Chapter 24 Electric Currents and DC Circuits

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


The position and velocity of the electron are given by the kinematic equations

r = v0t + 1 at2 2

andv = 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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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

It

= (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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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

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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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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

VI

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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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


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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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

RA

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 0C and Mean Temperature Coefficient of Resistivity (α)

Material ρ, Ω m α 0C−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 × 1014

2 × 105 - 2 × 1014

1013 - 1014

1013 - 1016

1011 - 1015

108 - 1011

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

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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.

The cross-sectional area of the wire is given by

Solution

A = πd2

4

( )230.644 10 m

4

π −×=

= 3.26 × 10−7 m2

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

R = ρ l A

( )

× Ω 10 ×= −

−27

8

m 103.26m 500

m 1.72

= 26.4 Ω


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.

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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 m5.00 m

0.0200 m 0.0250 ml

RA

ρ= = × 10 Ω

= 3.00 × 1017 Ω


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

ρ − ρ0 = m(t − t0) Solving for the resistivity ρ, gives

ρ = ρ0[1 + m (t − t0)] (24.8) ρ0 Let us now define a new constant α, called the mean temperature coefficient of resistivity, as

α = m ρ0

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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.

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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

Solution

ρ = ρ0[1 + α (t − t0)]

= (1.72 × 10−8 Ω m)[1 + (3.93 × 10−3 0C−1)(200 − 20.0) 0C ] = 2.94 × 10−8 Ω m


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)] (24.10) A A

However, ρl/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 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.

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

Solution

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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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 (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

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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)

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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)

2P 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 = 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= (24.20)

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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 = 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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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


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

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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 (24.23)

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Dividing both sides of equation 24.23 by I gives

V = R1 + R2 + R3 (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 2R R R 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

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Multiplying each term by ρ/A gives

31 2 ll llA A A A

ρ ρ ρ ρ= + +

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.

The equivalent resistance, given by equation 24.26, is

Solution

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


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

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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.


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

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.


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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 (24.29) R1

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I2 = V2 (24.30) R2

I3 = V3 (24.31) R3

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

I = V1 + V2 + V3 (24.32) R1 R2 R3 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 (24.33)

V R1 R2 R3

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 1R 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 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 2 3

1 1 1 1 1...

nR R R R R= + + + + (24.36)

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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.

From equation 24.35 the equivalent resistance is

Solution

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.


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

http://www.farmingdale.edu/faculty/peter-nolan/pdf/univphys/UPEX24-12,13,14,15,16.XLS


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.


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

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.


Example 24.15

Power supplied to a parallel circuit. What is the power supplied to the circuit in example 24.12?

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The power supplied by the battery is

Solution

P = IV = (0.650 A)(6 V) = 3.90 W


Example 24.16

Power dissipated in a parallel circuit. What is the power dissipated in each resistor in example 24.14?


Solution


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 round-off error.


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




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,

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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 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


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


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.


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




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



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.

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 = 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

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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 VA across the ammeter, and the voltmeter reads the voltage drop

V = VR + VA The calculated resistance is

RV VR

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 VR across the resistor. At first glance, it might seem that the problem is

24-33



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

R V

VR

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


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,

V1 = V3

Therefore, from Ohm’s law, 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


2 2 4 4

1 1 3 3

I R I RI R I R

= (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

2 4

1 3

R RR R

=

Solving for the unknown resistor R2, we obtain

42

3

RR

R= 1R (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.

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

Solution

( )( ) ( )4

2 13

4.00 5.00 2.00

10.0 R

R RR

Ω= = Ω =

ΩΩ


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



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 junctionI 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 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


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 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


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 Numerically I2 = I1 + I3 I2 = I1 + I3 (24.51) E1 − I1R1 − I2R2 + E2 = 0 6.00 − 10.0I1 − 20.0I2 + 12.0 = 0 (24.56) − E3 − E2 + I2R2 + I3R3 = 0 − 6.00 − 12.0 + 20.0I2 + 30.0I3 = 0 (24.57)

24-39


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

18.0 − 30.0I1 − 20.0I3 = 0 (24.58)

and − 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,

(1.50)(− 18.0 + 20.0I1 + 50.0I3) = (1.50)(0)

gives −27.0 + 30.0I1 + 75.0I3 = 0 (24.60)

Adding this to equation 24.58 gives

18.0 − 30.0I1 − 20.0I3 = 0 (24.58) ADD − 27.0 + 30.0I1 + 75.0I3 = 0 (24.60)

−9.00 + 0 + 55.0I3 = 0 Solving for the current I3, we obtain

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

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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

6.00 − 10.0(0.491) − 20.0I2 + 12.0 = 0 Solving for I2,

I2 = 0.655 A

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

I2 = I1 + I3 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

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.

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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

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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. ).

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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. ).

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Summary of Important Equations Current I = q (24.1) t Ohm’s law I = V (24.3) R Resistance of a wire R = ρ l (24.7) A Work done within a battery W = qV (24.13) Power P = IV (24.15) Power dissipated in a resistor P =I2R (24.18)

Resistors in series

V = V1 + V2 + V3 (24.22) I = I1 = I2 = I3

R = R1 + R2 + R3 (24.26) P = P1 + P2 + P3

Resistors in parallel V = V1 = V2 = V3 (24.27) I = I1 + I2 + I3 (24.28)

1 = 1 + 1 + 1 (24.35) R R1 R2 R3

P = P1 + P2 + P3

Terminal voltage of a battery V = E − Ir (24.37) Batteries in series E = E1 + E2 + E3 + … + En (24.38) Identical batteries in parallel E = E1 = E2 = E3 (24.39)

I = I1 + I2 + I3 (24.40) Wheatstone bridge R2 = R4 R1 (24.49) R3 Kirchhoff’s first rule Σ Ientering junction = Σ Ileaving junction (24.50) Kirchhoff’s second rule ∆Vclosed loop = 0 (24.52)

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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

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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.

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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.

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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.


*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.


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*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.


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.

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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.


*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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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.


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.


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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.


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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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.


*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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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.


To go to these Interactive Tutorials click on this sentence.

To go to another chapter, return to the table of contents by clicking on this sentence.

http://www.farmingdale.edu/faculty/peter-nolan/pdf/univphys/UPTutCh24.xls

http://www.farmingdale.edu/faculty/peter-nolan/pdf/UPBTOC.pdf

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Chapter 24 Electric Currents and DC Circuits