858 CHAPTER 28 • Direct Current Circuits

Chapter 28

Direct Current Circuits

CHAPTER OUTLINE

28.1 Electromotive Force

28.2 Resistors in Series and

Parallel

28.3 Kirchhoff’s Rules

28.4 RC Circuits

28.5 Electrical Meters

28.6 Household Wiring and

Electrical 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, that

is, a current for which the drift velocity of the charges is always in the same direction.

(George Semple)

858This chapter is concerned with the analysis of simple electric circuits that contain

batteries, resistors, and capacitors in various combinations. We will see some circuits in

which resistors can be combined using simple rules. The analysis of more complicated

circuits is simpliﬁed using two rules known as Kirchhoff’s rules, which follow from the

laws 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 that

currents in the circuit are constant in magnitude and direction. A current that is

constant 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, we

describe electrical meters for measuring current and potential difference, and discuss

electrical circuits in the home.

28.1 Electromotive Force

In Section 27.6 we discussed a closed circuit in which a battery produces a potential

difference 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 the

potential difference at the battery terminals is constant in a particular circuit, the

current in the circuit is constant in magnitude and direction and is called direct

current. A battery is called either a source of electromotive force or, more commonly, a

source 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 battery

is the maximum possible voltage that the battery can provide between its

terminals. You can think of a source of emf as a “charge pump.” When an electric

potential 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 to

a resistor. We shall generally assume that the connecting wires have no resistance.

Battery

–

+

Figure 28.1 A circuit consisting of a resistor

Resistor connected to the terminals of a battery.

859&

860 CHAPTER 28 • Direct Current Circuits

The positive terminal of the battery is at a higher potential than the negative termi-

ε

r

a b

–+

nal. Because a real battery is made of matter, there is resistance to the ﬂow of charge

within 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 its

I I

terminal voltage) equals its emf. However, for a real battery, the terminal voltage is not

R

equal to the emf for a battery in a circuit in which there is a current. To understand

d c

why this is so, consider the circuit diagram in Figure 28.2a, where the battery of

(a)

Figure 28.1 is represented by the dashed rectangle containing an ideal, resistance-free

emf ! in series with an internal resistance r. Now imagine moving through the battery

from a to b and measuring the electric potential at various locations. As we pass from

V

ε

the negative terminal to the positive terminal, the potential increases by an amount !.

r R

However, as we move through the resistance r, the potential decreases by an amount

Ir, where I is the current in the circuit. Thus, the terminal voltage of the battery

ε 1

"V# V $ V is

b a

Ir

IR

"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—

ac b d

for example, the emf of a D cell is 1.5 V. The actual potential difference between the

terminals of the battery depends on the current in the battery, as described by Equa-

(b)

tion 28.1.

Active Figure 28.2 (a) Circuit

Figure 28.2b is a graphical representation of the changes in electric potential as the

diagram of a source of emf ! (in

circuit is traversed in the clockwise direction. By inspecting Figure 28.2a, we see that

this case, a battery), of internal

the terminal voltage "V must equal the potential difference across the external resis-

resistance r, connected to an

external resistor of resistance R.

tance R, often called the load resistance. The load resistor might be a simple resistive

(b) Graphical representation

circuit element, as in Figure 28.1, or it could be the resistance of some electrical device

showing how the electric potential

(such as a toaster, an electric heater, or a lightbulb) connected to the battery (or, in

changes as the circuit in part (a) is

the case of household devices, to the wall outlet). The resistor represents a load on the

traversed clockwise.

battery because the battery must supply energy to operate the device. The potential dif-

At the Active Figures link

ference across the load resistance is "V# IR. Combining this expression with Equation

at http://www.pse6.com, you

28.1, we see that

can adjust the emf and

resistances r and R to see the

!# IR% Ir (28.2)

effect on the current and on the

Solving for the current gives

graph in part (b).

!

# (28.3)

R% r

This equation shows that the current in this simple circuit depends on both the load

resistance R external to the battery and the internal resistance r. If R is much greater

than 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

2 2

I!# I R% I r (28.4)

! PITFALL PREVENTION

This equation indicates that, because power !# I "V (see Eq. 27.22), the total power

28.1 What Is Constant in a

2

output I! of the battery is delivered to the external load resistance in the amount I R

Battery?

2

and to the internal resistance in the amount I r.

It is a common misconception

that a battery is a source of

constant current. Equation 28.3

Quick Quiz 28.1 In order to maximize the percentage of the power that is

clearly shows that this is not true.

The current in the circuit delivered from a battery to a device, the internal resistance of the battery should be

depends on the resistance

(a) as low as possible (b) as high as possible (c) The percentage does not depend on

connected to the battery. It is also

the internal resistance.

not true that a battery is a source

of constant terminal voltage,

as shown by Equation 28.1. A 1

The terminal voltage in this case is less than the emf by an amount Ir. In some situations, the

battery is a source of constant

terminal voltage may exceed the emf by an amount Ir. This happens when the direction of the current is

emf.

opposite that of the emf, as in the case of charging a battery with another source of emf.SECTION 28.1 • Electromotive Force 861

Interactive

Example 28.1 Terminal Voltage of a Battery

A battery has an emf of 12.0 V and an internal resistance of Hence, the power delivered by the battery is the sum

0.05’. Its terminals are connected to a load resistance of of these quantities, or 47.1 W. You should check this result,

3.00’. using the expression !# I!.

(A) Find the current in the circuit and the terminal voltage

What If? As a battery ages, its internal resistance

of the battery.

increases. Suppose the internal resistance of this battery

rises to 2.00! toward the end of its useful life. How does

Solution Equation 28.3 gives us the current:

this alter the ability of the battery to deliver energy?

! 12.0 V

I# # # 3.93 A

Answer Let us connect the same 3.00-’ load resistor to the

R% r 3.05 ’

battery. The current in the battery now is

and from Equation 28.1, we ﬁnd the terminal voltage:

! 12.0 V

I# # # 2.40 A

R% r (3.00 ’% 2.00 ’)

"V#$ ! Ir# 12.0 V$ (3.93 A)(0.05’)# 11.8 V

and the terminal voltage is

To check this result, we can calculate the voltage across the

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

load resistance R:

Notice that the terminal voltage is only 60% of the emf. The

"V# IR# (3.93 A)(3.00’)# 11.8 V

powers delivered to the load resistor and internal resistance

are

(B) Calculate the power delivered to the load resistor, the

2 2

power delivered to the internal resistance of the battery, and

! # I R# (2.40 A) (3.00’)# 17.3 W

R

the power delivered by the battery.

2 2

! # I r# (2.40 A) (2.00’)# 11.5 W

r

Solution The power delivered to the load resistor is

Notice that 40% of the power from the battery is delivered

2 2

to the internal resistance. In part (B), this percentage is

! # I R# (3.93 A) (3.00’)# 46.3 W

R

1.6%. Consequently, even though the emf remains ﬁxed,

the increasing internal resistance signiﬁcantly reduces the

The power delivered to the internal resistance is

ability of the battery to deliver energy.

2 2

! # I r# (3.93 A) (0.05’)# 0.772 W

r

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.

Example 28.2 Matching the Load

Show that the maximum power delivered to the load resis- to zero, and solving for R. The details are left as a problem

tance R in Figure 28.2a occurs when the load resistance for you to solve (Problem 57).

matches the internal resistance—that is, when R# r.

Solution The power delivered to the load resistance is

2

!

equal to I R, where I is given by Equation 28.3:

2

! R

!

2

max

!# I R#

2

(R% r)

When ! is plotted versus R as in Figure 28.3, we ﬁnd that

2

! reaches a maximum value of ! /4r at R# r. When R is

2

large, there is very little current, so that the power I R

delivered to the load resistor is small. When R is small,

the current is large and there is signiﬁcant loss of power

R

2

I r as energy is delivered to the internal resistance. When

r 2r 3r

R# r, these effects balance to give a maximum transfer of

Figure 28.3 (Example 28.2) Graph of the power ! delivered

power.

by a battery to a load resistor of resistance R as a function of R.

We can also prove that the power maximizes at R# r by The power delivered to the resistor is a maximum when the

load resistance equals the internal resistance of the battery.

differentiating ! with respect to R, setting the result equal862 CHAPTER 28 • Direct Current Circuits

28.2 Resistors in Series and Parallel

Suppose that you and your friends are at a crowded basketball game in a sports arena and

decide to leave early. You have two choices: (1) your group can exit through a single door

and 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 member

of your group can exit through a separate door in the main hall of the arena, where each

will have to push his or her way through a single group of people standing by the door. In

which 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 separate

doors than down the hallway where each of you has to push through several groups of

people. We could describe the groups of people in the hallway as being in series, because

each of you must push your way through all of the groups. The groups of people

around the doors in the arena can be described as being in parallel. Each member of

your group must push through only one group of people, and each member pushes

through a different group of people. This simple analogy will help us understand the

behavior of currents in electric circuits containing more than one resistor.

When two or more resistors are connected together as are the lightbulbs in Figure

28.4a, 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 of

charge Q exits resistor R , charge Q must also enter the second resistor R . (This is

1 2

analogous to all members of your group pushing through each crowd in the single

hallway of the sports arena.) Otherwise, charge will accumulate on the wire between

the resistors. Thus, the same amount of charge passes through both resistors in a given

time interval. Hence,

for a series combination of two resistors, the currents are the same in both resis-

tors because the amount of charge that passes through R must also pass through

1

R in the same time interval.

2

The potential difference applied across the series combination of resistors will divide

2

between the resistors. In Figure 28.4b, because the voltage drop from a to b equals IR

1

and the voltage drop from b to c equals IR , the voltage drop from a to c is

2

"V# IR % IR # I(R % R )

1 2 1 2

R R

1 2

R = R + R

eq 1 2

I = I = I

1 2

R R

1 2

ab c ac

+

–

I I

I

ΔV ΔV

Battery

+ – + –

(a) (b) (c)

Active Figure 28.4 (a) A series connection of two lightbulbs with resistances R and

1

At the Active Figures link

R . (b) Circuit diagram for the two-resistor circuit. The current in R is the same as that

2 1

at http://www.pse6.com, you

in R . (c) The resistors replaced with a single resistor having an equivalent resistance

2

can adjust the battery voltage

R # R % R .

eq 1 2

and resistances R and R to

1 2

see the effect on the currents

2

The term voltage drop is synonymous with a decrease in electric potential across a resistor and is

and voltages in the individual

used often by individuals working with electric circuits.

resistors.SECTION 28.2 • Resistors in Series and Parallel 863

The potential difference across the battery is also applied to the equivalent resistance

R in Figure 28.4c:

eq

"V# IR

eq

where we have indicated that the equivalent resistance has the same effect on the

circuit 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 series

with a single equivalent resistance whose value is the sum of the individual resistances:

"V# IR # I(R % R ) 9: R # R % R (28.5)

eq 1 2 eq 1 2

The resistance R is equivalent to the series combination R % R in the sense that

eq 1 2

the circuit current is unchanged when R replaces R % R .

eq 1 2

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

R # R % R % R %((( (28.6) The equivalent resistance of

eq 1 2 3

several resistors in series

This relationship indicates that the equivalent resistance of a series connection of

resistors is the numerical sum of the individual resistances and is always

greater than any individual resistance.

Looking back at Equation 28.3, the denominator is the simple algebraic sum of the

external and internal resistances. This is consistent with the fact that internal and

external resistances are in series in Figure 28.2a.

Note that if the ﬁlament of one lightbulb in Figure 28.4 were to fail, the circuit

! PITFALL PREVENTION

would no longer be complete (resulting in an open-circuit condition) and the second

bulb would also go out. This is a general feature of a series circuit—if one device in the

28.2 Lightbulbs Don’t

series creates an open circuit, all devices are inoperative.

Burn

We will describe the end of the

life of a lightbulb by saying that

the ﬁlament fails, rather than by

Quick Quiz 28.2 In Figure 28.4, imagine positive charges pass ﬁrst through

saying that the lightbulb “burns

R and then through R . Compared to the current in R , the current in R is

1 2 1 2 out.” The word burn suggests a

(a) smaller, (b) larger, or (c) the same.

combustion process, which is not

what occurs in a lightbulb.

Quick Quiz 28.3 If a piece of wire is used to connect points b and c in Figure

28.4b, does the brightness of bulb R (a) increase, (b) decrease, or (c) remain the same?

1

Quick Quiz 28.4 With the switch in the circuit of Figure 28.5 closed (left),

there is no current in R , because the current has an alternate zero-resistance path

2

through the switch. There is current in R and this current is measured with the amme-

1

ter (a device for measuring current) at the right side of the circuit. If the switch is

opened (Fig. 28.5, right), there is current in R . What happens to the reading on the

2

ammeter when the switch is opened? (a) the reading goes up; (b) the reading goes

down; (c) the reading does not change.

R R

1 1

R A R A

2 2

Switch closed Switch open

Figure 28.5 (Quick Quiz 28.4) What happens when the switch is opened?864 CHAPTER 28 • Direct Current Circuits

At the Active Figures link

at http://www.pse6.com, you

R

1

can adjust the battery voltage

and resistances R and R to

1 2

see the effect on the currents

and voltages in the individual

resistors.

R

2

ΔV = ΔV = ΔV

1 2

11 1

= +

R R R R

1 eq 1 2

I

1

R

+ 2

–

a b

I I

2

I

ΔV ΔV

Battery

+ –

+ –

(a) (b) (c)

Active Figure 28.6 (a) A parallel connection of two lightbulbs with resistances R and

1

R . (b) Circuit diagram for the two-resistor circuit. The potential difference across R is

2 1

the same as that across R . (c) The resistors replaced with a single resistor having an

2

equivalent resistance given by Equation 28.7.

Now consider two resistors connected in parallel, as shown in Figure 28.6. When

! PITFALL PREVENTION

charges reach point a in Figure 28.6b, called a junction, they split into two parts, with some

going through R and the rest going through R . A junction is any point in a circuit

1 2

28.3 Local and Global

where a current can split (just as your group might split up and leave the sports arena

Changes

through several doors, as described earlier.) This split results in less current in each indi-

A local change in one part of a

vidual resistor than the current leaving the battery. Because electric charge is conserved,

circuit may result in a global

the current I that enters point a must equal the total current leaving that point:

change throughout the circuit.

For example, if a single resistance

I# I % I

1 2

is changed in a circuit containing

where I is the current in R and I is the current in R .

several resistors and batteries, the

1 1 2 2

currents in all resistors and batter- As can be seen from Figure 28.6, both resistors are connected directly across the

ies, the terminal voltages of all bat-

terminals of the battery. Therefore,

teries, and the voltages across all

resistors may change as a result.

when resistors are connected in parallel, the potential differences across the resis-

tors is the same.

! PITFALL PREVENTION Because the potential differences across the resistors are the same, the expression

"V# IR gives

28.4 Current Does Not

"V "V 1 1 "V

Take the Path of

I# I % I # % #"V % #

1 2

! "

Least Resistance R R R R R

1 2 1 2 eq

You may have heard a phrase like

where R is an equivalent single resistance which will have the same effect on the

eq

“current takes the path of least

circuit as the two resistors in parallel; that is, it will draw the same current from the

resistance” in reference to a par-

battery (Fig. 28.6c). From this result, we see that the equivalent resistance of two resis-

allel combination of current

tors in parallel is given by

paths, such that there are two or

more paths for the current to 1 1 1

# % (28.7)

take. The phrase is incorrect.

R R R

eq 1 2

The current takes all paths.

or

Those paths with lower resistance

1 R R

1 2

will have large currents, but even

R # #

eq

1 1 R % R

very high-resistance paths will 1 2

%

carry some of the current.

R R

1 2SECTION 28.2 • Resistors in Series and Parallel 865

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

1 1 1 1

# % % %((( (28.8) The equivalent resistance of

R R R R

eq 1 2 3

several resistors in parallel

We can see from this expression that the inverse of the equivalent resistance of two

or more resistors connected in parallel is equal to the sum of the inverses of the

individual resistances. Furthermore, the equivalent resistance is always less

than 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 operate

on the same voltage.

Quick Quiz 28.5 In Figure 28.4, imagine that we add a third resistor in series

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 28.6 In Figure 28.6, imagine that we add a third resistor in

parallel 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 R . There is current in R and this current is measured with the

2 1

ammeter at the right side of the circuit. If the switch is closed (Fig. 28.7, right), there is

current in R . What happens to the reading on the ammeter when the switch is closed?

2

(a) the reading goes up; (b) the reading goes down; (c) the reading does not change.

R R

2 2

R R

1 1

A A

Switch open Switch closed

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 lighting on the cable at 10-foot intervals, so the light ﬁxtures are in

in his back yard. To save money, he purchases inexpensive parallel. Because of the cable’s resistance, the brightness

18-gauge cable, which has a relatively high resistance per of the bulbs in the light ﬁxtures is not as desired. Which

unit length. This cable consists of two side-by-side wires problem does the homeowner have? (a) All of the bulbs

separated by insulation, like the cord on an appliance. glow equally less brightly than they would if lower-

He runs a 200-foot length of this cable from the power resistance cable had been used. (b) The brightness of the

supply to the farthest point at which he plans to position a bulbs decreases as you move farther from the power

light ﬁxture. He attaches light ﬁxtures across the two wires supply.866 CHAPTER 28 • Direct Current Circuits

Solution A circuit diagram for the system appears in ﬁxture R is less than the terminal voltage. There is a

C

Figure 28.8. The horizontal resistors (such as R and R ) further voltage drop across resistors R and R . Conse-

D E

A B

represent the resistance of the wires in the cable between quently, the voltage across light ﬁxture R is smaller than

F

the light ﬁxtures while the vertical resistors (such as R ) that across R . This continues on down the line of light

C

C

represent the resistance of the light ﬁxtures themselves. ﬁxtures, so the correct choice is (b). Each successive light

Part of the terminal voltage of the power supply is dropped ﬁxture has a smaller voltage across it and glows less brightly

across resistors R and R . Thus, the voltage across light than the one before.

A B

Resistance of Resistance in

light fixtures wires of cable

R R

A D

Power

R R

C F

supply

R R

B E

Figure 28.8 (Conceptual Example 28.3) The circuit diagram for a set of landscape

light ﬁxtures connected in parallel across the two wires of a two-wire cable. The

horizontal resistors represent resistance in the wires of the cable. The vertical resistors

represent the light ﬁxtures.

Example 28.4 Find the Equivalent Resistance

Four resistors are connected as shown in Figure 28.9a.

I # 2.0 A. We could have guessed this at the start by noting

2

that the current in the 3.0-’ resistor has to be twice that in

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

the 6.0-’ resistor, in view of their relative resistances and the

fact that the same voltage is applied to each of them.

Solution The combination of resistors can be reduced in

As a ﬁnal check of our results, note that "V #

bc

steps, as shown in Figure 28.9. The 8.0-’ and 4.0-’ resistors

(6.0’)I # (3.0’)I # 6.0 V and "V # (12.0’)I# 36 V;

1 2 ab

are in series; thus, the equivalent resistance between a and b

therefore, "V #"V %"V # 42 V, as it must.

ac ab bc

is 12.0 ’ (see Eq. 28.5). The 6.0-’ and 3.0-’ resistors are in

parallel, so from Equation 28.7 we ﬁnd that the equivalent

resistance from b to c is 2.0 ’. Hence, the equivalent resis-

6.0 Ω

tance from a to c is 14.0 ’.

(B) What is the current in each resistor if a potential differ-

I

1

8.0 Ω 4.0 Ω

ence of 42 V is maintained between a and c?

b

a c

I

Solution The currents in the 8.0-’ and 4.0-’ resistors are

I

2

(a)

the same because they are in series. In addition, this is the

same as the current that would exist in the 14.0-’ equivalent

3.0 Ω

resistor subject to the 42-V potential difference. Therefore,

using Equation 27.8 (R#"V/I ) and the result from part

(A), we obtain

12.0 Ω 2.0 Ω

"V 42 V

ac

I# # # 3.0 A

a b c

R 14.0 ’

eq

(b)

This is the current in the 8.0-’ and 4.0-’ resistors. When

this 3.0-A current enters the junction at b, however, it splits,

14.0 Ω

with part passing through the 6.0-’ resistor (I ) and part

1

a c

through the 3.0-’ resistor (I ). Because the potential differ-

2

(c)

ence is "V across each of these parallel resistors, we see

bc

that (6.0 ’)I # (3.0 ’)I , or I # 2I . Using this result and

Figure 28.9 (Example 28.4) The original network of resistors

1 2 2 1

the fact that I % I # 3.0 A, we ﬁnd that I # 1.0 A and is reduced to a single equivalent resistance.

1 2 1

Example 28.5 Finding R by Symmetry Arguments

eq

Consider ﬁve resistors connected as shown in Figure 28.10a.

connections. We can, however, assume a current entering

Find the equivalent resistance between points a and b.

junction a and then apply symmetry arguments. Because of

the symmetry in the circuit (all 1-’ resistors in the outside

Solution If we inspect this system of resistors, we realize that

loop), the currents in branches ac and ad must be equal;

we cannot reduce it by using our rules for series and parallel

hence, the electric potentials at points c and d must be equal.SECTION 28.2 • Resistors in Series and Parallel 867

This means that "V # 0 and there is no current between the remaining circuit then reduced as in Figures 28.10c and

cd

points c and d. As a result, points c and d may be connected d. From this reduction we see that the equivalent resistance

together without affecting the circuit, as in Figure 28.10b. of the combination is 1’. Note that the result is 1’ regard-

Thus, the 5-’ resistor may be removed from the circuit and less of the value of the resistor connected between c and d.

5 Ω

c

1 Ω 1 Ω

1 Ω 1 Ω

0.5 Ω 0.5 Ω 1 Ω

5 Ω

a b

a

b ac,d b a b

c,d

1 Ω 1 Ω

1 Ω 1 Ω

d

(a) (b) (c) (d)

Figure 28.10 (Example 28.5) Because of the symmetry in this circuit, the 5-’ resistor

does not contribute to the resistance between points a and b and therefore can be

disregarded when we calculate the equivalent resistance.

Interactive

Example 28.6 Three Resistors in Parallel

Three resistors are connected in parallel as shown in Figure (B) Calculate the power delivered to each resistor and the

28.11a. A potential difference of 18.0 V is maintained total power delivered to the combination of resistors.

between points a and b.

2

Solution We apply the relationship !# I R to each resis-

tor and obtain

(A) Find the current in each resistor.

Solution The resistors are in parallel, and so the potential

2 2

3.00-’: ! # I R # (6.00 A) (3.00 ’)# 108 W

1 1 1

difference across each must be 18.0 V. Applying the relation-

ship "V# IR to each resistor gives

2 2

6.00-’: ! # I R # (3.00 A) (6.00 ’)# 54.0 W

"V 18.0 V 2 2 2

I # # # 6.00 A

1

R 3.00 ’

1

2 2

9.00-’: ! # I R # (2.00 A) (9.00 ’)# 36.0 W

3 3 3

"V 18.0 V

I # # # 3.00 A

2

R 6.00 ’

2

This shows that the smallest resistor receives the most

"V 18.0 V

power. Summing the three quantities gives a total power of

I # # # 2.00 A

3

R 9.00 ’ 198 W.

3

I

a

I I I

1 2 3

18.0 V

3.00 Ω 6.00 Ω 9.00 Ω

b

(a)

a

I I I

1 2 3

I

Figure 28.11 (Example 28.6) (a) Three

18.0 V resistors connected in parallel. The

3.00 Ω 6.00 Ω 9.00 Ω

voltage across each resistor is 18.0 V.

(b) Another circuit with three resistors

b

and a battery. Is this equivalent to the

(b)

circuit in part (a) of the ﬁgure?868 CHAPTER 28 • Direct Current Circuits

(C) Calculate the equivalent resistance of the circuit. What If? What if the circuit is as shown in Figure 28.11b

instead of as in Figure 28.11a? How does this affect the

Solution We can use Equation 28.8 to ﬁnd R :

calculation?

eq

Answer There is no effect on the calculation. The physical

1 1 1 1

# % %

placement of the battery is not important. In Figure 28.11b,

R 3.00 ’ 6.00 ’ 9.00 ’

eq

the battery still applies a potential difference of 18.0 V

between points a and b, so the two circuits in Figure 28.11

18.0 ’

are electrically identical.

R # # 1.64 ’

eq

11.0

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

and resistors.

Conceptual Example 28.7 Operation of a Three-Way Lightbulb

Figure 28.12 illustrates how a three-way lightbulb is con- in parallel, if one of them (for example, the 75-W ﬁlament)

structed to provide three levels of light intensity. The socket breaks, the bulb will still operate in two of the switch posi-

of the lamp is equipped with a three-way switch for selecting tions as current exists in the other (100-W) ﬁlament.

different light intensities. The bulb contains two ﬁlaments.

When the lamp is connected to a 120-V source, one ﬁlament

receives 100 W of power, and the other receives 75 W.

Explain how the two ﬁlaments are used to provide three

100-W filament

different light intensities.

Solution The three light intensities are made possible by

75-W filament

applying the 120 V to one ﬁlament alone, to the other ﬁla-

ment alone, or to the two ﬁlaments in parallel. When switch

S is closed and switch S is opened, current exists only in

1 2

the 75-W ﬁlament. When switch S is open and switch S is

1 2

closed, current exists only in the 100-W ﬁlament. When both

switches are closed, current exists in both ﬁlaments, and the S

1

total power is 175 W.

120 V

If the ﬁlaments were connected in series and one of

S

2

them were to break, no charges could pass through the

bulb, and the bulb would give no illumination, regardless of

Figure 28.12 (Conceptual Example 28.7) A three-way

the switch position. However, with the ﬁlaments connected lightbulb.

Application Strings of Lights

Strings of lights are used for many ornamental purposes, a series-wired string. As a result, these bulbs are inherently

3

such as decorating Christmas trees. Over the years, both more dangerous (more likely to start a ﬁre, for instance),

parallel and series connections have been used for strings but if one bulb in a parallel-wired string fails or is removed,

of lights powered by 120 V. Series-wired bulbs are safer than the rest of the bulbs continue to glow. (A 25-bulb string of

parallel-wired bulbs for indoor Christmas-tree use because 4-W bulbs results in a power of 100 W; the total power

series-wired bulbs operate with less energy per bulb and at a becomes substantial when several strings are used.)

lower temperature. However, if the ﬁlament of a single bulb A new design was developed for so-called “miniature”

fails (or if the bulb is removed from its socket), all the lights wired in series, to prevent the failure of one bulb from

lights on the string go out. The popularity of series-wired causing the entire string to go out. This design creates a

light strings diminished because troubleshooting a failed connection (called a jumper) across the ﬁlament after it

bulb was a tedious, time-consuming chore that involved fails. When the ﬁlament breaks in one of these miniature

trial-and-error substitution of a good bulb in each socket lightbulbs, the break in the ﬁlament represents the largest

along the string until the defective bulb was found. resistance in the series, much larger than that of the intact

In a parallel-wired string, each bulb operates at 120 V. ﬁlaments. As a result, most of the applied 120 V appears

By design, the bulbs are brighter and hotter than those on across the bulb with the broken ﬁlament. Inside the

3

These and other household devices, such as the three-way lightbulb in Conceptual Example 28.7

and the kitchen appliances discussed in Section 28.6, actually operate on alternating current (AC), to

be introduced in Chapter 33.SECTION 28.3 • Kirchhoff’s Rules 869

lightbulb, a small jumper loop covered by an insulating jumper connection mentioned in the preceding paragraph.

material is wrapped around the ﬁlament leads. When the All the other bulbs not only stay on but glow more brightly

ﬁlament fails and 120 V appears across the bulb, an arc because the total resistance of the string is reduced and con-

burns the insulation on the jumper and connects the sequently the current in each bulb increases.

ﬁlament leads. This connection now completes the circuit Let us assume that the resistance of a bulb remains at

through the bulb even though its ﬁlament is no longer 16.9’ even though its temperature rises as a result of the

active (Fig. 28.13). increased current. If one bulb fails, the potential difference

Suppose that all the bulbs in a 50-bulb miniature-light across each of the remaining bulbs increases to 120 V/49#

string are operating. A 2.40-V potential drop occurs across 2.45 V, the current increases from 0.142 A to 0.145 A, and the

each bulb because the bulbs are in series. A typical power power increases to 0.355 W. As more bulbs fail, the current

input to this style of bulb is 0.340 W. The ﬁlament resis- keeps rising, the ﬁlament of each bulb operates at a higher

tance of each bulb at the operating temperature is temperature, and the lifetime of the bulb is reduced. For this

2

(2.40 V) /(0.340 W)# 16.9 ’. The current in each bulb is reason, you should check for failed (nonglowing) bulbs in

2.40 V/16.9 ’# 0.142 A. When a bulb fails, the resistance such a series-wired string and replace them as soon as possible,

across its terminals is reduced to zero because of the alternate in order to maximize the lifetimes of all the bulbs.

I I

Filament

I

Jumper

Glass insulator

(a)

(b) (c)

Figure 28.13 (a) Schematic diagram of a modern “miniature” holiday lightbulb, with a

jumper connection to provide a current path if the ﬁlament breaks. When the ﬁlament

is 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.

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. Very

often, however, it is not possible to reduce a circuit to a single loop. The procedure

for analyzing more complex circuits is greatly simpliﬁed if we use two principles called

Kirchhoff ’s rules:

1. Junction rule. The sum of the currents entering any junction in a circuit must

equal the sum of the currents leaving that junction:

I # I (28.9)

# in # out

2. Loop rule. The sum of the potential differences across all elements around any

closed circuit loop must be zero:

"V# 0 (28.10)

#

closed

loop

George Semple870 CHAPTER 28 • Direct Current Circuits

I

Kirchhoff’s ﬁrst rule is a statement of conservation of electric charge. All charges

2

I

1

that enter a given point in a circuit must leave that point because charge cannot build

up at a point. If we apply this rule to the junction shown in Figure 28.14a, we obtain

I

3

I # I % I

1 2 3

(a)

Figure 28.14b represents a mechanical analog of this situation, in which water ﬂows

through a branched pipe having no leaks. Because water does not build up anywhere

Flow in

in the pipe, the ﬂow rate into the pipe equals the total ﬂow rate out of the two

Flow out

branches on the right.

Kirchhoff’s second rule follows from the law of conservation of energy. Let us

imagine moving a charge around a closed loop of a circuit. When the charge

returns to the starting point, the charge–circuit system must have the same total

(b)

energy as it had before the charge was moved. The sum of the increases in energy as

Figure 28.14 (a) Kirchhoff’s

the charge passes through some circuit elements must equal the sum of the

junction rule. Conservation of

decreases in energy as it passes through other elements. The potential energy

charge requires that all charges

decreases whenever the charge moves through a potential drop $ IR across a resis-

entering a junction must leave that

tor or whenever it moves in the reverse direction through a source of emf. The

junction. Therefore, I # I % I .

1 2 3

(b) A mechanical analog of the

potential energy increases whenever the charge passes through a battery from the

junction rule: the amount of water

negative terminal to the positive terminal.

ﬂowing out of the branches on the

When applying Kirchhoff’s second rule in practice, we imagine traveling around the

right must equal the amount

loop and consider changes in electric potential, rather than the changes in potential energy

ﬂowing into the single branch on

described in the preceding paragraph. You should note the following sign conventions

the left.

when 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-

I

tial difference "V across the resistor is $ IR (Fig. 28.15a).

(a)

• If a resistor is traversed in the direction opposite the current, the potential differ-

ab

ΔV = –IR

ence "V across the resistor is % IR (Fig. 28.15b).

• If a source of emf (assumed to have zero internal resistance) is traversed in the

I

direction of the emf (from $ to %), the potential difference "V is %! (Fig.

(b)

28.15c). The emf of the battery increases the electric potential as we move

ab

ΔV = +IR

through it in this direction.

• If a source of emf (assumed to have zero internal resistance) is traversed in the

direction opposite the emf (from % to $ ), the potential difference "V is $!

ε

– +

(Fig. 28.15d). In this case the emf of the battery reduces the electric potential as

(c)

ab

we move through it.

ΔV = +εε

Limitations exist on the numbers of times you can usefully apply Kirchhoff’s rules

in analyzing a circuit. You can use the junction rule as often as you need, so long as

ε

each time you write an equation you include in it a current that has not been used in

+ –

(d)

a preceding junction-rule equation. In general, the number of times you can use the

ab

ε

ΔV = –ε junction rule is one fewer than the number of junction points in the circuit. You can

Figure 28.15 Rules for apply the loop rule as often as needed, as long as a new circuit element (resistor or

determining the potential

battery) or a new current appears in each new equation. In general, in order to

differences across a resistor and a

solve a particular circuit problem, the number of independent equations you

battery. (The battery is assumed to

need to obtain from the two rules equals the number of unknown currents.

have no internal resistance.) Each

Complex networks containing many loops and junctions generate great numbers

circuit element is traversed from

left to right. of independent linear equations and a correspondingly great number of unknowns.

Such situations can be handled formally through the use of matrix algebra. Computer

software can also be used to solve for the unknowns.

The following examples illustrate how to use Kirchhoff’s rules. In all cases, it is

assumed that the circuits have reached steady-state conditions—that is, the currents

in the various branches are constant. Any capacitor acts as an open branch in a

circuit; that is, the current in the branch containing the capacitor is zero under

steady-state conditions.SECTION 28.3 • Kirchhoff’s Rules 871

PROBLEM-SOLVING HINTS

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. Although

the assignment of current directions is arbitrary, you must adhere rigorously to

the assigned directions when applying Kirchhoff’s rules.

Apply the junction rule to any junctions in the circuit that provide new

•

relationships among the various currents.

Apply the loop rule to as many loops in the circuit as are needed to solve for

•

the unknowns. To apply this rule, you must correctly identify the potential

difference as you imagine crossing each element while traversing the closed

loop (either clockwise or counterclockwise). Watch out for errors in sign!

Gustav Kirchhoff

Solve the equations simultaneously for the unknown quantities. Do not be German Physicist (1824–1887)

•

alarmed if a current turns out to be negative; its magnitude will be correct and the

Kirchhoff, a professor at

direction is opposite to that which you assigned.

Heidelberg, and Robert Bunsen

invented the spectroscope and

founded the science of

spectroscopy, which we shall

study in Chapter 42. They

Quick Quiz 28.8 In using Kirchhoff’s rules, you generally assign a separate

discovered the elements cesium

unknown current to (a) each resistor in the circuit (b) each loop in the circuit (c) each

and rubidium and invented

astronomical spectroscopy. (AIP

branch in the circuit (d) each battery in the circuit.

ESVA/W.F. Meggers Collection)

Example 28.8 A Single-Loop Circuit

A single-loop circuit contains two resistors and two batteries,

d: a represents a potential difference of $ IR . Applying

2

as shown in Figure 28.16. (Neglect the internal resistances

Kirchhoff’s loop rule gives

of the batteries.)

"V# 0

#

(A) Find the current in the circuit.

! $ IR $! $ IR # 0

1 1 2 2

Solution We do not need Kirchhoff’s rules to analyze this Solving for I and using the values given in Figure 28.16, we

simple circuit, but let us use them anyway just to see how obtain

they are applied. There are no junctions in this single-loop

! $! 6.0 V$ 12 V

1 2

(1) I# # # $0.33 A

circuit; thus, the current is the same in all elements. Let us

R % R 8.0 ’% 10 ’

1 2

assume that the current is clockwise, as shown in Figure

28.16. Traversing the circuit in the clockwise direction, The negative sign for I indicates that the direction of the

starting at a, we see that a: b represents a potential differ- current is opposite the assumed direction. Notice that the

ence of %! , b: c represents a potential difference of emfs in the numerator subtract because the batteries have

1

$ IR , c: d represents a potential difference of $! , and opposite polarities in Figure 28.16. In the denominator, the

1 2

resistances add because the two resistors are in series.

(B) What power is delivered to each resistor? What power is

ε = 6.0 V

1

delivered by the 12-V battery?

I

– +

ab

Solution Using Equation 27.23,

2 2

! # I R # (0.33 A) (8.0 ’)# 0.87 W

R = 10 Ω R = 8.0 Ω 1 1

2 1

2 2

! # I R # (0.33 A) (10 ’)# 1.1 W

2 2

d c

– +

Hence, the total power delivered to the resistors is

ε = 12 V

2 ! %! # 2.0 W.

1 2

The 12-V battery delivers power I! # 4.0 W. Half of this

Figure 28.16 (Example 28.8) A series circuit containing two

2

batteries and two resistors, where the polarities of the batteries power is delivered to the two resistors, as we just calculated.

are in opposition.

The other half is delivered to the 6-V battery, which is being872 CHAPTER 28 • Direct Current Circuits

charged by the 12-V battery. If we had included the internal now in the same direction, the signs of ! and ! are the

1 2

resistances of the batteries in our analysis, some of the power same and Equation (1) becomes

would appear as internal energy in the batteries; as a result,

! %! 6.0 V% 12 V

1 2

we would have found that less power was being delivered to

I# # # 1.0 A

R % R 8.0 ’% 10 ’

1 2

the 6-V battery.

The new powers delivered to the resistors are

What If? What if the polarity of the 12.0-V battery were

2 2

! # I R # (1.0 A) (8.0’)# 8.0 W

reversed? How would this affect the circuit? 1 1

2 2

! # I R # (1.0 A) (10’)# 10 W

2 2

Answer While we could repeat the Kirchhoff’s rules

calculation, let us examine Equation (1) and modify it This totals 18 W, nine times as much as in the original circuit,

accordingly. Because the polarities of the two batteries are in which the batteries were opposing each other.

Interactive

Example 28.9 Applying Kirchhoff’s Rules

Find the currents I , I , and I in the circuit shown in Figure Note that in loop befcb we obtain a positive value when

1 2 3

28.17. traversing the 6.0-’ resistor because our direction of travel

is opposite the assumed direction of I . Expressions (1), (2),

1

Solution Conceptualize by noting that we cannot simplify

and (3) represent three independent equations with three

the circuit by the rules of adding resistances in series and

unknowns. Substituting Equation (1) into Equation (2)

in parallel. (If the 10.0-V battery were taken away, we could

gives

reduce the remaining circuit with series and parallel com-

10.0 V$ (6.0’)I $ (2.0’) (I % I )# 0

1 1 2

binations.) Thus, we categorize this problem as one in

which we must use Kirchhoff’s rules. To analyze the circuit,

(4) 10.0 V# (8.0’)I % (2.0’)I

1 2

we arbitrarily choose the directions of the currents as la-

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

beled in Figure 28.17. Applying Kirchhoff’s junction rule

gives

to junction c gives

(5) $ 12.0 V#$ (3.0’)I % (2.0’)I

1 2

(1) I % I # I

1 2 3

Subtracting Equation (5) from Equation (4) eliminates I ,

2

We now have one equation with three unknowns—I , I , and

1 2

giving

I . There are three loops in the circuit—abcda, befcb, and

3

22.0 V# (11.0’)I

1

aefda. We therefore need only two loop equations to deter-

mine the unknown currents. (The third loop equation

would give no new information.) Applying Kirchhoff’s loop I # 2.0 A

1

rule to loops abcda and befcb and traversing these loops

clockwise, we obtain the expressions

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

1 2

(2.0’)I # (3.0’)I $ 12.0 V

2 1

(2) abcda 10.0 V$ (6.0’)I $ (2.0’)I # 0

1 3

# (3.0’)(2.0 A)$ 12.0 V#$ 6.0 V

(3) befcb $14.0 V% (6.0’)I $ 10.0 V$ (4.0’) I # 0

1 2

I # $3.0 A

2

Finally,

14.0 V

e f

I # I % I #

$1.0 A

3 1 2

To ﬁnalize the problem, note that I and I are both nega-

4.0 Ω 2 3

I

2

tive. This indicates only that the currents are opposite the

I

direction we chose for them. However, the numerical values

– + 1

b c

are correct. What would have happened had we left the

6.0 Ω

I

10.0 V 3

current directions as labeled in Figure 28.17 but traversed

the loops in the opposite direction?

a d

2.0 Ω

Figure 28.17 (Example 28.9) A circuit containing different

branches.

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

– +SECTION 28.4 • RC Circuits 873

Example 28.10 A Multiloop Circuit

(A) Under steady-state conditions, ﬁnd the unknown currents From Equation (1) we see that I # I $ I , which, when

1 3 2

I , I , and I in the multiloop circuit shown in Figure 28.18. substituted into Equation (3), gives

1 2 3

(4) (8.00’)I $ (5.00’)I % 8.00 V# 0

2 3

Solution First note that because the capacitor represents

an open circuit, there is no current between g and b along Subtracting Equation (4) from Equation (2), we eliminate I

3

path ghab under steady-state conditions. Therefore, when and ﬁnd that

the charges associated with I reach point g, they all go

1

4.00 V

I #$ # $0.364 A

toward point b through the 8.00-V battery; hence, I # I .

2

gb 1

11.0 ’

Labeling the currents as shown in Figure 28.18 and applying

Because our value for I is negative, we conclude that the di-

2

Equation 28.9 to junction c, we obtain

rection of I is from c to f in the 3.00-’ resistor. Despite this

2

(1) I % I # I

1 2 3

interpretation of the direction, however, we must continue

to use this negative value for I in subsequent calculations

2

Equation 28.10 applied to loops defcd and cfgbc, traversed

because our equations were established with our original

clockwise, gives

choice of direction.

(2) defcd 4.00 V$ (3.00’)I $ (5.00’)I # 0

2 3

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

2

(3) cfgbc (3.00’)I $ (5.00’)I % 8.00 V# 0

2 1

I # 1.38 A I # 1.02 A

1 3

4.00 V

(B) What is the charge on the capacitor?

d e

– +

Solution We can apply Kirchhoff’s loop rule to loop bghab

I

I 5.00 Ω 3

3

(or any other loop that contains the capacitor) to ﬁnd the

3.00 Ω

potential difference "V across the capacitor. We use this

cap

c f

potential difference in the loop equation without reference

I

2 to a sign convention because the charge on the capacitor

I

5.00 Ω

I 1

1 depends only on the magnitude of the potential difference.

Moving clockwise around this loop, we obtain

b

g

$ 8.00 V%"V $ 3.00 V# 0

cap

8.00 V

I = 0

"V # 11.0 V

cap

Because Q# C "V (see Eq. 26.1), the charge on the

cap

a – + h

capacitor is

6.00 µ F

3.00 V

Figure 28.18 (Example 28.10) A multiloop circuit. Kirchhoff’s

Q# (6.00)F)(11.0 V)# 66.0)C

loop rule can be applied to any closed loop, including the one

containing the capacitor.

Why is the left side of the capacitor positively charged?

28.4 RC Circuits

So far we have analyzed direct current circuits in which the current is constant. In DC

circuits containing capacitors, the current is always in the same direction but may vary

in time. A circuit containing a series combination of a resistor and a capacitor is called

an RC circuit.

Charging a Capacitor

Figure 28.19 shows a simple series RC circuit. Let us assume that the capacitor in this

circuit 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 in

4

the circuit, and the capacitor begins to charge. Note that during charging, charges do

4

In previous discussions of capacitors, we assumed a steady-state situation, in which no current was

present in any branch of the circuit containing a capacitor. Now we are considering the case before the

steady-state condition is realized; in this situation, charges are moving and a current exists in the wires

connected to the capacitor.

– +

– +874 CHAPTER 28 • Direct Current Circuits

Resistor

Capacitor

R R

q

–

+

–

C I

+ q

Switch

S S

ε ε

Battery

At the Active Figures link

(a) (b) t < 0 (c) t > 0

at http://www.pse6.com, you

can adjust the values of R and Active Figure 28.19 (a) A capacitor in series with a resistor, switch, and battery.

C to see the effect on the (b) Circuit diagram representing this system at time t* 0, before the switch is closed.

charging of the capacitor. (c) Circuit diagram at time t+ 0, after the switch has been closed.

not jump across the capacitor plates because the gap between the plates represents an

open circuit. Instead, charge is transferred between each plate and its connecting wires

due to the electric ﬁeld established in the wires by the battery, until the capacitor is

fully charged. As the plates are being charged, the potential difference across the

capacitor increases. The value of the maximum charge on the plates depends on the

voltage of the battery. Once the maximum charge is reached, the current in the circuit

is zero because the potential difference across the capacitor matches that supplied by

the battery.

To analyze this circuit quantitatively, let us apply Kirchhoff’s loop rule to the circuit

after the switch is closed. Traversing the loop in Fig. 28.19c clockwise gives

q

!$ $ IR# 0 (28.11)

C

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 the

signs on ! and IR. For the capacitor, notice that we are traveling in the direction from

the positive plate to the negative plate; this represents a decrease in potential. Thus, we

use a negative sign for this potential difference in Equation 28.11. Note that q and I are

instantaneous values that depend on time (as opposed to steady-state values) as the

capacitor 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 on

the capacitor is zero, and from Equation 28.11 we ﬁnd that the initial current I in the

0

circuit is a maximum and is equal to

!

I # (current at t# 0) (28.12)

0

R

At this time, the potential difference from the battery terminals appears entirely across

the resistor. Later, when the capacitor is charged to its maximum value Q , charges

cease to ﬂow, the current in the circuit is zero, and the potential difference from the

battery terminals appears entirely across the capacitor. Substituting I# 0 into Equation

28.11 gives the charge on the capacitor at this time:

Q# C ! (maximum charge) (28.13)

To determine analytical expressions for the time dependence of the charge and

current, we must solve Equation 28.11—a single equation containing two variables, q

and I. The current in all parts of the series circuit must be the same. Thus, the current

in the resistance R must be the same as the current between the capacitor plates and theSECTION 28.4 • RC Circuits 875

wires. This current is equal to the time rate of change of the charge on the capacitor

plates. Thus, we substitute I# dq/dt into Equation 28.11 and rearrange the equation:

dq ! q

# $

dt R RC

To ﬁnd an expression for q, we solve this separable differential equation. We ﬁrst

combine the terms on the right-hand side:

dq C! q q$ C!

# $ #$

dt RC RC RC

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

dq 1

#$ dt

q$ C ! RC

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

q t

dq 1

#$ dt

$ $

(q$ C!) RC

0 0

q$ C! t

ln #$

! "

$C! RC

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

Charge as a function of time for

$t/RC $t/RC

q(t )# C!(1$ e )# Q(1$ e ) (28.14)

a capacitor being charged

where e is the base of the natural logarithm and we have made the substitution from

Equation 28.13.

We can ﬁnd an expression for the charging current by differentiating Equation

28.14 with respect to time. Using I# dq/dt, we ﬁnd that

!

Current as a function of time for

$t/RC

I(t)# e (28.15)

a capacitor being charged

R

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 I #!/R at t# 0 and decays exponentially to zero

0

as t:,. The quantity RC, which appears in the exponents of Equations 28.14 and

28.15, is called the time constant - of the circuit. It represents the time interval

during which the current decreases to 1/e of its initial value; that is, in a time interval

$1 $ 2

-, I# e I # 0.368I . In a time interval 2-, I# e I # 0.135I , and so forth.

0 0 0 0

$1

Likewise, in a time interval -, the charge increases from zero to C![1$ e ]# 0.632C!.

q I

Cε

ε

I I =

0 0

R

0.632Cε

0.368I

0

τ =RC

t t

τ τ

(a) (b)

Figure 28.20 (a) Plot of capacitor charge versus time for the circuit shown in Figure

28.19. 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 approaches

inﬁnity. (b) Plot of current versus time for the circuit shown in Figure 28.19. The

current has its maximum value I #!/R at t# 0 and decays to zero exponentially as t

0

approaches inﬁnity. After a time interval equal to one time constant - has passed, the

current is 36.8% of its initial value.-

876 CHAPTER 28 • Direct Current Circuits

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

"V Q Q

–Q

[ ]# [RC ]# . # # ["t]# T

% & % &

C R

I "V Q / "t

+Q

Because -# RC has units of time, the combination -/RC is dimensionless, as it must be

in order to be an exponent of e in Equations 28.14 and 28.15.

S

2

The energy output of the battery as the capacitor is fully charged is Q!# C! .

t < 0

1 1

2

After the capacitor is fully charged, the energy stored in the capacitor is Q!# C! ,

2 2

(a)

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.

–q

C R I

+q

Discharging a Capacitor

Now consider the circuit shown in Figure 28.21, which consists of a capacitor carrying

S an initial charge Q , a resistor, and a switch. When the switch is open, a potential differ-

t > 0

ence Q /C exists across the capacitor and there is zero potential difference across the

(b)

resistor because I# 0. If the switch is closed at t# 0, the capacitor begins to discharge

through the resistor. At some time t during the discharge, the current in the circuit is I

Active Figure 28.21 (a) A

charged capacitor connected to a

and the charge on the capacitor is q (Fig. 28.21b). The circuit in Figure 28.21 is the

resistor and a switch, which is open

same as the circuit in Figure 28.19 except for the absence of the battery. Thus, we elim-

for t* 0. (b) After the switch is

inate the emf ! from Equation 28.11 to obtain the appropriate loop equation for the

closed at t# 0, a current that

circuit in Figure 28.21:

decreases in magnitude with time is

q

set up in the direction shown, and

$ $ IR# 0 (28.16)

the charge on the capacitor

C

decreases exponentially with time.

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

At the Active Figures link

dq q

at http://www.pse6.com, you

$R #

dt C

can adjust the values of R and

C to see the effect on the

dq 1

#$ dt

discharging of the capacitor.

q RC

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

q t

dq 1

#$ dt

$ $

q RC

Q 0

q t

ln #$

! "

Q RC

$t/RC

q(t )# Qe (28.17)

Charge as a function of time for

a discharging capacitor

Differentiating this expression with respect to time gives the instantaneous current as a

function of time:

dq d Q

$t/RC $t/RC

Current as a function of time for I(t )# # (Qe )#$ e (28.18)

dt dt RC

a discharging capacitor

where Q /RC# I is the initial current. The negative sign indicates that as the capaci-

0

tor discharges, the current direction is opposite its direction when the capacitor was

being charged. (Compare the current directions in Figs. 28.19c and 28.21b.) We see

that both the charge on the capacitor and the current decay exponentially at a rate

characterized by the time constant -# RC.

Quick Quiz 28.9 Consider the circuit in Figure 28.19 and assume that the

battery 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 the

following is equal to the emf of the battery? (d) C (e) R (f) neither C nor R.)

)

SECTION 28.4 • RC Circuits 877

Quick Quiz 28.10 Consider the circuit in Figure 28.22 and assume that the

battery 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 very

long time, the current in the battery is (f) zero (g) !/2R (h) 2!/R (i) !/R (j) impos-

sible to determine.

C

ε

RR

Figure 28.22 (Quick Quiz 28.10) How does the current

vary after the switch is closed?

Conceptual Example 28.11 Intermittent Windshield Wipers

Many automobiles are equipped with windshield wipers that through a multiposition switch. As it increases with time, the

can operate intermittently during a light rainfall. How does voltage across the capacitor reaches a point at which it

the operation of such wipers depend on the charging and triggers the wipers and discharges, ready to begin another

discharging of a capacitor? charging cycle. The time interval between the individual

sweeps of the wipers is determined by the value of the time

Solution The wipers are part of an RC circuit whose time

constant.

constant can be varied by selecting different values of R

Interactive

Example 28.12 Charging a Capacitor in an RC Circuit

An uncharged capacitor and a resistor are connected in

q(µ µC)

series to a battery, as shown in Figure 28.23. If !# 12.0 V,

Q = 60.0 µ µC

60

5

C# 5.00)F, and R# 8.00. 10 ’, ﬁnd the time constant

50

of the circuit, the maximum charge on the capacitor, the

40

maximum current in the circuit, and the charge and current

as functions of time.

30

20

Solution The time constant of the circuit is -# RC#

5 $6 10

(8.00. 10 ’)(5.00. 10 F)# 4.00 s. The maximum

t(s)

0

charge on the capacitor is Q# C!# (5.00)F)(12.0 V)#

01234567

60.0)C. The maximum current in the circuit is I #

0

(a)

5

!/R # (12.0 V)/(8.00. 10 ’)# 15.0)A. Using these

values and Equations 28.14 and 28.15, we ﬁnd that

t =τ

$t/4.00 s

I(µ µA)

q(t )# (60.0 C)(1$ e )

I = 15.0 µ µA

15

0

$t/4.00 s

I(t )#

(15.0 A)e

10

Graphs of these functions are provided in Figure 28.24.

R

5

C

t(s)

0

01234567

ε

(b)

+ –

S

Figure 28.24 (Example 28.12) Plots of (a) charge versus time

Figure 28.23 (Example 28.12) The switch in this series RC and (b) current versus time for the RC circuit shown in Figure

5

circuit, open for times t* 0, is closed at t# 0. 28.23, with !# 12.0 V, R# 8.00. 10 ’, and C# 5.00)F.

At the Interactive Worked Example link at http://www.pse6.com, you can vary R, C, and ! and observe the charge and

current as functions of time while charging or discharging the capacitor.878 CHAPTER 28 • Direct Current Circuits

Example 28.13 Discharging a Capacitor in an RC Circuit

2

Consider a capacitor of capacitance C that is being dis- where U # Q /2C is the initial energy stored in the capaci-

0

charged through a resistor of resistance R, as shown in tor. As in part (A), we now set U# U /4 and solve for t:

0

Figure 28.21.

U

0

$2t/RC

# U e

0

4

(A) After how many time constants is the charge on the

capacitor one-fourth its initial value? 1

$2t/RC

# e

4

Solution The charge on the capacitor varies with time

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

$t/RC

according to Equation 28.17, q(t)# Qe . To ﬁnd the time

interval during which q drops to one-fourth its initial value,

1

t# RC ln 4# 0.693RC# 0.693-

we substitute q(t)# Q /4 into this expression and solve for t: 2

Q

What If? What if we wanted to describe the circuit in terms

$t/RC

# Qe

4 of the time interval required for the charge to fall to one-half

its original value, rather than by the time constant "? This

1

$t/RC

# e

4 would give a parameter for the circuit called its half-life t .

1/2

How is the half-life related to the time constant?

Taking logarithms of both sides, we ﬁnd

Answer After one half-life, the charge has fallen from Q to

t

$ln 4 #$ Q /2. Thus, from Equation 28.17,

RC

Q

$t /RC

1/ 2

# Qe

2

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

1

$t /RC

1/2

# e

2

(B) The energy stored in the capacitor decreases with time

leading to

as the capacitor discharges. After how many time constants

is this stored energy one-fourth its initial value?

t # 0.693-

1/2

2

The concept of half-life will be important to us when we

Solution Using Equations 26.11 (U# Q /2C) and 28.17, we

study nuclear decay in Chapter 44. The radioactive decay of

can express the energy stored in the capacitor at any time t as

an unstable sample behaves in a mathematically similar

2 2

q Q

$2t/RC $2t/RC manner to a discharging capacitor in an RC circuit.

U# # e # U e

0

2C 2C

Example 28.14 Energy Delivered to a Resistor

the energy delivered to the resistor must equal the time inte-

A 5.00-)F capacitor is charged to a potential difference of

2

gral of I R dt:

800 V and then discharged through a 25.0-kV resistor. How

much energy is delivered to the resistor in the time interval

, ,

required to fully discharge the capacitor? 2 $t/RC 2

Energy# I R dt# ($I e ) R dt

$ $

0

0 0

Solution We shall solve this problem in two ways. The ﬁrst

To evaluate this integral, we note that the initial current I is

0

way is to note that the initial energy in the circuit equals the

2 equal to !/R and that all parameters except t are constant.

energy stored in the capacitor, C! /2 (see Eq. 26.11). Once

Thus, we ﬁnd

the capacitor is fully discharged, the energy stored in it is

zero. Because energy in an isolated system is conserved, the

2 ,

!

initial energy stored in the capacitor is transformed into in-

$2t/RC

(1) Energy# e dt

$

ternal energy in the resistor. Using the given values of C and

R

0

!, we ﬁnd

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

1 1

2 $6 2

Energy# C! # (5.00. 10 F)(800 V) # we ﬁnd

1.60 J

2 2

1

2

Energy# C!

2

The second way, which is more difﬁcult but perhaps more

which agrees with the result we obtained using the simpler

instructive, is to note that as the capacitor discharges

approach, as it must. Note that we can use this second

through the resistor, the rate at which energy is delivered to

2

approach to ﬁnd the total energy delivered to the resistor at

the resistor is given by I R, where I is the instantaneous

any time after the switch is closed by simply replacing the

current given by Equation 28.18. Because power is deﬁned

upper limit in the integral with that speciﬁc value of t.

as the rate at which energy is transferred, we conclude thatSECTION 28.5 • Electrical Meters 879

Scale

28.5 Electrical Meters

The Galvanometer

The galvanometer is the main component in analog meters for measuring current and

voltage. (Many analog meters are still in use although digital meters, which operate on a

different principle, are currently in wide use.) Figure 28.25 illustrates the essential

N S

features of a common type called the D’Arsonval galvanometer. It consists of a coil of wire

mounted 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 acts

on 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 greater

Spring Coil

the torque and the more the coil rotates before the spring tightens enough to stop the

Figure 28.25 The principal compo-

rotation. Hence, the deﬂection of a needle attached to the coil is proportional to the

nents of a D’Arsonval galvanometer.

current. Once the instrument is properly calibrated, it can be used in conjunction with

When the coil situated in a magnetic

other circuit elements to measure either currents or potential differences.

ﬁeld carries a current, the magnetic

torque causes the coil to twist. The

angle through which the coil rotates

is proportional to the current in the

The Ammeter

coil because of the counteracting

torque of the spring.

A device that measures current is called an ammeter. The charges constituting the

current to be measured must pass directly through the ammeter, so the ammeter

R R

1 2

must be connected in series with other elements in the circuit, as shown in Figure

28.26. When using an ammeter to measure direct currents, you must connect it so

–

that charges enter the instrument at the positive terminal and exit at the negative

A

terminal.

Ideally, an ammeter should have zero resistance so that the current being +

measured is not altered. In the circuit shown in Figure 28.26, this condition requires

that the resistance of the ammeter be much less than R % R . Because any ammeter

1 2

ε

always has some internal resistance, the presence of the ammeter in the circuit slightly

Figure 28.26 Current can be mea-

reduces the current from the value it would have in the meter’s absence.

sured with an ammeter connected in

A typical off-the-shelf galvanometer is often not suitable for use as an ammeter,

series with the elements in which the

primarily because it has a resistance of about 60 ’. An ammeter resistance this great

measurement of a current is desired.

considerably alters the current in a circuit. You can understand this by considering An ideal ammeter has zero resistance.

the following example. The current in a simple series circuit containing a 3-V

Galvanometer

battery and a 3-’ resistor is 1 A. If you insert a 60-’ galvanometer in this circuit to

measure the current, the total resistance becomes 63 ’ and the current is reduced

to 0.048 A!

60 Ω

A second factor that limits the use of a galvanometer as an ammeter is the fact that

a typical galvanometer gives a full-scale deﬂection for currents on the order of 1 mA or

less. Consequently, such a galvanometer cannot be used directly to measure currents

greater than this value. However, it can be converted to a useful ammeter by placing a

R

shunt resistor R in parallel with the galvanometer, as shown in Figure 28.27. The value p

p

of R must be much less than the galvanometer resistance so that most of the current

p

to be measured is directed to the shunt resistor.

Active Figure 28.27 A galva-

nometer is represented here by its

internal resistance of 60 ’. When a

The Voltmeter

galvanometer is to be used as an

ammeter, a shunt resistor R is

p

A device that measures potential difference is called a voltmeter. The potential differ-

connected in parallel with the

ence between any two points in a circuit can be measured by attaching the terminals of

galvanometer.

the voltmeter between these points without breaking the circuit, as shown in Figure

At the Active Figures link

28.28. The potential difference across resistor R is measured by connecting the volt-

2

at http://www.pse6.com, you can

meter in parallel with R . Again, it is necessary to observe the polarity of the instru-

2

predict the value of R needed

p

ment. The positive terminal of the voltmeter must be connected to the end of the

to cause full-scale deﬂection in

resistor that is at the higher potential, and the negative terminal to the end of the resis-

the circuit of Figure 28.26, and

tor at the lower potential.

test your result.880 CHAPTER 28 • Direct Current Circuits

Galvanometer

V

R

s

60 Ω

R R

1 2

Active Figure 28.29 When the galvanometer

is used as a voltmeter, a resistor R is

s

connected in series with the galvanometer.

ε

Figure 28.28 The potential difference across

At the Active Figures link at

a resistor can be measured with a voltmeter

http://www.pse6.com, you can predict

connected in parallel with the resistor. An

the value of R needed to cause full-

s

ideal voltmeter has inﬁnite resistance.

scale deﬂection in the circuit of Figure

28.28, and test your result.

An ideal voltmeter has inﬁnite resistance so that no current exists in it. In

Figure 28.28, this condition requires that the voltmeter have a resistance much greater

than R . In practice, if this condition is not met, corrections should be made for the

2

known resistance of the voltmeter.

A galvanometer can also be used as a voltmeter by adding an external resistor R in

s

series with it, as shown in Figure 28.29. In this case, the external resistor must have a

value 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 in

this chapter. In our world of electrical appliances, it is useful to understand the power

requirements and limitations of conventional electrical systems and the safety

measures 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 to

120 V 5

these wires. One wire is called the live wire, as illustrated in Figure 28.30, and the other

Live

is called the neutral wire. The neutral wire is grounded; that is, its electric potential is

Meter

taken to be zero. The potential difference between the live and neutral wires is about

Neutral

120 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 companies

Circuit

breaker

supply to businesses and households. (Alternating voltage and current are discussed in

Chapter 33.)

A meter is connected in series with the live wire entering the house to record the

household’s energy consumption. After the meter, the wire splits so that there are

several separate circuits in parallel distributed throughout the house. Each circuit

contains a circuit breaker (or, in older installations, a fuse). The wire and circuit

breaker for each circuit are carefully selected to meet the current demands for that

R R R

1 2 3

circuit. If a circuit is to carry currents as large as 30 A, a heavy wire and an appropriate

circuit breaker must be selected to handle this current. A circuit used to power only

0 V

lamps and small appliances often requires only 20 A. Each circuit has its own circuit

breaker to provide protection for that part of the entire electrical system of the house.

Figure 28.30 Wiring diagram for a

household circuit. The resistances

represent appliances or other

5

electrical devices that operate with

Live wire is a common expression for a conductor whose electric potential is above or below ground

an applied voltage of 120 V.

potential.SECTION 28.6 • Household Wiring and Electrical Safety 881

+120 V –120 V

(a) (b)

Figure 28.31 (a) An outlet for connection to a 240-V supply. (b) The connections for

each of the openings in a 240-V outlet.

As an example, consider a circuit in which a toaster oven, a microwave oven, and a

coffee maker are connected (corresponding to R , R , and R in Fig. 28.30). We can

1 2 3

calculate the current in each appliance by using the expression !# I"V. The toaster

oven, rated at 1 000 W, draws a current of 1 000 W/120 V# 8.33 A. The microwave

oven, rated at 1 300 W, draws 10.8 A, and the coffee maker, rated at 800 W, draws

6.67 A. If the three appliances are operated simultaneously, they draw a total current of

25.8 A. Therefore, the circuit should be wired to handle at least this much current. If

the rating of the circuit breaker protecting the circuit is too small—say, 20 A—the

breaker will be tripped when the third appliance is turned on, preventing all three

appliances from operating. To avoid this situation, the toaster oven and coffee maker

can 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 V

for their operation. The power company supplies this voltage by providing a third wire

that is 120 V below ground potential (Fig. 28.31). The potential difference between

this 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 to

operating it at 120 V; therefore, smaller wires can be used in the higher-voltage circuit

without overheating.

Electrical Safety

When the live wire of an electrical outlet is connected directly to ground, the circuit is

completed and a short-circuit condition exists. A short circuit occurs when almost zero

resistance exists between two points at different potentials; this results in a very large

current. When this happens accidentally, a properly operating circuit breaker opens

the circuit and no damage is done. However, a person in contact with ground can be

electrocuted by touching the live wire of a frayed cord or other exposed conductor. An

exceptionally effective (and dangerous!) ground contact is made when the person

either touches a water pipe (normally at ground potential) or stands on the ground

with wet feet. The latter situation represents effective ground contact because normal,

nondistilled water is a conductor due to the large number of ions associated with

impurities. 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 the

magnitude of the current, the length of time it acts, the part of the body touched by

the live wire, and the part of the body in which the current exists. Currents of 5 mA or

less cause a sensation of shock but ordinarily do little or no damage. If the current is

larger than about 10 mA, the muscles contract and the person may be unable to

release the live wire. If a current of about 100 mA passes through the body for only a

few seconds, the result can be fatal. Such a large current paralyzes the respiratory

George Semple882 CHAPTER 28 • Direct Current Circuits

“Ouch!”

Motor

“Neutral”

I

120 V

I

Wall Circuit

outlet breaker

“Live”

I

Ground

(a)

“Neutral”

Motor

I

“Ground”

I

120 V

I

3-wire Circuit

outlet breaker

“Live”

I

Ground

(b)

Figure 28.32 (a) A diagram of the circuit for an electric drill with only two connecting

wires. The normal current path is from the live wire through the motor connections

and back to ground through the neutral wire. In the situation shown, the live wire has

come into contact with the drill case. As a result, the person holding the drill acts as a

current path to ground and receives an electric shock. (b) This shock can be avoided

by 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 produce

serious (and sometimes fatal) burns. In practice, no contact with live wires is regarded

as safe whenever the voltage is greater than 24 V.

Many 120-V outlets are designed to accept a three-pronged power cord. (This

feature is required in all new electrical installations.) One of these prongs is the live

wire 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 that

normally carries no current but is both grounded and connected directly to the casing

of 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 the

appliance experiences an electric shock because the body provides a low-resistance

path to ground.

Special power outlets called ground-fault interrupters (GFIs) are now used in

kitchens, bathrooms, basements, exterior outlets, and other hazardous areas of new

homes. These devices are designed to protect persons from electric shock by sensing

small currents (’ 5 mA) leaking to ground. (The principle of their operation is

described in Chapter 31.) When an excessive leakage current is detected, the current is

shut off in less than 1 ms.Summary 883

SUM MARY

The emf of a battery is equal to the voltage across its terminals when the current is Take a practice test for

this chapter by clicking on

zero. That is, the emf is equivalent to the open-circuit voltage of the battery.

the Practice Test link at

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

http://www.pse6.com.

…

R # R % R % R % (28.6)

eq 1 2 3

The equivalent resistance of a set of resistors connected in parallel is found

from the relationship

1 1 1 1

# % % %((( (28.8)

R R R R

eq 1 2 3

If it is possible to combine resistors into series or parallel equivalents, the preceding

two equations make it easy to determine how the resistors inﬂuence the rest of the

circuit.

Circuits involving more than one loop are conveniently analyzed with the use of

Kirchhoff’s rules:

1. Junction rule. The sum of the currents entering any junction in an electric circuit

must equal the sum of the currents leaving that junction:

I # I (28.9)

# in # out

2. Loop rule. The sum of the potential differences across all elements around any

circuit loop must be zero:

"V# 0 (28.10)

#

closed

loop

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

statement 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 the

current, "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 a

source 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 you

to analyze electric circuits.

If a capacitor is charged with a battery through a resistor of resistance R, the

charge on the capacitor and the current in the circuit vary in time according to the

expressions

$t/RC

q(t )# Q(1$ e ) (28.14)

!

$t/RC

I(t )# e (28.15)

R

where Q# C! is the maximum charge on the capacitor. The product RC is called the

time constant- of the circuit. If a charged capacitor is discharged through a resistor

of resistance R, the charge and current decrease exponentially in time according to

the expressions

$t/RC

q(t )# Qe (28.17)

$t/RC

I(t )#$I e (28.18)

0

where Q is the initial charge on the capacitor and I # Q /RC is the initial current in

0

the circuit.884 CHAPTER 28 • Direct Current Circuits

QUESTIONS

1. Explain the difference between load resistance in a circuit 14. What is the internal resistance of an ideal ammeter? Of

and internal resistance in a battery. an ideal voltmeter? Do real meters ever attain these

ideals?

2. Under what condition does the potential difference across

the terminals of a battery equal its emf? Can the terminal 15. A “short circuit” is a path of very low resistance in a circuit

voltage ever exceed the emf? Explain. in parallel with some other part of the circuit. Discuss the

effect of the short circuit on the portion of the circuit it

3. Is the direction of current through a battery always from

parallels. Use a lamp with a frayed cord as an example.

the negative terminal to the positive terminal? Explain.

16. If electric power is transmitted over long distances, the

4. How would you connect resistors so that the equivalent

resistance of the wires becomes signiﬁcant. Why? Which

resistance is larger than the greatest individual resistance?

method of transmission would result in less energy

Give an example involving three resistors.

wasted—high current and low voltage or low current and

5. How would you connect resistors so that the equivalent

high voltage? Explain your answer.

resistance is smaller than the least individual resistance?

17. Are the two headlights of a car wired in series or in paral-

Give an example involving three resistors.

lel? How can you tell?

6. Given three lightbulbs and a battery, sketch as many differ-

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

ent electric circuits as you can.

What are they?

7. When resistors are connected in series, which of the follow-

19. Figure Q28.19 shows a series combination of three

ing would be the same for each resistor: potential differ-

lightbulbs, all rated at 120 V with power ratings of 60 W,

ence, current, power?

75 W, and 200 W. Why is the 60-W lamp the brightest and

8. When resistors are connected in parallel, which of the fol-

the 200-W lamp the dimmest? Which bulb has the greatest

lowing would be the same for each resistor: potential dif-

resistance? How would their intensities differ if they were

ference, current, power?

connected in parallel?

9. What advantage might there be in using two identical

resistors in parallel connected in series with another

identical parallel pair, rather than just using a single

resistor?

10. An incandescent lamp connected to a 120-V source with a

short extension cord provides more illumination than the

same lamp connected to the same source with a very long

extension cord. Explain.

11. Why is it possible for a bird to sit on a high-voltage wire

without being electrocuted?

12. When can the potential difference across a resistor be

positive?

13. Referring to Figure Q28.13, describe what happens to the

lightbulb after the switch is closed. Assume that the capaci-

tor has a large capacitance and is initially uncharged, and

assume that the light illuminates when connected directly

Figure Q 28.19

across the battery terminals.

20. A student claims that the second lightbulb in series is less

bright than the ﬁrst, because the ﬁrst bulb uses up some of

the current. How would you respond to this statement?

21. Is a circuit breaker wired in series or in parallel with the

device it is protecting?

C

22. So that your grandmother can listen to A Prairie Home Com-

panion, you take her bedside radio to the hospital where

she is staying. You are required to have a maintenance

worker test it for electrical safety. Finding that it develops

120 V on one of its knobs, he does not let you take it up to

your grandmother’s room. She complains that she has had

the radio for many years and nobody has ever gotten

+ –

a shock from it. You end up having to buy a new plastic

Switch

Battery

radio. Is this fair? Will the old radio be safe back in her

Figure Q28.13 bedroom?

Henry Leap and Jim LehmanProblems 885

23. Suppose you fall from a building and on the way down A B C

grab a high-voltage wire. If the wire supports you as you

hang from it, will you be electrocuted? If the wire then

breaks, should you continue to hold onto an end of the

wire as you fall?

24.

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

What disadvantages?

25. When electricians work with potentially live wires, they

often use the backs of their hands or ﬁngers to move wires.

S

ε

Why do you suppose they use this technique?

26. What procedure would you use to try to save a person who

is “frozen” to a live high-voltage wire without endangering

your own life? Figure Q28.29

27. If it is the current through the body that determines

how serious a shock will be, why do we see warnings of

30. If your car’s headlights are on when you start the ignition,

high voltage rather than high current near electrical

why do they dim while the car is starting?

equipment?

31. A ski resort consists of a few chair lifts and several inter-

28. Suppose you are ﬂying a kite when it strikes a high-voltage

connected downhill runs on the side of a mountain, with a

wire. What factors determine how great a shock you

lodge at the bottom. The lifts are analogous to batteries,

receive?

and the runs are analogous to resistors. Describe how two

29. A series circuit consists of three identical lamps connected runs can be in series. Describe how three runs can be in

to a battery as shown in Figure Q28.29. When the switch S parallel. Sketch a junction of one lift and two runs. State

is closed, what happens (a) to the intensities of lamps A Kirchhoff’s junction rule for ski resorts. One of the skiers

and B; (b) to the intensity of lamp C; (c) to the current in happens to be carrying a skydiver’s altimeter. She never

the circuit; and (d) to the voltage across the three lamps? takes the same set of lifts and runs twice, but keeps passing

(e) Does the power delivered to the circuit increase, you at the ﬁxed location where you are working. State

decrease, or remain the same? Kirchhoff’s loop rule for ski resorts.

PROBLEMS

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

Section 28.1 Electromotive Force (b) when the starter motor is operated, taking an addi-

tional 35.0 A from the battery?

A battery has an emf of 15.0 V. The terminal voltage

1.

of the battery is 11.6 V when it is delivering 20.0 W of

Section 28.2 Resistors in Series and Parallel

power to an external load resistor R. (a) What is the value

5. The current in a loop circuit that has a resistance of R is

of R? (b) What is the internal resistance of the battery?

1

2.00 A. The current is reduced to 1.60 A when an addi-

2. (a) What is the current in a 5.60-’ resistor connected to a

tional resistor R # 3.00 ’ is added in series with R .

2 1

battery that has a 0.200-’ internal resistance if the termi-

What is the value of R ?

1

nal voltage of the battery is 10.0 V? (b) What is the emf of

6. (a) Find the equivalent resistance between points a and b

the battery?

in Figure P28.6. (b) A potential difference of 34.0 V is

3. Two 1.50-V batteries — with their positive terminals in the

applied between points a and b. Calculate the current in

same direction — are inserted in series into the barrel of a

each resistor.

ﬂashlight. One battery has an internal resistance of 0.255 ’,

the other an internal resistance of 0.153’. When the

switch is closed, a current of 600 mA occurs in the lamp. 7.00 Ω

(a) What is the lamp’s resistance? (b) What fraction of the

4.00 Ω 9.00 Ω

chemical energy transformed appears as internal energy in

the batteries?

10.0 Ω

4. An automobile battery has an emf of 12.6 V and an inter-

nal resistance of 0.080 0’. The headlights together

present equivalent resistance 5.00’ (assumed constant).

a b

What is the potential difference across the headlight bulbs

(a) when they are the only load on the battery and Figure P28.6886 CHAPTER 28 • Direct Current Circuits

7. A lightbulb marked “75 W [at] 120 V” is screwed into a 11. Three 100-’ resistors are connected as shown in Figure

socket at one end of a long extension cord, in which each P28.11. The maximum power that can safely be deliv-

of the two conductors has resistance 0.800’. The other ered to any one resistor is 25.0 W. (a) What is the maxi-

end of the extension cord is plugged into a 120-V outlet. mum voltage that can be applied to the terminals a and

Draw a circuit diagram and ﬁnd the actual power deliv- b? For the voltage determined in part (a), what is the

ered to the bulb in this circuit. power delivered to each resistor? What is the total power

delivered?

8. Four copper wires of equal length are connected in

2 2

series. Their cross-sectional areas are 1.00 cm , 2.00 cm ,

2 2

3.00 cm , and 5.00 cm . A potential difference of 120 V is

100 Ω

applied across the combination. Determine the voltage

2

across the 2.00-cm wire.

100 Ω

a b

9. Consider the circuit shown in Figure P28.9. Find

(a) the current in the 20.0-’ resistor and (b) the potential

difference between points a and b.

100 Ω

Figure P28.11

10.0 Ω

25.0 V

10.0 Ω

12. Using only three resistors—2.00 ’, 3.00 ’, and 4.00 ’—

a b

ﬁnd 17 resistance values that may be obtained by various

combinations of one or more resistors. Tabulate the com-

binations in order of increasing resistance.

5.00 Ω 20.0 Ω

5.00 Ω

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 the

Figure P28.9

500-’ resistor.

10. For the purpose of measuring the electric resistance of 14. A 6.00-V battery supplies current to the circuit shown in

shoes through the body of the wearer to a metal ground Figure P28.14. When the double-throw switch S is open, as

plate, the American National Standards Institute (ANSI) shown in the ﬁgure, the current in the battery is 1.00 mA.

speciﬁes the circuit shown in Figure P28.10. The potential When the switch is closed in position 1, the current in the

difference "V across the 1.00-M’ resistor is measured with battery is 1.20 mA. When the switch is closed in position 2,

a high-resistance voltmeter. (a) Show that the resistance of the current in the battery is 2.00 mA. Find the resistances

the footwear is given by R , R , and R .

1 2 3

50.0 V$"V

R # 1.00 M’

shoes

! "

"V R R

1 2

(b) In a medical test, a current through the human body R

2

should not exceed 150)A. Can the current delivered by

1

the ANSI-speciﬁed circuit exceed 150)A? To decide,

6.00 V S

consider a person standing barefoot on the ground plate.

2

R

3

Figure P28.14

1.00 MΩ

15.

Calculate the power delivered to each resistor in the

circuit shown in Figure P28.15.

V

2.00 Ω

50.0 V

3.00 Ω 1.00 Ω

18.0 V

4.00 Ω

Figure P28.15

Figure P28.10Problems 887

16. Two resistors connected in series have an equivalent resis- 7.00 Ω 15.0 V

tance of 690 ’. When they are connected in parallel, their

equivalent resistance is 150 ’. Find the resistance of each

resistor.

I

1

5.00 Ω

17. An electric teakettle has a multiposition switch and two

A

heating coils. When only one of the coils is switched on, the

well-insulated kettle brings a full pot of water to a boil over

I

2

the time interval "t. When only the other coil is switched

2.00 Ω ε

on, it requires a time interval of 2"t to boil the same

amount of water. Find the time interval required to boil the

same amount of water if both coils are switched on (a) in a

Figure P28.20

parallel connection and (b) in a series connection.

18. In Figures 28.4 and 28.6, let R # 11.0 ’, R # 22.0 ’,

1 2

and let the battery have a terminal voltage of 33.0 V. (a) In

21. Determine the current in each branch of the circuit

the parallel circuit shown in Figure 28.6, to which resistor

shown in Figure P28.21.

is more power delivered? (b) Verify that the sum of the

2

power (I R) delivered to each resistor equals the power

supplied by the battery (!# I"V ). (c) In the series

3.00 Ω

circuit, which resistor uses more power? (d) Verify that the

2

sum of the power (I R) used by each resistor equals the

power supplied by the battery (!# I"V ). (e) Which

5.00 Ω

circuit conﬁguration uses more power?

1.00 Ω

19. Four resistors are connected to a battery as shown in

Figure P28.19. The current in the battery is I, the battery

8.00 Ω 1.00 Ω

emf is !, and the resistor values are R # R, R # 2R,

1 2

+

R # 4R, R # 3R. (a) Rank the resistors according to the

3 4

12.0 V

potential difference across them, from largest to smallest.

+

$

4.00 V

Note any cases of equal potential differences. (b) Deter-

$

mine the potential difference across each resistor in terms

of !. (c) Rank the resistors according to the current in

them, from largest to smallest. Note any cases of equal

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

currents. (d) Determine the current in each resistor in

terms of I. (e) What If? If R is increased, what happens to

3

22. In Figure P28.21, show how to add just enough ammeters

the current in each of the resistors? (f) In the limit that

to measure every different current. Show how to add just

R :,, what are the new values of the current in each

3

enough voltmeters to measure the potential difference

resistor in terms of I, the original current in the battery?

across each resistor and across each battery.

23. The circuit considered in Problem 21 and shown in Figure

R = 2R

2

P28.21 is connected for 2.00 min. (a) Find the energy

delivered by each battery. (b) Find the energy delivered to

each resistor. (c) Identify the types of energy transformations

R = R

1 that occur in the operation of the circuit and the total

amount of energy involved in each type of transformation.

R = 3R

4

24. Using Kirchhoff’s rules, (a) ﬁnd the current in each resistor

in Figure P28.24. (b) Find the potential difference between

ε

points c and f. Which point is at the higher potential?

I

R = 4R 4.00 kΩ

3

c

b d

Figure P28.19

εε εε εε

1 2 3

R

3

60.0 V 80.0 V

70.0 V

Section 28.3 Kirchhoff’s Rules

R

Note: The currents are not necessarily in the direction 3.00 kΩ

2

shown for some circuits.

2.00 kΩ

a e

f

R

1

20. The ammeter shown in Figure P28.20 reads 2.00 A. Find

I , I , and !. Figure P28.24

1 2888 CHAPTER 28 • Direct Current Circuits

25. Taking R# 1.00 k’ and !# 250 V in Figure P28.25, 12.0 V

4.00 Ω

determine the direction and magnitude of the current in

the horizontal wire between a and e.

b

2.00 Ω

a

R 2R

c

b d

+

6.00 Ω

2 8.00 V

ε

ε

4R 3R –

Figure P28.29

a e

30. Calculate the power delivered to each resistor shown in

Figure P28.25

Figure P28.30.

26. In the circuit of Figure P28.26, determine the current in

2.0 Ω

each resistor and the voltage across the 200-’ resistor.

40 V 360 V 80 V 50 V 20 V

4.0 Ω 4.0 Ω

2.0 Ω

200 Ω 80 Ω 20 Ω 70 Ω

Figure P28.30

Section 28.4 RC Circuits

31. Consider a series RC circuit (see Fig. 28.19) for which

Figure P28.26

R# 1.00 M’, C# 5.00)F, and !# 30.0 V. Find (a) the

time constant of the circuit and (b) the maximum charge

27. A dead battery is charged by connecting it to the live battery

on the capacitor after the switch is closed. (c) Find the

of another car with jumper cables (Fig. P28.27). Determine

current in the resistor 10.0 s after the switch is closed.

the current in the starter and in the dead battery.

32. A 2.00-nF capacitor with an initial charge of 5.10)C is

discharged through a 1.30-k’ resistor. (a) Calculate the

current in the resistor 9.00)s after the resistor is con-

nected across the terminals of the capacitor. (b) What

0.01 Ω 1.00 Ω

charge remains on the capacitor after 8.00)s? (c) What is

0.06 Ω

the maximum current in the resistor?

Starter

33. A fully charged capacitor stores energy U . How much

0

+ +

12 V 10 V

energy remains when its charge has decreased to half its

– –

original value?

Live Dead

34. A capacitor in an RC circuit is charged to 60.0% of its

battery battery

maximum value in 0.900 s. What is the time constant of

Figure P28.27

the circuit?

35. Show that the integral in Equation (1) of Example 28.14

28. For the network shown in Figure P28.28, show that the

has the value RC/2.

resistance R # (27/17) ’.

ab

36. In the circuit of Figure P28.36, the switch S has been open

for a long time. It is then suddenly closed. Determine the

time constant (a) before the switch is closed and (b) after

1.0 Ω 1.0 Ω

a b

the switch is closed. (c) Let the switch be closed at t# 0.

Determine the current in the switch as a function of time.

1.0 Ω

3.0 Ω 5.0 Ω

50.0 kΩ

Figure P28.28

10.0 V

10.0 µF

S

29. For the circuit shown in Figure P28.29, calculate (a) the

current in the 2.00-’ resistor and (b) the potential differ- 100 kΩ

ence between points a and b. Figure P28.36

–

+Problems 889

37. The circuit in Figure P28.37 has been connected for a long Section 28.5 Electrical Meters

time. (a) What is the voltage across the capacitor? (b) If the

41. Assume that a galvanometer has an internal resistance of

battery is disconnected, how long does it take the capacitor

60.0 ’ and requires a current of 0.500 mA to produce full-

to discharge to one tenth of its initial voltage?

scale deﬂection. What resistance must be connected in

parallel with the galvanometer if the combination is to

serve as an ammeter that has a full-scale deﬂection for

a current of 0.100 A?

1.00 Ω

8.00 Ω

42. A typical galvanometer, which requires a current of

1.00 µ µF

1.50 mA for full-scale deﬂection and has a resistance of

10.0 V

75.0 ’, may be used to measure currents of much greater

values. To enable an operator to measure large currents

4.00 Ω

2.00 Ω

without damage to the galvanometer, a relatively small

shunt resistor is wired in parallel with the galvanometer,

Figure P28.37

as suggested in Figure 28.27. Most of the current then

goes through the shunt resistor. Calculate the value of the

shunt resistor that allows the galvanometer to be used to

38. In places such as a hospital operating room and a factory

measure a current of 1.00 A at full-scale deﬂection.

for electronic circuit boards, electric sparks must be

(Suggestion: use Kirchhoff’s rules.)

avoided. A person standing on a grounded ﬂoor and

43. The same galvanometer described in the previous problem

touching nothing else can typically have a body capaci-

may be used to measure voltages. In this case a large resis-

tance of 150 pF, in parallel with a foot capacitance of

tor is wired in series with the galvanometer, as suggested in

80.0 pF produced by the dielectric soles of his or her

Figure 28.29. The effect is to limit the current in the

shoes. The person acquires static electric charge from

galvanometer when large voltages are applied. Most of the

interactions with furniture, clothing, equipment, packag-

potential drop occurs across the resistor placed in series.

ing materials, and essentially everything else. The static

Calculate the value of the resistor that allows the

charge is conducted to ground through the equivalent

galvanometer to measure an applied voltage of 25.0 V at

resistance of the two shoe soles in parallel with each other.

full-scale deﬂection.

A pair of rubber-soled street shoes can present an equiva-

lent resistance of 5 000 M’. A pair of shoes with special 44. Meter loading. Work this problem to ﬁve-digit precision.

static-dissipative soles can have an equivalent resistance of Refer to Figure P28.44. (a) When a 180.00-’ resistor is

1.00 M’. Consider the person’s body and shoes as connected across a battery of emf 6.000 0 V and internal

forming an RC circuit with the ground. (a) How long does resistance 20.000 ’, what is the current in the resistor?

it take the rubber-soled shoes to reduce a 3 000-V static What is the potential difference across it? (b) Suppose

charge to 100 V? (b) How long does it take the static- now an ammeter of resistance 0.500 00 ’ and a voltmeter

dissipative shoes to do the same thing? of resistance 20 000 ’ are added to the circuit as shown in

Figure P28.44b. Find the reading of each. (c) What If?

39. A 4.00-M’ resistor and a 3.00-)F capacitor are connected

Now one terminal of one wire is moved, as shown in

in series with a 12.0-V power supply. (a) What is the time

Figure P28.44c. Find the new meter readings.

constant for the circuit? (b) Express the current in the

circuit and the charge on the capacitor as functions of 45. Design a multirange ammeter capable of full-scale deﬂec-

time. tion for 25.0 mA, 50.0 mA, and 100 mA. Assume the meter

movement is a galvanometer that has a resistance of

40. Dielectric materials used in the manufacture of capacitors

25.0’ and gives a full-scale deﬂection for 1.00 mA.

are characterized by conductivities that are small but not

zero. Therefore, a charged capacitor slowly loses its charge 46. Design a multirange voltmeter capable of full-scale

by “leaking” across the dielectric. If a capacitor having deflection for 20.0 V, 50.0 V, and 100 V. Assume the

capacitance C leaks charge such that the potential differ- meter movement is a galvanometer that has a resistance

ence has decreased to half its initial (t# 0) value at a time of 60.0’ and gives a full-scale deflection for a current of

t, what is the equivalent resistance of the dielectric? 1.00 mA.

6.000 0 V

20.000 Ω

A A

V V

180.00 Ω

(a) (b) (c)

Figure P28.44890 CHAPTER 28 • Direct Current Circuits

4.00 V

47. A particular galvanometer serves as a 2.00-V full-scale volt-

2.00 Ω

meter when a 2 500-’ resistor is connected in series with

a

it. It serves as a 0.500-A full-scale ammeter when a 0.220-’

resistor is connected in parallel with it. Determine the

internal resistance of the galvanometer and the current

required to produce full-scale deﬂection.

4.00 Ω

12.0 V

Section 28.6 Household Wiring and Electrical

Safety

b

48. An 8.00-ft extension cord has two 18-gauge copper wires,

10.0 Ω

each having a diameter of 1.024 mm. At what rate is

Figure P28.54

energy delivered to the resistance in the cord when it is

carrying a current of (a) 1.00 A and (b) 10.0 A?

49.

An electric heater is rated at 1 500 W, a toaster at

55. Assume you have a battery of emf ! and three identical

750 W, and an electric grill at 1 000 W. The three

lightbulbs, each having constant resistance R. What is the

appliances are connected to a common 120-V household

total power delivered by the battery if the bulbs are con-

circuit. (a) How much current does each draw? (b) Is a

nected (a) in series? (b) in parallel? (c) For which connec-

circuit with a 25.0-A circuit breaker sufﬁcient in this

tion will the bulbs shine the brightest?

situation? Explain your answer.

56. A group of students on spring break manages to reach a

50. Aluminum wiring has sometimes been used instead of

deserted island in their wrecked sailboat. They splash

copper for economy. According to the National Electrical

ashore with fuel, a European gasoline-powered 240-V gen-

Code, the maximum allowable current for 12-gauge

erator, a box of North American 100-W 120-V lightbulbs, a

copper wire with rubber insulation is 20 A. What should

500-W 120-V hot pot, lamp sockets, and some insulated

be the maximum allowable current in a 12-gauge

wire. While waiting to be rescued, they decide to use the

aluminum wire if the power per unit length delivered to

generator to operate some lightbulbs. (a) Draw a diagram

the resistance in the aluminum wire is the same as that

of a circuit they can use, containing the minimum number

delivered in the copper wire?

of lightbulbs with 120 V across each bulb, and no higher

51. Turn on your desk lamp. Pick up the cord, with your

voltage. Find the current in the generator and its power

thumb and index ﬁnger spanning the width of the cord.

output. (b) One student catches a ﬁsh and wants to cook

(a) Compute an order-of-magnitude estimate for the

it in the hot pot. Draw a diagram of a circuit containing

current in your hand. You may assume that at a typical

the hot pot and the minimum number of lightbulbs with

instant the conductor inside the lamp cord next to your

120 V across each device, and not more. Find the current

2

thumb is at potential ( 10 V and that the conductor next

in the generator and its power output.

to your index ﬁnger is at ground potential (0 V). The

57. A battery has an emf ! and internal resistance r. A variable

resistance of your hand depends strongly on the thickness

load resistor R is connected across the terminals of the bat-

and the moisture content of the outer layers of your skin.

tery. (a) Determine the value of R such that the potential

Assume that the resistance of your hand between ﬁnger-

4 difference across the terminals is a maximum. (b) Deter-

tip and thumb tip is ( 10 ’. You may model the cord as

mine the value of R so that the current in the circuit is a

having rubber insulation. State the other quantities you

maximum. (c) Determine the value of R so that the power

measure or estimate and their values. Explain your rea-

delivered to the load resistor is a maximum. Choosing the

soning. (b) Suppose that your body is isolated from any

load resistance for maximum power transfer is a case of

other charges or currents. In order-of-magnitude terms

what is called impedance matching in general. Impedance

describe the potential of your thumb where it contacts the

matching is important in shifting gears on a bicycle, in

cord, and the potential of your ﬁnger where it touches

connecting a loudspeaker to an audio ampliﬁer, in con-

the cord.

necting a battery charger to a bank of solar photoelectric

cells, and in many other applications.

58. A 10.0-)F capacitor is charged by a 10.0-V battery through

Additional Problems

a resistance R. The capacitor reaches a potential difference

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

of 4.00 V in a time 3.00 s after charging begins. Find R.

sistor radio. If the batteries can move a charge of 240 C,

how long will they last if the radio has a resistance of

59. When two unknown resistors are connected in series with

200’?

a battery, the battery delivers 225 W and carries a total

current of 5.00 A. For the same total current, 50.0 W is

53. A battery has an emf of 9.20 V and an internal resistance

delivered when the resistors are connected in parallel.

of 1.20 ’. (a) What resistance across the battery will

Determine the values of the two resistors.

extract from it a power of 12.8 W? (b) a power of 21.2 W?

60. When two unknown resistors are connected in series with

54. Calculate the potential difference between points a and b

a battery, the battery delivers total power ! and carries a

in Figure P28.54 and identify which point is at the higher

s

total current of I. For the same total current, a total power

potential.Problems 891

! is delivered when the resistors are connected in paral-

p

R

1

lel. Determine the values of the two resistors.

61. A power supply has an open-circuit voltage of 40.0 V and R

ΔV

2

Voltage–

an internal resistance of 2.00 ’. It is used to charge two

controlled

storage batteries connected in series, each having an emf

switch

C V ΔV

c

of 6.00 V and internal resistance of 0.300’. If the

charging current is to be 4.00 A, (a) what additional

resistance should be added in series? (b) At what rate does

(a)

the internal energy increase in the supply, in the batteries, ΔV (t)

c

and in the added series resistance? (c) At what rate does

ΔV

the chemical energy increase in the batteries?

2ΔV

62. Two resistors R and R are in parallel with each other. 3

1 2

Together they carry total current I. (a) Determine the ΔV

3

current in each resistor. (b) Prove that this division of the

T

total current I between the two resistors results in less

t

power delivered to the combination than any other

(b)

division. It is a general principle that current in a direct

Figure P28.66

current circuit distributes itself so that the total power delivered to

the circuit is a minimum.

67. Three 60.0-W, 120-V lightbulbs are connected across a

63. The value of a resistor R is to be determined using the

120-V power source, as shown in Figure P28.67. Find

ammeter–voltmeter setup shown in Figure P28.63. The

(a) the total power delivered to the three bulbs and

ammeter has a resistance of 0.500’, and the voltmeter has

(b) the voltage across each. Assume that the resistance of

a resistance of 20 000’. Within what range of actual

each bulb is constant (even though in reality the resis-

values of R will the measured values be correct to within

tance might increase markedly with current).

5.00% if the measurement is made using the circuit shown

in (a) Figure P28.63a and (b) Figure P28.63b?

R

A

R

1

R R

120 V

2 3

V

(a)

Figure P28.67

68. Switch S has been closed for a long time, and the electric

R

A circuit shown in Figure P28.68 carries a constant current.

Take C # 3.00)F, C # 6.00)F, R # 4.00 k’, and

1 2 1

R # 7.00 k’. The power delivered to R is 2.40 W.

2 2

V

(a) Find the charge on C . (b) Now the switch is opened.

1

After many milliseconds, by how much has the charge on

(b)

C changed?

2

Figure P28.63

64. A battery is used to charge a capacitor through a resistor,

as shown in Figure 28.19. Show that half the energy sup-

C R

1 1

plied by the battery appears as internal energy in the resis-

S

tor and that half is stored in the capacitor.

65. The values of the components in a simple series RC circuit

containing a switch (Fig. 28.19) are C# 1.00)F, R# 2.00.

R C

2 2

6

10 ’, and !# 10.0 V. At the instant 10.0 s after the switch

is closed, calculate (a) the charge on the capacitor, (b) the

current in the resistor, (c) the rate at which energy is being

Figure P28.68

stored in the capacitor, and (d) the rate at which energy is

being delivered by the battery.

69. Four resistors are connected in parallel across a 9.20-V

66. The switch in Figure P28.66a closes when "V + 2"V/3

battery. They carry currents of 150 mA, 45.0 mA,

c

and opens when "V *"V/3. The voltmeter reads a

14.00 mA, and 4.00 mA. (a) If the resistor with the largest

c

voltage as plotted in Figure P28.66b. What is the period T

resistance is replaced with one having twice the resis-

of the waveform in terms of R , R , and C ?

tance, what is the ratio of the new current in the battery

1 2892 CHAPTER 28 • Direct Current Circuits

to the original current? (b) What If? If instead the resis- 72. A regular tetrahedron is a pyramid with a triangular base.

tor with the smallest resistance is replaced with one Six 10.0-’ resistors are placed along its six edges, with

having twice the resistance, what is the ratio of the new junctions at its four vertices. A 12.0-V battery is connected

total current to the original current? (c) On a February to any two of the vertices. Find (a) the equivalent

night, energy leaves a house by several heat leaks, includ- resistance of the tetrahedron between these vertices and

ing the following: 1 500 W by conduction through the (b) the current in the battery.

ceiling; 450 W by inﬁltration (air ﬂow) around the

73. The circuit shown in Figure P28.73 is set up in the

windows; 140 W by conduction through the basement

laboratory to measure an unknown capacitance C with the

wall above the foundation sill; and 40.0 W by conduction

use of a voltmeter of resistance R# 10.0 M’ and a battery

through the plywood door to the attic. To produce the

whose emf is 6.19 V. The data given in the table are the

biggest saving in heating bills, which one of these energy

measured voltages across the capacitor as a function of

transfers should be reduced ﬁrst?

time, where t# 0 represents the instant at which the switch

70. Figure P28.70 shows a circuit model for the transmission is opened. (a) Construct a graph of ln(!/"V ) versus t, and

of an electrical signal, such as cable TV, to a large perform a linear least-squares ﬁt to the data. (b) From the

number of subscribers. Each subscriber connects a load slope of your graph, obtain a value for the time constant of

resistance R between the transmission line and the the circuit and a value for the capacitance.

L

ground. The ground is assumed to be at zero potential

and able to carry any current between any ground ε

S

connections with negligible resistance. The resistance of

the transmission line itself between the connection

points of different subscribers is modeled as the constant

C

resistance R . Show that the equivalent resistance across

T

the signal source is

R

1 2 1/2

R # [(4R R % R ) % R ]

eq T L T T

2

Suggestion: Because the number of subscribers is large,

Voltmeter

the equivalent resistance would not change noticeably if

Figure P28.73

the ﬁrst subscriber cancelled his service. Consequently, the

equivalent resistance of the section of the circuit to the

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

right of the ﬁrst load resistor is nearly equal to R .

eq

6.19 0

R R R

T T T

5.55 4.87

4.93 11.1

4.34 19.4

3.72 30.8

Signal

R R R

L L L

source

3.09 46.6

2.47 67.3

1.83 102.2

Figure P28.70

74. The student engineer of a campus radio station wishes to

verify the effectiveness of the lightning rod on the antenna

71. In Figure P28.71, suppose the switch has been closed for a

mast (Fig. P28.74). The unknown resistance R is between

x

time sufﬁciently long for the capacitor to become fully

points C and E. Point E is a true ground but is inaccessible

charged. Find (a) the steady-state current in each resistor

for direct measurement since this stratum is several meters

and (b) the charge Q on the capacitor. (c) The switch is

below the Earth’s surface. Two identical rods are driven

now opened at t# 0. Write an equation for the current I

R

2

into the ground at A and B, introducing an unknown resis-

through R as a function of time and (d) ﬁnd the time

2

tance R . The procedure is as follows. Measure resistance

y

interval required for the charge on the capacitor to fall to

R between points A and B, then connect A and B with a

1

one-ﬁfth its initial value.

heavy conducting wire and measure resistance R between

2

S 12.0 kΩ

10.0 µ µF

AB C

R =15.0 kΩ

2

9.00 V

3.00 kΩ

R R R

y x y

E

Figure P28.71 Figure P28.74Answers to Quick Quizzes 893

points A and C. (a) Derive an equation for R in terms of and that the voltage across the capacitor as a function of

x

the observable resistances, R and R . (b) A satisfactory time is

1 2

ground resistance would be R * 2.00’. Is the grounding

x

r

$t/R C

eq

V # !(1$ e )

of the station adequate if measurements give R # 13.0 ’ C

1

r% R

and R # 6.00’?

2

(c) What If? If the capacitor is fully charged, and the

75. The circuit in Figure P28.75 contains two resistors,

switch is then opened, how does the voltage across the

R # 2.00 k’ and R # 3.00 k’, and two capacitors,

1 2

capacitor behave in this case?

C # 2.00)F and C # 3.00)F, connected to a battery with

1 2

emf !# 120 V. No charge is on either capacitor before

switch S is closed. Determine the charges q and q on

1 2

Answers to Quick Quizzes

capacitors C and C , respectively, after the switch is closed.

1 2

28.1 (a). Power is delivered to the internal resistance of a bat-

(Suggestion: First reconstruct the circuit so that it becomes a

tery, so decreasing the internal resistance will decrease

simple RC circuit containing a single resistor and single

this “lost” power and increase the percentage of the

capacitor in series, connected to the battery, and then deter-

power delivered to the device.

mine the total charge q stored in the equivalent circuit.)

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 pass

C

R

1

1

through a resistor.

bc de

28.3 (a). Connecting b to c “shorts out” bulb R and changes

2

the total resistance of the circuit from R % R to just R .

1 2 1

R

2 C Because the resistance of the circuit has decreased (and

2

the emf supplied by the battery does not change), the

ε

current in the circuit increases.

S

+ –

28.4 (b). When the switch is opened, resistors R and R are in

a f 1 2

series, so that the total circuit resistance is larger than when

Figure P28.75

the switch was closed. As a result, the current decreases.

28.5 (b), (d). Adding another series resistor increases the total

6

76. This problem illustrates how a digital voltmeter affects the

resistance of the circuit and thus reduces the current in

voltage across a capacitor in an RC circuit. A digital

the circuit. The potential difference across the battery ter-

voltmeter of internal resistance r is used to measure the

minals increases because the reduced current results in a

voltage across a capacitor after the switch in Figure P28.76

smaller voltage decrease across the internal resistance.

is closed. Because the meter has ﬁnite resistance, part of

28.6 (a), (e). If the second resistor were connected in paral-

the current supplied by the battery passes through the

lel, the total resistance of the circuit would decrease,

meter. (a) Apply Kirchhoff’s rules to this circuit, and use

and the current in the battery would increase. The po-

the fact that i # dq/dt to show that this leads to the

C

tential difference across the terminals would decrease

differential equation

because the increased current results in a greater volt-

dq q r age drop across the internal resistance.

R % # !

eq

dt C r% R

28.7 (a). When the switch is closed, resistors R and R are

1 2

in parallel, so that the total circuit resistance is smaller

where R # rR/(r% R). (b) Show that the solution to this

eq

than when the switch was open. As a result, the current

differential equation is

increases.

r

$t/R C

eq

q# C! (1$ e ) 28.8 (c). A current is assigned to a given branch of a circuit.

r% R

There may be multiple resistors and batteries in a given

branch.

Voltmeter

28.9 (b), (d). Just after the switch is closed, there is no

charge on the capacitor, so there is no voltage across it.

r

Charges begin to ﬂow in the circuit to charge up the ca-

R

pacitor, so that all of the voltage "V# IR appears across

C the resistor. After a long time, the capacitor is fully

charged 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 charge

on the capacitor. Current exists in both branches of the

circuit as the capacitor begins to charge, so the right

S ε

half of the circuit is equivalent to two resistances R in

1

Figure P28.76

parallel for an equivalent resistance of R. After a long

2

time, the capacitor is fully charged and the current in

6

the right-hand branch drops to zero. Now, current

After Joseph Priest, "Meter Resistance: Don't Forget It!" The Physics

Teacher, January 2003, p. 40. exists only in a resistance R across the battery.

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