T his chapter is concerned with the analysis of simple electric circuits that contain

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)

T his 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.

CHAPTER 28• Direct Current Circuits

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

I I tery with zero internal resistance, the potential difference across the battery (called its terminal voltage ) equals its emf. However, for a real battery, the terminal voltage is not

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

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

However, as we move through the resistance r, the potential decreases by an amount Ir , where I is the current in the circuit. Thus, the terminal voltage of the battery

"V # V b a is Ir 1 $V IR

"V # ! $ Ir

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—

a b c d for example, the emf of a D cell is 1.5 V. The actual potential difference between the (b)

terminals of the battery depends on the current in the battery, as described by Equa- tion 28.1.

Active Figure 28.2 (a) Circuit diagram of a source of emf ! (in

Figure 28.2b is a graphical representation of the changes in electric potential as the this case, a battery), of internal

circuit is traversed in the clockwise direction. By inspecting Figure 28.2a, we see that resistance r, connected to an

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

changes as the circuit in part (a) is (such as a toaster, an electric heater, or a lightbulb) connected to the battery (or, in traversed clockwise.

the case of household devices, to the wall outlet). The resistor represents a load on the battery because the battery must supply energy to operate the device. The potential dif-

At the Active Figures link at http://www.pse6.com, you

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

can adjust the emf and

28.1, we see that

resistances r and R to see the

! # IR % Ir

effect on the current and on the graph in part (b).

Solving for the current gives

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

I PITFALL PREVENTION #I ! 2 R%I 2 r

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

28.1 What Is Constant in a

output I ! of the battery is delivered to the external load resistance in the amount I 2 R

Battery?

and to the internal resistance in the amount I 2 r .

It is a common misconception that a battery is a source of constant current. Equation 28.3 clearly shows that this is not true.

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

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

Example 28.1 Terminal Voltage of a Battery Interactive

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

Solution Equation 28.3 gives us the current: rises to 2.00 ! toward the end of its useful life. How does this alter the ability of the battery to deliver energy?

12.0 V

I# #

Answer Let us connect the same 3.00-' load resistor to the R%r

3.93 A

battery. The current in the battery now is and from Equation 28.1, we find the terminal voltage:

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

R%r

(3.00 ' % 2.00 ')

11.8 V

and the terminal voltage is

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

"V # ! $ I r # 12.0 V $ (2.40 A) (2.00 ') # 7.2 V 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 power delivered to the internal resistance of the battery, and

17.3 W the power delivered by the battery.

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

! r #I 2 r # (2.40 A) 2 (2.00 ') # 11.5 W Solution The power delivered to the load resistor is Notice that 40% of the power from the battery is delivered

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

46.3 W

to the internal resistance. In part (B), this percentage is 1.6%. Consequently, even though the emf remains fixed,

The power delivered to the internal resistance is the increasing internal resistance significantly reduces the ability of the battery to deliver energy.

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

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

equal to I 2 R, where I is given by Equation 28.3:

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

! reaches a maximum value of ! 2 /4r at R # r. When R is

large, there is very little current, so that the power I 2 R

delivered to the load resistor is small. When R is small, the current is large and there is significant loss of power

I 2 r as energy is delivered to the internal resistance. When R

3r R # r, these effects balance to give a maximum transfer of

2r

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 differentiating ! with respect to R, setting the result equal

load resistance equals the internal resistance of the battery.

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 1 , charge Q must also enter the second resistor R 2 . (This is 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 1 must also pass through

R 2 in the same time interval. The potential difference applied across the series combination of resistors will divide

between the resistors. In Figure 28.4b, because the voltage drop 2 from a to b equals IR 1 and the voltage drop from b to c equals IR 2 , the voltage drop from a to c is

(c) Active Figure 28.4 (a) A series connection of two lightbulbs with resistances R 1 and

At the Active Figures link

R 2 . (b) Circuit diagram for the two-resistor circuit. The current in R 1 is the same as that

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

in R 2 . (c) The resistors replaced with a single resistor having an equivalent resistance

can adjust the battery voltage

R eq #R 1 %R 2 .

and resistances R 1 and R 2 to

see the effect on the currents and voltages in the individual

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

resistors.

used often by individuals working with electric circuits.

SECTION 28.2• Resistors in Series and Parallel

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

"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 eq # I(R 1 %R 2 ) 9: R eq #R 1 %R 2 (28.5) The resistance R eq is equivalent to the series combination R 1 %R 2 in the sense that

the circuit current is unchanged when R eq replaces R 1 %R 2 .

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

R eq #R 1 %R 2 %R 3 %(((

The equivalent resistance of 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

Quick Quiz 28.2 the filament fails, rather than by

In Figure 28.4, imagine positive charges pass first through

saying that the lightbulb “burns

out.” The word burn suggests a (a) smaller, (b) larger, or (c) the same.

R 1 and then through R 2 . Compared to the current in R 1 , the current in R 2 is

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 1 (a) increase, (b) decrease, or (c) remain the same?

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

there is no current in R 2 , because the current has an alternate zero-resistance path through the switch. There is current in R 1 and this current is measured with the amme-

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 2 . What happens to the reading on the

ammeter when the switch is opened? (a) the reading goes up; (b) the reading goes down; (c) the reading does not change.

Switch closed

Switch open

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

CHAPTER 28• Direct Current Circuits

At the Active Figures link at http://www.pse6.com, you

can adjust the battery voltage

and resistances R 1 and R 2 to

see the effect on the currents and voltages in the individual resistors.

(c) Active Figure 28.6 (a) A parallel connection of two lightbulbs with resistances R 1 and

R 2 . (b) Circuit diagram for the two-resistor circuit. The potential difference across R 1 is the same as that across R 2 . (c) The resistors replaced with a single resistor having an

equivalent resistance given by Equation 28.7.

▲ PITFALL PREVENTION

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

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

28.3 Local and Global

going through R 1 and the rest going through R 2 . A junction is any point in a circuit

Changes

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

A local change in one part of a through several doors, as described earlier.) This split results in less current in each indi- circuit may result in a global

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

the current I that enters point a must equal the total current leaving that point: For example, if a single resistance

I#I 1 %I 2

is changed in a circuit containing several resistors and batteries, the

where I 1 is the current in R 1 and I 2 is the current in R 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 Take the Path of

"V

"V

R 1 R 2 ! R 1 R 2 " R eq

Least Resistance

You may have heard a phrase like where R is an equivalent single resistance which will have the same effect on the “current takes the path of least

eq

resistance” in reference to a par- circuit as the two resistors in parallel; that is, it will draw the same current from the

allel combination of current battery (Fig. 28.6c). From this result, we see that the equivalent resistance of two resis- paths, such that there are two or

tors in parallel is given by

more paths for the current to

take. The phrase is incorrect.

R eq

The current takes all paths.

or

Those paths with lower resistance will have large currents, but even

R eq

very high-resistance paths will

1 %R 2

carry some of the current.

SECTION 28.2• Resistors in Series and Parallel

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

The equivalent resistance of

R eq

R 1 R 2 R 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 2 . There is current in R 1 and this current is measured with the

ammeter at the right side of the circuit. If the switch is closed (Fig. 28.7, right), there is current in R 2 . What happens to the reading on the ammeter when the switch is closed? (a) the reading goes up; (b) the reading goes down; (c) the reading does not change.

Switch open

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.

CHAPTER 28• Direct Current Circuits

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

further voltage drop across resistors R D and R E . Conse- represent the resistance of the wires in the cable between

Figure 28.8. The horizontal resistors (such as R A and R B )

quently, the voltage across light fixture R F is smaller than

that across R C . This continues on down the line of light represent the resistance of the light fixtures themselves.

the light fixtures while the vertical resistors (such as R C )

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 A and R B . Thus, the voltage across light

than the one before.

Resistance of

Resistance in

light fixtures

wires of cable

Power supply

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.

2.0 A. We could have guessed this at the start by noting (A) Find the equivalent resistance between points a and c.

that the current in the 3.0-' resistor has to be twice that in the 6.0-' resistor, in view of their relative resistances and the

Solution The combination of resistors can be reduced in fact that the same voltage is applied to each of them. steps, as shown in Figure 28.9. The 8.0-' and 4.0-' resistors

As a final check of our results, note that "V bc # are in series; thus, the equivalent resistance between a and b

6.0 V and "V ab # (12.0 ')I # 36 V; is 12.0 ' (see Eq. 28.5). The 6.0-' and 3.0-' resistors are in

(6.0 ')I 1 # (3.0 ')I 2 #

42 V, as it must. parallel, so from Equation 28.7 we find that the equivalent

therefore, "V ac # "V ab % "V bc #

resistance from b to c is 2.0 '. Hence, the equivalent resis- tance from a to c is 14.0 '.

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

ence of 42 V is maintained between a and c?

Solution The currents in the 8.0-' and 4.0-' resistors are I c

the same because they are in series. In addition, this is the same as the current that would exist in the 14.0-' equivalent

(a)

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

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,

with part passing through the 6.0-' resistor (I 1 ) and part

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

through the 3.0-' resistor (I 2 ). Because the potential differ-

(c)

Figure 28.9 (Example 28.4) The original network of resistors the fact that I 1 %I 2 #

that (6.0 ')I 1 # (3.0 ')I 2 , or I 2 # 2I 1 . Using this result and

3.0 A, we find that I 1 #

1.0 A and

is reduced to a single equivalent resistance.

Example 28.5 Finding R eq by Symmetry Arguments

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

the remaining circuit then reduced as in Figures 28.10c and points c and d. As a result, points c and d may be connected

This means that "V cd #

0 and there is no current between

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.

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

(b)

(c)

does not contribute to the resistance between points a and b and therefore can be disregarded when we calculate the equivalent resistance.

Example 28.6 Three Resistors in Parallel Interactive

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. Solution We apply the relationship ! # I 2 R to each resis- (A) Find the current in each resistor.

tor and obtain

Solution The resistors are in parallel, and so the potential

1 #I 1 2 R 1 # (6.00 A) 2 (3.00 ') # 108 W ship "V # IR to each resistor gives

3.00-': !

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

6.00-': ! 2 #I 2 2 R 2 # (3.00 A) 2 (6.00 ') # 54.0 W

R 1 3.00 '

"V 2 18.0 V 9.00-': ! 3 #I 3 R 3 # (2.00 A) 2 (9.00 ') # 36.0 W

3.00 A

R 2 6.00 '

This shows that the smallest resistor receives the most "V

18.0 V

power. Summing the three quantities gives a total power of

3 R 2.00 A

Figure 28.11 (Example 28.6) (a) Three

resistors connected in parallel. The voltage across each resistor is 18.0 V. b (b) Another circuit with three resistors and a battery. Is this equivalent to the

(b)

circuit in part (a) of the figure?

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 eq :

calculation?

1 % Answer There is no effect on the calculation. The physical 1 % 1 R eq

placement of the battery is not important. In Figure 28.11b, the battery still applies a potential difference of 18.0 V between points a and b, so the two circuits in Figure 28.11

are electrically identical.

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 1 is closed and switch S 2 is opened, current exists only in the 75-W filament. When switch S 1 is open and switch S 2 is

closed, current exists only in the 100-W filament. When both switches are closed, current exists in both filaments, and the

total power is 175 W. If the filaments were connected in series and one of

120 V

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

more dangerous (more likely to start a fire, for instance), parallel and series connections have been used for strings

such as decorating Christmas trees. 3 Over the years, both

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

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 (2.40 V) 2 /(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.

Filament I

Jumper Glass insulator

George Semple

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 in # # I out

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

closed #

"V # 0

loop

CHAPTER 28• Direct Current Circuits

I 2 Kirchhoff’s first rule is a statement of conservation of electric charge. All charges

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 1 #I 2 %I 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

charge requires that all charges decreases in energy as it passes through other elements. The potential energy

entering a junction must leave that decreases whenever the charge moves through a potential drop $ IR across a resis-

tor or whenever it moves in the reverse direction through a source of emf. The (b) A mechanical analog of the

junction. Therefore, I 1 #I 2 %I 3 .

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 right must equal the amount

When applying Kirchhoff’s second rule in practice, we imagine traveling around the flowing into the single branch on

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

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

V = – IR

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

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

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

(b) a V = +IR

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

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 $ !

(c) – + (Fig. 28.15d). In this case the emf of the battery reduces the electric potential as a ∆

V= + ε ε

we move through it. 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

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

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

junction rule is one fewer than the number of junction points in the circuit. You can Figure 28.15 Rules for

∆ V= – ε ε

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

have no internal resistance.) Each need to obtain from the two rules equals the number of unknown currents. circuit element is traversed from

Complex networks containing many loops and junctions generate great numbers 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

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

Kirchhoff’s Rules

• Draw a circuit diagram, and label all the known and unknown quantities. You

must assign a direction to the current in each branch of the circuit. 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

Quick Quiz 28.8 study in Chapter 42. They

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

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

and rubidium and invented branch in the circuit (d) each battery in the circuit.

astronomical spectroscopy. (AIP ESVA/W.F. Meggers Collection)

Example 28.8

A Single-Loop Circuit

d : a represents a potential difference of $ IR 2 . Applying as shown in Figure 28.16. (Neglect the internal resistances

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

Kirchhoff’s loop rule gives

of the batteries.)

# "V # 0

(A) Find the current in the circuit.

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

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

! 2 6.0 V $ 12 V

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

0.33 A 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 % ! 1 , b : c represents a potential difference of emfs in the numerator subtract because the batteries have $ IR 1 , c : d represents a potential difference of $ ! 2 , and opposite polarities in Figure 28.16. In the denominator, the

resistances add because the two resistors are in series. (B) What power is delivered to each resistor? What power is

ε 1 = 6.0 V

delivered by the 12-V battery?

Solution Using Equation 27.23,

R 2 = 10 Ω

R 1 = 8.0 Ω

! 1 #I 2 R 1 # (0.33 A) 2 (8.0 ') #

0.87 W ! 2 #I 2 R 2 # (0.33 A) 2 (10 ') #

1.1 W

Hence, the total power delivered to the resistors is

ε 2 = 12 V

2.0 W.

The 12-V battery delivers power I ! 2 # 4.0 W. Half of this

Figure 28.16 (Example 28.8) A series circuit containing two

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 being

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 ! 1 and ! 2 are the

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, we would have found that less power was being delivered to

6.0 V % 12 V #

1.0 A the 6-V battery.

I#

R 1 %R 2 8.0 ' % 10 ' The new powers delivered to the resistors are

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

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

Answer While we could repeat the Kirchhoff’s rules (10 ') # 10 W 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.

Example 28.9 Applying Kirchhoff’s Rules Interactive

Find the currents I 1 ,I 2 , and I 3 in the circuit shown in Figure

Note that in loop befcb we obtain a positive value when

28.17. traversing the 6.0-' resistor because our direction of travel is opposite the assumed direction of I 1 . Expressions (1), (2),

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

10.0 V $ (6.0 ')I 1 $ (2.0 ') (I 1 %I 2 )#0 which we must use Kirchhoff’s rules. To analyze the circuit,

10.0 V # (8.0 ')I 1 % (2.0 ')I 2 we arbitrarily choose the directions of the currents as la-

beled in Figure 28.17. Applying Kirchhoff’s junction rule Dividing each term in Equation (3) by 2 and rearranging to junction c gives

gives

12.0 V # $ (3.0 ')I % (2.0 ')I (1)

Subtracting Equation (5) from Equation (4) eliminates I 2 ,

We now have one equation with three unknowns—I 1 ,I 2 , and

giving

I 3 . There are three loops in the circuit—abcda, befcb, and aefda. We therefore need only two loop equations to deter-

22.0 V # (11.0 ')I 1 mine the unknown currents. (The third loop equation

would give no new information.) Applying Kirchhoff’s loop

2.0 A rule to loops abcda and befcb and traversing these loops clockwise, we obtain the expressions

Using this value of I 1 in Equation (5) gives a value for I 2 : (2)

abcda

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

10.0 V $ (6.0 ')I 1 $ (2.0 ')I 3 # 0 (2.0 ')I 2 # (3.0 ')I 1 $

(3) befcb $14.0 V % (6.0 ')I 1 $

10.0 V $ (4.0 ') I 2 # 0

3.0 A

Finally, 14.0 V

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

tive. This indicates only that the currents are opposite the b – +

c direction we chose for them. However, the numerical values

10.0 V 6.0 Ω

are correct. What would have happened had we left the

I 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

Example 28.10

A Multiloop Circuit

(A) Under steady-state conditions, find the unknown currents From Equation (1) we see that I 1 #I 3 $I 2 , which, when

I 1 ,I 2 , and I 3 in the multiloop circuit shown in Figure 28.18.

substituted into Equation (3), gives

(8.00 ')I 2 $ (5.00 ')I 3 Solution First note that because the capacitor represents %

8.00 V # 0 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 1 reach point g, they all go

4.00 V

toward point b through the 8.00-V battery; hence, I gb #I 1 .

Labeling the currents as shown in Figure 28.18 and applying Equation 28.9 to junction c, we obtain

Because our value for I 2 is negative, we conclude that the di- rection of I 2 is from c to f in the 3.00-' resistor. Despite this

I 1 %I 2 #I 3 interpretation of the direction, however, we must continue Equation 28.10 applied to loops defcd and cfgbc, traversed

to use this negative value for I 2 in subsequent calculations clockwise, gives