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, resistancefree
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 opencircuit 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 realworld 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 tworesistor 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 opencircuit 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 zeroresistance 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 tworesistor 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 highresistance 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 12volt landscape lighting on the cable at 10foot 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
18gauge 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 sidebyside 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 200foot 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 twowire 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 42V 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 3.0A 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 ThreeWay Lightbulb
Figure 28.12 illustrates how a threeway lightbulb is con in parallel, if one of them (for example, the 75W 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 threeway switch for selecting tions as current exists in the other (100W) ﬁlament.
different light intensities. The bulb contains two ﬁlaments.
When the lamp is connected to a 120V source, one ﬁlament
receives 100 W of power, and the other receives 75 W.
Explain how the two ﬁlaments are used to provide three
100W filament
different light intensities.
Solution The three light intensities are made possible by
75W 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 75W ﬁlament. When switch S is open and switch S is
1 2
closed, current exists only in the 100W 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 threeway
the switch position. However, with the ﬁlaments connected lightbulb.
Application Strings of Lights
Strings of lights are used for many ornamental purposes, a serieswired 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 parallelwired string fails or is removed,
of lights powered by 120 V. Serieswired bulbs are safer than the rest of the bulbs continue to glow. (A 25bulb string of
parallelwired bulbs for indoor Christmastree use because 4W bulbs results in a power of 100 W; the total power
serieswired 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 socalled “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 serieswired 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, timeconsuming chore that involved fails. When the ﬁlament breaks in one of these miniature
trialanderror 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 parallelwired 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 threeway 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 50bulb miniaturelight across each of the remaining bulbs increases to 120 V/49#
string are operating. A 2.40V 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 serieswired 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 Christmastree 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 highpotential 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 junctionrule 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 steadystate 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
steadystate conditions.SECTION 28.3 • Kirchhoff’s Rules 871
PROBLEMSOLVING 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 SingleLoop Circuit
A singleloop 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 singleloop
! $! 6.0 V$ 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 12V 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 12V 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 6V battery, which is being872 CHAPTER 28 • Direct Current Circuits
charged by the 12V 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 6V battery.
The new powers delivered to the resistors are
What If? What if the polarity of the 12.0V 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 10.0V 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 steadystate 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 steadystate conditions. Therefore, when and ﬁnd that
the charges associated with I reach point g, they all go
1
4.00 V
I #$ # $0.364 A
toward point b through the 8.00V battery; hence, I # I .
2
gb 1
11.0 ’
Labeling the currents as shown in Figure 28.18 and applying
Because our value for I is negative, we conclude that the di
2
Equation 28.9 to junction c, we obtain
rection of I is from c to f in the 3.00’ resistor. Despite this
2
(1) I % I # I
1 2 3
interpretation of the direction, however, we must continue
to use this negative value for I in subsequent calculations
2
Equation 28.10 applied to loops defcd and cfgbc, traversed
because our equations were established with our original
clockwise, gives
choice of direction.
(2) defcd 4.00 V$ (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 steadystate situation, in which no current was
present in any branch of the circuit containing a capacitor. Now we are considering the case before the
steadystate 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 steadystate 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 righthand 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 onefourth 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 onefourth 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 onehalf
its original value, rather than by the time constant "? This
1
$t/RC
# e
4 would give a parameter for the circuit called its halflife t .
1/2
How is the halflife related to the time constant?
Taking logarithms of both sides, we ﬁnd
Answer After one halflife, 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 onefourth its initial value?
t # 0.693
1/2
2
The concept of halflife 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 25.0kV 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 offtheshelf 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 3V
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 fullscale 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 fullscale 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 constantemf 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 240V supply. (b) The connections for
each of the openings in a 240V 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 20A circuit and the microwave oven on a separate 20A circuit.
Many heavyduty 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 240V line requires half as much current compared to
operating it at 120 V; therefore, smaller wires can be used in the highervoltage circuit
without overheating.
Electrical Safety
When the live wire of an electrical outlet is connected directly to ground, the circuit is
completed and a shortcircuit 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
3wire 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 120V outlets are designed to accept a threepronged 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 lowresistance
path to ground.
Special power outlets called groundfault 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 opencircuit 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 60W lamp the brightest and
8. When resistors are connected in parallel, which of the fol
the 200W 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 120V 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 highvoltage 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 highvoltage 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 120V 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 highvoltage 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 highvoltage
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 1.50V 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 120V 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 crosssectional 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 2.00cm 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 6.00V battery supplies current to the circuit shown in
shoes through the body of the wearer to a metal ground Figure P28.14. When the doublethrow 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 1.00M’ resistor is measured with battery is 1.20 mA. When the switch is closed in position 2,
a highresistance 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 ANSIspeciﬁed 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
wellinsulated 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 2.00nF capacitor with an initial charge of 5.10)C is
discharged through a 1.30k’ 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 fullscale 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 fullscale 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 fullscale 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.
fullscale deﬂection.
A pair of rubbersoled 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 ﬁvedigit precision.
staticdissipative 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 rubbersoled shoes to reduce a 3 000V 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 4.00M’ resistor and a 3.00)F capacitor are connected
Now one terminal of one wire is moved, as shown in
in series with a 12.0V 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 fullscale 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 fullscale 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 fullscale
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 fullscale 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 2.00V fullscale volt
2.00 Ω
meter when a 2 500’ resistor is connected in series with
a
it. It serves as a 0.500A fullscale 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 fullscale deﬂection.
4.00 Ω
12.0 V
Section 28.6 Household Wiring and Electrical
Safety
b
48. An 8.00ft extension cord has two 18gauge 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 120V 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 25.0A 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 gasolinepowered 240V gen
Code, the maximum allowable current for 12gauge
erator, a box of North American 100W 120V lightbulbs, a
copper wire with rubber insulation is 20 A. What should
500W 120V hot pot, lamp sockets, and some insulated
be the maximum allowable current in a 12gauge
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 orderofmagnitude 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 orderofmagnitude 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 10.0V battery through
Additional Problems
a resistance R. The capacitor reaches a potential difference
52. Four 1.50V 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 opencircuit 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 60.0W, 120V lightbulbs are connected across a
63. The value of a resistor R is to be determined using the
120V 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 9.20V
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 12.0V 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 leastsquares 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 steadystate 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
oneﬁfth 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 righthand 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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