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

r = v

0

t +

1

at

2

2

and

v = v

0

+ 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 v

d

, the drift velocity, in the expected

24-2

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

q

I

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 × 10

18

electronic charges, the

ampere is the flow of 6.242 × 10

18

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.

24-3

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 V

1

, a current I

1

is

recorded by the ammeter. When a larger battery of potential difference V

2

replaces

battery V

1

, a larger current I

2

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,

24-5

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

V

I

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

24-6

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

To go to this Interactive Example click on this sentence.

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?

The resistance of the circuit, found from the rearrangement of Ohm’s law, is

Solution

R =

V

=

12.0 V

= 40.0 Ω

I 0.300 A

To go to this Interactive Example click on this sentence.

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,

24-7

Chapter 24 Electric Currents and DC Circuits

R ∝

1

(24.5)

A

We can combine proportionalities 24.4 and 24.5 into the one proportionality

R ∝

l

(24.6)

A

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

l

R

A

ρ

=

(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

0

C

and Mean Temperature Coefficient of Resistivity (α)

Material

ρ, Ω m

α

0

C

−1

Aluminum

Brass

Carbon (graphite)

Copper

Gold

Iron

Lead

Mercury

Platinum

Silver

Tungsten

Amber

Bakelite

Glass

Hard Rubber

Mica

Wood

2.82 × 10

−8

7.00 × 10

−8

3500 × 10

−8

1.72 × 10

−8

2.44 × 10

−8

9.71 × 10

−8

20.6 × 10

−8

98.4 × 10

−8

10.6 × 10

−8

1.59 × 10

−8

5.51 × 10

−8

5.00 × 10

14

2 × 10

5

- 2 × 10

14

10

13

- 10

14

10

13

- 10

16

10

11

- 10

15

10

8

- 10

11

3.9 × 10

−3

2.00 × 10

−3

−0.5 × 10

−3

3.93 × 10

−3

3.40 × 10

−3

5.20 × 10

−3

4.30 × 10

−3

8.90 × 10

−3

3.90 × 10

−3

3.80 × 10

−3

4.50 × 10

−3

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 10

13

-10

15

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.

The cross-sectional area of the wire is given by

Solution

A =

πd

2

4

(

)

2

3

0.644 10 m

4

π

−

×

=

= 3.26 × 10

−7

m

2

The resistance of the wire spool, found from equation 24.7, is

R = ρ

l

A

( )

×

Ω 10 ×=

−

−

27

8

m 103.26

m 500

m 1.72

= 26.4 Ω

To go to this Interactive Example click on this sentence.

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.

The resistance from one end of the rod to the other, found from equation 24.7, is

Solution

( )

(

)

( )( )

14

0.300 m

5.00 m

0.0200 m 0.0250 m

l

R

A

ρ= = × 10 Ω

= 3.00

×

10

17

Ω

To go to this Interactive Example click on this sentence.

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 − ρ

0

∆

t

t

−

t

0

where ρ is the resistivity of the material at the temperature

t

and ρ

0

is the

resistivity of the material at the temperature

t

0

. Rearranging the equation we get

ρ − ρ

0

=

m

(

t

−

t

0

)

Solving for the resistivity ρ, gives

ρ = ρ

0

[1 +

m

(

t

−

t

0

)] (24.8)

ρ

0

Let us now define a new constant α, called the

mean temperature coefficient of

resistivity,

as

α =

m

ρ

0

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 ρ

0

at the reference temperature

t

0

, 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

0

C

−

1

. With this new temperature coefficient of resistivity, we

can write equation 24.8 as

ρ = ρ

0

[1 + α (

t

−

t

0

)] (24.9)

Equation 24.9 gives the resistivity of a material at the temperature t when the

resistivity of the material ρ

0

is known at the reference temperature t

0

.

Example 24.5

Temperature dependence of resistivity.

The resistivity of copper at 20.0

0

C is 1.72

×

10

−8

Ω m. Find its resistivity at 200

0

C.

24-11

Chapter 24 Electric Currents and DC Circuits

The temperature coefficient of resistivity for copper, found from table 24.1, is 3.93

×

10

−3

0

C

−

1

. The resistivity of copper at 200

0

C, found from equation 24.9, is

Solution

ρ = ρ

0

[1 + α (

t

−

t

0

)]

= (1.72

×

10

−8

Ω m)[1 + (3.93

×

10

−3

0

C

−

1

)(200 − 20.0)

0

C ]

= 2.94

×

10

−8

Ω m

To go to this Interactive Example click on this sentence.

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

= ρ

0

l

[1 + α(

t

−

t

0

)] (24.10)

A

A

However, ρ

l

/

A

is equal to the resistance

R

from equation 24.7. Therefore, we can

write equation 24.10 as

(

)

0

1R R t tα

0

= + −

(24.11)

Equation 24.11 gives the resistance of a resistor

R

, at the temperature

t

, if the

resistance

R

0

, at the reference temperature

t

0

, is known.

Example 24.6

Temperature dependence of resistance.

If the resistance of a copper resistor is 50.0 Ω

at 20.0

0

C, find its resistance at 200

0

C.

The temperature coefficient of resistivity for copper, found from table 24.1, is 3.93

×

10

−3

0

C

−1

. The resistance of the copper resistor at 200

0

C, found from equation 24.11,

is

Solution

R

=

R

0

[1 + α(

t

−

t

0

)]

= (50.0 Ω)[1 + (3.93

×

10

−3

0

C

−1

)(200 − 20.0)

0

C]

= 85.4 Ω

To go to this Interactive Example click on this sentence.

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

T

c

, 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

I

2

R

. 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

(24.12)

t

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

P

=

W

=

qV

(24.14)

t

t

But

I

=

q

(24.1)

t

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 (24.17)

24-14

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

)

2

P

I R=

(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

=

V

2

(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

=

(24.20)

24-15

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?

a.

Because the power rating of the bulb is known, we can find the current from

equation 24.15 as

Solution

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

=

I

2

R

= (0.500 A)

2

(240 Ω)

= 60.0 W

Note that the power supplied is equal to the power dissipated as expected.

To go to this Interactive Example click on this sentence.

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?

We find the current from

Solution

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

To go to this Interactive Example click on this sentence.

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 PE

1

. 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 PE

2

. 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 PE

3

. 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,

PE

top

= PE

1

+ PE

2

+ PE

3

(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

V

1

. As it falls

through the second resistor it drops in potential by

V

2

. The charge experiences

another potential drop,

V

3

, as it falls through

R

3

.

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

=

V

1

+

V

2

+

V

3

(24.22)

The voltage drop across each resistor in figure 24.11, given by Ohm’s law, equation

24.20, is

V

1

=

IR

1

V

2

=

IR

2

V

3

=

IR

3

Replacing these equations into equation 24.22 gives

V

=

IR

1

+

IR

2

+

IR

3

(24.23)

24-18

Chapter 24 Electric Currents and DC Circuits

Dividing both sides of equation 24.23 by

I

gives

V

=

R

1

+

R

2

+

R

3

(24.24)

I

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

1 2

R R R R

3

=

+ +

(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

R

1

,

R

2

, and

R

3

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

l

1

,

l

2

, and

l

3

, respectively, then the length of the wire when the

three resistors are connected in series is just

l

=

l

1

+

l

2

+

l

3

24-19

Chapter 24 Electric Currents and DC Circuits

Multiplying each term by ρ/

A

gives

31 2

ll l

l

A A A

A

ρ

ρ ρ ρ= + +

But using equation 24.7,

R

= ρ

l

/

A

gives

R

=

R

1

+

R

2

+

R

3

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

R

1

= 20.0 Ω,

R

2

= 30.0 Ω, and

R

3

= 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.

The equivalent resistance, given by equation 24.26, is

Solution

R

=

R

1

+

R

2

+

R

3

= 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.

The power supplied by the battery is

Solution

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

P

1

=

I

2

R

1

= (0.0667 A)

2

(20.0 Ω) = 0.0890 W

P

2

=

I

2

R

2

= (0.0667 A)

2

(30.0 Ω) = 0.133 W

P

3

=

I

2

R

3

= (0.0667 A)

2

(40.0 Ω) = 0.178 W

The total power dissipated in all resistors is

P

=

P

1

+

P

2

+

P

3

= 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.

The potential drop across each resistor, found by Ohm’s law, is

Solution

V

1

=

IR

1

= (0.0667 A)(20.0 Ω) = 1.33 V

V

2

=

IR

2

= (0.0667 A)(30.0 Ω) = 2.00 V

V

3

=

IR

3

= (0.0667 A)(40.0 Ω) = 2.67 V

Notice that the sum of the potential drops,

V

1

+

V

2

+

V

3

, 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

=

V

1

=

V

2

=

V

3

(24.27)

When the total current

I

from the battery reaches the junction

A

, it divides into

three parts;

I

1

goes through resistor

R

1

,

I

2

goes through resistor

R

2

, and

I

3

goes

through

R

3

. Because none of the charge disappears,

I

=

I

1

+

I

2

+

I

3

(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,

I

1

=

V

1

(24.29)

R

1

24-22

Chapter 24 Electric Currents and DC Circuits

I

2

=

V

2

(24.30)

R

2

I

3

=

V

3

(24.31)

R

3

Replacing these values of the currents back into the current equation, 24.28, gives

I

=

V

1

+

V

2

+

V

3

(24.32)

R

1

R

2

R

3

But, the potentials are equal by equation 24.27 (i.e.,

V

=

V

1

=

V

2

=

V

3

). Hence,

equation 24.32 becomes

I

=

V

+

V

+

V

R

1

R

2

R

3

Dividing both sides of the equation by

V

, gives

I

=

1

+

1

+

1

(24.33)

V R

1

R

2

R

3

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

(24.34)

V

R

where

R

is the total resistance of the entire circuit. Equating 24.34 to equation

24.33 gives

1 2

1 1 1 1

R R R R

= + +

3

(24.35)

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

R

1

,

R

2

, and

R

3

, 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 2 3

1 1 1 1 1

...

n

R R R R R

= + + + +

(24.36)

24-23

Chapter 24 Electric Currents and DC Circuits

Example 24.12

Equivalent resistance of resistors in parallel.

Three resistors

R

1

= 20.0 Ω,

R

2

= 30.0

Ω, and

R

3

= 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.

From equation 24.35 the equivalent resistance is

Solution

1

=

1

+

1

+

1

R R

1

R

2

R

3

=

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.

The total current in the circuit, found from Ohm’s law with

R

as the equivalent

resistance of the circuit, is

Solution

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.

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,

Solution

I

1

=

V

=

6.00 V

= 0.300 A

R

1

20.0 Ω

I

2

=

V

=

6.00 V

= 0.200 A

R

2

30.0 Ω

I

3

=

V

=

6.00 V

= 0.150 A

R

3

40.0 Ω

Note that the current through each resistor is different, but the sum of

I

1

+

I

2

+

I

3

=

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

The power supplied by the battery is

Solution

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?

The power dissipated in each resistor is

Solution

P

1

=

I

1

2

R

1

= (0.300 A)

2

(20.0 Ω) = 1.80 W

P

2

=

I

2

2

R

2

= (0.200 A)

2

(30.0 Ω) = 1.20 W

P

3

=

I

3

2

R

3

= (0.150 A)

2

(40.0 Ω) = 0.900 W

Again note that the sum of the powers dissipated in each resistor

P

1

+

P

2

+

P

3

= 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 round-

off 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

R

2

and

R

3

are connected in parallel. Their equivalent

resistance

R

23

, 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

R

23

R

2

R

3

=

1

+

1

=

0.0583

30.0 Ω 40.0 Ω Ω

R

23

= 17.2 Ω

The circuit of figure 24.14(a) can be replaced by its equivalent circuit, figure

24.14(b), where

R

23

is in series with

R

1

. The equivalent resistance of

R

1

and

R

23

in

series, found from equation 24.26, is

R

123

=

R

1

+

R

23

= 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

R

123

= 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

R

123

37.2 Ω

Since the resistor

R

1

is in series with the battery, this same current flows through it

(i.e.,

I

1

=

I

= 0.161 A). However, this is not the current through

R

2

and

R

3

because

they are in parallel and the current divides as it enters the two paths. In order to

determine

I

2

and

I

3

we must first determine the voltage drop across

R

2

and

R

3

.

The voltage drop across

R

1

, found from Ohm’s law, is

V

1

=

I

1

R

1

= (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

=

V

1

+

V

2

The potential drop

V

2

across the parallel resistors is

V

2

=

V

−

V

1

= 6.00 V − 3.22 V

= 2.78 V

The potential drop

V

3

is also equal to 2.78 V since

R

2

and

R

3

are connected in

parallel. With the potential drop across the parallel resistors known, the current in

each resistor, found from Ohm’s law, is

I

2

=

V

2

=

2.78 V

= 0.0927 A

R

2

30.0 Ω

I

3

=

V

3

=

2.78 V

= 0.0695 A

R

3

40.0 Ω

Note that the sum of the currents

I

2

+

I

3

= 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

P

1

=

I

1

2

R

1

= (0.161 A)

2

(20.0 Ω) = 0.518 W

P

2

=

I

2

2

R

2

= (0.0927 A)

2

(30.0 Ω) = 0.258 W

P

3

=

I

3

2

R

3

= (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 Force and the Internal

Resistance of a Battery

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

The potential difference across the battery, given by equation 24.37, is

Solution

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.

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

Solution

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

=

E

1

+

E

2

+

E

3

+ … +

E

n

(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

=

E

1

=

E

2

=

E

3

(24.39)

At junction

A

, the three battery currents combine to give the circuit current

I

=

I

1

+

I

2

+

I

3

(24.40)

Because we assume the batteries to be identical (

I

1

=

I

2

=

I

3

), 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.

The total current

I

in the circuit, found from Ohm’s law, is

Solution

I

=

E

=

6.00 V

= 0.0120 A

R

500 Ω

This total current is supplied by three batteries so that,

I

= 3

I

0

, where

I

0

is the

actual current from any one battery. Therefore, the battery current is

I

0

=

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

(24.41)

I

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

V

A

across the ammeter, and the voltmeter reads the voltage drop

V

=

V

R

+

V

A

The calculated resistance is

R

V V

R

I

+

=

A

(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

V

R

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

I

R

flows

through the resistor while a current

I

v

flows through the voltmeter. The ammeter

reads the current

I

=

I

R

+

I

v

The computed resistance

R

therefore becomes

R

R V

V

R

I I

=

+

(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

R

1

,

R

3

, and

R

4

, in the circuit are assumed to be known, whereas

R

2

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,

I

1

and

I

3

.

I

=

I

1

+

I

3

When current

I

1

reaches the junction

C

it splits into two currents,

I

2

through

R

2

and

I

g

the current through the galvanometer. That is,

I

1

=

I

2

+

I

g

(24.44)

The galvanometer needle deflects indicating that there is a current in it. At junction

D

the galvanometer current

I

g

combines with current

I

3

to form current

I

4

, which

passes through resistor

R

4

. That is,

I

3

+

I

g

=

I

4

(24.45)

At junction

B

, currents

I

2

and

I

4

combine to form the original current

I

.

A current

I

g

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

R

1

,

R

3

, and

R

4

are varied until the galvanometer reads zero (i.e.,

I

g

= 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

R

1

is now in parallel with

R

3

and

therefore the voltages across them are equal, that is,

V

1

=

V

3

Therefore, from Ohm’s law,

I

1

R

1

=

I

3

R

3

(24.46)

It is also obvious that

R

2

is now in parallel with

R

4

and hence the voltages across

them are also equal, that is,

V

2

=

V

4

I

2

R

2

=

I

4

R

4

(24.47)

Dividing equation 24.47 by equation 24.46, we obtain

24-35

Chapter 24 Electric Currents and DC Circuits

2 2 4 4

1 1 3 3

I R I R

I R I R

=

(24.48)

However, since the bridge is balanced,

I

g

= 0, and equation 24.44 reduces to

I

1

=

I

2

and equation 24.45 to

I

3

=

I

4

Therefore, the currents in equation 24.48 cancel, leaving the simple ratio of the

resistors

2 4

1 3

R R

R R

=

Solving for the unknown resistor

R

2

, we obtain

4

2

3

R

R

R

=

1

R

(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

R

1

= 5.00 Ω,

R

3

=

10.0 Ω, and

R

4

= 4.00 Ω. Find the unknown resistance

R

2

.

The value of the unknown resistor, found from equation 24.49, is

Solution

(

)

( )

( )

4

2 1

3

4.00

5.00 2.00

10.0

R

R R

R

Ω

= = Ω =

Ω

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,

entering junction leaving junction

I I=

∑ ∑

(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

I

1

emanates from battery

E

1

,

I

2

from

E

2

, and

I

3

from

E

3

, 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

I

2

=

I

1

+

I

3

(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,

2

We 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

closed loop

0V

∆

=

(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

E

1

.

The sum of the

potential rises and drops as the loop is traversed is

E

1

−

I

1

R

1

−

I

2

R

2

+

E

2

= 0 (24.53)

Note that

E

1

and

E

2

are positive since the charge emanates from the positive

terminals in traversing the loop counterclockwise, and that the products,

I

1

R

1

and

24-38

Chapter 24 Electric Currents and DC Circuits

I

2

R

2

, 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

−

E

1

−

E

2

+

I

2

R

2

+

I

1

R

1

= 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

IR

s 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

I

1

,

I

2

,

and

I

3

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

E

3

. The

sum of the potential rises and drops around the right-hand loop is

−

E

3

−

E

2

+

I

2

R

2

+

I

3

R

3

= 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

IR

s 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,

I

1

,

I

2

, and

I

3

,

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

Numerically

I

2

=

I

1

+

I

3

I

2

=

I

1

+

I

3

(24.51)

E

1

−

I

1

R

1

−

I

2

R

2

+

E

2

= 0 6.00

−

10.0

I

1

−

20.0

I

2

+ 12.0 = 0 (24.56)

−

E

3

−

E

2

+

I

2

R

2

+

I

3

R

3

= 0

−

6.00

−

12.0 + 20.0

I

2

+ 30.0

I

3

= 0 (24.57)

24-39

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

I

2

from the equations by substituting

equation 24.51 into both equations 24.56 and 24.57, that is,

6.00

−

10.0

I

1

−

20.0(

I

1

+

I

3

) + 12.0 = 0

−

6.00

−

12.0 + 20.0(

I

1

+

I

3

) + 30.0

I

3

= 0

Combining terms, the loop equations become

18.0

−

30.0

I

1

−

20.0

I

3

= 0 (24.58)

and

−

18.0 + 20.0

I

1

+ 50.0

I

3

= 0 (24.59)

The problem is now reduced to solving two equations for the two unknowns

I

1

and

I

3

. 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 +30.0

I

1

, and when this equation is added to equation 24.58, the

I

1

term is eliminated. That is,

(1.50)(

−

18.0 + 20.0

I

1

+ 50.0

I

3

) = (1.50)(0)

gives

−

27.0 + 30.0

I

1

+ 75.0

I

3

= 0 (24.60)

Adding this to equation 24.58 gives

18.0

−

30.0

I

1

−

20.0

I

3

= 0 (24.58)

ADD

−

27.0 + 30.0

I

1

+ 75.0

I

3

= 0 (24.60)

−

9.00 + 0 + 55.0

I

3

= 0

Solving for the current

I

3

, we obtain

55.0

I

3

= 9.00

I

3

= 0.164 A

Substituting this value for the current

I

3

back into equation 24.58 allows us to solve

for the current

I

1

. That is,

18.0

−

30.0

I

1

−

20.0(0.164) = 0

I

1

= 0.491 A

At this point the currents

I

1

and

I

3

could be substituted back into equation 24.51 to

find the current

I

2

. However, we will not do this. Instead we will evaluate the

current

I

2

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

I

1

just found, back into equation 24.56, yields

6.00

−

10.0(0.491)

−

20.0

I

2

+ 12.0 = 0

Solving for

I

2

,

I

2

= 0.655 A

As a check, place the values of the currents just found back into equation 24.51:

I

2

=

I

1

+

I

3

0.655 A = 0.491 A + 0.164 A = 0.655 A

Thus,

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

E

1

−

I

1

R

1

−

I

2

R

2

+

E

2

= 0

6.00

−

10.0

I

1

−

20.0

I

2

+ 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

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