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Chapter 17Current and Resistance

17.1 ELECTRIC CURRENT

568

17CHAPTER

Current and ResistanceO U T L I N E

17.1 Electric Current17.2 A Microscopic View:

Current and Drift Speed17.3 Current and Voltage

Measurements in Circuits17.4 Resistance and Ohm’s Law17.5 Resistivity17.6 Temperature Variation

of Resistance17.7 Superconductors17.8 Electrical Energy and

Power17.9 Electrical Activity in the

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These power lines transfer energyfrom the power company to homesand businesses. The energy is trans-ferred at a very high voltage, possiblyhundreds of thousands of volts insome cases. The high voltage resultsin less loss of power due to resistancein the wires, so it is used despite thefact that it makes power lines verydangerous.

Many practical applications and devices are based on the principles of static electricity, butelectricity was destined to become an inseparable part of our daily lives when scientistslearned how to produce a continuous flow of charge for relatively long periods of time usingbatteries. The battery or voltaic cell was invented in 1800 by the Italian physicist AlessandroVolta. Batteries supplied a continuous flow of charge at low potential, in contrast to earlierelectrostatic devices that produced a tiny flow of charge at high potential for brief periods.This steady source of electric current allowed scientists to perform experiments to learn howto control the flow of electric charges in circuits. Today, electric currents power our lights,radios, television sets, air conditioners, computers, and refrigerators. They ignite the gasolinein automobile engines, travel through miniature components making up the chips ofmicrocomputers, and provide the power for countless other invaluable tasks.

In this chapter we define current and discuss some of the factors that contribute to theresistance to the flow of charge in conductors. We also discuss energy transformations inelectric circuits. These topics will be the foundation for additional work with circuits in laterchapters.

17.1 ELECTRIC CURRENTIn Figure 17.1, charges move in a direction perpendicular to a surface of area A.(The area could be the cross-sectional area of a wire, for example.) The current isthe rate at which charge flows through this surface.

Suppose !Q is the amount of charge that flows through an area A in atime interval !t and that the direction of flow is perpendicular to the area.Then the current I is equal to the amount of charge divided by the timeinterval:

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17.1 Electric Current 569

A

I

+

+

++

+

Figure 17.1 Charges in motionthrough an area A. The time rate offlow of charge through the area is de-fined as the current I. The directionof the current is the direction of flowof positive charges.

[17.1]

SI unit: coulomb/second (C/s), or the ampere (A).

One ampere of current is equivalent to one coulomb of charge passing throughthe cross-sectional area in a time interval of 1 s.

When charges flow through a surface as in Figure 17.1, they can be positive, nega-tive, or both. The direction of conventional current used in this book is the directionpositive charges flow. (This historical convention originated about 200 years ago,when the ideas of positive and negative charges were introduced.) In a common con-ductor such as copper, the current is due to the motion of negatively charged elec-trons, so the direction of the current is opposite the direction of motion of the elec-trons. On the other hand, for a beam of positively-charged protons in an accelerator,the current is in the same direction as the motion of the protons. In some cases—gases and electrolytes, for example—the current is the result of the flows of both pos-itive and negative charges. Moving charges, whether positive or negative, are referredto as charge carriers. In a metal, for example, the charge carriers are electrons.

In electrostatics, where charges are stationary, the electric potential is the sameeverywhere in a conductor. This is no longer true for conductors carrying current:as charges move along a wire, the electric potential is continually decreasing (ex-cept in the special case of superconductors).

I ! !Q!t

EXAMPLE 17.1 Turn on the LightGoal Apply the concept of current.

Problem The amount of charge that passes through the filament of a certain lightbulb in 2.00 s is 1.67 C. Find(a) the current in the bulb and (b) the number of electrons that pass through the filament in 5.00 s.

Strategy Substitute into Equation 17.1 for part (a), then multiply the answer by the time given in part (b) to getthe total charge that passes in that time. The total charge equals the number N of electrons going through the circuittimes the charge per electron.

Solution(a) Compute the current in the lightbulb.

Substitute the charge and time into Equation 17.1:

(b) Find the number of electrons passing through thefilament in 5.00 s.

The total number N of electrons times the charge perelectron equals the total charge, I !t :

Substitute and solve forN : N(1.60 " 10#19 C/electron) $ (0.835 A)(5.00 s)

N $ 2.61 " 1019 electrons

(1) Nq $ I !t

0.835 AI $!Q!t

$1.67 C2.00 s

$

Remarks In developing the solution, it was important to use units to ensure the correctness of equations such asEquation (1). Notice the enormous number of electrons passing through a given point in a typical circuit.

Exercise 17.1Suppose 6.40 " 1021 electrons pass through a wire in 2.00 min. Find the current.

Answer 8.53 A

TIP 17.1 Current Flow is RedundantThe phrases flow of current and currentflow are commonly used, but here theword flow is redundant because cur-rent is already defined as a flow (ofcharge). Avoid this construction!

! Direction of current

44920_17_p568-591 1/5/05 1:41 PM Page 56917.1 Electric Current 569

A

I

+

+

++

+

Figure 17.1 Charges in motionthrough an area A. The time rate offlow of charge through the area is de-fined as the current I. The directionof the current is the direction of flowof positive charges.

[17.1]

SI unit: coulomb/second (C/s), or the ampere (A).

One ampere of current is equivalent to one coulomb of charge passing throughthe cross-sectional area in a time interval of 1 s.

When charges flow through a surface as in Figure 17.1, they can be positive, nega-tive, or both. The direction of conventional current used in this book is the directionpositive charges flow. (This historical convention originated about 200 years ago,when the ideas of positive and negative charges were introduced.) In a common con-ductor such as copper, the current is due to the motion of negatively charged elec-trons, so the direction of the current is opposite the direction of motion of the elec-trons. On the other hand, for a beam of positively-charged protons in an accelerator,the current is in the same direction as the motion of the protons. In some cases—gases and electrolytes, for example—the current is the result of the flows of both pos-itive and negative charges. Moving charges, whether positive or negative, are referredto as charge carriers. In a metal, for example, the charge carriers are electrons.

In electrostatics, where charges are stationary, the electric potential is the sameeverywhere in a conductor. This is no longer true for conductors carrying current:as charges move along a wire, the electric potential is continually decreasing (ex-cept in the special case of superconductors).

I ! !Q!t

EXAMPLE 17.1 Turn on the LightGoal Apply the concept of current.

Problem The amount of charge that passes through the filament of a certain lightbulb in 2.00 s is 1.67 C. Find(a) the current in the bulb and (b) the number of electrons that pass through the filament in 5.00 s.

Strategy Substitute into Equation 17.1 for part (a), then multiply the answer by the time given in part (b) to getthe total charge that passes in that time. The total charge equals the number N of electrons going through the circuittimes the charge per electron.

Solution(a) Compute the current in the lightbulb.

Substitute the charge and time into Equation 17.1:

(b) Find the number of electrons passing through thefilament in 5.00 s.

The total number N of electrons times the charge perelectron equals the total charge, I !t :

Substitute and solve forN : N(1.60 " 10#19 C/electron) $ (0.835 A)(5.00 s)

N $ 2.61 " 1019 electrons

(1) Nq $ I !t

0.835 AI $!Q!t

$1.67 C2.00 s

$

Remarks In developing the solution, it was important to use units to ensure the correctness of equations such asEquation (1). Notice the enormous number of electrons passing through a given point in a typical circuit.

Exercise 17.1Suppose 6.40 " 1021 electrons pass through a wire in 2.00 min. Find the current.

Answer 8.53 A

TIP 17.1 Current Flow is RedundantThe phrases flow of current and currentflow are commonly used, but here theword flow is redundant because cur-rent is already defined as a flow (ofcharge). Avoid this construction!

! Direction of current

44920_17_p568-591 1/5/05 1:41 PM Page 569

17.2 A MICROSCOPIC VIEW: CURRENT AND DRIFT SPEED

570 Chapter 17 Current and Resistance


Macroscopic currents can be related to the motion of the microscopic charge car-riers making up the current. It turns out that current depends on the averagespeed of the charge carriers in the direction of the current, the number of chargecarriers per unit volume, and the size of the charge carried by each.

Consider identically charged particles moving in a conductor of cross-sectionalarea A (Fig. 17.3). The volume of an element of length !x of the conductor is A !x. If n represents the number of mobile charge carriers per unit volume, thenthe number of carriers in the volume element is nA !x. The mobile charge !Q inthis element is therefore

!Q " number of carriers # charge per carrier " (nA !x)q

where q is the charge on each carrier. If the carriers move with a constant averagespeed called the drift speed vd, the distance they move in the time interval !t is !x " vd !t. We can therefore write

!Q " (nAvd !t)q

If we divide both sides of this equation by !t, we see that the current in the con-ductor is

[17.2]

To understand the meaning of drift speed, consider a conductor in which thecharge carriers are free electrons. If the conductor is isolated, these electrons un-dergo random motion similar to the motion of the molecules of a gas. The driftspeed is normally much smaller than the free electrons’ average speed betweencollisions with the fixed atoms of the conductor. When a potential difference is ap-plied between the ends of the conductor (say, with a battery), an electric field is setup in the conductor, creating an electric force on the electrons and hence a cur-rent. In reality, the electrons don’t simply move in straight lines along the conduc-tor. Instead, they undergo repeated collisions with the atoms of the metal, and theresult is a complicated zigzag motion with only a small average drift speed alongthe wire (Active Fig. 17.4). The energy transferred from the electrons to the metalatoms during a collision increases the vibrational energy of the atoms and causesa corresponding increase in the temperature of the conductor. Despite thecollisions, however, the electrons move slowly along the conductor in a directionopposite with the drift velocity .v:dE

:

I "!Q!t

" nqvdA

Consider positive and negative charges moving horizontally through the four re-gions in Figure 17.2. Rank the magnitudes of the currents in these four regionsfrom lowest to highest. (Ia is the current in Figure 17.2a, Ib the current in Figure17.2b, etc.) (a) Id, Ia, Ic, Ib (b) Ia, Ic, Ib, Id (c) Ic, Ia, Id, Ib (d) Id, Ib, Ic, Ia(e) Ia, Ib, Ic, Id (f) none of these

Quick Quiz 17.1

(a)

++

+

++

++

+

(b) (c) (d)

–

––

––

–

Figure 17.2 (Quick Quiz 17.1)∆x

Aq

vd

vd∆t

Figure 17.3 A section of a uniformconductor of cross-sectional area A.The charge carriers move with a speedvd , and the distance they travel in time!t is given by !x " vd!t. The numberof mobile charge carriers in the section of length !x is given bynAvd!t, where n is the number of mobile carriers per unit volume.

vd

E

–

ACTIVE FIGURE 17.4A schematic representation of thezigzag motion of a charge carrier ina conductor. The sharp changes in direction are due to collisions withatoms in the conductor. Note that thenet motion of electrons is oppositethe direction of the electric field.

Log into PhysicsNow atwww.cp7e.com and go to Active Figure 17.4, where you can observethe random zigzag motion of acharge carrier and see how themotion is affected by an electric field.

TIP 17.2 Electrons are Everywhere in the CircuitElectrons don’t have to travel fromthe light switch to the light for thelight to operate. Electrons already inthe filament of the lightbulb move in response to the electric field set up bythe battery. Also, the battery does notprovide electrons to the circuit; it provides energy to the existing electrons.

44920_17_p568-591 1/5/05 1:41 PM Page 570







!Q " (nAvd !t)q


[17.2]


:

I "!Q!t

" nqvdA


Quick Quiz 17.1

(a)

++

+

++

++

+

(b) (c) (d)

–

––

––

–


Aq

vd

vd∆t


vd

E

–




44920_17_p568-591 1/5/05 1:41 PM Page 570







!Q " (nAvd !t)q


[17.2]


:

I "!Q!t

" nqvdA


Quick Quiz 17.1

(a)

++

+

++

++

+

(b) (c) (d)

–

––

––

–


Aq

vd

vd∆t


vd

E

–




44920_17_p568-591 1/5/05 1:41 PM Page 570







!Q " (nAvd !t)q


[17.2]


:

I "!Q!t

" nqvdA


Quick Quiz 17.1

(a)

++

+

++

++

+

(b) (c) (d)

–

––

––

–

Figure 17.2 (Quick Quiz 17.1)!x

Aq

vd

vd!t


vd

E

–




44920_17_p568-591 1/5/05 1:41 PM Page 570







!Q " (nAvd !t)q


[17.2]


:

I "!Q!t

" nqvdA


Quick Quiz 17.1

(a)

++

+

++

++

+

(b) (c) (d)

–

––

––

–


Aq

vd

vd∆t


vd

E

–




44920_17_p568-591 1/5/05 1:41 PM Page 570

drift speed







!Q " (nAvd !t)q


[17.2]


:

I "!Q!t

" nqvdA


Quick Quiz 17.1

(a)

++

+

++

++

+

(b) (c) (d)

–

––

––

–

Figure 17.2 (Quick Quiz 17.1)!x

Aq

vd

vd!t


vd

E

–




44920_17_p568-591 1/5/05 1:41 PM Page 570

17.2 A Microscopic View: Current and Drift Speed 571

Example 17.2 shows that drift speeds are typically very small. In fact, the driftspeed is much smaller than the average speed between collisions. Electrons travel-ing at 2.46 ! 10"4 m/s, as in the example, would take about 68 min to travel 1 m!

EXAMPLE 17.2 Drift Speed of ElectronsGoal Calculate a drift speed and compare it with the rms speed of an electron gas.

Problem A copper wire of cross-sectional area 3.00 ! 10"6 m2 carries a current of 10.0 A. (a) Assuming that eachcopper atom contributes one free electron to the metal, find the drift speed of the electrons in this wire. (b) Use theideal gas model to compare the drift speed with the random rms speed an electron would have at 20.0°C. The den-sity of copper is 8.92 g/cm3, and its atomic mass is 63.5 u.

Strategy All the variables in Equation 17.2 are known except for n, the number of free charge carriers per unit vol-ume. We can find n by recalling that one mole of copper contains an Avogadro’s number (6.02 ! 1023) of atoms andeach atom contributes one charge carrier to the metal. The volume of one mole can be found from copper’s knowndensity and atomic mass. The atomic mass is the same, numerically, as the number of grams in a mole of the sub-stance.

Solution(a) Find the drift speed of the electrons.

Calculate the volume of one mole of copper from itsdensity and its atomic mass:

V #m$

#63.5 g

8.92 g/cm3 # 7.12 cm3

Convert the volume from cm3 to m3: 7.12 cm3 ! 1 m102 cm "3

# 7.12 ! 10"6 m3

Divide Avogadro’s number (the number of electrons inone mole) by the volume per mole to obtain the num-ber density: # 8.46 ! 1028 electrons/m3

n #6.02 ! 1023 electrons/mole

7.12 ! 10"6 m3/mole

Solve Equation 17.2 for the drift speed, and substitute:

vd # 2.46 ! 10"4 m/s

#10.0 C/s

(8.46 ! 1028 electrons/m3)(1.60 ! 10"19 C)(3.00 ! 10"6 m2)

vd #I

nqA

(b) Find the rms speed of a gas of electrons at 20.0°C.

Apply Equation 10.18: vrms # √ 3kBTme

Convert the temperature to the Kelvin scale, and substi-tute values:

# 1.15 ! 105 m/s

vrms # √ 3(1.38 ! 10"23 J/K)(293 K )9.11 ! 10"31 kg

Remarks The drift speed of an electron in a wire is very small—only about one-billionth of its random thermalspeed.

Exercise 17.2What current in a copper wire with a cross-sectional area of 7.50 ! 10"7 m2 would result in a drift speed of 5.00 ! 10"4 m/s?

Answer 5.08 A

44920_17_p568-591 1/5/05 1:41 PM Page 571

17.2 A Microscopic View: Current and Drift Speed 571

Example 17.2 shows that drift speeds are typically very small. In fact, the driftspeed is much smaller than the average speed between collisions. Electrons travel-ing at 2.46 ! 10"4 m/s, as in the example, would take about 68 min to travel 1 m!

EXAMPLE 17.2 Drift Speed of ElectronsGoal Calculate a drift speed and compare it with the rms speed of an electron gas.

Problem A copper wire of cross-sectional area 3.00 ! 10"6 m2 carries a current of 10.0 A. (a) Assuming that eachcopper atom contributes one free electron to the metal, find the drift speed of the electrons in this wire. (b) Use theideal gas model to compare the drift speed with the random rms speed an electron would have at 20.0°C. The den-sity of copper is 8.92 g/cm3, and its atomic mass is 63.5 u.

Strategy All the variables in Equation 17.2 are known except for n, the number of free charge carriers per unit vol-ume. We can find n by recalling that one mole of copper contains an Avogadro’s number (6.02 ! 1023) of atoms andeach atom contributes one charge carrier to the metal. The volume of one mole can be found from copper’s knowndensity and atomic mass. The atomic mass is the same, numerically, as the number of grams in a mole of the sub-stance.

Solution(a) Find the drift speed of the electrons.

Calculate the volume of one mole of copper from itsdensity and its atomic mass:

V #m$

#63.5 g

8.92 g/cm3 # 7.12 cm3

Convert the volume from cm3 to m3: 7.12 cm3 ! 1 m102 cm "3

# 7.12 ! 10"6 m3

Divide Avogadro’s number (the number of electrons inone mole) by the volume per mole to obtain the num-ber density: # 8.46 ! 1028 electrons/m3

n #6.02 ! 1023 electrons/mole

7.12 ! 10"6 m3/mole

Solve Equation 17.2 for the drift speed, and substitute:

vd # 2.46 ! 10"4 m/s

#10.0 C/s

(8.46 ! 1028 electrons/m3)(1.60 ! 10"19 C)(3.00 ! 10"6 m2)

vd #I

nqA

(b) Find the rms speed of a gas of electrons at 20.0°C.

Apply Equation 10.18: vrms # √ 3kBTme

Convert the temperature to the Kelvin scale, and substi-tute values:

# 1.15 ! 105 m/s

vrms # √ 3(1.38 ! 10"23 J/K)(293 K )9.11 ! 10"31 kg

Remarks The drift speed of an electron in a wire is very small—only about one-billionth of its random thermalspeed.

Exercise 17.2What current in a copper wire with a cross-sectional area of 7.50 ! 10"7 m2 would result in a drift speed of 5.00 ! 10"4 m/s?

Answer 5.08 A

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17.3 CURRENT AND VOLTAGE MEASUREMENTS IN CIRCUITS


In view of this low speed, you might wonder why a light turns on almost instanta-neously when a switch is thrown. Think of the flow of water through a pipe. If adrop of water is forced into one end of a pipe that is already filled with water, adrop must be pushed out the other end of the pipe. Although it may take an indi-vidual drop a long time to make it through the pipe, a flow initiated at one endproduces a similar flow at the other end very quickly. Another familiar analogy isthe motion of a bicycle chain. When the sprocket moves one link, the other linksall move more or less immediately, even though it takes a given link some time tomake a complete rotation. In a conductor, the electric field driving the free elec-trons travels at a speed close to that of light, so when you flip a light switch, themessage for the electrons to start moving through the wire (the electric field)reaches them at a speed on the order of 108 m/s!

17.3 CURRENT AND VOLTAGE MEASUREMENTS IN CIRCUITS

To study electric current in circuits, we need to understand how to measure cur-rents and voltages.

The circuit shown in Figure 17.5a is a drawing of the actual circuit necessary formeasuring the current in Example 17.1. Figure 17.5b shows a stylized figure calleda circuit diagram which represents the actual circuit of Figure 17.5a. This circuitconsists of only a battery and a lightbulb. The word “circuit” means “a closed loop

Suppose a current-carrying wire has a cross-sectional area that gradually becomessmaller along the wire, so that the wire has the shape of a very long cone. Howdoes the drift speed vary along the wire? (a) It slows down as the cross section be-comes smaller. (b) It speeds up as the cross section becomes smaller. (c) It doesn’tchange. (d) More information is needed.

Quick Quiz 17.2

+–

A+–

+–V

Battery

Ammeter

Voltmeter

Bulb

(b)(a)

I

I

II

0.835 A

0.0 V

+ –+ –

+–

Figure 17.5 (a) A sketch of an actual circuit used to measure the current in a flashlight bulb and thepotential difference across it. (b) A schematic diagram of the circuit shown in part (a). (c) A digitalmultimeter can be used to measure both currents and potential differences. Here, the meter is measur-ing the potential difference across a 9-V battery.

c, M

ichae

l Dal

ton,

Fund

amen

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17.4 RESISTANCE AND OHM’S LAW

17.4 Resistance and Ohm’s Law 573

of some sort around which current circulates.” The battery pumps charge throughthe bulb and around the loop. No charge would flow without a complete conduct-ing path from the positive terminal of the battery into one side of the bulb, out theother side, and through the copper conducting wires back to the negative termi-nal of the battery. The most important quantities that characterize how the bulbworks in different situations are the current I in the bulb and the potential differ-ence !V across the bulb. To measure the current in the bulb, we place an amme-ter, the device for measuring current, in line with the bulb so there is no path forthe current to bypass the meter; all of the charge passing through the bulb mustalso pass through the ammeter. The voltmeter measures the potential difference,or voltage, between the two ends of the bulb’s filament. If we use two meters simul-taneously as in Figure 17.5a, we can remove the voltmeter and see if its presenceaffects the current reading. Figure 17.5c shows a digital multimeter—a convenientdevice, with a digital readout, that can be used to measure voltage, current, or re-sistance. An advantage of using a digital multimeter as a voltmeter is that it willusually not affect the current, since a digital meter has enormous resistance to theflow of charge in the voltmeter mode.

At this point, you can measure the current as a function of voltage (an I – !Vcurve) of various devices in the lab. All you need is a variable voltage supply (anadjustable battery) capable of supplying potential differences from about " 5 Vto # 5 V, a bulb, a resistor, some wires and alligator clips, and a couple of multi-meters. Be sure to always start your measurements using the highest multimeterscales (say, 10 A and 1 000 V), and increase the sensitivity one scale at a time toobtain the highest accuracy without overloading the meters. (Increasing thesensitivity means lowering the maximum current or voltage that the scalereads.) Note that the meters must be connected with the proper polarity withrespect to the voltage supply, as shown in Figure 17.5b. Finally, follow your in-structor’s directions carefully to avoid damaging the meters and incurring asoaring lab fee.

17.4 RESISTANCE AND OHM’S LAWWhen a voltage (potential difference) !V is applied across the ends of a metallicconductor as in Figure 17.7, the current in the conductor is found to be propor-tional to the applied voltage; I $ !V. If the proportionality holds, we can write !V % IR , where the proportionality constant R is called the resistance of theconductor. In fact, we define the resistance as the ratio of the voltage across theconductor to the current it carries:

[17.3]R ! !VI

Look at the four “circuits” shown in Figure 17.6 and select those that will light thebulb.

Quick Quiz 17.3

– +

(b)

– +

(c) (d)

AMPS

+ –

+ –+–

(a)

Figure 17.6 (Quick Quiz 17.3)

l

E

Vb Va

IA

Figure 17.7 A uniform conductorof length l and cross-sectional area A.The current I in the conductor is proportional to the applied voltage !V % Vb " Va. The electric field

set up in the conductor is also proportional to the current.E:

! Resistance

44920_17_p568-591 1/5/05 1:41 PM 73

17.4 Resistance and Ohm’s Law 573

of some sort around which current circulates.” The battery pumps charge throughthe bulb and around the loop. No charge would flow without a complete conduct-ing path from the positive terminal of the battery into one side of the bulb, out theother side, and through the copper conducting wires back to the negative termi-nal of the battery. The most important quantities that characterize how the bulbworks in different situations are the current I in the bulb and the potential differ-ence !V across the bulb. To measure the current in the bulb, we place an amme-ter, the device for measuring current, in line with the bulb so there is no path forthe current to bypass the meter; all of the charge passing through the bulb mustalso pass through the ammeter. The voltmeter measures the potential difference,or voltage, between the two ends of the bulb’s filament. If we use two meters simul-taneously as in Figure 17.5a, we can remove the voltmeter and see if its presenceaffects the current reading. Figure 17.5c shows a digital multimeter—a convenientdevice, with a digital readout, that can be used to measure voltage, current, or re-sistance. An advantage of using a digital multimeter as a voltmeter is that it willusually not affect the current, since a digital meter has enormous resistance to theflow of charge in the voltmeter mode.

At this point, you can measure the current as a function of voltage (an I – !Vcurve) of various devices in the lab. All you need is a variable voltage supply (anadjustable battery) capable of supplying potential differences from about " 5 Vto # 5 V, a bulb, a resistor, some wires and alligator clips, and a couple of multi-meters. Be sure to always start your measurements using the highest multimeterscales (say, 10 A and 1 000 V), and increase the sensitivity one scale at a time toobtain the highest accuracy without overloading the meters. (Increasing thesensitivity means lowering the maximum current or voltage that the scalereads.) Note that the meters must be connected with the proper polarity withrespect to the voltage supply, as shown in Figure 17.5b. Finally, follow your in-structor’s directions carefully to avoid damaging the meters and incurring asoaring lab fee.

17.4 RESISTANCE AND OHM’S LAWWhen a voltage (potential difference) !V is applied across the ends of a metallicconductor as in Figure 17.7, the current in the conductor is found to be propor-tional to the applied voltage; I $ !V. If the proportionality holds, we can write !V % IR , where the proportionality constant R is called the resistance of theconductor. In fact, we define the resistance as the ratio of the voltage across theconductor to the current it carries:

[17.3]R ! !VI

Look at the four “circuits” shown in Figure 17.6 and select those that will light thebulb.

Quick Quiz 17.3

– +

(b)

– +

(c) (d)

AMPS

+ –

+ –+–

(a)

Figure 17.6 (Quick Quiz 17.3)

l

E

Vb Va

IA

Figure 17.7 A uniform conductorof length l and cross-sectional area A.The current I in the conductor is proportional to the applied voltage !V % Vb " Va. The electric field

set up in the conductor is also proportional to the current.E:

! Resistance

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17.5 RESISTIVITY

17.5 Resistivity 575

17.5 RESISTIVITYElectrons don’t move in straight-line paths through a conductor. Instead, they un-dergo repeated collisions with the metal atoms. Consider a conductor with a volt-age applied across its ends. An electron gains speed as the electric force associatedwith the internal electric field accelerates it, giving it a velocity in the directionopposite that of the electric field. A collision with an atom randomizes the elec-tron’s velocity, reducing it in the direction opposite the field. The process thenrepeats itself. Together, these collisions affect the electron somewhat as a force ofinternal friction would. This is the origin of a material’s resistance.

The resistance of an ohmic conductor increases with length, which makes sensebecause the electrons going through it must undergo more collisions in a longerconductor. A smaller cross-sectional area also increases the resistance of a conduc-tor, just as a smaller pipe slows the fluid moving through it. The resistance, then, isproportional to the conductor’s length l and inversely proportional to its cross-sectional area A,

[17.5]

where the constant of proportionality, !, is called the resistivity of the material.1Every material has a characteristic resistivity that depends on its electronic struc-ture and on temperature. Good electric conductors have very low resistivities, andgood insulators have very high resistivities. Table 17.1 lists the resistivities of vari-ous materials at 20°C. Because resistance values are in ohms, resistivity values mustbe in ohm-meters (" # m).

R $ ! lA

An assortment of resistors used for avariety of applications in electroniccircuits.

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In Figure 17.8b, does the resistance of the diode (a) increase or (b) decrease asthe positive voltage %V increases?

Quick Quiz 17.4

EXAMPLE 17.3 Resistance of a Steam IronGoal Use Ohm’s law to calculate a resistance.

Problem All electric devices are required to have identifying plates that specify their electrical characteristics. The plate on a certain steam iron states that the iron carries a current of 6.40 A when connected to a source of 1.20 & 102 V. What is the resistance of the steam iron?

Strategy Substitute into Ohm’s law.

SolutionApply Equation 17.3: 18.8 "R $

%VI

$1.20 & 102 V

6.40 A $

Exercise 17.3The resistance of a hot plate is 48.0 ". How much current does the plate carry when connected to a 1.20 & 102-Vsource?

Answer 2.50 A

1The symbol ! used for resistivity shouldn’t be confused with the same symbol used earlier in the book for density. Of-ten, a single symbol is used to represent different quantities.

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resistivity

17.6 TEMPERATURE VARIATION OF RESISTANCE


room when the people are in motion than when they are standing still. The in-creased electron scattering with increasing temperature results in increased resis-tivity. Technically, thermal expansion also affects resistance; however, this is a verysmall effect.

Over a limited temperature range, the resistivity of most metals increases lin-early with increasing temperature according to the expression

! " !0[1 # $(T % T0)] [17.6]

where ! is the resistivity at some temperature T (in Celsius degrees), !0 is the resis-tivity at some reference temperature T0 (usually taken to be 20°C), and $ is a pa-rameter called the temperature coefficient of resistivity. Temperature coefficientsfor various materials are provided in Table 17.1. The interesting negative values of$ for semiconductors arise because these materials possess weakly bound chargecarriers that become free to move and contribute to the current as the tempera-ture rises.

Because the resistance of a conductor with a uniform cross section is propor-tional to the resistivity according to Equation 17.5 (R " !l/A), the temperaturevariation of resistance can be written

R " R0[1 # $(T % T0)] [17.7]

Precise temperature measurements are often made using this property, as shownby the following example.

EXAMPLE 17.5 A Platinum Resistance ThermometerGoal Apply the temperature dependence of resistance.

Problem A resistance thermometer, which measures temperature by measuring the change in resistance of a con-ductor, is made of platinum and has a resistance of 50.0 & at 20.0°C. (a) When the device is immersed in a vessel con-taining melting indium, its resistance increases to 76.8 &. From this information, find the melting point of indium.(b) The indium is heated further until it reaches a temperature of 235°C. What is the ratio of the new current in theplatinum to the current Imp at the melting point?

Strategy In part (a), solve Equation 17.7 for T % T0 and get $ for platinum from Table 17.1, substituting knownquantities. For part (b), use Ohm’s law in Equation 17.7.

Solution(a) Find the melting point of indium.

Solve Equation 17.7 for T % T0:

" 137'C

T % T0 "R % R0

$R 0"

76.8 & % 50.0 &[3.92 ( 10%3 ('C)%1][50.0 &]

Substitute T0 " 20.0°C and obtain the melting point ofindium:

T " 157°C

(b) Find the ratio of the new current to the old whenthe temperature rises from 157°C to 235°C.

Write Equation 17.7, with R0 and T0 replaced by Rmp andTmp, the resistance and temperature at the melting point.

R " R mp[1 # $(T % Tmp)]

According to Ohm’s law, R " )V/I and R mp " )V/Imp.Substitute these expressions into Equation 17.7:

)VI

")VImp

[1 # $(T % Tmp)]

Cancel the voltage differences, invert the two expres-sions, and then divide both sides by Imp:

IImp

"1

1 # $(T % Tmp)

In an old-fashioned carbon filamentincandescent lamp, the electrical re-sistance is typically 10 &, but changeswith temperature.

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room when the people are in motion than when they are standing still. The in-creased electron scattering with increasing temperature results in increased resis-tivity. Technically, thermal expansion also affects resistance; however, this is a verysmall effect.

Over a limited temperature range, the resistivity of most metals increases lin-early with increasing temperature according to the expression

! " !0[1 # $(T % T0)] [17.6]

where ! is the resistivity at some temperature T (in Celsius degrees), !0 is the resis-tivity at some reference temperature T0 (usually taken to be 20°C), and $ is a pa-rameter called the temperature coefficient of resistivity. Temperature coefficientsfor various materials are provided in Table 17.1. The interesting negative values of$ for semiconductors arise because these materials possess weakly bound chargecarriers that become free to move and contribute to the current as the tempera-ture rises.

Because the resistance of a conductor with a uniform cross section is propor-tional to the resistivity according to Equation 17.5 (R " !l/A), the temperaturevariation of resistance can be written

R " R0[1 # $(T % T0)] [17.7]

Precise temperature measurements are often made using this property, as shownby the following example.

EXAMPLE 17.5 A Platinum Resistance ThermometerGoal Apply the temperature dependence of resistance.

Problem A resistance thermometer, which measures temperature by measuring the change in resistance of a con-ductor, is made of platinum and has a resistance of 50.0 & at 20.0°C. (a) When the device is immersed in a vessel con-taining melting indium, its resistance increases to 76.8 &. From this information, find the melting point of indium.(b) The indium is heated further until it reaches a temperature of 235°C. What is the ratio of the new current in theplatinum to the current Imp at the melting point?

Strategy In part (a), solve Equation 17.7 for T % T0 and get $ for platinum from Table 17.1, substituting knownquantities. For part (b), use Ohm’s law in Equation 17.7.

Solution(a) Find the melting point of indium.

Solve Equation 17.7 for T % T0:

" 137'C

T % T0 "R % R0

$R 0"

76.8 & % 50.0 &[3.92 ( 10%3 ('C)%1][50.0 &]

Substitute T0 " 20.0°C and obtain the melting point ofindium:

T " 157°C

(b) Find the ratio of the new current to the old whenthe temperature rises from 157°C to 235°C.

Write Equation 17.7, with R0 and T0 replaced by Rmp andTmp, the resistance and temperature at the melting point.

R " R mp[1 # $(T % Tmp)]

According to Ohm’s law, R " )V/I and R mp " )V/Imp.Substitute these expressions into Equation 17.7:

)VI

")VImp

[1 # $(T % Tmp)]

Cancel the voltage differences, invert the two expres-sions, and then divide both sides by Imp:

IImp

"1

1 # $(T % Tmp)

In an old-fashioned carbon filamentincandescent lamp, the electrical re-sistance is typically 10 &, but changeswith temperature.

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temperature coefficient of resistivity


TABLE 17.1Resistivities and Temperature Coefficients of Resistivity for Various Materials (at 20°C)

Temperature CoefficientResistivity of Resistivity

Material (! " m) [(°C)#1]

Silver 1.59 $ 10#8 3.8 $ 10#3

Copper 1.7 $ 10#8 3.9 $ 10#3

Gold 2.44 $ 10#8 3.4 $ 10#3

Aluminum 2.82 $ 10#8 3.9 $ 10#3

Tungsten 5.6 $ 10#8 4.5 $ 10#3

Iron 10.0 $ 10#8 5.0 $ 10#3

Platinum 11 $ 10#8 3.92 $ 10#3

Lead 22 $ 10#8 3.9 $ 10#3

Nichromea 150 $ 10#8 0.4 $ 10#3

Carbon 3.5 $ 105 # 0.5 $ 10#3

Germanium 0.46 # 48 $ 10#3

Silicon 640 # 75 $ 10#3

Glass 1010–1014

Hard rubber ! 1013

Sulfur 1015

Quartz (fused) 75 $ 1016

aA nickel–chromium alloy commonly used in heating elements.

As a lightbulb ages, why does it gives off less light thanwhen new?

Explanation There are two reasons for the lightbulb’sbehavior, one electrical and one optical, but both arerelated to the same phenomenon occurring within thebulb. The filament of an old lightbulb is made of atungsten wire that has been kept at a high temperaturefor many hours. High temperatures evaporate tung-sten from the filament, decreasing its radius. FromR % &l/A, we see that a decreased cross-sectional arealeads to an increase in the resistance of the filament.

This increasing resistance with age means that thefilament will carry less current for the same appliedvoltage. With less current in the filament, there is lesslight output, and the filament glows more dimly.

At the high operating temperature of the filament,tungsten atoms leave its surface, much as water mole-cules evaporate from a puddle of water. The atomsare carried away by convection currents in the gas inthe bulb and are deposited on the inner surface of theglass. In time, the glass becomes less transparentbecause of the tungsten coating, which decreases theamount of light that passes through the glass.

Applying Physics 17.1 Dimming of Aging Lightbulbs

INTERACTIVE EXAMPLE 17.4 The Resistance of Nichrome WireGoal Combine the concept of resistivity with Ohm’s law.

Problem (a) Calculate the resistance per unit length of a 22-gauge nichrome wire of radius 0.321 mm. (b) If a po-tential difference of 10.0 V is maintained across a 1.00-m length of the nichrome wire, what is the current in thewire? (c) The wire is melted down and recast with twice its original length. Find the new resistance RN as a multiple ofthe old resistance RO .

Strategy Part (a) requires substitution into Equation 17.5, after calculating the cross-sectional area, while part (b) isa matter of substitution into Ohm’s law. Part (c) requires some algebra. The idea is to take the expression for the newresistance and substitute expressions for lN and AN , the new length and cross-sectional area, in terms of the old lengthand cross-section. For the area substitution, use the fact that the volumes of the old and new wires are the same.

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17.8 ELECTRICAL ENERGY AND POWER o


power efficiently is also receiving serious consideration. Modern superconductingelectronic devices consisting of two thin-film superconductors separated by athin insulator have been constructed. Among these devices are magnetometers(magnetic-field measuring devices) and various microwave devices.

17.8 ELECTRICAL ENERGY AND POWERIf a battery is used to establish an electric current in a conductor, chemical energystored in the battery is continuously transformed into kinetic energy of the chargecarriers. This kinetic energy is quickly lost as a result of collisions between thecharge carriers and fixed atoms in the conductor, causing an increase in the tem-perature of the conductor. In this way, the chemical energy stored in the battery iscontinuously transformed into thermal energy.

In order to understand the process of energy transfer in a simple circuit, con-sider a battery with terminals connected to a resistor (Active Fig. 17.11; rememberthat the positive terminal of the battery is always at the higher potential). Nowimagine following a quantity of positive charge !Q around the circuit from pointA, through the battery and resistor, and back to A. Point A is a reference point thatis grounded (the ground symbol is ), and its potential is taken to be zero. Asthe charge !Q moves from A to B through the battery, the electrical potentialenergy of the system increases by the amount !Q !V, and the chemical potentialenergy in the battery decreases by the same amount. (Recall from Chapter 16 that!PE " q !V.) However, as the charge moves from C to D through the resistor, itloses this electrical potential energy during collisions with atoms in the resistor. Inthe process, the energy is transformed to internal energy corresponding to in-creased vibrational motion of those atoms. Because we can ignore the very smallresistance of the interconnecting wires, no energy transformation occurs for pathsBC and DA. When the charge returns to point A, the net result is that some of thechemical energy in the battery has been delivered to the resistor and has caused itstemperature to rise.

The charge !Q loses energy !Q !V as it passes through the resistor. If !t is thetime it takes the charge to pass through the resistor, then the rate at which it loseselectric potential energy is

where I is the current in the resistor and !V is the potential difference across it. Ofcourse, the charge regains this energy when it passes through the battery, at theexpense of chemical energy in the battery. The rate at which the system losespotential energy as the charge passes through the resistor is equal to the rate atwhich the system gains internal energy in the resistor. Therefore, the power !,representing the rate at which energy is delivered to the resistor, is

[17.8]

While this result was developed by considering a battery delivering energy to a re-sistor, Equation 17.8 can be used to determine the power transferred from a volt-age source to any device carrying a current I and having a potential difference !Vbetween its terminals.

Using Equation 17.8 and the fact that !V " IR for a resistor, we can express thepower delivered to the resistor in the alternate forms

[17.9]

When I is in amperes, !V in volts, and R in ohms, the SI unit of power is the watt(introduced in Chapter 5). The power delivered to a conductor of resistance R is

! " I 2R "!V 2

R

! " I!V

!Q!t

!V " I!V

Figure 17.10 A small permanentmagnet floats freely above a ceramicdisk made of the superconductorYBa2Cu3O7, cooled by liquid nitrogenat 77 K. The superconductor has zeroelectric resistance at temperatures below 92 K and expels any appliedmagnetic field.

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ACTIVE FIGURE 17.11A circuit consisting of a battery and aresistance R. Positive charge flowsclockwise from the positive to thenegative terminal of the battery. PointA is grounded.

Log into PhysicsNow at www.cp7e.comand go to Active Figure 17.11, whereyou can adjust the battery voltage andthe resistance, and see the resultingcurrent in the circuit and the powerdissipated as heat by the resistor.

Power !

Power delivered to a resistor !

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[17.8]



[17.9]


! " I 2R "!V 2

R

! " I!V

!Q!t

!V " I!V


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Power !


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[17.8]



[17.9]


! " I 2R "!V 2

R

! " I!V

!Q!t

!V " I!V


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Power !


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[17.8]



[17.9]


! " I 2R "!V 2

R

! " I!V

!Q!t

!V " I!V


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Power !


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17.8 Electrical Energy and Power 581

often referred to as an I 2R loss. Note that Equation 17.9 applies only to resistorsand not to nonohmic devices such as lightbulbs and diodes.

Regardless of the ways in which you use electrical energy in your home, you ulti-mately must pay for it or risk having your power turned off. The unit of energyused by electric companies to calculate consumption, the kilowatt-hour, is definedin terms of the unit of power and the amount of time it’s supplied. One kilowatt-hour (kWh) is the energy converted or consumed in 1 h at the constant rate of1 kW. It has the numerical value

1 kWh ! (103 W)(3600 s) ! 3.60 " 106 J [17.10]

On an electric bill, the amount of electricity used in a given period is usually statedin multiples of kilowatt-hours.

Why do lightbulbs fail so often right after they’returned on?

Explanation Once the switch is closed, the line voltageis applied across the bulb. As the voltage is appliedacross the cold filament when the bulb is first turnedon, the resistance of the filament is low, the current is

high, and a relatively large amount of power is deliv-ered to the bulb. This current spike at the beginningof operation is the reason why lightbulbs often fail justafter they are turned on. As the filament warms, its resistance rises and the current decreases. As a result,the power delivered to the bulb decreases, and thebulb is less likely to burn out.

Applying Physics 17.2 Lightbulb Failures

A voltage #V is applied across the ends of a nichrome heater wire having a cross-sectional area A and length L . The same voltage is applied across the ends of a sec-ond heater wire having a cross-sectional area A and length 2L. Which wire getshotter? (a) the shorter wire, (b) the longer wire, or (c) more information isneeded.

Quick Quiz 17.6

For the two resistors shown inFigure 17.12, rank the currentsat points a through f fromlargest to smallest.

(a) Ia ! Ib $ Ie ! If $ Ic ! Id(b) Ia ! Ib $ Ic ! Id $ Ie ! If(c) Ie ! If $ Ic ! Id $ Ia ! Ib

Quick Quiz 17.7

Two resistors, A and B, are connected in a series circuit with a battery. The resist-ance of A is twice that of B. Which resistor dissipates more power? (a) resistor A(b) resistor B (c) More information is needed.

Quick Quiz 17.8

#V

30 W

60 W

e f

c d

a b

A

B

Figure 17.12 (QuickQuiz 17.7)

TIP 17.3 MisconceptionAbout CurrentCurrent is not “used up” in a resistor.Rather, some of the energy thecharges have received from the voltage source is delivered to the resistor, making it hot and causing itto radiate. Also, the current doesn’tslow down when going through theresistor: it’s the same throughout thecircuit.

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