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 and r = v 0 t + 1 at 2 2 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 vd, the drift velocity, in the expected 24-2

direction as seen in figure 24.2(c). Because of the constant interaction with the atoms of the lattice structure, it takes almost 30 s in some metals for the electron to drift about 1 cm. This is extremely slow as compared to the motion that would be experienced by the electron in figure 24.1. Because of this complexity of the motion of the electron within the solid wire, we will make no further analysis of the motion of the electron by Newton s laws and the kinematic equations; we will adopt a much simpler procedure. When a potential difference is applied across the ends of a wire and electrons start to drift within the wire, we will say that an electric current exists in the wire. The electric current is defined as the amount of electric charge that flows through a cross section of the wire per unit time. We write this mathematically as I q = (24.1) t 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

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

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

I V (24.2) To make an equality out of this proportionality, we introduce a constant of proportionality (1/R) and the proportionality 24.2 becomes the equation I V = (24.3) R 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

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? Solution The resistance of the circuit, found from the rearrangement of Ohm s law, is R = V = 12.0 V = 40.0 Ω I 0.300 A 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

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 = ρ (24.7) A 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

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. Solution The cross-sectional area of the wire is given by A = πd 2 4 ( 0.644 10 3 m ) 2 π = 4 = 3.26 10 7 m 2 The resistance of the wire spool, found from equation 24.7, is R = ρ l A 8 500 m ( 1.72 10 Ω m) = 7 = 26.4 Ω 3.26 10 To go to this Interactive Example click on this sentence. m 2 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

Figure 24.8 The resistance of an amber rod. Solution The resistance from one end of the rod to the other, found from equation 24.7, is ( ) ( )( ) l 14 0.300 m R = ρ = ( 5.00 10 Ω m) A 0.0200 m 0.0250 m = 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 Solving for the resistivity ρ, gives ρ ρ 0 = m(t t 0 ) 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

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 ρ0 is known at the reference temperature t0. 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

Solution The temperature coefficient of resistivity for copper, found from table 24.1, is 3.93 10 3 0C 1. The resistivity of copper at 200 0 C, found from equation 24.9, is ρ = ρ 0 [1 + α (t t 0 )] = (1.72 10 8 Ω m)[1 + (3.93 10 3 0C 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, ρl/a is equal to the resistance R from equation 24.7. Therefore, we can write equation 24.10 as R = R 1 0 + α ( t t 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. Solution The temperature coefficient of resistivity for copper, found from table 24.1, is 3.93 10 3 0C 1. The resistance of the copper resistor at 200 0 C, found from equation 24.11, is R = R 0 [1 + α(t t 0 )] = (50.0 Ω)[1 + (3.93 10 3 0C 1 )(200 20.0) 0 C] = 85.4 Ω To go to this Interactive Example click on this sentence. 24-12

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 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

Figure 24.10 Conservation of energy in an electric circuit. the point of zero potential to the point of higher potential V within the battery. Recall from the definition of the potential that the potential is the potential energy per unit charge, V = PE = W (21.19) q q where W is the work that must be done to give the charge its potential energy. From equation 21.19, the work done within the battery is simply W = qv (24.13) The power supplied by the battery is found from equations 24.12 and 24.13 as But P = W = qv (24.14) t t 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

Therefore, the power consumed in the circuit is Power consumed = Power supplied = IV If the circuit contains only a resistor, then the voltage across the resistor is given by Ohm s law as V = IR. Therefore, the power consumed or dissipated in the resistor is P = IV = I(IR) P 2 = 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

It is sometimes convenient to mark the position of the resistor that is at the highest potential with a plus (+) sign, and the point of the resistor that is at the lowest potential with a negative sign ( ) to remind ourselves that the charge will fall from a plus (+) to a minus ( ) potential. This is shown in figure 24.10(b). The mechanical equivalent of the circuit of figure 24.10(a) is shown in figure 24.10(c). A ball of mass m is lifted to a height h by a person. The person who does the work lifting the mass is equivalent to the battery. The ball has potential energy at the top. We assume that the medium that the ball will fall through is resistive. Therefore, the ball will not fall freely, continually accelerating, but rather will be slowed down by the friction of the medium until the ball moves at a constant velocity, its terminal velocity. This is similar to the charge moving at the constant drift velocity. As the ball falls and loses potential energy, the lost potential energy shows up as heat generated by the frictional forces between the ball and the medium. This is similar to the loss of energy of the charge through Joule heating as it falls through the resistor. Example 24.7 A 60-W light bulb. A 60.0-W light bulb is screwed into a 120-V lamp outlet. (a) What current will flow through the bulb and what is the resistance of the bulb? (b) What is the power dissipated in the Joule heating of the wire in the bulb? Solution a. Because the power rating of the bulb is known, we can find the current from equation 24.15 as P = IV I = P = 60.0 W = 0.500 A V V 120 V V = 0.500 A The resistance of the bulb, found from Ohm s law, is R = V = 120 V = 240 Ω I 0.500 A b. The power dissipated in the Joule heating of the wire in the bulb, found from equation 24.18, is P = 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

Example 24.8 A 120-W light bulb. What is the current and resistance of a 120-W light bulb connected to a 120-V source? Solution We find the current from I = P = 120 W = 1.00 A V 120 V We determine the resistance of the bulb by Ohm s law as R = V = 120 V = 120 Ω I 1.00 A 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

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, PEtop = 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

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 R = R1 + R2 + 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

Multiplying each term by ρ/a gives Chapter 24 Electric Currents and DC Circuits l l1 l2 l ρ = ρ + ρ + ρ 3 A A A A But using equation 24.7, R = ρl/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. Solution The equivalent resistance, given by equation 24.26, is 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

The power dissipated in each resistor is P = VI = (6.00 V)(0.0667 A) = 0.400 W 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. Solution The potential drop across each resistor, found by Ohm s law, is 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

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

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 1 1 1 = + + (24.35) R R R R 1 2 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 3 1 1 1 1 1 = + + +... + (24.36) R R1 R2 R3 R n 24-23

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. Solution From equation 24.35 the equivalent resistance is 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. Solution The total current in the circuit, found from Ohm s law with R as the equivalent resistance of the circuit, is 24-24

I = V = 6.00 V = 0.650 A R 9.23 Ω Notice that the current from the battery is almost ten times greater when the resistors are connected in parallel then when connected in series. This is easily explained, since the resistance in parallel is only about one-tenth of the resistance when the resistors are in series. To go to this Interactive Example click on this sentence. Example 24.14 The current in each parallel resistor. Find the current through each resistor in example 24.12. Solution Because the voltage across each resistor is 6.00 V, we can find the current from Ohm s law applied to each resistor as given in equations 24.29, 24.30, and 24.31. Namely, 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

Solution The power supplied by the battery is P = IV = (0.650 A)(6 V) = 3.90 W To go to this Interactive Example click on this sentence. Example 24.16 Power dissipated in a parallel circuit. What is the power dissipated in each resistor in example 24.14? The power dissipated in each resistor is Solution P 1 = I 12 R 1 = (0.300 A) 2 (20.0 Ω) = 1.80 W P 2 = I 22 R 2 = (0.200 A) 2 (30.0 Ω) = 1.20 W P 3 = I3 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 roundoff error. To go to this Interactive Example click on this sentence. 24.8 Combinations of Resistors in Series and Parallel A typical circuit showing a simple combination of resistors connected in series and parallel is shown in figure 24.14. Let us determine the current through each resistor and the voltage drop across it. To do so, we use the techniques of sections 24.6 and 24.7. Resistors R 2 and R 3 are connected in parallel. Their equivalent resistance R 23, found from equation 24.36, is 24-26

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

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 Note that the sum of the currents 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 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 The power dissipated in each resistor is P = IV = (0.161 A)(6.00 V) = 0.966 W P 1 = I 12 R 1 = (0.161 A) 2 (20.0 Ω) = 0.518 W P 2 = I2 2 R 2 = (0.0927 A) 2 (30.0 Ω) = 0.258 W P 3 = I3 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

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 which is, of course, less than the emf of the battery. 1 V = E Ir (24.37) 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

Solution The potential difference across the battery, given by equation 24.37, is V = E Ir = 1.50 V (0.0100 A)(2.00 Ω) = 1.48 V Notice that when the current in the circuit is relatively small, the potential drop across the internal resistance is also quite small, and the terminal voltage is very close to the emf of the battery. If the current in the circuit is much larger, as for example I = 0.500 A then the terminal voltage is V = E Ir = 1.50 V (0.500 A)(2.00 Ω) = 0.500 V which is very much less than the emf of the battery. To go to this Interactive Example click on this sentence. Example 24.18 The emf of a battery connected to a circuit. A battery has an emf of 1.50 V and an internal resistance of 3.00 Ω. It is connected as shown in figure 24.15 to a resistance of 500 Ω. Find the current in the circuit, and the terminal voltage of the battery. Solution The internal resistance r is in series with the load resistor R, so the total resistance in the circuit is just the sum of the two of them. Using Ohm s law to determine the current we have I = E = 1.50 V = 2.98 10 3 A R + r 500 Ω + 3.00 Ω The terminal voltage across the battery is V = E Ir = 1.50 V (2.98 10 3 A)(3.00 Ω) = 1.49 V 24-30

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 + + 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 = 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. Solution The total current I in the circuit, found from Ohm s law, is I = E = 6.00 V = 0.0120 A R 500 Ω This total current is supplied by three batteries so that, I = 3I 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

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 VA across the ammeter, and the voltmeter reads the voltage drop The calculated resistance is V = VR + VA R V + V I R 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 V R R = I I R + V (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