Chapter 24 Electric Currents and DC Circuits
“What we call physics comprises that group of natural
sciences which base their concepts on measurements;
and whose concepts and propositions lend themselves
to mathematical formulation. Its realm is accordingly
defined as that part of the sum total of our knowledge
which is capable of being expressed in mathematical
terms.”
Albert Einstein
24.1 Electric Current
Before 1800 the study of electricity was limited to electrostatics, the study of
charges at rest. It was impossible to obtain large amounts of electric charge for any
continuous period of time. In 1800, Alessandro Volta (1745-1827), an Italian
physicist, invented the “Voltaic pile,” later to be called an electric battery. The
battery converted chemical energy into electrical energy thereby supplying a
potential difference and relatively large quantities of charge that could flow from
the battery for relatively long periods of time. With the invention of the battery,
applications of electricity grew by leaps and bounds.
Let us first consider the motion of electric charges when a battery is used to
supply a potential difference between two parallel plates. The space between the
plates is either a vacuum or is filled with air. The negative plate has a small hole in
it to allow the introduction of an electron into the uniform electric field, as shown in
figure 24.1. The electron immediately experiences a force in the electric field, given
by
F = qE = −eE
Figure 24.1 Motion of an electron in air or a vacuum.
where e is the charge on the electron. The motion of the electron can be determined
by the techniques discussed in section 21.10. The electron will experience the
acceleration:
a = F = − eE
m
m
Chapter 24 Electric Currents and DC Circuits
The position and velocity of the electron are given by the kinematic equations
and
r = v0t + 1 at2
2
v = v0 + at
If a battery is now connected across the ends of a length of wire, the flow of
electrons is much more complicated because of the molecular structure of the wire
itself. The atoms of the metal wire are arranged in a lattice structure, as shown in
figure 24.2(a). Metals can be thought of as an array of positive ions surrounded by a
Figure 24.2 Motion of an electron in a wire.
more or less uniform sea of negatively charged electrons that help hold the ions in
place. The outer electron of the atom in the metal wire is loosely bound, and in the
lattice structure moves about freely. The free electron of each atom moves about in
a random manner within the lattice structure due to the constant interaction
between the moving electron and each fixed positive ionized atom it sees as it moves
about. These electrons resemble the random motion of gas molecules and are
sometimes referred to as the electron gas. They have no net directed motion along
the wire. When a potential difference is applied across the ends of the wire, as in
figure 24.2(b), an electric field is set up within the wire. The electron in this field
starts to move, but the motion is not the simple motion of the electron in figure 24.1
because of the constant interactions with the positive ions of the lattice. The
electron slowly drifts with an average velocity vd, the drift velocity, in the expected
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Chapter 24 Electric Currents and DC Circuits
direction as seen in figure 24.2(c). Because of the constant interaction with the
atoms of the lattice structure, it takes almost 30 s in some metals for the electron to
drift about 1 cm. This is extremely slow as compared to the motion that would be
experienced by the electron in figure 24.1.
Because of this complexity of the motion of the electron within the solid wire,
we will make no further analysis of the motion of the electron by Newton’s laws and
the kinematic equations; we will adopt a much simpler procedure. When a potential
difference is applied across the ends of a wire and electrons start to drift within the
wire, we will say that an electric current exists in the wire. The electric current is
defined as the amount of electric charge that flows through a cross section of the wire
per unit time. We write this mathematically as
I=
q
t
(24.1)
The SI unit of current is the ampere, named after the French physicist, André
Marie Ampère (1775-1836). An ampere of current is a flow of one coulomb of charge
per second. That is,
1 ampere = 1 coulomb
second
We abbreviate this as
1 A = 1 C/s
Because one coulomb of charge represents 6.242 × 1018 electronic charges, the
ampere is the flow of 6.242 × 1018 electrons per second. Sometimes we use a smaller
unit of current, the milliampere, abbreviated mA.
1 mA = 10−3 A
Because electric current was defined before any knowledge of the existence of
electrons, the original current was assumed to be a flow of positive charges. (It is
interesting to note that it was Benjamin Franklin [1706-1790] an American
statesman and scientist who first called the charges positive and negative. He
assumed it was the positive charges that moved.) Although it is now known that
this is incorrect for the conduction through a solid, it is usual to continue with this
historical convention and define the current to be a flow of positive charges from a
position of high potential to one of low potential. This current is called
conventional current to distinguish it from the actual electron flow. The flow of
electrons is called the electron current. In all the cases considered, a flow of
positive charges in one direction (conventional current) is completely equivalent to a
flow of negative charges in the opposite direction (electron current). In gases and
electrolytic liquids, both positive and negative charges flow in opposite directions,
but both contribute to the positive current.
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Chapter 24 Electric Currents and DC Circuits
The flow of charges in a wire is represented in what is called a circuit
diagram, and the simplest one is shown in figure 24.3. This circuit consists of the
Figure 24.3 A circuit diagram.
battery and a very long wire, which is connected to both terminals of the battery.
The battery, the supplier of a potential difference, is shown as the two horizontal
lines. The longer line represents the positive terminal of the battery, the point of
highest potential, and is labeled with a positive sign (+). The shorter line represents
the negative terminal of the battery and is labeled with a negative sign in the
diagram (−). This negative terminal is arbitrarily chosen to be the point of zero
potential. The line from the positive terminal to the negative terminal is the
external circuit and here represents the length of wire connected to the two
terminals of the battery. The potential difference maintained by the battery is
labeled V in the figure.
The flow of charge in the circuit is analogous to dropping a ball from a height
h and is shown in figure 24.4. At the height h, the ball has a positive potential
Figure 24.4 Analogy of mechanical and electrical motion.
energy with respect to the ground, which is at zero potential energy. The ball falls
from a position of high potential energy to the ground, as shown in figure 24.4(a).
The electrical circuit of figure 24.3 is pulled apart in figure 24.4(b) to show the
analogy more clearly. In the electrical circuit, a positive charge leaves the positive
terminal of the battery where it has the potential V, and ‘’falls’’ through the
connecting wire to the ‘’ground’’ or zero of potential. This analogy between the
mechanical case and the electrical case is even clearer if you recall that we defined
the potential as the potential energy per unit charge. So when a positive charge
flows or ‘’falls’’ from high potential to low potential it is falling from a position of
high potential energy to a position of low potential energy. Whenever a potential
difference exists between any two points in a circuit, positive charge flows from the
24-4
Chapter 24 Electric Currents and DC Circuits
point of high potential to the point of low potential. The resulting current is called a
direct current (DC) since the charges flow in only one direction.
24.2 Ohm’s Law
If a wire is connected to both terminals of a battery, charges will flow. To determine
the current produced by these flowing charges, a device called an ammeter is placed
in the circuit and is shown as A in figure 24.5(a). The ammeter is a device that
displays the amount of current in a circuit. Another device, called a voltmeter, is
Figure 24.5 Experimental determination of Ohm’s law.
connected across the battery and is shown as V in figure 24.5(a). The voltmeter
reads the potential difference between any two points in an electrical circuit. In
figure 24.5(a) the voltmeter reads the potential difference across the terminals of
the battery. A picture of a typical laboratory meter is shown in figure 24.5(c).
For a particular battery maintaining a potential difference V1, a current I1 is
recorded by the ammeter. When a larger battery of potential difference V2 replaces
battery V1, a larger current I2 is observed in the circuit. By successively using
different batteries, we observe different currents. If we plot the current recorded by
the ammeter in the circuit against the applied battery voltage, we obtain a linear
relationship, as shown in figure 24.6. This implies that the current in the circuit is
Figure 24.6 Ohm’s law.
directly proportional to the applied voltage, that is,
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Chapter 24 Electric Currents and DC Circuits
I∝V
(24.2)
To make an equality out of this proportionality, we introduce a constant of
proportionality (1/R) and the proportionality 24.2 becomes the equation
I=
V
R
(24.3)
where R is called the resistance of the wire. Equation 24.3 is called Ohm’s law after
Georg Simon Ohm (1787-1854), a German physicist who discovered the relation in
1827. Ohm’s law states that the current in a circuit is directly proportional to the
applied potential difference V and inversely proportional to the resistance R of the
circuit. The SI unit of resistance is the ohm, designated by the Greek letter Ω,
where
1 ohm = 1 volt
= 1 V/A
ampere
(For completeness, we should note that there are some materials that do not obey
Ohm’s law. That is, a graph of the current versus the voltage is not a straight line.
For such materials, the resistance is not a constant, but varies with voltage.
However, the resistance for any voltage can still be defined as the ratio of the
voltage to the current. We will not be concerned with such materials in this book.)
Every wire has some resistance and the resistance of a circuit is shown in the
circuit diagram in figure 24.5(b) as a saw tooth symbol that is labeled as R. The
wire, or any other resistive material, is called a resistor. Ohm’s law enables us to
determine the current in a circuit when the resistance of the circuit and the applied
voltage are known.
Example 24.1
Finding the current by Ohm’s law. A 6.00-V battery is applied to a circuit having a
resistance of 10.0 Ω. Find the current in the circuit (see figure 24.7).
Figure 24.7 Simple application of Ohm’s law.
Solution
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Chapter 24 Electric Currents and DC Circuits
The current in the circuit, found by Ohm’s law, equation 24.3, is
I = V = 6.00 V = 0.600 V
R 10.0 Ω
V/A
= 0.600 A
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Example 24.2
Finding the resistance by Ohm’s law. A potential difference of 12.0 V is applied to a
circuit and a current of 0.300 A is observed. What is the resistance of the circuit?
Solution
The resistance of the circuit, found from the rearrangement of Ohm’s law, is
R = V = 12.0 V = 40.0 Ω
I
0.300 A
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24.3 Resistivity
The concept of the resistance of a wire can be more easily understood by looking at
the lattice structure of the wire as in figure 24.2(c). When a potential difference is
applied to the ends of the wire the electrons slowly drift along the wire. The greater
the length of the wire, the longer it takes for the electrons to drift along the wire.
Since the current is the flow of charge per unit time, a longer period of time implies
a smaller current in the wire. This reduced current is viewed from Ohm’s law as an
increase in the resistance of the wire. Therefore, the resistance of a wire is directly
proportional to the length l of the wire, that is,
R∝l
(24.4)
The larger the cross-sectional area of the wire the greater are the number of
electrons that can pass through it per unit time. Therefore, the greater the area, the
larger the current in the wire. Viewed from Ohm’s law, equation 24.3, this
increased current implies a smaller value of resistance. Hence, the resistance of a
wire is inversely proportional to the cross-sectional area of the wire, that is,
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Chapter 24 Electric Currents and DC Circuits
R∝ 1
A
(24.5)
We can combine proportionalities 24.4 and 24.5 into the one proportionality
R∝ l
A
(24.6)
To make an equation out of this, we need a constant of proportionality. The
constant of proportionality should depend on the material that the wire is made of,
because the electrons in the wire constantly interact with the atoms of the lattice.
Different atoms exert different forces on the free electrons, and hence affect the
drift velocity of the electrons. The proportionality constant is called the resistivity ρ,
and proportionality 24.6 becomes the equation
R=ρ
l
A
(24.7)
Equation 24.7 says that the resistance of a wire is directly proportional to its
resistivity and its length, and inversely proportional to its cross-sectional area. A
table of resistivities for various materials is shown in table 24.1. The SI unit of
resistivity is an ohm meter, abbreviated Ω m. Everything else being equal,
Table 24.1
Resistivities (ρ) of Some Different Materials at 20 0C
and Mean Temperature Coefficient of Resistivity (α)
Material
ρ, Ω m
α 0C−1
Aluminum
2.82 × 10−8
3.9 × 10−3
Brass
7.00 × 10−8
2.00 × 10−3
−8
Carbon (graphite)
3500 × 10
−0.5 × 10−3
Copper
1.72 × 10−8
3.93 × 10−3
Gold
2.44 × 10−8
3.40 × 10−3
−8
Iron
9.71 × 10
5.20 × 10−3
Lead
20.6 × 10−8
4.30 × 10−3
Mercury
98.4 × 10−8
8.90 × 10−3
−8
Platinum
10.6 × 10
3.90 × 10−3
Silver
1.59 × 10−8
3.80 × 10−3
Tungsten
5.51 × 10−8
4.50 × 10−3
Amber
5.00 × 1014
Bakelite
2 × 105 - 2 × 1014
Glass
1013 - 1014
Hard Rubber
1013 - 1016
Mica
1011 - 1015
Wood
108 - 1011
24-8
Chapter 24 Electric Currents and DC Circuits
materials with relatively small values of resistivity ρ, such as metals, have small
resistances and hence make good conductors of electricity. Materials with large
values of ρ, the nonmetals, have large resistances and therefore make poor
conductors. Poor conductors are, however, good insulators. A good conductor has a
resistivity of the order of 10−8 Ω m, whereas a good insulator has a resistivity value
of the order of 1013-1015 Ω m. Good conductors of electricity are also good conductors
of heat. This is because the highly mobile electrons are also carriers of thermal
energy in conductors. For the same reason, good electrical insulators are also good
thermal insulators.
Example 24.3
The resistance of a spool of wire. Find the resistance of a spool of copper wire, 500 m
long with a diameter d of 0.644 mm.
Solution
The cross-sectional area of the wire is given by
A = πd2
4
=
π ( 0.644 × 10−3 m )
2
4
= 3.26 × 10−7 m2
The resistance of the wire spool, found from equation 24.7, is
R=ρl
A
(
)
500 m


= 1.72 × 10 −8 Ω m 
−7
2 
 3.26 × 10 m 
= 26.4 Ω
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Example 24.4
The resistance of an amber rod. Find the resistance of an amber rod 30.0 cm long by
2.00 cm high and 2.50 cm thick, as shown in figure 24.8.
24-9
Chapter 24 Electric Currents and DC Circuits
Figure 24.8 The resistance of an amber rod.
Solution
The resistance from one end of the rod to the other, found from equation 24.7, is
R=ρ
( 0.300 m )
l
= 5.00 × 1014 Ω m
A
( 0.0200 m )( 0.0250 m )
(
)
= 3.00 × 1017 Ω
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24.4 The Variation of Resistance with Temperature
It is found experimentally that the resistivity of a material is not constant but
rather varies with temperature. A graph of the variation of the resistivity with
temperature for a metal is shown in figure 24.9(a). Notice that the graph is a curve,
which means that the variation is not linear. However, a straight line can be drawn
that approximates the curve for a range of temperatures, as shown in figure 24.9(b).
The slope m of this straight line is given as
m = ∆ρ = ρ − ρ0
∆t t − t0
where ρ is the resistivity of the material at the temperature t and ρ0 is the
resistivity of the material at the temperature t0. Rearranging the equation we get
Solving for the resistivity ρ, gives
ρ − ρ0 = m(t − t0)
ρ = ρ0[1 + m (t − t0)]
ρ0
(24.8)
Let us now define a new constant α, called the mean temperature coefficient of
resistivity, as
α= m
ρ0
24-10
Chapter 24 Electric Currents and DC Circuits
Figure 24.9 The variation of resistivity with temperature.
Thus, by measuring the slope of the curve in the area of interest, and dividing by
the resistivity ρ0 at the reference temperature t0, we obtain the mean temperature
coefficient of resistivity α. The value of α for various materials is shown in table
24.1. Notice that since the slope m is a ratio of resistivity to temperature, the
temperature coefficient of resistivity α , which is equal to that slope divided by the
resistivity, has units of 0C−1. With this new temperature coefficient of resistivity, we
can write equation 24.8 as
ρ = ρ0[1 + α (t − t0)]
(24.9)
Equation 24.9 gives the resistivity of a material at the temperature t when the
resistivity of the material ρ0 is known at the reference temperature t0.
Example 24.5
Temperature dependence of resistivity. The resistivity of copper at 20.0 0C is 1.72 ×
10−8 Ω m. Find its resistivity at 200 0C.
24-11
Chapter 24 Electric Currents and DC Circuits
Solution
The temperature coefficient of resistivity for copper, found from table 24.1, is 3.93 ×
10−3 0C−1. The resistivity of copper at 200 0C, found from equation 24.9, is
ρ = ρ0[1 + α (t − t0)]
= (1.72 × 10 Ω m)[1 + (3.93 × 10−3 0C−1)(200 − 20.0) 0C ]
= 2.94 × 10−8 Ω m
−8
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Because the resistivity of a wire changes with temperature, the resistance of that
wire also changes with temperature. If we multiply both sides of equation 24.9 by
the length of the wire and divide by the cross-sectional area of the wire, we obtain
ρ l = ρ0 l [1 + α(t − t0)]
A
A
(24.10)
However, ρl/A is equal to the resistance R from equation 24.7. Therefore, we can
write equation 24.10 as
R = R0 1 + α (t − t0 ) 
(24.11)
Equation 24.11 gives the resistance of a resistor R, at the temperature t, if the
resistance R0, at the reference temperature t0, is known.
Example 24.6
Temperature dependence of resistance. If the resistance of a copper resistor is 50.0 Ω
at 20.0 0C, find its resistance at 200 0C.
Solution
The temperature coefficient of resistivity for copper, found from table 24.1, is 3.93 ×
10−3 0C−1. The resistance of the copper resistor at 200 0C, found from equation 24.11,
is
R = R0[1 + α(t − t0)]
= (50.0 Ω)[1 + (3.93 × 10−3 0C−1)(200 − 20.0) 0C]
= 85.4 Ω
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24-12
Chapter 24 Electric Currents and DC Circuits
We should note that not all materials have the same kind of variation of
resistivity with temperature. A group of materials known as semiconductors have a
temperature variation as shown in figure 24.9(c).
When the resistivity of metals is plotted against the absolute temperature, a
very strange phenomenon occurs at very low temperatures, as shown in figure
24.9(d). At a certain temperature, known as the critical temperature Tc, the
resistivity, and hence the resistance of the material, drops to zero. The material is
then said to be superconducting, and the material is called a superconductor. This
phenomenon was first discovered by Kamerlingh Onnes in the Netherlands in 1911.
He found that for mercury, the resistivity effectively dropped to zero when the
temperature was reduced to 0.05 K. Researchers have since found newer materials
that become superconducting at much higher temperatures. By the 1960s the
critical temperature for superconduction was found to be as high as 20 K. Steady
research into newer materials has raised the critical temperature even further. By
the end of 1986 the critical temperature of 40 to 50 K had been reached with an
oxide of barium, lanthanum, and copper. By March 1987 superconductivity had
been made to occur at temperatures above 90 K.
The reason for the great importance of superconductivity lies in the fact that
once the resistance of a material drops to zero, the energy dissipated in the
resistance also drops to zero and the current continues to exist forever. (We will see
that the energy dissipated in a resistor is given by I2R. If R is zero, no energy is
lost.) If newer superconducting materials can be found that have still higher critical
temperatures, a time will come when superconductivity will be a practical new
technology that will revolutionize the entire electrical industry. The cost of
delivering electricity will be greatly reduced by making essentially lossless
transmission lines. Some superconducting thin films will be used in new devices for
electronic components in computers. Electric motors will run with a minimum of
energy. The change will be as great as it was with the discovery of the transistor
and the subsequent use of the microchip in electrical components.
24.5 Conservation of Energy and the Electric Circuit -Power Expended in a Circuit
Consider the simple resistive circuit in figure 24.10. The power supplied by the
battery is the work it does per unit time, that is
P=W
t
(24.12)
But the work done within the battery is done by chemical means, and its final
result is to move a positive charge q from the negative terminal inside the battery to
the positive terminal within the battery. That is, a charge q had to be ‘’lifted’’ from
24-13
Chapter 24 Electric Currents and DC Circuits
Figure 24.10 Conservation of energy in an electric circuit.
the point of zero potential to the point of higher potential V within the battery.
Recall from the definition of the potential that the potential is the potential energy
per unit charge,
V = PE = W
(21.19)
q
q
where W is the work that must be done to give the charge its potential energy. From
equation 21.19, the work done within the battery is simply
W = qV
(24.13)
The power supplied by the battery is found from equations 24.12 and 24.13 as
But
P = W = qV
t
t
(24.14)
I=q
t
(24.1)
the current coming out of the battery. Combining equation 24.1 with equation 24.14
gives the power supplied by the battery as
P = IV
(24.15)
The conservation of energy applied to the circuit can be expressed as
Energy supplied to the circuit = Energy consumed in the circuit
(24.16)
Since the power is energy per unit time, if we divide both sides of equation 24.16 by
the time t, we get
Power supplied to the circuit = Power consumed in the circuit
24-14
(24.17)
Chapter 24 Electric Currents and DC Circuits
Therefore, the power consumed in the circuit is
Power consumed = Power supplied = IV
If the circuit contains only a resistor, then the voltage across the resistor is given by
Ohm’s law as V = IR. Therefore, the power consumed or dissipated in the resistor is
P = IV = I(IR)
P = I 2R
(24.18)
Equation 24.18 gives the rate at which energy is dissipated in the resistor with
time. This energy that is lost as the charge “falls” through the resistor shows up as
heat in the resistor, and is usually referred to as Joule heat. We can also express
equation 24.18 as
P = V2
(24.19)
R
by substituting I = V/R into equation 24.18.
Recall from chapter 7, the unit of power is the watt, where
1 watt = 1 joule = 1 J/s
second
Checking the units for the power supplied to a circuit we get
P = IV = ampere volt = (coulomb)(joule) = joule = watt
second coulomb second
which we can abbreviate as
1 W = 1 A V = (1 C/s)(1 J/C) = 1 J/s = 1 W
Thus, in electrical circuits, we represent the unit for power as
watt = ampere volt = A V
because it is equivalent to the previous definition.
When the charge q leaves the battery, it is at the potential V. As it ‘’falls’’
through the resistor, figure 24.10(b), it loses energy as it falls to a position of lower
potential. Therefore, we say that there is a potential drop across the resistor.
The potential drop across the resistor is given by Ohm’s law as
V = IR
24-15
(24.20)
Chapter 24 Electric Currents and DC Circuits
It is sometimes convenient to mark the position of the resistor that is at the highest
potential with a plus (+) sign, and the point of the resistor that is at the lowest
potential with a negative sign (−) to remind ourselves that the charge will fall from
a plus (+) to a minus (−) potential. This is shown in figure 24.10(b).
The mechanical equivalent of the circuit of figure 24.10(a) is shown in figure
24.10(c). A ball of mass m is lifted to a height h by a person. The person who does
the work lifting the mass is equivalent to the battery. The ball has potential energy
at the top. We assume that the medium that the ball will fall through is resistive.
Therefore, the ball will not fall freely, continually accelerating, but rather will be
slowed down by the friction of the medium until the ball moves at a constant
velocity, its terminal velocity. This is similar to the charge moving at the constant
drift velocity. As the ball falls and loses potential energy, the lost potential energy
shows up as heat generated by the frictional forces between the ball and the
medium. This is similar to the loss of energy of the charge through Joule heating as
it ‘’falls’’ through the resistor.
Example 24.7
A 60-W light bulb. A 60.0-W light bulb is screwed into a 120-V lamp outlet. (a) What
current will flow through the bulb and what is the resistance of the bulb? (b) What
is the power dissipated in the Joule heating of the wire in the bulb?
Solution
a. Because the power rating of the bulb is known, we can find the current from
equation 24.15 as
P = IV
I = P = 60.0 W = 0.500 A V
V
120 V
V
= 0.500 A
The resistance of the bulb, found from Ohm’s law, is
R = V = 120 V = 240 Ω
I
0.500 A
b. The power dissipated in the Joule heating of the wire in the bulb, found from
equation 24.18, is
P = I2R = (0.500 A)2(240 Ω)
= 60.0 W
Note that the power supplied is equal to the power dissipated as expected.
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24-16
Chapter 24 Electric Currents and DC Circuits
Example 24.8
A 120-W light bulb. What is the current and resistance of a 120-W light bulb
connected to a 120-V source?
Solution
We find the current from
I = P = 120 W = 1.00 A
V
120 V
We determine the resistance of the bulb by Ohm’s law as
R = V = 120 V = 120 Ω
I
1.00 A
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Up to now, we studied a circuit that contained only one resistor. Suppose
there are several resistors in the circuit. Is there a difference in the circuit if these
resistors are connected in different ways? The answer is yes and we will now study
different combinations of these resistors -- in particular, resistors in series, resistors
in parallel, and combinations of resistors in series and parallel.
24.6 Resistors in Series
A typical circuit having resistors in series is shown in figure 24.11(a). The
characteristic of a series circuit is that the same current that flows from the battery
flows through each resistor. That is, the current I is the same everywhere in the series
circuit. Figure 24.11(b) displays the circuit unfolded, showing how the positive
charge will fall from high potential to low potential. Figure 24.11(c) shows a
mechanical analogue of the circuit of resistors in series. A ball is picked up from the
ground and placed at the top of a three-step stairway. At the top step the ball has
its maximum potential energy. If the ball is given a slight push it rolls off the top
step dropping down to the first step, losing an amount of potential energy PE1. This
lost potential energy is first converted to kinetic energy as the ball falls, and the
kinetic energy is then converted to thermal energy in the collision of the ball with
the step. The ball then rolls off the first step dropping down to the second step,
losing an amount of potential energy PE2. It then rolls off the second step, falling
24-17
Chapter 24 Electric Currents and DC Circuits
Figure 24.11 Resistors in Series
down to the third and last step at the ground level. This time it loses an amount of
potential energy PE3. The law of conservation of energy says that the total energy
given to the ball to place it at the top step must be equal to the total energy that it
loses as it falls from step to step to the ground. That is,
PEtop = PE1 + PE2 + PE3
(24.21)
The electrical circuit, figure 24.11(b), is analogous to this mechanical staircase. The
charge q is raised to the potential V by the chemical action of the battery. As the
charge ‘’falls’’ through the first resistor, it drops in potential by V1. As it falls
through the second resistor it drops in potential by V2. The charge experiences
another potential drop, V3, as it falls through R3. By the law of conservation of
energy the potential supplied to the charge by the battery must be equal to the
potential that the charge loses as it falls through the resistors. Therefore,
V = V1 + V2 + V3
(24.22)
The voltage drop across each resistor in figure 24.11, given by Ohm’s law, equation
24.20, is
V1 = IR1
V2 = IR2
V3 = IR3
Replacing these equations into equation 24.22 gives
V = IR1 + IR2 + IR3
24-18
(24.23)
Chapter 24 Electric Currents and DC Circuits
Dividing both sides of equation 24.23 by I gives
V = R1 + R2 + R3
I
(24.24)
where V is the total potential applied to the circuit and I is the total current in the
circuit, so V/I should, by Ohm’s law, equal R, the total resistance of the circuit, that
is
V =R
(24.25)
I
From equations 24.24 and 24.25, it is obvious that
R = R1 + R2 + R3
(24.26)
That is, the sum of the three resistances in the series circuit is equivalent to one
resistance R called the equivalent resistance. Figure 24.12(a) is equivalent to the
simpler circuit shown in figure 24.12(b), with R, the equivalent resistance, given by
equation 24.26. That is, the three resistors R1, R2, and R3 could be replaced by the
one equivalent resistor R without detecting any electrical change in the circuit.
Figure 24.12 Equivalent circuit for resistors in series.
Although equation 24.26 was derived for three resistors it is obvious that it holds
for the sum of any number of resistors in series.
The fact that the equivalent resistance of resistors in series is just the sum of
the individual resistances should not be too surprising. Because, if each of the
resistors were a wire of the same material, same cross-sectional area, but with
different lengths l1, l2, and l3, respectively, then the length of the wire when the
three resistors are connected in series is just
l = l1 + l2 + l3
24-19
Chapter 24 Electric Currents and DC Circuits
Multiplying each term by ρ/A gives
ρ
l
l
l
l
=ρ 1 +ρ 2 +ρ 3
A
A
A
A
But using equation 24.7, R = ρl/A gives
R = R1 + R2 + R3
which is the same result as before, equation 24.26. This derivation may be easier to
see but it does not give the same insight as is obtained by using the law of
conservation of energy.
Example 24.9
Resistors in series. Resistors R1 = 20.0 Ω, R2 = 30.0 Ω, and R3 = 40.0 Ω are connected
in series to a 6.00-V battery. Find the equivalent resistance of the circuit and the
current in this series circuit.
Solution
The equivalent resistance, given by equation 24.26, is
R = R1 + R2 + R3
= 20.0 Ω + 30.0 Ω + 40.0 Ω
= 90.0 Ω
The current is found by Ohm’s law, with R the equivalent resistance, that is,
I = V = 6.00 V = 0.0667 A
R 90.0 Ω
To go to this Interactive Example click on this sentence.
Example 24.10
Power dissipated in resistors in series. Find the power supplied and the power
dissipated in each resistor in example 24.9.
Solution
The power supplied by the battery is
24-20
Chapter 24 Electric Currents and DC Circuits
P = VI = (6.00 V)(0.0667 A) = 0.400 W
The power dissipated in each resistor is
P1 = I2R1 = (0.0667 A)2(20.0 Ω) = 0.0890 W
P2 = I2R2 = (0.0667 A)2(30.0 Ω) = 0.133 W
P3 = I2R3 = (0.0667 A)2(40.0 Ω) = 0.178 W
The total power dissipated in all resistors is
P = P1 + P2 + P3
= 0.0890 W + 0.133 W + 0.178 W
= 0.400 W
Note that the power supplied to the circuit is equal to the power dissipated in the
resistors.
To go to this Interactive Example click on this sentence.
Example 24.11
Potential drop across resistors in series. Find the potential drop across each resistor
of examples 24.9 and 24.10.
Solution
The potential drop across each resistor, found by Ohm’s law, is
V1 = IR1 = (0.0667 A)(20.0 Ω) = 1.33 V
V2 = IR2 = (0.0667 A)(30.0 Ω) = 2.00 V
V3 = IR3 = (0.0667 A)(40.0 Ω) = 2.67 V
Notice that the sum of the potential drops, V1 + V2 + V3, is equal to 6.00 V, which is
equal to the applied voltage of 6.00 V.
To go to this Interactive Example click on this sentence.
24-21
Chapter 24 Electric Currents and DC Circuits
24.7 Resistors in Parallel
A typical circuit with resistors connected in parallel is shown in figure 24.13(a).
Resistors in a parallel circuit are connected such that the top of each resistor is
connected to the same point A, whereas the bottom of each resistor is connected to
Figure 24.13 Resistors in parallel.
the same point B. Therefore, the potential difference between A and B is the same
as the potential difference across each resistor. We assume that the resistance of
the connecting wires is negligible compared to the resistors in the circuit, and can
be ignored. Therefore, the potential across AB is the same as the potential V
supplied by the battery. Consequently, the characteristic of resistors connected in
parallel is that the potential difference is the same across every resistor. That is,
V = V1 = V2 = V3
(24.27)
When the total current I from the battery reaches the junction A, it divides into
three parts; I1 goes through resistor R1, I2 goes through resistor R2, and I3 goes
through R3. Because none of the charge disappears,
I = I1 + I2 + I3
(24.28)
Equation 24.28 is a statement of the law of conservation of electric charge.
Electric charge can neither be created nor destroyed and hence the electric charges
entering a junction must be equal to the electric charges leaving a junction. Thus, the
electric current entering a junction is equal to the electric current leaving a junction.
At the junction B these three currents again combine to form the same total current
I that entered the junction A. The current in each resistor can be found by applying
Ohm’s law to that resistor. That is,
I1 = V1
R1
24-22
(24.29)
Chapter 24 Electric Currents and DC Circuits
I2 = V2
R2
I3 = V3
R3
(24.30)
(24.31)
Replacing these values of the currents back into the current equation, 24.28, gives
I = V1 + V2 + V3
R1 R2 R3
(24.32)
But, the potentials are equal by equation 24.27 (i.e., V = V1 = V2 = V3). Hence,
equation 24.32 becomes
I= V + V + V
R1 R2 R3
Dividing both sides of the equation by V, gives
I = 1 + 1 + 1
V R1 R2 R3
(24.33)
But the left-hand side of equation 24.33 contains the total current I in the circuit,
divided by the total voltage V applied to the circuit, and by Ohm’s law is simply
I = 1
V
R
(24.34)
where R is the total resistance of the entire circuit. Equating 24.34 to equation
24.33 gives
1
1
1
1
=
+
+
(24.35)
R R1 R2 R3
Equation 24.35 says that the reciprocal of the total resistance of the circuit is
equivalent to the sum of the reciprocals of each parallel resistance. We call R the
equivalent resistance of the resistors in parallel. The resistor R could replace the
three resistors R1, R2, and R3, and the circuit would still behave the same way
electrically.
Although we derived equation 24.35 from a circuit with only three resistors
in parallel, the form is completely general. If there were n resistors in parallel, the
equivalent resistance would be found from
1
1
1
1
1
=
+
+
+ ... +
R R1 R2 R3
Rn
24-23
(24.36)
Chapter 24 Electric Currents and DC Circuits
Example 24.12
Equivalent resistance of resistors in parallel. Three resistors R1 = 20.0 Ω, R2 = 30.0
Ω, and R3 = 40.0 Ω are connected in parallel to a 6.00-V battery, as shown in figure
24.13. Find the equivalent resistance of the circuit.
Solution
From equation 24.35 the equivalent resistance is
1 = 1 + 1 + 1
R
R1 R2 R3
= 1
+
1 +
1 = 0.108
20.0 Ω 30.0 Ω 40.0 Ω
Ω
R = 9.23 Ω
Notice that in this calculation 1/R = 0.108, and to get the actual value of R we need
to take the reciprocal of 0.108, which yields the correct value of the resistance as
9.23 Ω. Failure to take the final reciprocal is a very common student error.
To go to this Interactive Example click on this sentence.
Compare this result with example 24.9 in which the same three resistors
were connected in series. There, the total equivalent resistance was 90.0 Ω when
the resistors were connected in series, whereas here the same resistors connected in
parallel give an equivalent resistance of only 9.23 Ω. It is clear from these two
examples that the way the resistors are connected in a circuit makes a great deal of
difference in their effect on the circuit. Note that when the resistors are connected in
parallel, the equivalent resistance is always less than the smallest of the original
resistances. In this case the total resistance, 9.23 Ω, is less than the smallest
resistance of 20.0 Ω. The equivalent circuit of the parallel circuit in figure 24.13(a)
is shown in figure 24.13(b), where R is the equivalent resistance, found in equation
24.35.
Example 24.13
The total current in the parallel circuit. Find the total current coming from the
battery in the example 24.12.
Solution
The total current in the circuit, found from Ohm’s law with R as the equivalent
resistance of the circuit, is
24-24
Chapter 24 Electric Currents and DC Circuits
I = V = 6.00 V = 0.650 A
R 9.23 Ω
Notice that the current from the battery is almost ten times greater when the
resistors are connected in parallel then when connected in series. This is easily
explained, since the resistance in parallel is only about one-tenth of the resistance
when the resistors are in series.
To go to this Interactive Example click on this sentence.
Example 24.14
The current in each parallel resistor. Find the current through each resistor in
example 24.12.
Solution
Because the voltage across each resistor is 6.00 V, we can find the current from
Ohm’s law applied to each resistor as given in equations 24.29, 24.30, and 24.31.
Namely,
I1 = V = 6.00 V = 0.300 A
R1 20.0 Ω
I2 = V = 6.00 V = 0.200 A
R2 30.0 Ω
I3 = V = 6.00 V = 0.150 A
R3 40.0 Ω
Note that the current through each resistor is different, but the sum of I1 + I2 + I3 =
0.650 A is equal, to the total current of 0.650 A flowing from the battery in the
circuit. Also note that the resistor with the smallest value of resistance has the
largest value of current passing through it. This is sometimes stated as: The current
always takes the path of least resistance.
To go to this Interactive Example click on this sentence.
Example 24.15
Power supplied to a parallel circuit. What is the power supplied to the circuit in
example 24.12?
24-25
Chapter 24 Electric Currents and DC Circuits
Solution
The power supplied by the battery is
P = IV = (0.650 A)(6 V) = 3.90 W
To go to this Interactive Example click on this sentence.
Example 24.16
Power dissipated in a parallel circuit. What is the power dissipated in each resistor
in example 24.14?
Solution
The power dissipated in each resistor is
P1 = I12R1 = (0.300 A)2(20.0 Ω) = 1.80 W
P2 = I22R2 = (0.200 A)2(30.0 Ω) = 1.20 W
P3 = I32R3 = (0.150 A)2(40.0 Ω) = 0.900 W
Again note that the sum of the powers dissipated in each resistor
P1 + P2 + P3 = 1.80 W + 1.20 W + 0.900 W
= 3.90 W
is the same as the total power supplied to the circuit by the battery, within roundoff error.
To go to this Interactive Example click on this sentence.
24.8 Combinations of Resistors in Series and Parallel
A typical circuit showing a simple combination of resistors connected in series and
parallel is shown in figure 24.14. Let us determine the current through each
resistor and the voltage drop across it. To do so, we use the techniques of sections
24.6 and 24.7. Resistors R2 and R3 are connected in parallel. Their equivalent
resistance R23, found from equation 24.36, is
24-26
Chapter 24 Electric Currents and DC Circuits
Figure 24.14 Combinations of resistors in a circuit.
1 = 1 + 1
R23 R2 R3
= 1 + 1
= 0.0583
30.0 Ω 40.0 Ω
Ω
R23 = 17.2 Ω
The circuit of figure 24.14(a) can be replaced by its equivalent circuit, figure
24.14(b), where R23 is in series with R1. The equivalent resistance of R1 and R23 in
series, found from equation 24.26, is
R123 = R1 + R23
= 20.0 Ω + 17.2 Ω
= 37.2 Ω
The circuit of figure 24.14(b) can now be replaced by the equivalent circuit, figure
24.14(c), containing only one resistor, the equivalent resistor R123 = 37.2 Ω.
The current from the battery is now easily determined by Ohm’s law as
I = V = 6.00 V = 0.161 A
R123 37.2 Ω
Since the resistor R1 is in series with the battery, this same current flows through it
(i.e., I1 = I = 0.161 A). However, this is not the current through R2 and R3 because
they are in parallel and the current divides as it enters the two paths. In order to
determine I2 and I3 we must first determine the voltage drop across R2 and R3.
The voltage drop across R1, found from Ohm’s law, is
V1 = I1R1 = (0.161 A)(20.0 Ω) = 3.22 V
The total applied potential is equal to the sum of the potential drops, that is,
24-27
Chapter 24 Electric Currents and DC Circuits
V = V1 + V2
The potential drop V2 across the parallel resistors is
V2 = V − V1
= 6.00 V − 3.22 V
= 2.78 V
The potential drop V3 is also equal to 2.78 V since R2 and R3 are connected in
parallel. With the potential drop across the parallel resistors known, the current in
each resistor, found from Ohm’s law, is
I2 = V2 = 2.78 V = 0.0927 A
R2 30.0 Ω
I3 = V3 = 2.78 V = 0.0695 A
R3 40.0 Ω
Note that the sum of the currents
I2 + I3 = 0.0927 A + 0.0695 A = 0.162 A
is equal, within round-off errors of our calculations, to the total current in the
circuit, 0.161 A.
The power delivered to the circuit is
P = IV = (0.161 A)(6.00 V) = 0.966 W
The power dissipated in each resistor is
P1 = I12R1 = (0.161 A)2(20.0 Ω) = 0.518 W
P2 = I22R2 = (0.0927 A)2(30.0 Ω) = 0.258 W
P3 = I32R3 = (0.0695 A)2(40.0 Ω) = 0.193 W
The sum of the power dissipated in the resistors (0.969 W) is equal, within round-off
error, to the supplied power of 0.966 W.
24.9 The Electromotive
Resistance of a Battery
Force
and
the
Internal
In the early days of electricity, in order to keep an analogy with Newtonian
mechanics, it was assumed that if a charge moved through a wire, there must be
some force pushing on the charge to cause it to move through the wire. This force
acting on the charge was called an electromotive force. The battery, the supplier
24-28
Chapter 24 Electric Currents and DC Circuits
of this force, was called a seat of electromotive force, abbreviated emf, and
pronounced as the individual letters e-m-f. The emf is usually written as the script
letter E. Today of course, we know that this emf is really a misnomer. The battery is
supplying a potential difference. However, the name emf still remains. The emf is
measured in volts, since the emf is a potential difference. You will hear comments
such as, a battery has an emf of 6 V, or the generator supplies an emf of 120 V, and
the like.
If there is no current being drawn from the battery, the emf of the battery
and the potential difference between its two terminals are the same. When a
current exists, however, the potential difference between the terminals is always
less than the emf of the battery. The reason for this is that every battery has an
internal resistance, which is a characteristic of the battery and cannot be
eliminated. The internal resistance of the battery acts as though it were a resistor
in series with the battery itself. It is usually represented in a circuit diagram as the
lower case r in figure 24.15. When a current exists in the circuit there is a potential
drop, equal to the product Ir, across the internal resistance. The potential difference
between the terminals of the battery, becomes
V = E − Ir
(24.37)
which is, of course, less than the emf of the battery.1
Figure 24.15 The internal resistance of a battery.
Example 24.17
The potential and emf of a battery. A battery has an emf of 1.50 V and when
connected to a circuit supplies a current of 0.0100 A. If the internal resistance of the
battery is 2.00 Ω, what is the potential difference across the battery?
1
We should note that if the battery is being charged by an external source, then the terminal
voltage would be given by V = E + Ir.
24-29
Chapter 24 Electric Currents and DC Circuits
Solution
The potential difference across the battery, given by equation 24.37, is
V = E − Ir
= 1.50 V − (0.0100 A)(2.00 Ω)
= 1.48 V
Notice that when the current in the circuit is relatively small, the potential drop
across the internal resistance is also quite small, and the terminal voltage is very
close to the emf of the battery.
If the current in the circuit is much larger, as for example I = 0.500 A then
the terminal voltage is
V = E − Ir
= 1.50 V − (0.500 A)(2.00 Ω)
= 0.500 V
which is very much less than the emf of the battery.
To go to this Interactive Example click on this sentence.
Example 24.18
The emf of a battery connected to a circuit. A battery has an emf of 1.50 V and an
internal resistance of 3.00 Ω. It is connected as shown in figure 24.15 to a resistance
of 500 Ω. Find the current in the circuit, and the terminal voltage of the battery.
Solution
The internal resistance r is in series with the load resistor R, so the total resistance
in the circuit is just the sum of the two of them. Using Ohm’s law to determine the
current we have
I= E =
1.50 V
= 2.98 × 10−3 A
R + r 500 Ω + 3.00 Ω
The terminal voltage across the battery is
V = E − Ir
= 1.50 V − (2.98 × 10−3 A)(3.00 Ω)
= 1.49 V
24-30
Chapter 24 Electric Currents and DC Circuits
To go to this Interactive Example click on this sentence.
Because the value of r is usually very small, relative to the resistance R of
the circuit, we neglect it in many problems. In such cases, we assume that the
terminal voltage and the emf are the same. Whether we can make this assumption
or not, depends on the values of r and R and the accuracy that we are willing to
accept in the solution to the problem. In example 24.18, neglecting r gives a current
of 3.00 × 10−3 A, an error of about 0.7%. If that amount of error is acceptable in the
problem, the internal resistance can be ignored. If that error is not acceptable then
the internal resistance must be taken into account.
Batteries in Series
Since the emf of most batteries is only 1.50 V, we need to place many of them in
series to get a higher emf. If three 1.50-V batteries are connected in series, as in
figure 24.16, with the positive terminal of one battery connected to the negative
terminal of the next battery and so on, the emf of the combination is 3(1.50 V) =
Figure 24.16 Batteries in series.
4.50 V. In general, when batteries are connected in series, the total emf is the sum of
the emf’s of each battery. That is,
E = E1 + E2 + E3 + … + En
(24.38)
By connecting any number of batteries in series, any larger emf may be obtained.
Because each battery is in series, the current through each battery is the same, and
the total internal resistance of the combination is just the sum of the internal
resistance of each battery.
Batteries in Parallel
Three identical batteries, of negligible internal resistance, are connected in parallel
in figure 24.17. The negative terminals of all the batteries are all connected
together, and the positive terminals of all the batteries are all connected to one
another. Since the batteries are in parallel, the potential difference across the
combination is the same as the emf of each battery, that is,
24-31
Chapter 24 Electric Currents and DC Circuits
Figure 24.17 Identical batteries in parallel.
E = E1 = E2 = E3
(24.39)
At junction A, the three battery currents combine to give the circuit current
I = I1 + I2 + I3
(24.40)
Because we assume the batteries to be identical (I1 = I2 = I3), the total circuit
current available is three times the current available from any one battery. By a
suitable combination of batteries in series and parallel, batteries of any voltage and
current can be made.
Example 24.19
Batteries in parallel. Three identical 6.00-V batteries are connected in parallel to a
resistance of 500 Ω, as in figure 24.17. Find the current in the circuit and the
current coming from each battery. Assume the internal resistance of the batteries to
be zero.
Solution
The total current I in the circuit, found from Ohm’s law, is
I = E = 6.00 V = 0.0120 A
R 500 Ω
This total current is supplied by three batteries so that, I = 3I0, where I0 is the
actual current from any one battery. Therefore, the battery current is
I0 = I = 0.0120 A = 4.00 × 10−3 A
3
3
24-32
Chapter 24 Electric Currents and DC Circuits
To go to this Interactive Example click on this sentence.
24.10 The Wheatstone Bridge
A standard technique to determine the value of a particular resistor is to connect it
into a circuit, such as in figure 24.18(a). It is usually assumed that the resistance of
the ammeter is negligible, and the resistance of the voltmeter is so large that
negligible current flows through it. Therefore, the current I in the resistor is
measured, and the voltage V across it is also measured. The value of the resistance
R is easily determined from Ohm’s law as
R=V
I
(24.41)
Figure 24.18 Techniques for measuring voltage and current in a simple circuit.
As long as our assumptions hold, equation 24.41 is valid. Now let’s look a little
closer at our assumptions. If the resistance of the ammeter is anywhere near the
order of magnitude of the resistance R, the assumptions break down. Although the
ammeter does measure the current I through R, there is now a voltage drop VA
across the ammeter, and the voltmeter reads the voltage drop
The calculated resistance is
V = VR + VA
R=
VR + VA
I
(24.42)
which gives a value for R greater than the exact value found by equation 24.41
when the effects of the ammeter can be neglected.
To remedy this, we might say, let’s change the measuring technique to the
one shown in figure 24.18(b). With this second technique, the voltmeter reads only
the voltage VR across the resistor. At first glance, it might seem that the problem is
24-33
Chapter 24 Electric Currents and DC Circuits
solved. But if we look carefully we note that the current I in the circuit splits at
junction A, where the resistor and voltmeter are in parallel. A current IR flows
through the resistor while a current Iv flows through the voltmeter. The ammeter
reads the current
I = IR + Iv
The computed resistance R therefore becomes
R=
VR
I R + IV
(24.43)
But this is less than that computed from the exact value of equation 24.41, when
the effects of the voltmeter can be neglected in the circuit.
This analysis of the measurements of a simple circuit again points out the
importance of the assumptions made in deriving any equation. As long as the initial
assumptions hold, the derived equations hold. When there is a breakdown in the
assumptions, the derived equations are no longer valid. Note that when the
assumptions of a negligible resistance for the ammeter, and a negligible current in
the voltmeter (very high resistance voltmeter), are correct, equations 24.42 and
24.43 reduce to 24.41.
When the previous assumptions do not hold, it is still possible to make
accurate measurements of resistance by means of a circuit that is called the
Wheatstone bridge, figure 24.19(a). Its virtue is that it is a null detection method
Figure 24.19 The Wheatstone bridge.
24-34
Chapter 24 Electric Currents and DC Circuits
that does not depend on the characteristics of the galvanometer, but uses it merely
to be a sensitive detector of the condition of current versus no current. The values of
resistors R1, R3, and R4, in the circuit are assumed to be known, whereas R2 is the
unknown resistor. A battery of emf E is applied to the circuit and a current I
emanates from the battery. At the junction A, the current I splits into two currents,
I1 and I3.
I = I1 + I3
When current I1 reaches the junction C it splits into two currents, I2 through R2 and
Ig the current through the galvanometer. That is,
I1 = I2 + Ig
(24.44)
The galvanometer needle deflects indicating that there is a current in it. At junction
D the galvanometer current Ig combines with current I3 to form current I4, which
passes through resistor R4. That is,
I3 + Ig = I4
(24.45)
At junction B, currents I2 and I4 combine to form the original current I.
A current Ig exists in the galvanometer, because there is a difference of
potential between points C and D in the circuit. This difference in potential is
caused by the different currents in the different resistors. The values of these
resistors R1, R3, and R4 are varied until the galvanometer reads zero (i.e., Ig = 0).
When this occurs the bridge is said to be balanced. When the galvanometer reads
zero, it means that there is no longer a difference in potential between points C and
D. (Remember a current exists in a conductor when there is a difference in
potential. If there is no current, there is no difference in potential.) The circuit in
figure 24.19(a) can now be replaced by the one in figure 24.19(b). The points C and
D have coalesced electrically into one point, because they are at the same potential.
It is now obvious from figure 24.19(b) that R1 is now in parallel with R3 and
therefore the voltages across them are equal, that is,
Therefore, from Ohm’s law,
V1 = V3
I1R1 = I3R3
(24.46)
It is also obvious that R2 is now in parallel with R4 and hence the voltages across
them are also equal, that is,
V2 = V4
I2R2 = I4R4
(24.47)
Dividing equation 24.47 by equation 24.46, we obtain
24-35
Chapter 24 Electric Currents and DC Circuits
I 2 R2 I 4 R4
=
I1 R1 I 3 R3
(24.48)
However, since the bridge is balanced, Ig = 0, and equation 24.44 reduces to
I1 = I2
and equation 24.45 to
I3 = I4
Therefore, the currents in equation 24.48 cancel, leaving the simple ratio of the
resistors
R2 R4
=
R1 R3
Solving for the unknown resistor R2, we obtain
R2 =
R4
R1
R3
(24.49)
The Wheatstone bridge is a very accurate means of determining the resistance of an
unknown resistor. A picture of a typical commercial Wheatstone bridge is shown in
figure 24.19(c).
Example 24.20
The Wheatstone bridge. A Wheatstone bridge is balanced when R1 = 5.00 Ω, R3 =
10.0 Ω, and R4 = 4.00 Ω. Find the unknown resistance R2.
Solution
The value of the unknown resistor, found from equation 24.49, is
R2 =
( 4.00 Ω ) 5.00 Ω = 2.00 Ω
R4
R1 =
(
)
R3
(10.0 Ω )
To go to this Interactive Example click on this sentence.
24.11 Kirchhoff’s Rules
Quite often we find circuits in which the resistors are not simply in series or
parallel, or their combinations. An example of such a circuit is shown in figure
24-36
Chapter 24 Electric Currents and DC Circuits
24.20. The locations of the different batteries in the circuit prevents the resistors
from being in parallel. In order to solve this more complicated circuit problem, two
rules established by the German physicist Gustav Robert Kirchhoff (1824-1887) are
used. Kirchhoff’s first rule is: The sum of the currents entering a junction is equal
to the sum of the currents leaving the junction. That is,
∑I
entering junction
=∑ I leaving junction
(24.50)
Figure 24.20 A circuit problem solved by Kirchhoff’s rules.
This first rule is a statement of the law of conservation of electric charge.2
In order to apply the first rule, we must make an assumption about the
currents in the circuit. We assume that current I1 emanates from battery E 1, I2
from E2, and I3 from E3, all in the directions shown in figure 24.20. If the actual
directions of the currents are as assumed, the numerical values of the currents
found by the equations will be positive. If the direction of the currents is in the
opposite direction from those assumed, the numerical values of the currents found
by the equations will be negative, indicating that the direction of the currents is
really reversed. Using Kirchhoff’s first rule at junction A, gives
I2 = I1 + I3
(24.51)
Kirchhoff’s second rule states that the change in potential around a closed loop is
equal to zero. This can also be stated in the form, around any closed loop, the sum of
the potential rises plus the sum of the potential drops is equal to zero. That is,
2We
should note that Kirchhoff’s first rule is sometimes stated in the equivalent form that the
algebraic sum of the currents toward a junction is zero. That is, currents entering the junction are
positive and currents leaving the junction are negative. In this notation, equation 24.50 would take
the equivalent form ΣI = 0.
24-37
Chapter 24 Electric Currents and DC Circuits
∆Vclosed loop = 0
(24.52)
Kirchhoff’s second rule is really another statement of the law of conservation of
energy. It says that if a positive charge is carried completely around the loop of the
circuit, in either a clockwise or counterclockwise direction, on returning to its
original point, it will have the same value of V with which it started. Hence, there is
no change in potential when the charge returns to the same place where it started
from. The mechanical analogue is that of a person standing on, let’s say, the fifth
floor of a ten-story building. The person has a certain potential energy at that point.
The person now goes for a walk up and down the stairway. If the person goes up the
stairs he increases his potential energy. If he goes down the stairs he decreases his
potential energy. But when he returns to the fifth floor, from whence he started, he
has the same potential energy that he started with.
Because charge flows from a point of high potential to one at a lower
potential, it is convenient to place a plus (+) sign on the side of a resistor that the
current enters and a negative sign (−) on the side of the resistor that the current
leaves, as shown in figure 24.20. If the charge that is carried around the circuit
traverses the loop of the circuit in the direction of the current, then as it passes the
resistor it goes from a position of higher potential to one of lower potential, and
thereby loses potential as it crosses the resistor. That is, there is a potential drop
across the resistor, which we will designate as −IR. If the carried charge passes
through a resistor in the direction that is opposite to the direction of the current,
then it is going from a position of lower potential to one of higher potential and
hence it experiences a potential rise across the resistor, which we will designate as
+IR.
In traversing a battery, if the carried charge comes out of the positive
terminal of the battery, it is at a positive potential, and this we designate as + E. If
it comes out of the negative terminal of the battery, when traversing the circuit, it
has dropped in potential and this we designate as − E. A good mechanical analogy
here would be that the battery is like an elevator in a ten-story building. When the
person enters at the ground floor, he is carried up to the fifth floor by the elevator.
The elevator has caused his potential energy to increase. The stairs are like the
resistors. When the person walks up and down the stairs from the fifth floor, he
gains or loses potential energy. If he enters the elevator at any floor along his
excursion and rides to another floor, he can gain or lose additional potential energy.
With the help of these conventions, we can now apply Kirchhoff’s second rule
to figure 24.20. Let us carry a charge around the left-hand loop in a
counterclockwise direction, starting at the positive terminal of E1. The sum of the
potential rises and drops as the loop is traversed is
E1 − I1R1 − I2R2 + E2 = 0
(24.53)
Note that E1 and E2 are positive since the charge emanates from the positive
terminals in traversing the loop counterclockwise, and that the products, I1R1 and
24-38
Chapter 24 Electric Currents and DC Circuits
I2R2, are both negative since the charge dropped in potential as it traversed these
resistors in the counterclockwise direction. The decision to traverse the loop
counterclockwise was completely arbitrary. If we wanted to traverse the loop in a
clockwise direction, the sum of the potential drops and rises would yield
− E1 − E2 + I2R2 + I1R1 = 0
(24.54)
In this case the emf’s are negative because in traversing the loop in a clockwise
direction, the carried charge emerges from the negative terminals of the batteries.
The products of the IRs across the resistors are now positive, and are called
potential rises, because in traversing the loop in a clockwise direction, the carried
charge entered the resistors at their lowest potential (−) and emerged at their
highest potential (+). No new information is obtained in this way, however, because
if we multiply each term in equation 24.54 by a negative one, we get back equation
24.53. Therefore, it does not matter whether the loop is traversed in a clockwise or
counterclockwise direction.
Recall one of the fundamental principles of algebra that in the solution of
equations there must be as many equations as there are unknowns in order to solve
for all the unknowns. At this point, there are three unknowns, the currents I1, I2,
and I3 and only two equations 24.51 and 24.53. We need one additional equation to
solve the problem. We can obtain the third equation by applying Kirchhoff’s second
rule to the right-hand loop in figure 24.20. This loop will be traversed in a
counterclockwise direction starting at the negative terminal of the battery E3. The
sum of the potential rises and drops around the right-hand loop is
− E3 − E2 + I2R2 + I3R3 = 0
(24.55)
Again note that by our convention, in traversing the loop in a counterclockwise
direction, the carried charge emerged from the negative terminals of the batteries,
and hence the emf’s are negative. The products of the IRs are positive and are
potential rises because the resistors were entered at the position of their lowest
potential and exited at the position of their highest potential.
There are now three equations in the three unknown currents, I1, I2, and I3,
and we must solve them simultaneously. It is more convenient to place the
numerical values of the emf’s and the resistors into the equations and then solve
them simultaneously. The equations are
Algebraically
I2 = I1 + I3
E1 − I1R1 − I2R2 + E2 = 0
− E3 − E2 + I2R2 + I3R3 = 0
Numerically
I2 = I1 + I3
6.00 − 10.0I1 − 20.0I2 + 12.0 = 0
− 6.00 − 12.0 + 20.0I2 + 30.0I3 = 0
24-39
(24.51)
(24.56)
(24.57)
Chapter 24 Electric Currents and DC Circuits
We will find the solution to the three equations by the method of
substitution. We eliminate the current I2 from the equations by substituting
equation 24.51 into both equations 24.56 and 24.57, that is,
6.00 − 10.0I1 − 20.0(I1 + I3) + 12.0 = 0
− 6.00 − 12.0 + 20.0(I1 + I3) + 30.0I3 = 0
Combining terms, the loop equations become
and
18.0 − 30.0I1 − 20.0I3 = 0
(24.58)
− 18.0 + 20.0I1 + 50.0I3 = 0
(24.59)
The problem is now reduced to solving two equations for the two unknowns I1
and I3. The easiest way to solve these equations is to multiply one of the equations
by a factor that will make one current term in equation 24.59 equal to a current
term in equation 24.58. Then the two equations are either added or subtracted to
eliminate the common term, leaving one equation, in one unknown current, which is
then easily solved. For example, if we multiply both sides of equation 24.59 by 1.50,
one term becomes +30.0I1, and when this equation is added to equation 24.58, the I1
term is eliminated. That is,
gives
(1.50)(− 18.0 + 20.0I1 + 50.0I3) = (1.50)(0)
−27.0 + 30.0I1 + 75.0I3 = 0
(24.60)
Adding this to equation 24.58 gives
ADD
18.0 − 30.0I1 − 20.0I3 = 0
− 27.0 + 30.0I1 + 75.0I3 = 0
−9.00 +
Solving for the current I3, we obtain
0
(24.58)
(24.60)
+ 55.0I3 = 0
55.0I3 = 9.00
I3 = 0.164 A
Substituting this value for the current I3 back into equation 24.58 allows us to solve
for the current I1. That is,
18.0 − 30.0I1 − 20.0(0.164) = 0
I1 = 0.491 A
At this point the currents I1 and I3 could be substituted back into equation 24.51 to
find the current I2. However, we will not do this. Instead we will evaluate the
current I2 from one of the loop equations and use equation 24.51 as a check on our
24-40
Chapter 24 Electric Currents and DC Circuits
arithmetic. The reason for doing this check is that it is very easy to make a mistake
in the algebra and/or the arithmetic in the problem.
Substituting the value of I1 just found, back into equation 24.56, yields
Solving for I2,
6.00 − 10.0(0.491) − 20.0I2 + 12.0 = 0
I2 = 0.655 A
As a check, place the values of the currents just found back into equation 24.51:
Thus,
I2 = I1 + I3
0.655 A = 0.491 A + 0.164 A = 0.655 A
0.655 A = 0.655 A
CHECK
The current equation thus acts as a good check on our calculations. The problem is
now solved, since we know the three currents.
It is interesting and informative to plot the change in potential that a
positive charge would encounter as it traversed each battery and resistor as it
moves around the loop. Considering the first loop
E1 − I1R1 − I2R2 + E2 = 0
6.00 − 10.0I1 − 20.0I2 + 12.0 = 0
6.00 − 10.0(0.491) − 20.0(0.655) + 12.0 = 0
6.00 V − 4.91 V − 13.09 V + 12.0 V = 0
(24.53)
These terms are plotted in figure 24.21. Emerging from the positive terminal of the
Figure 24.21 Plot of the potential drops and rises as the left loop of figure 24.20 is
traversed.
24-41
Chapter 24 Electric Currents and DC Circuits
first battery the positive charge is at the potential E1, which is +6.00 V above
ground, the zero of potential. As the first resistor is traversed, there is a potential
drop VR1 of 4.91 V, leaving the charge at a potential of 6.00 V − 4.91 V = 1.09 V
above ground. This is the value of the potential as the charge enters resistor R2.
Across R2, the charge experiences a potential drop of 13.09 V. Thus, on leaving R2
the potential is at a negative value of 1.09 V − 13.09 V = −12.0 V. At that point, the
charge enters the negative terminal of battery 2 at a potential of −12 V. As battery 2
is traversed, chemical energy is converted into electrical energy until, when the
charge emerges from the positive terminal of the battery, it is 12.0 V higher in
potential then when it entered and it is now at a zero potential. The charge now
enters the negative terminal of battery 1, − E1, at a 0.00 V potential and is raised
6.00 V by chemical means as it emerges from the positive terminal of E1. The charge
is now back to where it started. As we can see from figure 24.21, the net change in
potential as the charge traversed the closed loop is zero.
A plot of the potential drops and rises experienced by a charge as it traverses
the right-hand loop is shown in figure 24.22. The loop equation is
− E3 − E2 + I2R2 + I3R3 = 0
− 6.00 − 12.00 + 20.0I2 + 30.0I3 = 0
− 6.00 V − 12.00 V + 13.09 V + 4.91 V = 0
(24.55)
Figure 24.22 Plot of potential drops and rises around the right hand loop of figure
24.20.
Starting at the negative terminal of battery 3, the charge is 6.00 V below ground.
The charge now enters the positive terminal of battery 2. When it emerges from the
negative terminal of battery 2 it is 12.0 V more negative, or 18.0 V below ground. As
24-42
Chapter 24 Electric Currents and DC Circuits
the charge enters resistor R2 it is at the end of the resistor’s lowest potential, and
when it emerges at the other end of the resistor it is at the resistor’s highest
potential. The charge, therefore, experienced a potential rise of 13.09 V, leaving it at
a potential of −18.0 V + 13.09 V = −4.91 V. The charge now enters resistor R3, where
it experiences a potential rise of +4.91 V, leaving it at the zero of potential and the
positive terminal of battery E3. As the battery is traversed a potential drop of 6.00 V
is experienced as the charge emerges from the negative terminal of E3. The charge
is back where it started and, as we can see from figure 24.22, the total change in
potential in traversing the closed loop is zero. It is not necessary to draw these
potential diagrams for every Kirchhoff’s rules problem. The student should,
however, draw at least one potential diagram to help in the understanding of
Kirchhoff’s second rule.
The Language of Physics
Electric current
Electric current is defined as the amount of electric charge that flows through a
cross section of a wire per unit time (p. ).
Ampere
The SI unit of current that is equal to a flow of one coulomb of charge per second (p.
).
Conventional current
A flow of positive charges in a circuit from a position of high potential to one of low
potential (p. ).
Electron current
The actual current in a circuit; it is a flow of electrons from a position of low
potential to one of high potential (p. ).
Direct current
An electric current in which the electric charges flow in only one direction in a
circuit (p. ).
Ohm’s law
For metallic conductors, the current in a circuit is directly proportional to the
applied potential difference and inversely proportional to the resistance in the
circuit (p. ).
Joule heat
The energy that is lost as a charge ‘’falls’’ through a resistor shows up as heat in the
resistor (p. ).
24-43
Chapter 24 Electric Currents and DC Circuits
Series circuit
A circuit in which each element of the circuit is connected to an adjacent element of
the circuit such that the same amount of charge flows through each and every
circuit element. For a resistive circuit, the current is the same through each resistor
(p.).
Equivalent resistance
A single resistor, whose value is equal to the combined resistance of the individual
resistors in the circuit. For resistors in series, the equivalent resistance is equal to
the sum of the individual resistances. For resistors in parallel, the reciprocal of the
equivalent resistance is equal to the sum of the reciprocals of the individual
resistances (p. ).
Parallel circuit
A circuit in which the circuit elements are connected in such a way that the
potential difference across all the elements of the circuit is the same. For a resistive
circuit, the potential difference across the resistors is equal (p. ).
Law of conservation of electric charge
Electric charge can neither be created nor destroyed (p. ).
Electromotive force (emf)
A potential difference that is supplied by a battery. If the internal resistance of the
battery is relatively small, then the emf is equal to the terminal voltage of the
battery (p. ).
Galvanometer
A device that indicates that there is a current in a circuit. By appropriate
construction, a galvanometer can be made into an ammeter or a voltmeter (p. ).
Wheatstone bridge
An electric circuit that is used to measure the value of an unknown resistor very
accurately (p. ).
Kirchhoff’s first rule
The sum of the currents in a circuit entering a junction is equal to the sum of the
currents leaving the junction (p. ).
Kirchhoff’s second rule
The change in potential around a closed loop of a circuit is equal to zero. This can
also be stated in the form, the sum of the potential rises and potential drops around
any closed loop of a circuit is equal to zero (p. ).
24-44
Chapter 24 Electric Currents and DC Circuits
Summary of Important Equations
Current
Ohm’s law
Resistance of a wire
Work done within a battery
I= q
t
I= V
R
R=ρl
A
W = qV
(24.13)
P = IV
(24.15)
Power
Power
dissipated
in
a
(24.1)
(24.3)
(24.7)
P =I2R
resistor
(24.18)
Resistors in series
Resistors in parallel
V = V1 + V2 + V3
I = I1 = I2 = I3
R = R1 + R2 + R3
P = P1 + P2 + P3
(24.22)
V = V1 = V2 = V3
I = I1 + I2 + I3
1 = 1 + 1 + 1
R R1 R2 R3
P = P1 + P2 + P3
(24.27)
(24.28)
(24.35)
V = E − Ir
(24.37)
Terminal voltage of a battery
(24.26)
E = E1 + E2 + E3 + … + En
Batteries in series
(24.38)
Identical batteries in parallel
E = E1 = E2 = E3
I = I1 + I2 + I3
(24.39)
(24.40)
Wheatstone bridge
R2 = R4 R1
R3
junction = Σ Ileaving
(24.49)
Kirchhoff’s first rule
Kirchhoff’s second rule
Σ Ientering
∆Vclosed loop = 0
24-45
junction
(24.50)
(24.52)
Chapter 24 Electric Currents and DC Circuits
Questions for Chapter 24
1. Can the flow of electric charge in a circuit be compared to the flow of a
fluid in a pipe? Point out where it is a good analogy and where it is a bad analogy.
2. Can a resistor be used as a thermometer?
*3. If a metal consists of positive ions surrounded by a sea of electrons, how
does the metal stay together?
4. If the drift velocity of an electron in a wire is so small that it takes almost
30 s for it to move about 1 cm, why is no delay observed on an ammeter when a
battery is connected to a circuit?
5. Describe the flow of the positive charges of a conventional current with the
flow of electrons in an electron current in a circuit. What are the advantages and
disadvantages of using conventional current?
6. Why would you not want to hook up the simple circuit shown in figure
24.3? Estimate the value of the resistance of the wire. Using the concept of Ohm’s
law, would a very large current exist? Would this be called ‘’shorting’’ the battery?
7. Compare a positive charge falling through a potential difference with a
mass falling in a gravitational field. Why is this a pretty good analogy? Try to make
a similar analogy with the motion of an electron in a circuit.
*8. A black box is a device that is contained in a closed box, so that you
cannot see inside the box to see what the device is. There is an input wire to the box
and an output wire from the box. All you can do is to infer characteristics about the
device from the information obtained from the input and output connections. A
range of different potentials is applied to this black box, and the current coming out
of the box is measured. A plot of the potential versus the current is a curve rather
than a straight line. What are some of the characteristics of this black box?
9. A copper block is 35 cm long, 2 cm thick, and 2 cm wide. Does the block
have the same resistance from end to end as it does from side to side? Does it
depend in any way on the size of the connection to the block?
*10. In the 1970s when copper became very expensive, some new homes were
built with aluminum wires rather than copper wires. If the aluminum wire is
connected to a copper outlet, and the coefficient of linear expansion for aluminum
and copper are different, what hazard could this produce?
11. At very low temperatures some materials become superconductors, that
is, their electrical resistance becomes essentially zero. What are the advantages and
disadvantages of superconductors?
12. How would you hook up three resistors in a circuit to get (a) the
maximum current and (b) the minimum current?
Problems for Chapter 24
24.1 Electric Current
24-46
Chapter 24 Electric Currents and DC Circuits
1. How many electrons are associated with a current of 5.00 A?
2. A wire carries a current of 7.50 A for a period of 30.0 min. How much
charge flows through a cross section of the wire? How many electrons does this
represent?
3. Suppose that the region between two oppositely charged parallel plates is
occupied by both singly charged positive ions and by electrons. If 5 × 106 electrons
flow from the negative plate to the positive plate in 4.00 s, while 6 × 106 ions flow
from the positive plate to the negative plate in the same time, find the value of the
net current flowing from the positive plate to the negative plate.
24.2 Ohm’s Law
4. A 500-Ω resistor is connected to a 12.0-V battery. Find the current through
the resistor.
5. What value of resistance is necessary to get a current of 2.50 A when it is
connected to a 120-V source?
6. If a current of 8.75 mA passes through a resistor of 550 Ω, find the voltage
drop across the resistor.
24.3 Resistivity
7. Find the resistance of a 100-m spool of #22 B&S gauge copper wire
(diameter 6.44 × 10−4 m).
8. The connecting wires in an electrical experiment are 1.00 m long and are
made of copper. Their diameter is 6.44 × 10−4 m. What is the resistance of the
connecting wires? Is it reasonable to neglect their effect in analyzing a circuit?
9. Does doubling both the length and the diameter of a wire have any effect
on its resistance?
10. What length of #22 B&S gauge copper wire is needed to make a resistance
of 500 Ω?
11. Find the resistance of an aluminum bar 1.00 m long, 2.00 cm wide, and
1.00 cm high from (a) end to end and (b) from the 1.00-cm side to the other 1.00-cm
side.
12. A wire 2.00 m long and 1.00 mm in diameter has resistance of 0.500 Ω. A
second wire 4.00 m long and 2.00 mm in diameter has twice the resistivity of the
first wire. Find the resistance of the second wire.
24.4 The Variation of Resistance with Temperature
13. If the resistance of a resistor is 20.00 Ω at 20.00 0C, find its resistance at
200.00 0C. The temperature coefficient of resistance α for copper is 3.93 × 10−3 /0C.
14. If the resistance of a copper wire is 500 Ω at 20 0C, find its resistance at
100 0C.
15. A copper wire has a resistance of 20.00 Ω at room temperature. The wire
is placed in an oven and the resistance is then measured as 30.0 Ω. Find the
temperature of the oven.
24-47
Chapter 24 Electric Currents and DC Circuits
16. The resistance of a copper wire changes from 200 Ω to 210 Ω. What was
the change in temperature of the wire?
17. A copper resistor has a resistance of 500 Ω at a temperature of 20.0 0C.
What is the resistance if the temperature doubles?
24.5 Conservation of Energy and the Electric Circuit -- Power Expended in
a Circuit
18. An electric toaster is rated at 1200 W. If it is connected to a 120-V line,
what current does it draw? What is the resistance of the toaster? How much energy
is used if the toaster is ‘’on’’ for 1.00 min?
19. A stereo amplifier is rated at 60.0 W. If it is connected to a 120-V line,
how much current does it draw? If the stereo is ‘’on’’ for 5 hr, how much energy is
used?
20. A 60.0-W light bulb is connected to a 120-V outlet. What is the current
through the bulb? How many electrons flow through the bulb per second?
21. A power line carries 1000 A and has a resistance of 20.0 Ω. How much
energy is lost per second in terms of Joule heat?
22. How much power is consumed in starting a car if 200 A is drawn from a
12.0-V battery?
24.6 Resistors in Series
23. Find the equivalent resistance of three resistors of 100, 200, and 300 Ω
when connected in series.
24. Find the equivalent resistance of four resistors of 250, 400, 186, and 375
Ω when connected in series.
24.7 Resistors in Parallel
25. Find the equivalent resistance of three resistors of 100, 200, and 300 Ω
when connected in parallel.
26. Find the equivalent resistance of four resistors of 250, 400, 186, and 375
Ω when connected in parallel.
24.8 Combinations of Resistors in Series and Parallel
27. Find the equivalent resistance of the circuit in the diagram.
Diagram for problem 27.
28. Find the equivalent resistance of the circuit in the diagram if R1 = 30.0 Ω,
R2 = 40.0 Ω, R3 = 50.0 Ω, R4 = 70.0 Ω, R5 = 300 Ω, R6 = 200 Ω, and R7 = 100 Ω.
24-48
Chapter 24 Electric Currents and DC Circuits
Diagram for problem 28.
Diagram for problem 29.
*29. In the diagram find (a) the equivalent resistance of the circuit, (b) the
current flowing from the battery, (c) the voltage drop across R1, (d) the voltage drop
across R2 and R3, (e) the current through each resistor, (f) the power supplied to the
circuit, and (g) the power dissipated in each resistor. The emf of the battery is 12.0
V.
30. Find the current through each resistor in the diagram.
Diagram for problem 30.
Diagram for problem 31.
*31. In the diagram, find (a) the equivalent resistance of the resistors in
parallel, (b) the equivalent resistance of the circuit, (c) the current from the battery,
(d) the voltage drop across R1, (e) the voltage drop across R2, R3, and R4, (f) the
current through each resistor, and (g) the power dissipated in each resistor.
32. Find the voltage drop across R3 in the diagram.
33. Find (a) the equivalent resistance of the circuit shown in the diagram if
R1 = 50.0 Ω, R2 = 80.0 Ω, R3 = 150 Ω, R4 = 30.0 Ω, R5 = 200 Ω, and R6 = 300 Ω and
(b) the current through each resistor.
Diagram for problem 32.
Diagram for problem 33.
24-49
Chapter 24 Electric Currents and DC Circuits
*34. In the diagram find (a) the equivalent resistance, (b) the current from
the battery, (c) the current through each resistor, (d) the voltage drop across each
resistor, (e) the power supplied to the circuit, and (f) the power dissipated in each
resistor.
Diagram for problem 34.
24.9 The Electromotive Force and the Internal Resistance of a Battery
35. A 6.00-V battery is connected to a 100-Ω resistor. A voltmeter placed
across the resistor measures 5.60 V. Find the internal resistance of the battery.
36. A 12-V battery has an internal resistance of 5.5 Ω. Find the voltage
applied to a circuit of 50 Ω.
37. Six 1.50-V batteries are connected in series, with the positive terminal of
one battery connected to the negative terminal of the next. What is the emf of the
combination? If the internal resistance of each battery is 2.00 Ω, find the terminal
voltage when there is a current of 200 mA in the circuit.
38. Six 1.50-V batteries are connected in series to a resistance of 50.0 Ω. Find
the current in the circuit (a) neglecting the internal resistance of the battery and (b)
taking the 2.00 Ω resistance of each battery into account.
39. Three 12-V batteries are connected in series to a 35-Ω resistor. If the
internal resistance of each battery is 10 Ω, 20 Ω, and 30 Ω, respectively, find the
voltage drop across the 35-Ω resistor.
40. When a battery is connected to an external resistance of 10.0 Ω, the
current through the external resistance is 0.100 A. When the same battery is
connected to an external resistance of 5 Ω, the current is 0.150 A. Find the emf of
the battery and the internal resistance of the battery.
24.10 The Wheatstone Bridge
41. The voltmeter reads 12.0 V and the ammeter reads a current of 5.45 ×
10−2 A in the circuit shown. The ammeter has a resistance of 20.0 Ω and the
voltmeter has a resistance of 5000 Ω. Taking the resistance of the ammeter and
voltmeter into account, find the value of the resistor R.
42. Find the value of R2 in the diagram if the bridge is balanced.
24-50
Chapter 24 Electric Currents and DC Circuits
Diagram for problem 41.
Diagram for problem 42.
43. Find the equivalent resistance of the Wheatstone bridge in problem 42
when the bridge is balanced. Find the current through each resistor.
24.11 Kirchhoff’s Rules
*44. In the accompanying diagram (a) find the current through each resistor
and (b) plot the potential as you traverse each loop.
Diagram for problem 44.
Diagram for problem 45.
*45. Two batteries, with emf’s and internal resistances shown, are connected
in parallel. Using Kirchhoff’s rules, find the current through R and the voltage drop
across R.
*46. Find the current through each resistor in the diagram.
*47. Using Kirchhoff’s rules, find the current through each resistor in the
diagram.
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Chapter 24 Electric Currents and DC Circuits
Diagram for problem 46.
Diagram for problem 47.
48. Three 12.0-V batteries with different internal resistances are connected
in parallel as shown in the diagram. If r1 = 10.0 Ω, r2 = 20.0 Ω, and r3 = 30.0 Ω, find
the voltage drop across AB.
Diagram for problem 48.
Additional Problems
*49. Find the potential difference across AB in the accompanying diagram if
R1 = 10.0 Ω, R2 = 20.0 Ω, R3 = 30.0 Ω, and the current through R3 is 0.300 A.
50. Find the voltage drop across R2 if R1 = 20.0 Ω, R2 = 30.0 Ω, R3 = 50.0 Ω,
and the emf is 6.00 V in the diagram.
Diagram for problem 49.
Diagram for problem 50.
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Chapter 24 Electric Currents and DC Circuits
51. Find the current through each of the seven resistors, and the potential
drop across each resistor in the circuit in the diagram.
Diagram for problem 51.
Diagram for problem 52.
*52. Find the equivalent resistance of the circuit shown in the diagram if R1 =
10.0 Ω, R2 = 20.0 Ω, R3 = 30.0 Ω, R4 = 40.0 Ω, R5 = 50.0 Ω, R6 = 60.0 Ω, and R7 = 70.0
Ω.
*53. If R1 = 10.0 Ω, R2 = 20.0 Ω, R3 = 30.0 Ω, and R4 = 40.0 Ω in the circuit
shown, find (a) the equivalent resistance of the circuit and, if E1 = 12.0 V, (b) the
current through each resistor.
Diagram for problem 53.
54. (a) What is the resistance of 100 resistors of 10.0 Ω each if they are all in
series? (b) What is the resistance of 100 resistors of 10.0 Ω each if they are all in
parallel?
*55. A 200-W immersion heater is placed in a beaker of water containing
0.500 liters at room temperature. How long will it take for the water to boil?
*56. An electrically insulated coil of wire is immersed in 100.0 g of water in a
calorimeter cup at room temperature. The wire is connected to a 120-V source and a
current of 2.00 A is observed in the wire. Find the temperature of the water after
60.0 s.
*57. A voltage divider is shown in the diagram. The resistor R is a variable
slide wire resistor. Show that by sliding the arrow contact in the figure along the
slide wire any fraction of the original applied voltage can be obtained. In particular
show that the voltage out of the divider is given by
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Chapter 24 Electric Currents and DC Circuits
Vout = lout Vin
lin
where lin is the original length of the wire resistor and lout is the shorter length of
the wire resistor that the output connector is attached to.
*58. The Wheatstone bridge shown in the diagram is not balanced. Using
Kirchhoff’s rules find the current through each resistor if R1 = 10.0 Ω, R2 = 20.0 Ω,
R3 = 30.0 Ω, R4 = 40.0 Ω, Rg = 50.0 Ω, and E1 = 6.00 V. Find the equivalent
resistance of the circuit.
Diagram for problem 57.
Diagram for problem 58.
*59. In problem 58, find the potential differences across points (a) AC, (b) AD,
(c) CD, (d) CB, and (e) DB.
Interactive Tutorials
60. Resistors in series. Three resistors, R1 = 25.0 Ω, R2 = 45.5 Ω, and R3 = 83.5
Ω, are connected in series to a 50.0-V battery. Find (a) the equivalent resistance R,
(b) the current coming from the battery, (c) the current through each resistor,
(d) the voltage drop across each resistor, (e) the power lost across each resistor, and
(f) the power supplied to the circuit.
61. Resistors in parallel. Three resistors, R1 = 25.0 Ω, R2 = 45.5 Ω, and R3 =
83.5 Ω, are connected in parallel to a 50.0-V battery. Find (a) the equivalent
resistance R, (b) the current coming from the battery, (c) the voltage drop across
each resistor, (d) the current through each resistor, (e) the power lost across each
resistor, and (f) the power supplied to the circuit.
62. Combination of resistors in series and parallel. Resistor R1 = 25.0 Ω, is in
series with the parallel resistors R2 = 45.5 Ω and R3 = 83.5 Ω, as in figure 24.14(a).
The resistors are connected to a 50.0-V battery. Find (a) the equivalent resistance
R, (b) the current coming from the battery, (c) the voltage drop across each resistor,
(d) the current through each resistor, (e) the power lost across each resistor, and
(f) the power supplied to the circuit.
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Chapter 24 Electric Currents and DC Circuits
63. Kirchhoff’s Rules. Three resistors R1 = 25.0 Ω, R2 = 45.5 Ω, and R3 = 83.5
Ω are connected to three batteries E1 = 6 V, E2 = 9 V, E3 = 12 V, as shown in the
diagram. Using determinants, find the current I1, I2, I3 through each resistor.
Assume that the currents are in the direction indicated. If a current is actually in
the opposite direction, the answer will give a negative value for the current.
Diagram for problem 63.
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