with negligible resistance. The resistance ofthe transmission line itself between the connectionpoints of different subscribers is modeled as the constantresistance RT. Show that the equivalent resistance acrossthe signal source is

Suggestion: Because the number of subscribers is large,the equivalent resistance would not change noticeably ifthe first subscriber cancelled his service. Consequently, theequivalent resistance of the section of the circuit to theright of the first load resistor is nearly equal to Req.

Req # 12 [(4RTRL % RT

2)1/2 % RT]

In Figure P28.71, suppose the switch has been closed for atime sufficiently long for the capacitor to become fullycharged. Find (a) the steady-state current in each resistorand (b) the charge Q on the capacitor. (c) The switch isnow opened at t # 0. Write an equation for the current through R2 as a function of time and (d) find the timeinterval required for the charge on the capacitor to fall toone-fifth its initial value.

IR 2

71.

72. A regular tetrahedron is a pyramid with a triangular base.Six 10.0-' resistors are placed along its six edges, withjunctions at its four vertices. A 12.0-V battery is connectedto any two of the vertices. Find (a) the equivalentresistance of the tetrahedron between these vertices and(b) the current in the battery.

73. The circuit shown in Figure P28.73 is set up in thelaboratory to measure an unknown capacitance C with theuse of a voltmeter of resistance R # 10.0 M' and a batterywhose emf is 6.19 V. The data given in the table are themeasured voltages across the capacitor as a function oftime, where t # 0 represents the instant at which the switchis opened. (a) Construct a graph of ln( /"V ) versus t, andperform a linear least-squares fit to the data. (b) From theslope of your graph, obtain a value for the time constant ofthe circuit and a value for the capacitance.

!

74. The student engineer of a campus radio station wishes toverify the effectiveness of the lightning rod on the antennamast (Fig. P28.74). The unknown resistance Rx is betweenpoints C and E. Point E is a true ground but is inaccessiblefor direct measurement since this stratum is several metersbelow the Earth’s surface. Two identical rods are driveninto the ground at A and B, introducing an unknown resis-tance Ry . The procedure is as follows. Measure resistanceR1 between points A and B, then connect A and B with aheavy conducting wire and measure resistance R2 between

3.00 kΩ

S

R2 =15.0 kΩ

12.0 kΩ

10.0 µF

9.00 V

µ

Figure P28.71

S

C

R

Voltmeter

ε

Figure P28.73

Ry Rx

A BC

Ry

E

Figure P28.74

RT RT RT

RL RL RLSignalsource

Figure P28.70

"V (V) t (s) ln( /"V )

6.19 05.55 4.874.93 11.14.34 19.43.72 30.83.09 46.62.47 67.31.83 102.2

!

Answers to Quick Quizzes 893

6 After Joseph Priest, "Meter Resistance: Don't Forget It!" The PhysicsTeacher, January 2003, p. 40.

points A and C. (a) Derive an equation for Rx in terms ofthe observable resistances, R1 and R2. (b) A satisfactoryground resistance would be Rx * 2.00 '. Is the groundingof the station adequate if measurements give R1 # 13.0 'and R2 # 6.00 '?

75. The circuit in Figure P28.75 contains two resistors,R1 # 2.00 k' and R 2 # 3.00 k', and two capacitors,C1 # 2.00 )F and C 2 # 3.00 )F, connected to a battery withemf # 120 V. No charge is on either capacitor beforeswitch S is closed. Determine the charges q1 and q2 oncapacitors C1 and C2, respectively, after the switch is closed.(Suggestion: First reconstruct the circuit so that it becomes asimple RC circuit containing a single resistor and singlecapacitor in series, connected to the battery, and then deter-mine the total charge q stored in the equivalent circuit.)

!

and that the voltage across the capacitor as a function oftime is

(c) What If? If the capacitor is fully charged, and theswitch is then opened, how does the voltage across thecapacitor behave in this case?

Answers to Quick Quizzes28.1 (a). Power is delivered to the internal resistance of a bat-

tery, so decreasing the internal resistance will decreasethis “lost” power and increase the percentage of thepower delivered to the device.

28.2 (c). In a series circuit, the current is the same in all resis-tors in series. Current is not “used up” as charges passthrough a resistor.

28.3 (a). Connecting b to c “shorts out” bulb R2 and changesthe total resistance of the circuit from R1 % R2 to just R1.Because the resistance of the circuit has decreased (andthe emf supplied by the battery does not change), thecurrent in the circuit increases.

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

28.5 (b), (d). Adding another series resistor increases the totalresistance of the circuit and thus reduces the current inthe circuit. The potential difference across the battery ter-minals increases because the reduced current results in asmaller voltage decrease across the internal resistance.

28.6 (a), (e). If the second resistor were connected in paral-lel, the total resistance of the circuit would decrease,and the current in the battery would increase. The po-tential difference across the terminals would decreasebecause the increased current results in a greater volt-age drop across the internal resistance.

28.7 (a). When the switch is closed, resistors R1 and R2 arein parallel, so that the total circuit resistance is smallerthan when the switch was open. As a result, the currentincreases.

28.8 (c). A current is assigned to a given branch of a circuit.There may be multiple resistors and batteries in a givenbranch.

28.9 (b), (d). Just after the switch is closed, there is nocharge on the capacitor, so there is no voltage across it.Charges begin to flow in the circuit to charge up the ca-pacitor, so that all of the voltage "V # IR appears acrossthe resistor. After a long time, the capacitor is fullycharged and the current drops to zero. Thus, the bat-tery voltage is now entirely across the capacitor.

28.10 (c), (i). Just after the switch is closed, there is no chargeon the capacitor. Current exists in both branches of thecircuit as the capacitor begins to charge, so the righthalf of the circuit is equivalent to two resistances R inparallel for an equivalent resistance of R . After a longtime, the capacitor is fully charged and the current inthe right-hand branch drops to zero. Now, currentexists only in a resistance R across the battery.

12

VC #r

r % R !(1 $ e$t/R eqC)

ε+ –

R2

R1 C1

C2

a

b c

f

S

d e

Figure P28.75

Voltmeter

S

R

C

r

εFigure P28.76

76. This problem6 illustrates how a digital voltmeter affects thevoltage across a capacitor in an RC circuit. A digitalvoltmeter of internal resistance r is used to measure thevoltage across a capacitor after the switch in Figure P28.76is closed. Because the meter has finite resistance, part ofthe current supplied by the battery passes through themeter. (a) Apply Kirchhoff’s rules to this circuit, and usethe fact that iC # dq/dt to show that this leads to thedifferential equation

where Req # rR/(r % R). (b) Show that the solution to thisdifferential equation is

q #r

r % R C! (1 $ e$t/ReqC)

R eq dqdt

%qC

#r

r % R !