CHAPTER - God and · PDF fileIf a conducting wire is connected to the terminals of a ......

Fig. 19–1 An electric field inside aconducting wire.

493

Electric Current

H ave you ever wondered exactly what happens when you turn on an electric

light—what causes the bulb to glow? Or have you wondered what lightning is and

what causes it? Unfortunately, electricity remains a great mystery even to most educated

people today. In this chapter we shall remove some of that mystery. We shall see what

forces act on the electrons in a light-bulb filament, causing it to light up. We shall describe

lightning, see how it balances other charge-flow processes on earth (Problem 32), and

even estimate our chances of being struck by it (Problem 33). We shall also describe how

batteries supply energy to an electric circuit and how high-temperature superconductors

may radically change electrical technology in the future.

In the two preceding chapters we studied electrostatics—electric charges at rest.

In this chapter we study charges in motion. We begin by introducing the concept of

electric current, a concept used to describe the motion of charges.

Electric CurrentIf a conducting wire is connected to the terminals of a battery, an electric field isproduced inside the wire, directed along its length (Fig. 19–1). The free electrons in thewire experience a force and move in the direction opposite the field.

19–1

C.HAPTER 19

Return to Table of Contents

During an electrical storm hugeamounts of electric charge aretransferred between the earth’ssurface and the atmosphere.

Fig. 19–2 Finding the current through asurface.

(a)

(b)

Fig. 19–3 (a) A proton moves througha surface to the right and produces apositive current to the right. (b) Anelectron moves through a surface tothe left and produces a positive currentto the right.

494 CHAPTER 19 Electric Current

The rate of flow of charge through a cross section of the wire is called electriccurrent. This term applies to the motion of any charge through any cross-sectionalarea. We define the electric current I to be the charge �Q per unit time passingthrough an area.

I � (average current) (19–1)

If the rate of flow of charge is not constant, the definition above represents only theaverage flow rate. It is then useful to define an instantaneous current as the limitingvalue of the expression above as �t approaches zero:

I � (instantaneous current) (19–2)

The SI unit for electric current is the coulomb per second, which is called an “ampere”(amp for short) and denoted by A.

1 A � 1 C/s (19–3)

In a lightning stroke, typically 20 C of charge may pass a point in 10�3 s. This meansthat the current through a cross section of the atmosphere perpendicular to the direc-tion of charge flow at that point is 20 C/10�3 s � 20,000 A.

Since the electric charge can be either positive or negative, electric current can alsobe positive or negative. Before we determine the sign of a current, we must specify thedirection in which we want to find the current. Suppose we want to find the current tothe right through a certain surface (Fig. 19–2). If positive charge flows to the right, �Qis positive on the right side of the surface, and so the current to the right is positive. Ifnegative charge flows to the right, �Q is negative on the right side and so the currentto the right is negative. For example, if �10 C of charge flows to the right through thesurface in 1 s, the current to the right is �10 A.

A negative current in one direction is equivalent to a positive current in theopposite direction. For example, if �10 C of charge flows to the right through thesurface in Fig. 19–2, the current to the right is negative, but the current to the left ispositive because the flow of negative charge from left to right means �Q on the leftside is positive. So a current of �10 A to the right is equivalent to a current of �10 Ato the left.

Fig. 19–3 shows a proton moving to the right through a surface and an electronmoving to the left through the same surface. In either case there is a positive current tothe right (or, equivalently, a negative current to the left). In most of our applicationsmoving electrons will produce the electric current. The direction of positive currentwill then be opposite the direction of the electrons’ motion. However, even though elec-trons produce a current, we shall often find it easier to think of the current as the flowof positive charge in the direction in which the current is positive.

�Q��t

�Q��t

limit�t → 0

Fig. 19–5 Electric current flows througha car’s headlights when the lights are on.

49519–2 Ohm’s Law

Ohm’s LawWhen you turn on a car’s headlights, electric current flows through the bulbs andconnecting wires (Fig. 19–5). When you turn on an electric heater or hairdryer, currentflows through a heating element. Suppose you want to know just how much currentflows in these situations. To solve such problems we shall often apply Ohm’s law—alaw relating the electric current I through some conducting medium to the potentialdifference V applied across the ends of the conductor. Ohm’s law states that currentis proportional to potential difference.

I � V

Many materials (metals, for example) are found to obey this simple relationship.Such materials are called “ohmic.” For other, “non-ohmic” materials (transistors, forexample), no such simple relationship between current and voltage exists.

ResistanceSince voltage and current are proportional, the ratio of voltage to current is a constantfor a particular piece of ohmic material. Thus another way to express Ohm’s law is tosay that the ratio of V to I is a constant. We call this ratio the electrical resistance ofthe piece of material and denote it by R:

R � (19–4)

Units of resistance are volts per amp, which we call ohms and abbreviate by theGreek letter �:

1 � � 1 V/A (19–5)

19–2

V�I

EXAMPLE 1 Current Charging a Capacitor

A wire is connected to one plate of a capacitor (Fig. 19–4). Findthe current to the right through the wire if 1.0 � 1019 electronsper second move from the plate through the wire to the left.

Fig. 19–4

SOLUTION The charge removed from the plate in 1.0 s is(1.0 � 1019)(�1.6 � 10�19 C) � �1.6 C. This charge passesthrough any cross section of the wire from right to left, meaningthat there is a negative current to the left of �1.6 C per second,or �1.6 A. Equivalently, there is a positive current to the right of�1.6 A. You can think of this current as a flow of 1.6 C ofpositive charge to the right through any cross section of the wireduring a 1.0 s interval. Such a positive charge flow would increasethe charge on the capacitor plate by 1.6 C each second—thesame effect produced by the actual process of removing �1.6 Cfrom the plate each second as the electrons are removed.

(a)

(b)

Fig. 19–6 The current through (a) acopper wire is much greater than thecurrent through (b) a strip of rubberwhen the same voltage is applied to each.

Fig. 19–7 A piece of material of length� and cross-sectional area A has resist-ance R � �/A, where is the resistivityof the material.


We can solve Eq. 19–4 for either V or I and express Ohm’s law either as

V � IR (19–6)

or

I � (19–7)

This last form of Ohm’s law can be used to find the current I through a piece of mate-rial when we know both its resistance R and the voltage V applied across it. For a givenapplied voltage, the greater the resistance of a piece of material, the less current willflow through it.

ResistivityDifferent kinds of materials can give radically different values of resistance. Forexample, you may apply a 1 V potential difference both across the ends of a copperwire and across the ends of a strip of rubber (Fig. 19–6). The wire might have aresistance of only 10�3 � and therefore carry a current

I � � � 103 A

The rubber, however, will have enormous resistance—perhaps 109 �, which wouldmean that the current through it would be

I � � � 10–9 A

Copper is a much better conductor than rubber, and so the copper wire has much lessresistance than the strip of rubber.

Resistivity is the physical property that indicates copper is a good conductor andrubber is a poor conductor. Experiment shows that the resistance R of a particular pieceof material is given by

R � (19–8)

where is the resistivity of the material, � is the length of the piece, and A is its cross-sectional area (Fig. 19–7).

V�R

1 V�10�3 �

V�R

V�R

1 V�109 �

��A

EXAMPLE 2 Current Through a Headlight

Find the current through an automobile headlight of resistance2.5 �, when the potential difference across the light is 12 V.

SOLUTION Applying Eq. 19–7, we find the current

I � � � 4.8 A12 V�2.5 �

V�R

(a)

(b)

Fig. 19–8 (a) Heavy copper wiredesigned to carry a large current.(b) Light copper wire designed tocarry a small current.

49719–2 Ohm’s Law

According to Eq. 19–8, for a given kind of material of resistivity , resistance willincrease if either � increases or A decreases. Thus a long, thin copper wire has greaterresistance than a short, thick copper wire. Thicker wires are used when large currentsmust be carried, for example, to connect appliances, such as electric stoves, that requirelarge currents (Fig. 19–8a). Thinner wires are used when the intended current issmaller, for example, to connect a light bulb (Fig. 19–8b).

The unit of resistivity is most easily expressed in terms of ohms. Solving Eq. 19–8for , we see that � RA/�, and so has units of (�-m2)/m � �-m.

Resistivity varies with temperature. For most metals increases linearly with temp-erature over a fairly broad temperature range. Thus we may express at temperatureT as

� 0[1 � (T � T0)] (19–9)

where 0 is the value of at the reference temperature T0 (usually 20� C) and is aconstant called the temperature coefficient of resistance. Since resistance is directlyproportional to resistivity, the resistance of a particular sample is related to its temper-ature by the same relationship:

R � R0[1 � (T � T0)] (19–10)

Table 19–1 gives resistivities and temperature coefficients for various materials. Thetable shows a tremendous range of values for , from about 10�8 �-m for good conduc-tors like silver, copper, and gold up to about108 to1016 �-m for glass, wood, and rubber.

Resistivities and temperature coefficients

*Resistivities of these semiconductors are strongly dependent on the presence and concentration of impu-rities. For example, doping silicon with aluminum can reduce the resistivity of the silicon by a factor of 10�6.

Table 19–1

Material 0 (�-m) at 20° C (°C�1)

SilverCopperGoldAluminumTungstenIronLeadMercuryNichrome (an alloy used in heating elements)CarbonBody fluidsGermanium*Silicon*WoodPolyethyleneGlassHard rubber

1.59\� 10 –8

1.72\� 10 –8

2.44\� 10 –8

2.82\� 10 –8

5.51\� 10 –8

10\� 10 –8

22\� 10 –8

96\� 10 –8

100\� 10 –8

3.5\� 10 –5

�\0.1�\0.5�\103

108\–\1012

2\� 1011

1010\–\1014

1013\–\1016

3.8\� 10 –3

3.9\� 10 –3

3.4\� 10 –3

3.9\� 10 –3

4.5\� 10 –3

5.0\� 10 –3

3.9\� 10 –3

0.9\� 10 –3

0.4\� 10 –3

–0.5\� 10 –3

–50\� 10 –3

Germanium and silicon are typical examples of materials known as semiconduc-tors, which have much higher resistivity than metallic conductors but much lowerresistivity than insulators. Semiconductors are the materials used to construct variouselectronic components, such as transistors, which we shall discuss in Chapter 29.

At very low temperatures certain materials have zero resistance. Such materials arecalled superconductors. Mercury is a superconductor below 4.2 K, and lead is asuperconductor below 7.2 K. Experiments have shown that once an electric current isproduced in a superconducting lead ring, the current will continue for more than a yearwithout any electric field or source of energy needed to produce the current, as long asthe lead’s temperature is kept below 7.2 K. Applications of superconductivity andrecent discoveries are discussed in an essay at the end of this section.

CHAPTER 19 Electric Current498

EXAMPLE 3 Current Through a Copper Wire with 1.50 V Across It

A potential difference of 1.50 V is applied to a 1.00 m longcopper wire of radius 0.500 mm (Fig. 19–9). Find the resistanceof the wire and the current through it at 20.0° C.

SOLUTION We apply Eq. 19–8 using the value of forcopper at 20° C given in Table 19–1 and using A � r 2 for thecircular cross section of the wire of radius r � 0.500 mm:

R � � �

� 2.19 � 10�2 �

We find the current by applying Ohm’s law:

I � � � 68.5 A

Because of copper’s low resistivity, the resistance of the wire isquite small and the current through it is large.

Fig. 19–9

Placing the wire directly across battery terminals the way wehave indicated here is not advisable, since it would soon wastethe battery’s energy by converting it to heat. Placing any goodconductor directly across the terminals of a larger battery can bedangerous because of the very large current produced.

The wire described here is relatively thin. A copper wire witha larger radius would have even less resistance and so wouldcarry even more current. If, however, the 0.500 mm radius wirewere made of Nichrome, it would have considerably moreresistance (about 1.3 �) and carry much less current for thesame applied voltage (1.5 V/1.3 � � 1.2 A).

1.50 V��2.19 � 10–2 �

V�R

�� r2

��A

(1.72 � 10–8 �-m)(1.00 m)��

(0.500 � 10–3 m)2

EXAMPLE 4 Resistance of a Hot Copper Wire

Find the resistance of the copper wire described in the lastexample at 100° C.

SOLUTION The resistance at 100° C is found when Eq.19–10 is applied:

R � R0[1 � (T � T0)]

The reference resistance R0 is the value of 2.19 � 10�2 � foundin the previous example for 20° C. Using the value of forcopper found in Table 19–1, we find that at 100° C,

R � (2.19 � 10�2 �)[1 � (3.9 � 10�3 (°C)�1)(100°C � 20° C)]

� 2.9 � 10�2 �

19–2 Ohm’s Law 499

The preceding examples show that copper wire with a diameter of 1 mm will haveresistance much less than 1 � per meter of length. Larger-diameter copper wire willhave even less resistance. Such small resistance means that copper wire can carry rela-tively large currents with only a very small voltage drop along its length. This propertymakes copper useful in connecting elements of an electric circuit. For example, we usecopper wire to connect an electric light bulb to a power supply. The same currentpasses through the wire and the light-bulb filament (an extremely thin tungsten wire),but the filament has much higher resistance than the wire. Therefore the voltage dropacross the filament is much greater than the voltage drop along the copper wire.Nearly the entire voltage drop applied to the connecting wires appears across thefilament, almost none of it across the wires (Fig. 19–10).

ResistorsAll kinds of electric devices—light bulbs, electric heaters, electric motors, and so on—have resistance, represented in circuit diagrams by the symbol . The samesymbol is used to represent the resistance of resistors, small devices designed toprovide electrical resistance. When inserted in an electric circuit, resistors are used tocontrol the amount of current in various parts of the circuit. Some resistors consist ofa very long coil of fine wire. Others are made of a composition material containingcarbon. The resistance of the resistors shown in Fig. 19–11 varies from 0.5 � to 106 �.

Fig. 19–11 Various resistors.

Connecting wires having negligible resistance are represented in a circuit diagramby straight lines. Thus two resistors connected by wire of negligible resistance arerepresented as shown in Fig. 19–12.

Fig. 19–12 Two resistors connected by a wire of negligible resistance.

Fig. 19–10 When 120 V is applied acrossthe ends of copper wire connected toa light bulb, there is a negligible voltagedrop along the connecting wires becauseof their negligible resistance. The entire120 V acts across the light-bulb filament,which has significant resistance.

EXAMPLE 5 Finding the Current Through a Resistor

Find the current through the resistor in Fig. 19–13.

SOLUTION Between the two ends of the resistor there is apotential difference V �15 V � 3 V �12 V. We find the currentby applying Ohm’s law:

I � � � 1.2 A

Fig. 19–13

Positive current is directed to the left, from the higher potentialto the lower potential—in other words, in the direction of theelectric field.

12 V�10 �

V�R

Fig. 19–15 The voltage drop across aresistor.

Fig. 19–16 Electrical energy is beingconverted to thermal energy in theheating element of this toaster.


Voltage Drop Across a ResistorThe preceding example illustrates an important rule: When a resistor is crossed froma to b, the voltage drop Vab is either �IR or �IR, depending on whether you aremoving in the indicated current direction or opposite that direction—�IR with thecurrent, �IR against it (Fig. 19–15):

Vab � �IR (�, with current; �, against current) (19–14)

This rule, which works for both positive and negative values of I, will prove useful insolving circuit problems in the next chapter.

Electric Power; Batteries and AC SourcesElectrical energy is used for a variety of purposes, for example, to provide mechanicalenergy to an electric motor or to provide thermal energy to an electric heater, anelectric stove, or a toaster (Fig. 19–16). In this section we shall see how to calculate theelectrical energy provided to such devices, and we shall also describe sources ofelectrical energy.

Consider the process that occurs as current flows through a resistor. We can think ofthis current as being produced by positive charges flowing from the higher potentialend to the lower potential end. This means that the charges lose electrical potential energyas they pass through the resistor. Conservation of energy requires that this loss be com -pensated by a gain in some other form of energy. The energy gained is not in the formof kinetic energy of the charges because the current is the same at both ends of theresistor, and so the charges must have the same average speed at both ends. The con -stant average speed is caused by continual collisions of charges with the lattice formedby the atoms of the resistor. These collisions cause the average kinetic energy of the latticeto increase, but the average velocity and the average kinetic energy of the currentproducing charges are unchanged. The increase in kinetic energy of the lattice meansthat the thermal energy of the resistor increases; that is, the resistor heats up. The rateof production of thermal energy is just the rate at which electrical energy is used.

Next we shall obtain an expression for the rate at which electrical potential energyis used, that is, for electric power consumption by devices such as resistors and elec-tric motors.

19–3

EXAMPLE 6 Finding the Voltage Drop Across a Resistor

A positive current of 2 A is directed from a to b through the 4 �

resistor in Fig. 19–14. Find the voltage drop from a to b.

Fig. 19–14

SOLUTION Positive current is in the direction of the electricfield and hence is directed from the end of the resistor at thehigher potential to the end at the lower potential. Thus thepotential at a is greater than the potential at b, and there is apositive voltage drop from a to b:

Vab � Va � Vb � �IR � �(2 A)(4 �) � �8 V

On the other hand, the voltage drop from b to a, Vba � Vb � Va,is �8 V.

Fig. 19–A Magnetic resonance imaging.

Fig. 19–B This experimental train iscapable of speeds up to 500 km/h.

Fig. 19–C A permanent magnet is levi-tated over a disk of YBa2Cu3O7, cooled byliquid nitrogen.

A Closer Look

The phenomenon of superconductivityhas been known since 1911, when Kam -merlingh Onnes discovered that the resist-ance of mer cury is exactly zero at temper-atures below 4.2 K. Recent developmentsin superconductivity and its applicationhave generated great worldwide interestin superconducting materials.

Superconducting wires used as coils inelectromagnets produce extremely strongmagnetic fields without the usual heatingproduced by current in conventional coils.Superconducting electromagnets are usedin physics high-energy accelerator labs andfor magnetic resonance imaging, a medicaldiagnostic technique (Fig. 19–A). In Japan andGermany new high-speed experimen taltrains are both levitated and propelled bymagnetic forces between the train’s super -conducting electromagnets and mag neticfields induced in the track below (Fig. 19–B).

Superconductors are now being used inelectronics. An invention used to measuremagnetic fields with great accuracy is the“superconducting quantum interferencedevice” (called “squid” for short), whichcontains very thin layers of superconduct -ing material.

Until the last few years, no known sup -er conductors exhibited their supercon-duct ing behavior above 20 K. At highertemperatures these superconductors havethe properties of ordinary conductors; thatis, they have nonzero resistance. Maintain -ing the superconducting state required cool -ing with expensive liquid helium to 4.2 K.Therefore the practical applications of sup -erconductivity have so far been lim ited tovery specialized and expensive devices.

However, in 1986 a new class of ceramic“high-temperature” superconductors wasdiscovered. The new materials are super-conducting at much higher temperaturesthan any previous materials. One com -pound (TlBaCaCuO) becomes supercon-ducting below 125 K, and some others be -come superconducting below about 90 K.

Of course these are still very low temp -eratures, �150� C or less! However, suchtemp eratures are not too difficult to attainusing liquid nitrogen at 77 K. Liquid nitro -gen is much more plentiful and much lessexpensive than liquid helium, costing lessthan some bottled water. Fig. 19–C showsa small permanent magnet suspended bymagnetic forces over a ceramic supercon-ductor cooled with liquid nitrogen. (Themag netic effects involved in this demon-stration are explained in Chapter 22.)

The discovery of “high-temperature”superconductors immediately gave rise tospeculation about wonderful new inven-tions—smaller and faster computers, pow -erful miniature electric motors, practicalelectric cars. However, there are two fun -damental problems with the new ceramicsuperconductors: (1) they are brittle anddifficult to shape into wires; (2) the super-conducting state is maintained only forvery low currents in bulk superconduct -ors, and practical applications require highcurrents. There is hope that these prob-lems can be overcome. Perhaps one dayeven room-temperature superconductorswill be developed.

Progress is difficult to predict , since thetheoretical understanding of superconduc-tivity in these new materials is very incom-plete. There was no satisfactory explanationof superconductivity at all until Bardeen,Cooper, and Schrieffer published the “BCS”theory in 1957, 46 years after KammerlinghOnnes’s discovery of superconductivity. TheBCS theory is not able to explain the mech -anism of superconductivity in the new high-temperature superconductors, however.Perhaps a better theoretical understandingwill provide the insight to come up withnew and better superconducting com -pounds. Or perhaps better superconduc-tors will be discovered without an under-standing of the mechanism. The history ofscientific discovery indicates that it couldgo either way.

Superconductivity

Fig. 19–17 Current through an elec-trical device with a potential differenceV � Va � Vb across it.


Electric Power LossConsider a device with terminals a and b maintained at constant potential Va and Vb

respectively and through which passes a constant positive current I from a to b (Fig.19–17). During a short time interval �t, a quantity of charge �Q � I �t enters thedevice at a and an equal quantity of charge leaves the device at b. If we assume thatVa � Vb, the effect of the device during this time interval is to lower the electricalpotential energy of the charge passing through it:

Electrical potential energy loss � �QVa � �QVb � �Q(Va � Vb) � (I �t)V

where we use V to denote the voltage drop Va � Vb.The electric power loss P is the rate of loss* of electrical potential energy and is

obtained when the expression above is divided by the time interval �t:

P � IV (19–15)

For example, if a current of 2 A passes through an electric motor that has a potentialdifference of 10 V across its terminals, the motor uses electric power at the rate P �IV � (2 A)(10 V) � 20 W. This is the rate at which the electric motor converts elec-trical potential energy to mechanical work.

The electric power loss in a resistor carrying a current I is found by applying Eq.19–15. In this case, we can utilize Ohm’s law to express the power loss in alternativeforms; substituting either V � IR or I � V/R in the expression P � IV, we get

P � IV � I2R � (for resistors) (19–16)

*In some devices—for example, a battery—positive current enters the low-potential terminal and leaves thehigh-potential terminal. There is then an increase in electrical potential energy as charge passes through thedevice. The rate of increase of this energy, the electric power gain, is also found by use of Eq. 19–15.

V 2

�R

EXAMPLE 7 Melting Ice with an Electric Window Defroster

The heating element in the rear-window defroster of a MazdaRX-7 has a resistance of 3.00 �. The element is connecteddirectly across the car’s 12.0 V battery. How much heat is pro -duced in the element in 10.0 min and how much ice will melt?

SOLUTION First we find the electric power used by theresistor. Since we know V and R, we use the third form ofEq. 19–16.

P � � � 48.0 W

Using the definition of power as energy per unit time and thedefinition of a watt as a J/s, we find that in a time interval of10.0 min, the electrical energy used, or the thermal energyproduced, is

P �t � (48.0 J/s)(10.0 min)� �� 2.88 � 104 J

Since 335 J/g is required to melt ice, if all the heat produced bythe defroster were absorbed by ice at 0° C, the mass of ice thatwould melt is

� 86.0 g2.88 � 104

��335 J/g

(12.0 V)2

�3.00 �

V2

�R

60 s�1 min

Fig. 19–18 Galvani’s experiment.

Fig. 19–19 Volta’s first battery.

Fig. 19–20 Common batteries.

50319–3 Electric Power; Batteries and AC Sources

BatteriesA current-carrying resistor continuously uses electrical energy and so must beconnected to a source that continuously produces electrical energy. A battery is acommon source of electrical energy.

The development of the modern electric battery began with a chance discovery byLuigi Galvani in 1780. While dissecting a frog, Galvani found that he could sometimesmake the muscles in the frog’s leg twitch by touching the leg simultaneously with hisdissecting knife and some other metal instrument (Fig. 19–18). Galvani believed thatthe effect was electrical and that the source was within the frog. He referred to thephenomenon as “animal electricity.” Volta, a contemporary of Galvani, performed hisown experiments and was the first to realize that the effect Galvani had observed wasproduced not by the frog but by the proximity of different metals in a conducting fluid(the frog’s body fluid).

Volta found that he could create a source of electricity by placing paper moistenedin a salt solution between two metallic disks, one made of zinc and the other of silver.When the ends of a metallic object were touched to the zinc and silver simultaneously,a small spark was observed. The effect was enhanced when he connected a series ofcells, each consisting of zinc, paper, and silver (Fig. 19–19). Thus Volta created the firstelectric battery.

The batteries shown in Fig. 19–20 are similar to Volta’s original battery. The dry-cell flashlight battery consists of an inner carbon rod and an outer zinc cylinder and isfilled with an acid absorbed in a solid material that separates the carbon from the zinc(Fig. 19–21).

The chemical reactions within the cell are complex, but the following simplifieddescription gives the essentials of what goes on. The acid slowly dissolves the zinc;that is, negative ions in the acid pull positive zinc ions from the metallic lattice struc-ture of the zinc cylinder. The removal of the zinc ions leaves the cylinder with anexcess of free electrons and therefore a net negative charge. An opposite kind ofreaction occurs on the carbon rod. Free electrons from the carbon are attracted to posi-tive ions in the acid and are pulled away from the rod, leaving it positively charged.The reactions continue until the potential difference across the terminals is 1.5 V, atwhich point there is a sufficiently strong electrostatic field within the battery to opposethe further interaction of ions, and so the reaction stops. Within the battery, there is adelicate equilibrium between the interionic forces favoring the chemical reaction andthe electrostatic force opposing the reaction. When the battery is connected in acircuit, electrons are removed from the negative terminal and travel around the circuit.The chemical reaction then proceeds at a rate sufficient to replace the electronsremoved and maintain the 1.5 V potential difference.*

*However, if the electrons are drawn away at a fast enough rate—that is, if the current through the batteryis large enough—there is a significant resistance to the motion of the ions. This means that the equilibriumelectric field and the terminal potential difference opposing the flow will be less. We shall return to a discus-sion of internal resistance at the end of this section.

Fig. 19–21 Schematic of a dry cell battery.The attraction between positive zinc ionsand negative ions in the acid is opposed bythe electric field within the cell.

Fig. 19–22 Chemical energy is con vert -ed to electrical potential energy inside abattery used as a source of energy.

Fig. 19–23 Electrical potential energyis being converted to chemical energy;in other words, the battery is beingenergized, or “charged.”

As charge passes through the battery, the charge gains electrical potential energy.This energy is supplied by the loss of internal energy, or “chemical energy,” of thesystem of electrodes and acid within the battery. Eventually the battery’s terminals aredissolved, and the chemicals are used up; the battery is dead, no longer capable ofconverting internal energy to electrical energy.

The voltage across the terminals of different batteries varies, depending on the chem-ical reactions and on the strength of interaction between terminals and acid. An auto-mobile battery contains six cells, each consisting of a lead electrode and a lead dioxideelectrode separated by sulfuric acid. The chemical reaction of sulfuric acid, lead, and leaddioxide results in a potential difference of 2 V across each cell. Therefore the entirebattery of six cells maintains a total potential difference of 12 V between its terminals.

A battery is represented in a circuit diagram by the symbol , with the longer linerepresenting the higher potential terminal, designated by a � on the battery. Fig.19–22 shows a battery and its circuit representation. The battery is being used as asource of electrical energy. Thus, inside the battery, positive current flows from thenegative terminal to the positive terminal as the battery transforms chemical energy toelectrical energy. Outside the battery positive current flows through the connectingwires away from the positive terminal and toward the negative terminal.

For some batteries, such as the automobile battery, the conversion of chemical energyto electrical energy can be reversed. Reversal occurs if the chemical reaction is re vers -ible and if there is another, stronger source that can be connected to the battery in sucha way that the current through the battery is reversed (Fig. 19–23). In passing through thebattery, positive charge loses electrical potential energy, and chemical energy is stored.The usual chemical process is reversed, and the original supply of chemicals (electrodesand acid) is replenished. When an automobile engine is running, the electric generator oralternator is the source that maintains the chemical energy of the bat tery. This processis loosely referred to as “charging” the battery. In the common flashlight battery, how -ever, irreversible processes occur, and they prevent the battery from being “recharged.”

EmfA battery is quantitatively described by its “emf” value.* The emf, denoted by E, isdefined to be the energy per unit charge provided by a source as charge crossesthe source’s terminals. Let �U denote the energy provided (chemical energy in thecase of a battery) to charge �Q crossing the terminals. Then we can express the defi-nition of emf as

E � (19–17)

Since emf is an energy per unit charge, the SI unit of emf is joules per coulomb, or volts,which is the same as the unit of electric potential. The rating of a battery or othersource is its emf, measured in volts. An automobile battery has an emf of 12 V, whereasa flashlight battery has an emf of 1.5 V. These values represent the energy supplied tocharge as the charge crosses the terminals. For example, when 1 C of charge crosses theterminals of a discharging 12 V battery, the battery supplies to the charge energy �U �(�Q)E � (1 C)(12 V) � 12 J. Thus the battery loses 12 J of its chemical energy.

The definition of emf applies not only to batteries and chemical energy but to otherenergy sources as well. For example, electric generators, solar cells, thermocouples, andeven electric eels are all sources of emf (Fig. 19–24). In each, electrical potential energyis created from some other form of energy. For example, a solar cell converts sunlightto electrical energy, and a thermocouple converts thermal energy to electrical energy.

*The name emf originated as an abbreviation for “electromotive force.” Since emf is not a force, we shallavoid using the term electromotive force.

�U��Q



Let P denote the power provided by a source of emf, the rate at which chemicalenergy or some other form of energy is used:

P �

Applying Eq. 19–17, we find

P �

or

P � IE (19–18)

In an ideal source of emf, some form of energy (chemical, solar, or other) is com -pletely converted to electrical energy. If positive charge �Q crosses a potential differ-ence V from the negative to the positive terminal of a source, the charge gains electricalpotential energy (�Q)V while the source loses energy (�Q)E. If the source is ideal, theenergy it loses equals the electrical energy gained by the charge, that is

(�Q)E � (�Q)V

orV � E

Thus in an ideal source of emf the voltage across the terminals equals the emf. Forexample, a normal automobile battery is a nearly ideal source of emf and so has aterminal voltage very close to its emf of 12 V.

(�Q)E�

�t

�U��t

EXAMPLE 8 Energy Conversion in a Flashlight Battery

A flashlight is powered by a 1.5 V battery that delivers power atthe rate of 3.0 W. Find (a) the current passing through the battery;(b) the charge crossing the battery terminals in 20 min; (c) thebattery’s loss of chemical energy during this interval.

SOLUTION (a) Applying Eq. 19–18, we find

I � � � 2.0 A

(b) We use the definition of current (I � �Q/�t) to find thecharge �Q passing through the battery during a time interval �t:

�Q � I �t � (2.0 A)(20 min)� � � 2400 C

(c) Solving Eq. 19–17 for �U, we obtain

�U � (�Q)E � (2400 C)(1.5 V) � 3600 J

We could also obtain this result by applying the definition ofpower (P � �U/�t):

�U � P �t � (3.0 W)(20 min)� � � 3600 J

This is the total energy delivered by the battery. This energy isconverted partly to radiant energy in the flashlight bulb andpartly to heat, both in the bulb and inside the battery.

60 s�1 min

60 s�1 min

3.0 W�1.5 V

P�E

Fig. 19–24A generator,solar cells, andan electric eel.

Fig. 19–25 The voltage drop across anideal source of emf.


Voltage Drop Across a Source of emfIn describing electric circuits, we often need to calculate the voltage drop across theterminals of a source. If one goes from a positive terminal a to a negative terminal b ofan ideal source, there is a positive drop in potential Vab � Va � Vb � �E. If one goesfrom a negative terminal a to a positive terminal b, there is a rise in potential, which isthe same as a negative voltage drop, Vab � Va � Vb � �E (Fig. 19–25).

ideal source Vab � �E (�, from � to �; �, from � to �) (19–19)

Fig. 19–26 shows some examples applying this result.

Fig. 19–26 Voltage drops across ideal sources of emf.

EXAMPLE 9 Energy Conversion in a Simple Circuit

A simple circuit consisting of an ideal battery, a resistor, andconnecting wires is shown in Fig. 19–27, along with its circuitdiagram. Describe the energy conversion that occurs as chargeflows through the circuit.

SOLUTION The battery maintains a certain potential differ-ence between its terminals, with a at the higher potential. Thereis no change in potential along the connecting wires, and hencepoints a and c are at the same potential and points b and d are atthe same potential. Thus the battery maintains the same poten-tial difference across the resistor as it does across its own termi-nals. Ohm’s law (V � IR) then predicts that the voltage dropacross the resistor produces a current through it from c to d.This same steady current exists throughout the circuit. (If thiswere not so, charge would pile up at some point and create alarge electric field. Such an electric field would immediatelycause the charge to disperse.)

As indicated in the figure, there is a positive counterclock-wise current. We can think of this current as positive chargemoving counterclockwise around the circuit. As the chargepasses from point b through the battery to point a, its electricalpotential energy increases while the chemical energy stored inthe battery decreases. Then the charge moves around the re -main der of the circuit, flowing through the resistor from c to dand losing electrical potential energy while producing thermalenergy in the resistor. Completing the circuit, the charge returnsto point b, where its electrical potential energy is the same aswhen it began at b.

Fig. 19–27

The charge then goes around the circuit again. The processis continuous. The current is the same at all points in the circuitat any instant. As some charge is passing through the battery, anequal amount of charge is moving through the resistor. Inside thebattery, chemical energy is continuously being used to pro duceelectrical potential energy, while inside the resistor, electricalpotential energy is continuously converted to thermal energy.

Fig. 19–29 A lightbulb connected to awall outlet and itscircuit representation.


Often in a battery the production of electrical potential energy is accompanied bysome internal heating of the battery. The battery can then be treated as an ideal source Econnected in series with a resistor r, where r represents the battery’s internal resistance(Fig. 19–28). When current flows through the battery, there will be either a positive ora negative voltage drop (�Ir) across this internal resistance, the sign depending on thedirection of current. The terminal voltage of the battery will then be either greater thanor less than E. If there is no current through the battery, then Ir � 0 and the terminalvoltage equals the emf. Thus you can measure the emf of a battery by measuring thepotential difference across its terminals when it is disconnected from a circuit.

Alternating Current SourceBatteries provide an energy source for devices such as flashlights, portable radios, andcalculators. The current through a battery flows in one direction. We call this directcurrent, or “DC.” The electrical energy provided by electrical outlets in your homeproduces alternating current, or “AC,” which means that the current alternates indirection back and forth. In Chapter 20 we shall discuss household electricity in moredetail, and in Chapter 22 we shall see how such electricity is generated at powerplants. For now, we note that the conducting wires connected to an electrical outlet aremaintained at a potential difference that varies with time. The same time-dependentvoltage is applied to any device plugged into an electrical outlet, an electric light, forexample (Fig. 19–29).

EXAMPLE 10 Terminal Voltage for a Non-Ideal Battery

Suppose that the battery in Fig. 19–28 has an emf E � 1.5 V, aninternal resistance r � 1 �, and a positive current I � 0.2 Adirected toward the left from c to a. Find the terminal voltagedrop Vac � Va � Vc.

SOLUTION We find the total voltage drop from a to c bysumming the voltage drops Vab and Vbc:

Va � Vc � Va � Vb � Vb � Vc

or

Vac � Vab � Vbc

Applying Eq. 19–19, we find Vab � �E. And since in crossing

the resistor from b to c we are going against the current, appli-cation of Eq. 19–14 gives Vbc � �Ir. Thus

Vac � �E � Ir � �1.5 V � (0.2 A)(1 �)

� �1.3 V

The physical significance of this result can be understood whenwe consider what happens to a charge of �1.0 C as it crossesthe battery terminals in the direction of the current from c to a.There is a loss of chemical energy of (1.0 C)(1.5 V) � 1.5 J, again in electrical potential energy of (1.0 C)(1.3 V) � 1.3 J,and a gain in thermal energy of (1.0 C)(0.2 V) � 0.2 J. Thusenergy is conserved. Chemical energy is converted to bothelectrical potential energy and thermal energy inside the battery.

Fig. 19–28 A real battery can be treatedas an ideal source E in series with aresistor r. This resistor represents theinternal resistance of the battery.

Fig. 19–30 The voltage provided by asource of alternating current.

Fig. 19–31 The average value ofsin2 (2 f t) equals �

12�.


The time dependence of the voltage is harmonic—a sine wave of frequency f andamplitude V0 (Fig. 19–30):

V � V0 sin (2 f t) (19–20)

Any harmonic voltage source provides a voltage of this form. For a standard householdoutlet in the United States, V0 � 170 V and f � 60 Hz. A resistor connected to an alter-nating voltage source will carry an alternating current, found by applying Ohm’s law.

I � � sin (2 f t)

orI � I0 sin (2 f t) (19–21)

where I0 � V0/R is the current amplitude.We can find the instantaneous power dissipated by a resistor by using a form of

Eq. 19–16:P � I2R

orP �

We are usually more interested in the time-averaged power dissipated by a resistor,rather than the instantaneous power. This average power P is determined by theaverage values of I 2 or V 2—that is, I 2 or V 2. Both I 2 and V 2 are proportional to sin2

(2 f t), the average value of which is (Fig. 19–31). Thus from Eqs. 19–20 and19–21 we find

I2 � I02 sin2 (2 f t) � I0

2

andV 2 � V0

2 sin2 (2 f t) � V02

It is convenient to express average power in terms of the “root-mean-square” of eithercurrent or voltage—defined as the square root of the mean current or voltage squared.From the preceding equation we see that the rms value of either I or V equals 1/�2�times the respective amplitude.

Irms � �I 2�� I0 (19–22)

Vrms � �V 2�� V0 (19–23)

For a standard electrical outlet in the United States, Vrms � 1/�2� (170 V) � 120 V.We usually refer to this simply as a 120 V outlet. In most of Europe, the rms voltageis 240 V.

We can use rms values of I and V to express average power. First we take theaverage of the expressions for instantaneous power:

P � I 2IR �

Then we use the definitions of Irms and Vrms to substitute I 2 � I 2rms and V 2 � V 2

rms andobtain

P � I 2rmsR � (19–24)

An rms current or voltage produces the same average power as a constant current orvoltage of the same value.

1�2

1�2

V 2

�R

V�R

V0�R

1�2

V 2rms

�R

V�2��R

1��2�

1��2�

Fig. 19–32 Moving charges produce acurrent through surface area A.

50919–4 Electric Current and Ohm’s Law on the Microscopic Level

Electric Current and Ohm’s Law on theMicroscopic Level

In this section we shall describe electric current in a conductor on the microscopic leveland obtain a more fundamental understanding of Ohm’s law.

Electric Current and Moving ChargesFirst we shall consider the various factors that determine the value of electric currentwhen charges flow through a region of space. For simplicity, suppose that identicalpositive charges q are uniformly spread over some region of space and are all movingwith the same velocity v to the right (Fig. 19–32). We want to find an expression forthe current I through a surface of area A. We shall see that the current depends on q, v,and A and also on the number of charges per unit volume, denoted by n.

What is the quantity of charge passing through the surface area A in time �t? Sinceall the charges move a distance v �t to the right during the time interval �t, the quan-tity of charge passing through A is all of the charge initially contained in the cylindricalregion of length v �t, to the left of A, shown in Fig. 19–32. The number of charges inthis region equals the product of n, the number of charges per unit volume, and thevolume of the cylinder, (v �t)A:

Number of charges passing through A in time �t � nvA �t

The quantity of charge �Q passing through A during �t equals the number of chargestimes the value q of each individual charge:

�Q � qnvA �t

Dividing by �t, we obtain an expression for the current through the surface:

I � � qnvA

If, instead of all charges having the same speed v, there is some distribution of speedswith an average value v�, the preceding equation is modified by use of this averagevalue v� instead of v. Furthermore, if we want to apply our result to negative chargecarriers as well as positive, we can do so by replacing q by its absolute value.

I � �q�nv�A (19–25)

*19–4

�Q��t

EXAMPLE 11 Rate at Which Heat is Produced by a Toaster

A Nichrome wire having a resistance of 10.0 � is the heatingelement in an electric toaster. Find the power used by the toasterwhen it is connected to a standard electrical outlet.

SOLUTION The outlet provides 120 V, rms. From Eq. 19–24we find

P � � � 1440 W

This is the rate at which electrical energy is being converted tothermal energy in the toaster.

(120 V)2

�10.0 �

V 2rms

�R

Fig. 19–33 In a metallic lattice, positiveions form a fixed, periodic pattern,whereas conduction electrons are freeto move over large distances.

Fig. 19–34 A mechanical model illustrat -ing the motion of conduction electronsthrough a metal.


Microscopic Form of Ohm’s LawTo get a better idea of the physical meaning of Ohm’s law, let us picture what happenswhen a potential difference is applied across the ends of a metallic conductor. Asmall section of a metal consists of atoms that form a regular array, called a “lattice.”In forming the lattice, each atom loses one or two electrons, and so the lattice consistsof positive ions rigidly bound in place, whereas the electrons given up are free to movethroughout the lattice (Fig. 19–33). These free electrons behave much like gas mole-cules, moving in random directions and colliding with positive ions.

An electric field may be set up within the conductor by applying a potential differ-ence across it, for example, by connecting it to the terminals of a battery. The electricfield then accelerates the free electrons in the direction opposite the field. The electronscontinue to collide frequently with the positive ions, and so the acceleration of the elec-trons by the field is interrupted frequently.

An analogous effect is the motion of marbles rolling down an incline and collidingwith a series of nails (Fig. 19–34). If there were no nails, a marble would experiencea constant acceleration down the incline, and its velocity would increase linearly withtime. Instead, a marble accelerates briefly, collides with a nail and stops or bouncesback and then begins to accelerate down the incline again. The continual alternationbetween acceleration and collision for each marble gives rise to an average downwardflow of marbles that is fairly constant. The steeper the incline, the greater will be theaverage speed of the marbles.

Similarly, conduction electrons are accelerated by an electric field for the brief timeinterval between collisions. This intermittent acceleration gives a low average electronvelocity directed opposite the electric field.

The average speed of marbles down an incline will increase as the steepness of theincline increases. So too does the average speed of conduction electrons increase as thestrength of the electric field increases. The electrons’ average speed v� is directlyproportional to the field strength E:

v� � E

EXAMPLE 12 How Long for a Light Bulb to Glow?

A copper wire with a cross-sectional area of 1.0 mm2 carries apositive current of 20 A to the right. The density of free elec-trons in copper is 8.5 � 1028/m3. Find the average speed of thefree electrons. Is this related to the time it takes for a light toturn on?

SOLUTION Recall from Chapter 17 that in metals it is elec-trons that are free to move. Thus the positive current to the rightis produced by the flow of negatively charged free electrons tothe left. We apply Eq. 19–25 to find the average speed of thefree electrons, whose motion produces the current:

I � �q�nv�A

v� � �

� 1.5 � 10�3 m /s � 1.5 mm/s

The electrons move to the left through the wire at an averagespeed of 1.5 mm/s. Thus on the average it requires 1 s for anelectron to move 1.5 mm along the wire, or 1000 s (17 min) tomove 1.5 m. This average electron speed should not be con fusedwith the speed at which electron motion is communicated fromone end of a current-carrying wire to the other. When a wire isconnected as part of a circuit, the current begins almost simul-taneously at all points in the wire. An electric field is veryquickly established along the length of the wire. This is an elec- tromagnetic wave phenomenon, and the field propagates at thespeed of light even though electrons move very slowly in re -sponse to the field. Thus when you turn on a light switch, thereis not a delay of minutes before the light bulb glows. Instead, anelectric field and electric current are established in the filamentof the light bulb and in the connecting wires almost instantly,and so the glow of the light bulb is practically immediate.

I��q�nA

20 A��(1.6 � 10–19 C)(8.5 � 10 28/m3)(1.0 � 10–6 m2)

Fig. 19–35 A wire carrying a current I.

51119–4 Electric Current and Ohm’s Law on the Microscopic Level

From Eq. 19–25 (I � �q�nv�A), we see that

v� �

Since v� is proportional to both E and I/A, it follows that E and I/A are proportional toeach other.

E �

The constant of proportionality is the resistivity of the metal, and so we may expressthis result as

E � (19–26)

This equation is the microscopic form of Ohm’s law, from which we shall now derivethe more familiar macroscopic form of Ohm’s law, V � IR.

Macroscopic Form of Ohm’s LawConsider a straight wire of length � and cross-sectional area A carrying a uniformcurrent I to the right (Fig. 19–35).

According to the microscopic form of Ohm’s law, there must also be a uniformelectric field to the right of strength

E �

This uniform electric field produces a potential difference V � Va � Vb, which,according to Eq. 18–14 (V � Ed), is the product of E and the length of the wire:

V � E�

Inserting the preceding expression for E into this equation gives

V � I

Defining the wire’s electrical resistance R as

R �

and inserting this into the preceding equation, we obtain the macroscopic form ofOhm’s law:

V � IR

The picture of metallic conduction we have presented here is greatly oversimplified.A treatment that is more complete and quantitative is provided in Appendix C.

I�A

I�A

I�A

I�A

��A

��A

512

Electric current is defined as the rate of flow of charge andis measured in amperes (A), coulombs per second:

I �

1 A � 1 C/s

Ohm’s law relates the voltage drop V across a particularpiece of material to the current I through it:

V � IR

where the resistance R is measured in ohms (�), or voltsper ampere:

1 � � 1 V/A

The resistance of a cylindrical conductor of length � andcross-sectional area A is related to the resistivity of themedium by the equation

R �

Resistivity depends on temperature and on the medium’stemperature coefficient of resistance :

� 0[1 � (T � T0)]

In crossing a resistor R carrying a current I, the voltagedrop Vab from point a to point b is given by either �IR or�IR, depending on the direction of the current relative to thepath from a to b:

Vab � �IR (�, with current; �, against current)

A battery is a source of energy that maintains an approxi-mately constant potential difference across its terminals. Asource’s emf E is the energy per unit charge provided by thesource to the charge crossing the terminals of the source.

E �

In an ideal battery, the terminal voltage equals the emf. Thevoltage drop Vab across an ideal source from a to b is either�E or �E, depending on direction:

Vab � �E (�, from � to �; �, from � to �)

A battery with internal resistance can be represented in acircuit by an ideal battery in series with a resistor. The rulesfor calculating voltage drops across resistors and idealsources may then be applied.

Electrical outlets maintain a potential difference betweenthe wires connected to the outlet. The potential difference

V � V0 sin (2 f t)

produces an alternating current I in a resistor connected tothe outlet, where

I � I0 sin (2 f t)

The rms value of current or voltage is defined as the squareroot of the mean current or voltage squared.

Irms � �I 2�� I0

Vrms � �V 2�� V0

Standard electrical outlets in the United States provide 120V, rms, at a frequency f � 60 Hz.

Electric power is the rate of conversion of electrical en -ergy to some other form. For a device with a voltage dropV across its terminals and current I through it, the electricpower is

P � IV

For a resistor, the power may also be expressed

P � I 2R

or

P �

These same expressions may be used for the average powerprovided by an alternating current source to a resistor if weuse rms values of I and V:

–P � I 2

rmsR �

In a source of emf E, the rate at which chemical energy (orsome other form of energy) is used is

P � IE

where I is the curren2t through the source.

V 2

�R

1��2�

1��2�

�U��Q

�Q��t

V 2rms�

R

��A

HAPTER SUMMARY19

C

513

1 Because of the earth’s atmospheric electric field, whichis directed downward during clear weather, positive ionsflow downward in the atmosphere and negative ionsflow upward. What is the sign of the current in the down -ward direction, resulting from the flow of the (a) posi -tive charges; (b) negative charges? What is the sign ofthe current in the upward direction resulting from theflow of the (c) positive charges; (d) negative charges?

2 Fig. 19–36 shows a glowing circle caused by a beam ofelectrons moving along a circular path in a magneticfield. (See Chapter 21 for a further discussion of suchphenomena.) The electrons move around the circle clock -wise. Is the current positive in the clockwise or coun ter -clockwise direction?

Fig. 19–36

3 A single strand of wire of resistance R is cut in half, andthe two halves are placed side by side to form parallelstrands. What is the resistance of the two-strand com -bination?

4 Two nails, one made of aluminum and the other made ofiron, have exactly the same size and shape. Supposethe same potential difference is applied across the endsof each nail.(a) In which nail will the current be greater?(b) In which nail will the thermal energy increase faster?

5 Does a 60 W light bulb or a 100 W light bulb havegreater electrical resistance?

6 Electrical potential energy is lost as an electric currentpasses through a resistor. Does this loss mean that thecurrent leaving the resistor is less than the currententering?

7 Conduction electrons moving through a copper wirepro duce an electric current. Does the electric field withinthe wire do positive, negative, or zero work on the elec-trons?

8 A lemon with a copper rod and a steel paperclip stuck inthe sides acts as a battery (Fig. 19–37).(a) What is the emf?(b) This battery is not at all practical as a source of energy

because of its very high internal resistance. Whatwould happen to the voltmeter’s reading if the lemonwere connected to a 1 � resistor to form a completecircuit?

Fig. 19–37

9 Suppose your car’s battery is dead. You wish to “charge”it by connecting it to a good battery with jumper cables—heavy copper cables used to connect the terminals of thetwo batteries. To energize your dead battery and avoid adangerously large current, should the cables connectthe terminals: (a) � to � and � to � or (b) � to � and� to �?

10 A lightweight electrical extension cord is intended tobe used to connect low-power electrical devices, such asradios and lights. Suppose you use such a cord to connecta heavyweight power tool to an electrical outlet. Whathappens and what should you do?

11 In a certain region of space, moving electrons produce acurrent through a surface. Suppose the electrons arereplaced by protons having the same number density nand the same average velocity. Would the current change?If so, how?

12 When you turn an electric heater on, which of the fol -lowing factors account for the time delay before theheating element begins to glow: (a) the time for elec-trons to move from one end of the element to the other;(b) the time for electrons to move from the power supplyto the heating element; (c) the time for electromagneticwaves to move through the wires; (d) the time for enoughcharge to pass through the element to provide sufficientthermal energy to increase the temperature?

Questions

Questions

Electric Current

1 A copper wire carries a positive current I to the right.Each second 1.00 � 1020 electrons pass through a crosssection of the wire.(a) In what direction are the electrons moving?(b) Calculate I.

2 A current of 2.50 A flows through a 1.00 mm diametercopper wire that is connected to a capacitor plate. Thecurrent is directed toward the plate. Find the chargetransferred to the plate in 1.00 � 10�3 s.

3 A copper wire carries a current of 10.0 A. How manyelectrons pass a point in the wire in 1 hour?

4 In what appears to be a single lightning bolt, there aretypically several distinct strokes over the same path, eachlasting on the order of 10�4 s, with an interval of about0.05 s between them. The duration of the several strokesis about 0.20 s. Thus, although the peak current mayreach a maximum of 20,000 A, the average current overthe 0.20 s is much less—typically about 120 A upward.Find the charge transferred to one point on the earth’ssurface during a typical 0.20 s discharge.

Ohm’s Law

5 A certain resistor is made with a 50.0 m length of finecopper wire, 5.00 � 10�2 mm in diameter, wound onto acylindrical form and having a fiber insulator separatingthe coils. Calculate the resistance.

6 There is a potential difference of 1.0 V between theends of a 10 cm long graphite rod that has a cross-sectional area of 1.0 mm2. The resistivity of graphite is7.5 � 10�6 �-m. Find (a) the resistance of the rod;(b) the current; (c) the electric field inside the rod.

7 A certain electric extension cord has a resistance of2.00 �. Suppose it is replaced by an extension cordmade of the same material. The replacement cord istwice the diameter of the original and twice as long.What is the resistance of the new cord?

8 A lead wire of cross-sectional area 1.00 mm2 is at atemperature of 2.00 K and carries a current of 0.500 A.(a) Find the voltage drop along a 10.0 m length of the

wire.(b) Will the temperature of the lead rise as a result of the

current?

19–1 19–2

514

13 Two copper wires of the same length but different diam-eters are connected in series in a circuit. Indicate foreach of the following quantities whether it is greater inthe larger or in the smaller wire, or whether it is the samein both: (a) resistance; (b) current; (c) temperature.

14 Two copper wires of the same length but different diam-eters are connected in parallel in a circuit. Indicate foreach of the following quantities whether it is greater inthe larger or in the smaller diameter wire, or whether itis the same in both: (a) resistance; (b) current; (c) tem -perature.

15 A copper wire and an aluminum wire, both havingthe same length and diameter, are connected in series.Indicate for each of the following quantities whether itis greater in the copper or in the aluminum wire, orwhether it is the same in both: (a) resistance; (b) current;(c) temp erature.

Answers to Odd-Numbered Questions1 (a) �; (b) �; (c) �; (d) �; 3 1⁄4R; 5 60 W; 7 positive;9 a; 11 Yes. It would have the same magnitude but oppositedirection; 13 (a) smaller; (b) same; (c) smaller; 15 (a) Al;(b) same; (c) Al

CHAPTER 19 Electric Current

Problems (listed by section)

515

9 When an electric heater is turned on, the Nichrome heat -ing element is at 0� C and draws a current of 16 A, witha potential difference of 120 V between the ends of theelement. The element heats up to 800� C, with the poten-tial difference constant. Find the final current.

10 Find the potential at b in Fig. 19–38.*11 In order to double its resistance, by how much would

you have to increase the temperature of a piece of(a) copper; (b) Nichrome?

Fig. 19–38

12 Find the value of resistance R in Fig. 19–39.

Fig. 19–39

13 (a) Find the current through the resistor in Fig. 19–40.(b) If the resistor is made of carbon, what will its resist-

ance be when its temperature is increased by 100 C�?

Fig. 19–40

Electric Power; Batteries and ACSources

14 A battery with an emf of 10.0 V and internal resistanceof 1.00 � carries a positive current of 0.500 A from itsnegative terminal to its positive terminal.(a) How long is required for 1.00 C of charge to pass

through the battery?(b) By how much will the electrical potential energy

of 1.00 C of charge increase in crossing the batteryterminals?

(c) By how much will the chemical energy decrease?(d) How much electrical power is supplied by the battery?(e) At what rate is chemical energy used?

15 A positive current of 2.00 A flows from the negative tothe positive terminal inside a battery that has an emf of10.0 V. How long does it take for this battery to lose100 J of chemical energy?

16 The terminal voltage of a certain battery is measuredwith the battery disconnected and is found to be 6.00 V.The internal resistance of the battery is 1.00 �. Find thevoltage across the battery terminals when there is a cur -rent of 2.00 A from the negative to the positive terminal.

17 (a) Find the potential at B in Fig. 19–41.(b) Find the emf of the battery.(c) Is the battery being “charged” or “discharged”; that

is, is electrical energy being converted to chemicalenergy or vice versa?

Fig. 19–41

*18 When a car’s starter is in use, it draws a large current.The car’s lights draw much less current. As a certain caris starting, the current through the battery is 60.0 A andthe potential difference across the battery terminals is9.00 V. When only the car’s lights are used, the currentthrough the battery is 2.00 A and the terminal potentialdifference is 11.9 V. Find the battery’s emf and internalresistance.

19–3

Problems

516

19 Car batteries are often rated in amp-hours. This ratingindicates the quantity of charge that can pass through thebattery before the battery’s chemical energy must berestored by the car’s generator or some other source. Theamp-hour unit is convenient for simple calculations. Forexample, a 60-amp-hour battery can supply a current of1.0 A for 60 h (or 60 A for 1.0 h) before going dead.Find the energy stored in (a) a 60 A-h, 12 V car battery;(b) a 1.0 A-h, 1.5 V flashlight battery.

20 A 2000 W air conditioner and a 500 W refrigerator eachoperate on 120 V, rms. Find the current through (a) theair conditioner; (b) the refrigerator.

21 Find the resistance of (a) a 100 W light bulb; (b) a2000 W heating element in a stove. Power ratings arebased on the assumption of a potential difference of120 V, rms across the terminals.

22 The total power input to all the electrical appliances in acertain household averages 500 W throughout the day.What will the electric bill be for a 30-day period if therate is 6 cents per kWh?

23 An electrical outlet for an electric stove provides a poten -tial difference of 240 V, rms. Find the amplitude of thepotential difference.

24 The instantaneous potential difference Vab across an elec-tric light bulb has a maximum value of �170 V and amin imum value of �170 V. What is the rms voltage?

25 If the instantaneous current through a heating element inan electric heater reaches a maximum value of 10 A tothe right at some instant, what is the current through theelement �

1

1

20� s later?

26 The rms voltage across a 5.00 � resistor is 35.0 V. Findthe maximum instantaneous current through the resistor.

*27 (a) What is the minimum length of time necessary tobring to a boil 1.00 L (1.00 � 103 cm3) of water, ini -tially at 20.0� C, using as a source of heat the 2000 Wheating element on a stove?

(b) What additional minimum time is necessary to boilthe water away?

28 Two ordinary 100 W, 120 V light bulbs are connected inseries, with 120 V, rms across the combination. Findthe total power used by the two bulbs.

Electric Current and Ohm’s Law onthe Microscopic Level

29 Find the average speed of the electrons in a 1.0 cmdiameter, copper power line, when it carries a current of20 A.

30 The peak instantaneous current in a certain lightningstroke is 20,000 A vertically upward. The current isapproximately constant at this value for 1.0 � 10 �4 s.

(a) Find the charge transferred to the ground during thistime interval.

(b) The current at one point in the lightning bolt isproduced by electrons moving at a speed of 1.0 �107 m/s through a circular cross section of radius1.0 cm. What is the number of electrons per m3 inthe discharge?

(c) Assuming this same electron density throughout,what is the total charge contained in a 1.0 km longlightning bolt that has an average cross-sectionalarea of 1.0 cm2?

*31 An automobile starter motor is an electric motor used tostart the gasoline engine. To operate the starter motor,the ignition switch is turned on, and a current of 50 A issupplied through a 2.0 mm radius copper wire thatconnects the motor to the automobile battery.(a) If electric charge had to travel the 0.50 m distance

from the battery to the motor before the motor wouldoperate, how long would it take?

(b) How far do the electrons move in the 2.0 s requiredto start the engine?

*32 There is a net charge of �5.9 � 105 C on the surface ofthe earth and a net positive charge spread throughout theearth’s atmosphere. This charge distribution produces anelectric field that, just above the earth’s surface, has an av -er age value of 130 N/C, downward, during clearweather. (During thunderstorms the field is muchstronger and oppositely directed.) Ions in the atmos-phere move in response to the field, producing anaverage downward current of 3.5 � 10�12 A per squaremeter of cross-sectional area.(a) Find the total ionic current into the earth’s surface.(b) If the ionic current remained constant and there were

no other currents at the earth’s surface, how longwould it take to remove the earth’s surface charge?

(c) Find the potential difference between the ground anda point 1.0 km above it, assuming the field remainsconstant.

(d) Find the resistance of a 1.0 km thick layer of theearth’s atmosphere just above the surface.

(e) Find the resistivity of the earth’s lower atmosphere.(f) What solid materials have roughly the same

resistivity?*33 Problem 32 shows that, at the earth’s surface, there is a

global, downward, ionic current of 1800 A. Lightningprovides an equal average upward current and therebymaintains the earth’s charge distribution. A lightningdischarge at one point lasts about 0.20 s, with an averageupward current of 120 A.(a) Find the average number of simultaneous lightning

discharges occurring somewhere on the earth.(b) If there were equal likelihood of lightning striking

anywhere on the earth, what would be the chance of

*19–4

CHAPTER 19 Electric Current

517

lightning striking at any instant the 0.10 m2 areawithin which you stand (Fig. 19–42)?

(c) What would be the chance of lightning striking suchan area one time in a 1-year interval?*

Fig. 19–42 What are the odds that you will be struck bylightning?

Additional Problems34 A “dead” but rechargeable car battery has an emf of

11.8 V and internal resistance of 1.00 �. To start the car,a good battery (E � 12.0 V, r � 0) is connected to thedead battery by jumper cables of negligible resistance(Fig. 19–43). The battery terminals are connected � to� and � to �. The weak battery is charged by thestrong battery when connected in this way.(a) How long does it take for the weak battery to receive

a 2.00 amp-hour charge? Assume that the emfs andresistances remain constant.

(b) How long would it take to acquire this same 2.00amp-hour charge if the engine were running and thegenerator were charging the battery with a current of4.00 A?

Fig. 19–43

*Each year deaths from lightning in the United States number about100 (about 1 out of every 2 million Americans). There are 10 times thisnumber of deaths caused by household electrocution and 500 times thisnumber in auto accidents.

*35 Fig. 19–44 shows copper and aluminum wires connectedtogether and carrying a current of 5.00 A to the right.The right end of the aluminum is at a potential of zero.Find (a) the resistance of the copper wire; (b) the resist-ance of the aluminum wire; (c) the potential at the leftend of the copper wire. (d) In which wire is the electricfield greater?

Fig. 19–44

*36 When the tungsten filament in a 100 W, 120 V lightbulb reaches 2000� C, 100 W of electric power is used topro duce light and heat.(a) Find the current through the filament under these

conditions, which are established very shortly afterthe light is turned on.

(b) Find the filament current and the power used whenthe light is first turned on, assuming an initial fila-ment temperature of 20� C.

(c) Estimate the time needed to heat the filament to2000� C. The filament has a mass of 0.040 g and aspecific heat of 0.032 cal/g-C�.

**37 The filament in an incandescent light bulb is a very thin,coiled tungsten wire. To radiate 100 W at 2000� C, thesurface area of the tungsten must equal 2.54 cm2 (seeProblem 42 in Chapter 13). Find the radius and thelength of the wire.

*38 An aluminum wire has the same resistance as a copperwire of the same length. Find the ratio of the aluminumwire’s diameter to the copper wire’s diameter.

*39 The electric field equals 1.00 � 106 N/C at the surfaceof the 20.0 cm radius dome of a Van de Graaf generator,described in Question 19 in Chapter 17. What is thecur rent carried by the belt if the power required tooperate the generator is 10.0 W? Assume negligiblefriction.

40 An electric heater draws an rms current of 10 A whenthere is a potential difference of 120 V, rms across theheating element. What does it cost to operate the heaterconstantly for 30 days if the power company charges6.0 cents per kWh?

41 Find the rms current in a 1.00 hp, 120 V electric motor,assuming that the mechanical power output of 1.00 hpequals the electrical power input.

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