power These lines transfer energy from the power company to homes and businesses. The energy is transferred at a very high voltage, possibly hundreds of thousands of volts in some cases. The high voltage results in less loss of power due to resistance in the wires, so it is used despite the fact that it makes power lines very dangerous.
17
O U T L I N E
CHAPTER
Telegraph Colour Library/FPG/Getty Images
17.1 17.2 17.3 17.4 17.5 17.6 17.7 17.8 17.9
Electric Current A Microscopic View: Current and Drift Speed Current and Voltage Measurements in Circuits Resistance and Ohm's Law Resistivity Temperature Variation of Resistance Superconductors Electrical Energy and Power Electrical Activity in the Heart
Current and Resistance
Many practical applications and devices are based on the principles of static electricity, but electricity was destined to become an inseparable part of our daily lives when scientists learned how to produce a continuous flow of charge for relatively long periods of time using batteries. The battery or voltaic cell was invented in 1800 by the Italian physicist Alessandro Volta. Batteries supplied a continuous flow of charge at low potential, in contrast to earlier electrostatic 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 how to 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 gasoline in automobile engines, travel through miniature components making up the chips of microcomputers, and provide the power for countless other invaluable tasks. In this chapter we define current and discuss some of the factors that contribute to the resistance to the flow of charge in conductors. We also discuss energy transformations in electric circuits. These topics will be the foundation for additional work with circuits in later chapters.
17.1 ELECTRIC CURRENT
In 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 is the rate at which charge flows through this surface. Suppose Q is the amount of charge that flows through an area A in a time 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 time interval:
568
17.1
Electric Current
569
I
Q t
[17.1]
TIP 17.1 Current Flow is Redundant
The phrases flow of current and current flow are commonly used, but here the word flow is redundant because current is already defined as a flow (of charge). Avoid this construction!
SI unit: coulomb/second (C/s), or the ampere (A). One ampere of current is equivalent to one coulomb of charge passing through the 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, negative, or both. The direction of conventional current used in this book is the direction positive charges flow. (This historical convention originated about 200 years ago, when the ideas of positive and negative charges were introduced.) In a common conductor such as copper, the current is due to the motion of negatively charged electrons, so the direction of the current is opposite the direction of motion of the electrons. 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 positive and negative charges. Moving charges, whether positive or negative, are referred to as charge carriers. In a metal, for example, the charge carriers are electrons. In electrostatics, where charges are stationary, the electric potential is the same everywhere in a conductor. This is no longer true for conductors carrying current: as charges move along a wire, the electric potential is continually decreasing (except in the special case of superconductors).
Direction of current
+ + + + A I
Figure 17.1 Charges in motion through an area A. The time rate of flow of charge through the area is defined as the current I. The direction of the current is the direction of flow of positive charges.
+
EXAMPLE 17.1 Turn on the Light
Goal 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 get the total charge that passes in that time. The total charge equals the number N of electrons going through the circuit times 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 the filament in 5.00 s. The total number N of electrons times the charge per electron equals the total charge, I t: Substitute and solve forN: (1) Nq I 10 t
19
I
Q t
1.67 C 2.00 s
0.835 A
N(1.60 N 2.61
C/electron)
(0.835 A)(5.00 s)
1019 electrons
Remarks In developing the solution, it was important to use units to ensure the correctness of equations such as Equation (1). Notice the enormous number of electrons passing through a given point in a typical circuit. Exercise 17.1 Suppose 6.40 1021 electrons pass through a wire in 2.00 min. Find the current.
Answer 8.53 A
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Quick Quiz 17.1
Consider positive and negative charges moving horizontally through the four regions in Figure 17.2. Rank the magnitudes of the currents in these four regions from lowest to highest. (Ia is the current in Figure 17.2a, Ib the current in Figure 17.2b, 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
Figure 17.2 Quiz 17.1) (Quick
x vd A q
+ +
+
+
+ +
+ +
(d)
(a)
(b)
(c)
vd t Figure 17.3 A section of a uniform conductor of cross-sectional area A. The charge carriers move with a speed v d , and the distance they travel in time t is given by x v d t. The number of mobile charge carriers in the section of length x is given by nAvd t, where n is the number of mobile carriers per unit volume.
17.2 A MICROSCOPIC VIEW: CURRENT AND DRIFT SPEED
Macroscopic currents can be related to the motion of the microscopic charge carriers making up the current. It turns out that current depends on the average speed of the charge carriers in the direction of the current, the number of charge carriers per unit volume, and the size of the charge carried by each. Consider identically charged particles moving in a conductor of cross-sectional area 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, then the number of carriers in the volume element is nA x. The mobile charge Q in this element is therefore Q number of carriers charge per carrier (nA x)q
vd
E
ACTIVE FIGURE 17.4 A schematic representation of the zigzag motion of a charge carrier in a conductor. The sharp changes in direction are due to collisions with atoms in the conductor. Note that the net motion of electrons is opposite the direction of the electric field.
where q is the charge on each carrier. If the carriers move with a constant average speed 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 t, we see that the current in the con-
If we divide both sides of this equation by ductor is I Q t
nqvd A
[17.2]
Log into PhysicsNow at www.cp7e.com and go to Active Figure 17.4, where you can observe the random zigzag motion of a charge carrier and see how the motion is affected by an electric field.
TIP 17.2 Electrons are Everywhere in the Circuit
Electrons don't have to travel from the light switch to the light for the light to operate. Electrons already in the filament of the lightbulb move in response to the electric field set up by the battery. Also, the battery does not provide electrons to the circuit; it provides energy to the existing electrons.
To understand the meaning of drift speed, consider a conductor in which the charge carriers are free electrons. If the conductor is isolated, these electrons undergo random motion similar to the motion of the molecules of a gas. The drift speed is normally much smaller than the free electrons' average speed between collisions with the fixed atoms of the conductor. When a potential difference is applied between the ends of the conductor (say, with a battery), an electric field is set up in the conductor, creating an electric force on the electrons and hence a current. In reality, the electrons don't simply move in straight lines along the conductor. Instead, they undergo repeated collisions with the atoms of the metal, and the result is a complicated zigzag motion with only a small average drift speed along the wire (Active Fig. 17.4). The energy transferred from the electrons to the metal atoms during a collision increases the vibrational energy of the atoms and causes a corresponding increase in the temperature of the conductor. Despite the collisions, however, the electrons move slowly along the conductor in a direction : opposite E with the drift velocity :d . v
17.2
A Microscopic View: Current and Drift Speed
571
EXAMPLE 17.2 Drift Speed of Electrons
Goal 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 each copper atom contributes one free electron to the metal, find the drift speed of the electrons in this wire. (b) Use the ideal gas model to compare the drift speed with the random rms speed an electron would have at 20.0C. The density 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 volume. We can find n by recalling that one mole of copper contains an Avogadro's number (6.02 1023) of atoms and each atom contributes one charge carrier to the metal. The volume of one mole can be found from copper's known density and atomic mass. The atomic mass is the same, numerically, as the number of grams in a mole of the substance. Solution (a) Find the drift speed of the electrons. Calculate the volume of one mole of copper from its density and its atomic mass: Convert the volume from cm3 to m3: V m 63.5 g 8.92 g/cm3 1m 102 cm
3
7.12 cm3
7.12 cm3
7.12
10
6
m3
Divide Avogadro's number (the number of electrons in one mole) by the volume per mole to obtain the number density: Solve Equation 17.2 for the drift speed, and substitute:
n
6.02 1023 electrons/mole 7.12 10 6 m3/mole 8.46 1028 electrons/m3
I nqA (8.46 10.0 C/s 1028 electrons/m3)(1.60 10
19
vd
C)(3.00
10
6
m2)
vd (b) Find the rms speed of a gas of electrons at 20.0C. Apply Equation 10.18: v rms
2.46
10
4
m/s
Convert the temperature to the Kelvin scale, and substitute values:
v rms
3k BT me 3(1.38 9.11 105 m/s 10
23
J/K)(293 K ) 10 31 kg
1.15
Remarks The drift speed of an electron in a wire is very small -- only about one-billionth of its random thermal speed. Exercise 17.2 What current in a copper wire with a cross-sectional area of 7.50 5.00 10 4 m/s? Answer 5.08 A
7
10
m2 would result in a drift speed of
Example 17.2 shows that drift speeds are typically very small. In fact, the drift speed is much smaller than the average speed between collisions. Electrons traveling at 2.46 10 4 m/s, as in the example, would take about 68 min to travel 1 m!
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In view of this low speed, you might wonder why a light turns on almost instantaneously when a switch is thrown. Think of the flow of water through a pipe. If a drop of water is forced into one end of a pipe that is already filled with water, a drop must be pushed out the other end of the pipe. Although it may take an individual drop a long time to make it through the pipe, a flow initiated at one end produces a similar flow at the other end very quickly. Another familiar analogy is the motion of a bicycle chain. When the sprocket moves one link, the other links all move more or less immediately, even though it takes a given link some time to make a complete rotation. In a conductor, the electric field driving the free electrons travels at a speed close to that of light, so when you flip a light switch, the message for the electrons to start moving through the wire (the electric field) reaches them at a speed on the order of 108 m/s!
Quick Quiz 17.2
Suppose a current-carrying wire has a cross-sectional area that gradually becomes smaller along the wire, so that the wire has the shape of a very long cone. How does the drift speed vary along the wire? (a) It slows down as the cross section becomes smaller. (b) It speeds up as the cross section becomes smaller. (c) It doesn't change. (d) More information is needed.
17.3 CURRENT AND VOLTAGE MEASUREMENTS IN CIRCUITS
To study electric current in circuits, we need to understand how to measure currents and voltages. The circuit shown in Figure 17.5a is a drawing of the actual circuit necessary for measuring the current in Example 17.1. Figure 17.5b shows a stylized figure called a circuit diagram which represents the actual circuit of Figure 17.5a. This circuit consists of only a battery and a lightbulb. The word "circuit" means "a closed loop
Battery +
c, Michael Dalton, Fundamental Photographs
+
I
0.0 V
Bulb A +
Ammeter I + I (a) I
0.835 A
+
V + Voltmeter (b)
(c) Figure 17.5 (a) A sketch of an actual circuit used to measure the current in a flashlight bulb and the potential difference across it. (b) A schematic diagram of the circuit shown in part (a). (c) A digital multimeter can be used to measure both currents and potential differences. Here, the meter is measuring the potential difference across a 9-V battery.
17.4
Resistance and Ohm's Law
573
of some sort around which current circulates." The battery pumps charge through the bulb and around the loop. No charge would flow without a complete conducting path from the positive terminal of the battery into one side of the bulb, out the other side, and through the copper conducting wires back to the negative terminal of the battery. The most important quantities that characterize how the bulb works in different situations are the current I in the bulb and the potential difference V across the bulb. To measure the current in the bulb, we place an ammeter, the device for measuring current, in line with the bulb so there is no path for the current to bypass the meter; all of the charge passing through the bulb must also 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 simultaneously as in Figure 17.5a, we can remove the voltmeter and see if its presence affects the current reading. Figure 17.5c shows a digital multimeter -- a convenient device, with a digital readout, that can be used to measure voltage, current, or resistance. An advantage of using a digital multimeter as a voltmeter is that it will usually not affect the current, since a digital meter has enormous resistance to the flow of charge in the voltmeter mode. At this point, you can measure the current as a function of voltage (an I V curve) of various devices in the lab. All you need is a variable voltage supply (an adjustable battery) capable of supplying potential differences from about 5 V to 5 V, a bulb, a resistor, some wires and alligator clips, and a couple of multimeters. Be sure to always start your measurements using the highest multimeter scales (say, 10 A and 1 000 V), and increase the sensitivity one scale at a time to obtain the highest accuracy without overloading the meters. (Increasing the sensitivity means lowering the maximum current or voltage that the scale reads.) Note that the meters must be connected with the proper polarity with respect to the voltage supply, as shown in Figure 17.5b. Finally, follow your instructor's directions carefully to avoid damaging the meters and incurring a soaring lab fee.
Quick Quiz 17.3
Look at the four "circuits" shown in Figure 17.6 and select those that will light the bulb.
+ + + +
AMPS
+
(a) Figure 17.6 (Quick Quiz 17.3)
(b)
(c)
(d)
l
A
I Va E
17.4 RESISTANCE AND OHM'S LAW
When a voltage (potential difference) V is applied across the ends of a metallic conductor as in Figure 17.7, the current in the conductor is found to be proportional 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 the conductor. In fact, we define the resistance as the ratio of the voltage across the conductor to the current it carries: R V I [17.3]
Vb
Figure 17.7 A uniform conductor of 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 : E set up in the conductor is also proportional to the current.
Resistance
574
Chapter 17
Current and Resistance
GEORG SIMON OHM (17871854)
A high school teacher in Cologne and later a professor at Munich, Ohm formulated the concept of resistance and discovered the proportionalities expressed in Equation 17.5.
Resistance has SI units of volts per ampere, called ohms ( ). If a potential difference of 1 V across a conductor produces a current of 1 A, the resistance of the conductor is 1 . For example, if an electrical appliance connected to a 120-V source carries a current of 6 A, its resistance is 20 . The concepts of electric current, voltage, and resistance can be compared with the flow of water in a river. As water flows downhill in a river of constant width and depth, the flow rate (water current) depends on the steepness of descent of the river and the effects of rocks, the riverbank, and other obstructions. The voltage difference is analogous to the steepness, and the resistance to the obstructions. Based on this analogy, it seems reasonable that increasing the voltage applied to a circuit should increase the current in the circuit, just as increasing the steepness of descent increases the water current. Also, increasing the obstructions in the river's path will reduce the water current, just as increasing the resistance in a circuit will lower the electric current. Resistance in a circuit arises due to collisions between the electrons carrying the current with fixed atoms inside the conductor. These collisions inhibit the movement of charges in much the same way as would a force of friction. For many materials, including most metals, experiments show that the resistance remains constant over a wide range of applied voltages or currents. This statement is known as Ohm's law, after Georg Simon Ohm (1789 1854), who was the first to conduct a systematic study of electrical resistance. Ohm's law is given by V IR [17.4]
Ohm's law
Bettmann/CORBIS
where R is understood to be independent of V, the potential drop across the resistor, and I, the current in the resistor. We will continue to use this traditional form of Ohm's law when discussing electrical circuits. A resistor is a conductor that provides a specified resistance in an electric circuit. The symbol for a resistor in circuit diagrams is a zigzag line: . Ohm's law is an empirical relationship valid only for certain materials. Materials that obey Ohm's law, and hence have a constant resistance over a wide range of voltages, are said to be ohmic. Materials having resistance that changes with voltage or current are nonohmic. Ohmic materials have a linear current voltage relationship over a large range of applied voltages (Fig. 17.8a). Nonohmic materials have a nonlinear current voltage relationship (Fig. 17.8b). One common semiconducting device that is nonohmic is the diode, a circuit element that acts like a one-way valve for current. Its resistance is small for currents in one direction (positive V ) and large for currents in the reverse direction (negative V ). Most modern electronic devices, such as transistors, have nonlinear current voltage relationships; their operation depends on the particular ways in which they violate Ohm's law.
I Slope = 1 R
Figure 17.8 (a) The current voltage curve for an ohmic material. The curve is linear, and the slope gives the resistance of the conductor. (b) A nonlinear current voltage curve for a semiconducting diode. This device doesn't obey Ohm's law.
I
V
V
(a)
(b)
17.5
Resistivity
575
Quick Quiz 17.4
Courtesy of Henry Leap and Jim Lehman
In Figure 17.8b, does the resistance of the diode (a) increase or (b) decrease as the positive voltage V increases?
An assortment of resistors used for a variety of applications in electronic circuits.
EXAMPLE 17.3 Resistance of a Steam Iron
Goal 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. Solution Apply Equation 17.3: Exercise 17.3 The resistance of a hot plate is 48.0 source? Answer 2.50 A V I 1.20 102 V 6.40 A
R
18.8
. How much current does the plate carry when connected to a 1.20
102-V
17.5 RESISTIVITY
Electrons don't move in straight-line paths through a conductor. Instead, they undergo repeated collisions with the metal atoms. Consider a conductor with a voltage applied across its ends. An electron gains speed as the electric force associated with the internal electric field accelerates it, giving it a velocity in the direction opposite that of the electric field. A collision with an atom randomizes the electron's velocity, reducing it in the direction opposite the field. The process then repeats itself. Together, these collisions affect the electron somewhat as a force of internal friction would. This is the origin of a material's resistance. The resistance of an ohmic conductor increases with length, which makes sense because the electrons going through it must undergo more collisions in a longer conductor. A smaller cross-sectional area also increases the resistance of a conductor, just as a smaller pipe slows the fluid moving through it. The resistance, then, is proportional to the conductor's length l and inversely proportional to its crosssectional area A, R l A [17.5]
where the constant of proportionality, , is called the resistivity of the material.1 Every material has a characteristic resistivity that depends on its electronic structure and on temperature. Good electric conductors have very low resistivities, and good insulators have very high resistivities. Table 17.1 lists the resistivities of various materials at 20C. Because resistance values are in ohms, resistivity values must be in ohm-meters ( m).
1The
symbol used for resistivity shouldn't be confused with the same symbol used earlier in the book for density. Often, a single symbol is used to represent different quantities.
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TABLE 17.1
Resistivities and Temperature Coefficients of Resistivity for Various Materials (at 20C)
Material Silver Copper Gold Aluminum Tungsten Iron Platinum Lead Nichromea Carbon Germanium Silicon Glass Hard rubber Sulfur Quartz (fused)
aA
Resistivity ( m) 1.59 10 8 1.7 10 8 2.44 10 8 2.82 10 8 5.6 10 8 10.0 10 8 11 10 8 22 10 8 150 10 8 3.5 105 0.46 640 101014 10 1013 1015 75 1016
Temperature Coefficient of Resistivity [(C) 1] 3.8 3.9 3.4 3.9 4.5 5.0 3.92 3.9 0.4 0.5 48 75 10 10 10 10 10 10 10 10 10 10 10 10
3 3 3 3 3 3 3 3 3 3 3 3
nickel chromium alloy commonly used in heating elements.
Applying Physics 17.1 Dimming of Aging Lightbulbs
As a lightbulb ages, why does it gives off less light than when new? Explanation There are two reasons for the lightbulb's behavior, one electrical and one optical, but both are related to the same phenomenon occurring within the bulb. The filament of an old lightbulb is made of a tungsten wire that has been kept at a high temperature for many hours. High temperatures evaporate tungsten from the filament, decreasing its radius. From R l/A, we see that a decreased cross-sectional area leads to an increase in the resistance of the filament. This increasing resistance with age means that the filament will carry less current for the same applied voltage. With less current in the filament, there is less light output, and the filament glows more dimly. At the high operating temperature of the filament, tungsten atoms leave its surface, much as water molecules evaporate from a puddle of water. The atoms are carried away by convection currents in the gas in the bulb and are deposited on the inner surface of the glass. In time, the glass becomes less transparent because of the tungsten coating, which decreases the amount of light that passes through the glass.
INTERACTIVE EXAMPLE 17.4 The Resistance of Nichrome Wire
Goal 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 potential difference of 10.0 V is maintained across a 1.00-m length of the nichrome wire, what is the current in the wire? (c) The wire is melted down and recast with twice its original length. Find the new resistance R N as a multiple of the old resistance R O . Strategy Part (a) requires substitution into Equation 17.5, after calculating the cross-sectional area, while part (b) is a matter of substitution into Ohm's law. Part (c) requires some algebra. The idea is to take the expression for the new resistance and substitute expressions for lN and AN , the new length and cross-sectional area, in terms of the old length and cross-section. For the area substitution, use the fact that the volumes of the old and new wires are the same.
17.6
Temperature Variation of Resistance
577
Solution (a) Calculate the resistance per unit length. Find the cross-sectional area of the wire: Obtain the resistivity of nichrome from Table 17.1, solve Equation 17.5 for R/l, and substitute: (b) Find the current in a 1.00-m segment of the wire if the potential difference across it is 10.0 V. Substitute given values into Ohm's law: (c) If the wire is melted down and recast with twice its original length, find the new resistance as a multiple of the old. Find the new area AN in terms of the old area A O , using the fact the volume doesn't change and lN 2l O : VN AN RN VO : AN lN A O (l O /2l O) lN AN AO lO : AN A O /2 4 lO AO 4R O AO(l O /l N) I V R 10.0 V 4.6 2.2 A A R l r2 (0.321 10
3
m)2
3.24 4.6
10 /m
7
m2
A
1.5 10 6 m 3.24 10 7 m2
Substitute into Equation 17.5:
(2l O) (AO/2)
Remarks From Table 17.1, the resistivity of nichrome is about 100 times that of copper, a typical good conductor. Therefore, a copper wire of the same radius would have a resistance per unit length of only 0.052 /m, and a 1.00-m length of copper wire of the same radius would carry the same current (2.2 A) with an applied voltage of only 0.115 V. Because of its resistance to oxidation, nichrome is often used for heating elements in toasters, irons, and electric heaters. Exercise 17.4 What is the resistance of a 6.0-m length of nichrome wire that has a radius 0.321 mm? How much current does it carry when connected to a 120-V source? Answer 28 ; 4.3 A
You can explore the resistance of different materials by logging into PhysicsNow at www.cp7e.com and going to Interactive Example 17.4.
Quick Quiz 17.5
Suppose an electrical wire is replaced with one having every linear dimension doubled (i.e. the length and radius have twice their original values). Does the wire now have (a) more resistance than before, (b) less resistance, or (c) the same resistance?
17.6 TEMPERATURE VARIATION OF RESISTANCE
The resistivity , and hence the resistance, of a conductor depends on a number of factors. One of the most important is the temperature of the metal. For most metals, resistivity increases with increasing temperature. This correlation can be understood as follows: as the temperature of the material increases, its constituent atoms vibrate with greater amplitudes. As a result, the electrons find it more difficult to get by those atoms, just as it is more difficult to weave through a crowded
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Courtesy of Central Scientific Company
room when the people are in motion than when they are standing still. The increased electron scattering with increasing temperature results in increased resistivity. Technically, thermal expansion also affects resistance; however, this is a very small effect. Over a limited temperature range, the resistivity of most metals increases linearly with increasing temperature according to the expression
0[1
(T
T0)]
[17.6]
In an old-fashioned carbon filament incandescent lamp, the electrical resistance is typically 10 , but changes with temperature.
where is the resistivity at some temperature T (in Celsius degrees), 0 is the resistivity at some reference temperature T0 (usually taken to be 20C), and is a parameter called the temperature coefficient of resistivity. Temperature coefficients for various materials are provided in Table 17.1. The interesting negative values of for semiconductors arise because these materials possess weakly bound charge carriers that become free to move and contribute to the current as the temperature rises. Because the resistance of a conductor with a uniform cross section is proportional to the resistivity according to Equation 17.5 (R l/A), the temperature variation of resistance can be written R R 0[1 (T T0)] [17.7]
Precise temperature measurements are often made using this property, as shown by the following example.
EXAMPLE 17.5 A Platinum Resistance Thermometer
Goal Apply the temperature dependence of resistance. Problem A resistance thermometer, which measures temperature by measuring the change in resistance of a conductor, is made of platinum and has a resistance of 50.0 at 20.0C. (a) When the device is immersed in a vessel containing 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 235C. What is the ratio of the new current in the platinum to the current I mp at the melting point? Strategy In part (a), solve Equation 17.7 for T T0 and get quantities. 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: T T0 R R0 R0 137 C 76.8 [3.92 10
3
for platinum from Table 17.1, substituting known
50.0 ( C) 1][50.0
]
Substitute T0 indium:
20.0C and obtain the melting point of
T
157C
(b) Find the ratio of the new current to the old when the temperature rises from 157C to 235C. Write Equation 17.7, with R 0 and T 0 replaced by R mp and T mp , the resistance and temperature at the melting point. According to Ohm's law, R V/I and R mp V/I mp. Substitute these expressions into Equation 17.7: Cancel the voltage differences, invert the two expressions, and then divide both sides by I mp : R R mp[1 (T Tmp )]
V I I I mp
V Imp
[1
(T
Tmp)]
1
1 (T
Tmp)
17.7
Superconductors
579
Substitute T 235C, Tmp 157C, and the value for , obtaining the desired ratio:
I Imp
0.766
Exercise 17.5 Suppose a wire made of an unknown alloy and having a temperature of 20.0C carries a current of 0.450 A. At 52.0C the current is 0.370 A for the same potential difference. Find the temperature coefficient of resistivity of the alloy. Answer 6.76 10
3
(C)
1
17.7 SUPERCONDUCTORS
There is a class of metals and compounds with resistances that fall virtually to zero below a certain temperature Tc called the critical temperature. These materials are known as superconductors. The resistance temperature graph for a superconductor follows that of a normal metal at temperatures above Tc (Fig. 17.9). When the temperature is at or below Tc , however, the resistance suddenly drops to zero. This phenomenon was discovered in 1911 by the Dutch physicist H. Kamerlingh Onnes as he and a graduate student worked with mercury, which is a superconductor below 4.1 K. Recent measurements have shown that the resistivities of superconductors below Tc are less than 4 10 25 m -- around 1017 times smaller than the resistivity of copper and in practice considered to be zero. Today thousands of superconductors are known, including such common metals as aluminum, tin, lead, zinc, and indium. Table 17.2 lists the critical temperatures of several superconductors. The value of Tc is sensitive to chemical composition, pressure, and crystalline structure. Interestingly, copper, silver, and gold, which are excellent conductors, don't exhibit superconductivity. One of the truly remarkable features of superconductors is the fact that once a current is set up in them, it persists without any applied voltage (because R 0). In fact, steady currents in superconducting loops have been observed to persist for years with no apparent decay! An important development in physics that created much excitement in the scientific community was the discovery of high-temperature copper-oxide-based superconductors. The excitement began with a 1986 publication by J. Georg Bednorz and K. Alex Mller, scientists at the IBM Zurich Research Laboratory in Switzerland, in which they reported evidence for superconductivity at a temperature near 30 K in an oxide of barium, lanthanum, and copper. Bednorz and Mller were awarded the Nobel Prize for physics in 1987 for their important discovery. The discovery was remarkable in view of the fact that the critical temperature was significantly higher than those of any previously known superconductors. Shortly thereafter a new family of compounds was investigated, and research activity in the field of superconductivity proceeded vigorously. In early 1987, groups at the University of Alabama at Huntsville and the University of Houston announced the discovery of superconductivity at about 92 K in an oxide of yttrium, barium, and copper (YBa2Cu3O7), shown as the gray disk in Figure 17.10. Late in 1987, teams of scientists from Japan and the United States reported superconductivity at 105 K in an oxide of bismuth, strontium, calcium, and copper. More recently, scientists have reported superconductivity at temperatures as high as 150 K in an oxide containing mercury. The search for novel superconducting materials continues, with the hope of someday obtaining a room-temperature superconducting material. This research is important both for scientific reasons and for practical applications. An important and useful application is the construction of superconducting magnets in which the magnetic field intensities are about ten times greater than those of the best normal electromagnets. Such magnets are being considered as a means of storing energy. The idea of using superconducting power lines to transmit
R() 0.15
0.10
0.05
Tc 0.00
4.0
4.2 T(K)
4.4
Figure 17.9 Resistance versus temperature for a sample of mercury (Hg). The graph follows that of a normal metal above the critical temperature Tc . The resistance drops to zero at the critical temperature, which is 4.2 K for mercury, and remains at zero for lower temperatures.
TABLE 17.2
Critical Temperatures for Various Superconductors
Material Zn Al Sn Hg Pb Nb Nb3Sn Nb3Ge YBa2Cu3O7 Bi Sr Ca Cu O Tl Ba Ca Cu O HgBa2Ca2Cu3O8 Tc (K) 0.88 1.19 3.72 4.15 7.18 9.46 18.05 23.2 90 105 125 134
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power efficiently is also receiving serious consideration. Modern superconducting electronic devices consisting of two thin-film superconductors separated by a thin insulator have been constructed. Among these devices are magnetometers (magnetic-field measuring devices) and various microwave devices.
Courtesy of IBM Research Laboratory
17.8 ELECTRICAL ENERGY AND POWER
If a battery is used to establish an electric current in a conductor, chemical energy stored in the battery is continuously transformed into kinetic energy of the charge carriers. This kinetic energy is quickly lost as a result of collisions between the charge carriers and fixed atoms in the conductor, causing an increase in the temperature of the conductor. In this way, the chemical energy stored in the battery is continuously transformed into thermal energy. In order to understand the process of energy transfer in a simple circuit, consider a battery with terminals connected to a resistor (Active Fig. 17.11; remember that the positive terminal of the battery is always at the higher potential). Now imagine following a quantity of positive charge Q around the circuit from point A, through the battery and resistor, and back to A. Point A is a reference point that is grounded (the ground symbol is ), and its potential is taken to be zero. As the charge Q moves from A to B through the battery, the electrical potential energy of the system increases by the amount Q V, and the chemical potential energy 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, it loses this electrical potential energy during collisions with atoms in the resistor. In the process, the energy is transformed to internal energy corresponding to increased vibrational motion of those atoms. Because we can ignore the very small resistance of the interconnecting wires, no energy transformation occurs for paths BC and DA. When the charge returns to point A, the net result is that some of the chemical energy in the battery has been delivered to the resistor and has caused its temperature to rise. The charge Q loses energy Q V as it passes through the resistor. If t is the time it takes the charge to pass through the resistor, then the rate at which it loses electric potential energy is Q t V I V
Figure 17.10 A small permanent magnet floats freely above a ceramic disk made of the superconductor YBa2Cu3O7, cooled by liquid nitrogen at 77 K. The superconductor has zero electric resistance at temperatures below 92 K and expels any applied magnetic field. I
B + A R
C
D
ACTIVE FIGURE 17.11 A circuit consisting of a battery and a resistance R. Positive charge flows clockwise from the positive to the negative terminal of the battery. Point A is grounded.
Log into PhysicsNow at www.cp7e.com and go to Active Figure 17.11, where you can adjust the battery voltage and the resistance, and see the resulting current in the circuit and the power dissipated as heat by the resistor.
where I is the current in the resistor and V is the potential difference across it. Of course, the charge regains this energy when it passes through the battery, at the expense of chemical energy in the battery. The rate at which the system loses potential energy as the charge passes through the resistor is equal to the rate at which the system gains internal energy in the resistor. Therefore, the power , representing the rate at which energy is delivered to the resistor, is I V [17.8]
Power
While this result was developed by considering a battery delivering energy to a resistor, Equation 17.8 can be used to determine the power transferred from a voltage source to any device carrying a current I and having a potential difference V between its terminals. Using Equation 17.8 and the fact that V IR for a resistor, we can express the power delivered to the resistor in the alternate forms
Power delivered to a resistor
I 2R
V2 R
[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
17.8
Electrical Energy and Power
581
often referred to as an I 2R loss. Note that Equation 17.9 applies only to resistors and not to nonohmic devices such as lightbulbs and diodes. Regardless of the ways in which you use electrical energy in your home, you ultimately must pay for it or risk having your power turned off. The unit of energy used by electric companies to calculate consumption, the kilowatt-hour, is defined in terms of the unit of power and the amount of time it's supplied. One kilowatthour (kWh) is the energy converted or consumed in 1 h at the constant rate of 1 kW. It has the numerical value 1 kWh (103 W)(3600 s) 3.60 106 J [17.10]
TIP 17.3 Misconception About Current
Current is not "used up" in a resistor. Rather, some of the energy the charges have received from the voltage source is delivered to the resistor, making it hot and causing it to radiate. Also, the current doesn't slow down when going through the resistor: it's the same throughout the circuit.
On an electric bill, the amount of electricity used in a given period is usually stated in multiples of kilowatt-hours.
Applying Physics 17.2 Lightbulb Failures
Why lightbulbs do fail so often right after they're turned on? Explanation Once the switch is closed, the line voltage is applied across the bulb. As the voltage is applied across the cold filament when the bulb is first turned on, the resistance of the filament is low, the current is high, and a relatively large amount of power is delivered to the bulb. This current spike at the beginning of operation is the reason why lightbulbs often fail just after 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 the bulb is less likely to burn out.
Quick Quiz 17.6
A voltage V is applied across the ends of a nichrome heater wire having a crosssectional area A and length L. The same voltage is applied across the ends of a second heater wire having a cross-sectional area A and length 2L. Which wire gets hotter? (a) the shorter wire, (b) the longer wire, or (c) more information is needed.
Quick Quiz 17.7
For the two resistors shown in Figure 17.12, rank the currents at points a through f from largest to smallest. (a) Ia (b) Ia (c) Ie Ib Ib If Ie Ic Ic If Id Id Ic Ie Ia Id If Ib
e
30 W A f
60 W c B d
a V
b
Figure 17.12 Quiz 17.7)
(Quick
Quick Quiz 17.8
Two resistors, A and B, are connected in a series circuit with a battery. The resistance of A is twice that of B. Which resistor dissipates more power? (a) resistor A (b) resistor B (c) More information is needed.
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Example 17.6 The Cost of Lighting Up Your Life
Goal Apply the electric power concept, and calculate the cost of power usage using kilowatt-hours. Problem A circuit provides a maximum current of 20.0 A at an operating voltage of 1.20 102 V. (a) How many 75 W bulbs can operate with this voltage source? (b) At $0.120 per kilowatt-hour, how much does it cost to operate these bulbs for 8.00 h? Strategy Find the necessary power with I V, then divide by 75.0 W per bulb to get the total number of bulbs. To find the cost, convert power to kilowatts and multiply by the number of hours, then multiply by the cost per kilowatt-hour. Solution (a) Find the number of bulbs that can be lighted. Substitute into Equation 17.8 to get the total power: Divide the total power by the power per bulb to get the number of bulbs. (b) Calculate the cost of this electricity for an 8.00-h day. Find the energy in kilowatt-hours: Energy t (2.40 103 W) 1.00 kW 1.00 103 W (8.00 h)
total
I V
(20.0 A)(1.20
total bulb
102 V)
2.40
103 W 32.0
Number of bulbs
2.40 103 W 75.0 W
19.2 kWh Multiply by the cost per kilowatt-hour: Cost (19.2 kWh)($0.12/kWh) $2.30
Remarks This amount of energy might correspond to what a small office uses in a working day, taking into account all power requirements (not just lighting). In general, resistive devices can have variable power output, depending on how the circuit is wired. Here, power outputs were specified, so such considerations were unnecessary. Exercise 17.6 (a) How many Christmas tree lights drawing 5.00 W of power each could be run on a circuit operating at 1.20 102 V and providing 15.0 A of current? (b) Find the cost to operate one such string 24.0 h per day for the Christmas season (two weeks), using the rate $0.12/kWh. Answers (a) 3.60 102 bulbs (b) $72.60
EXAMPLE 17.7 The Power Converted by an Electric Heater
Goal Calculate an electrical power output, and link to its effect on the environment through the first law of thermodynamics. Problem An electric heater is operated by applying a potential difference of 50.0 V to a nichrome wire of total resistance 8.00 . (a) Find the current carried by the wire and the power rating of the heater. (b) Using this heater, how long would it take to heat 2.50 103 moles of diatomic gas (e.g., a mixture of oxygen and nitrogen -- air) from a chilly 10.0C to 25.0C? Take the molar specific heat at constant volume of air to be 5 R . 2 Strategy For part (a), find the current with Ohm's law and substitute into the expression for power. Part (b) is an isovolumetric process, so the thermal energy provided by the heater all goes into the change in internal energy, U. Calculate this quantity using the first law of thermodynamics, and divide by the power to get the time. Solution (a) Compute the current and power output. Apply Ohm's law to get the current: I V R 50.0 V 8.00 6.25 A
17.9
Electrical Activity in The Heart
583
Substitute into Equation 17.9 to find the power: (b) How long does it take to heat the gas? Calculate the thermal energy transfer from the first law. Note that W 0 because the volume doesn't change. Q
I 2R
(6.25 A)2(8.00
)
313 W
U nCv T (2.50 103 mol)(5 8.31 J/mol K)(298 K 2 7.79 105 J Q 7.79 105 J 313 W 2.49 103 s
283 K)
Divide the thermal energy by the power, to get the time:
t
Remarks The number of moles of gas given here is approximately what would be found in a bedroom. Warming the air with this space heater requires only about forty minutes. However, the calculation doesn't take into account conduction losses. Recall that a 20-cm-thick concrete wall, as calculated in Chapter 11, permitted the loss of over two megajoules an hour by conduction! Exercise 17.7 A hot-water heater is rated at 4.50 103 W and operates at 2.40 102 V. (a) Find the resistance in the heating element, and the current. (b) How long does it take to heat 125 L of water from 20.0C to 50.0C, neglecting conduction and other losses? Answers (a) 12.8 , 18.8 A (b) 3.49 103 s
17.9 ELECTRICAL ACTIVITY IN THE HEART
Electrocardiograms
Every action involving the body's muscles is initiated by electrical activity. The voltages produced by muscular action in the heart are particularly important to physicians. Voltage pulses cause the heart to beat, and the waves of electrical excitation that sweep across the heart associated with the heartbeat are conducted through the body via the body fluids. These voltage pulses are large enough to be detected by suitable monitoring equipment attached to the skin. A sensitive voltmeter making good electrical contact with the skin by means of contacts attached with conducting paste can be used to measure heart pulses, which are typically of the order of 1 mV at the surface of the body. The voltage pulses can be recorded on an instrument called an electrocardiograph, and the pattern recorded by this instrument is called an electrocardiogram (EKG). In order to understand the information contained in an EKG pattern, it is necessary first to describe the underlying principles concerning electrical activity in the heart. The right atrium of the heart contains a specialized set of muscle fibers called the SA (sinoatrial) node that initiates the heartbeat (Fig. 17.13). Electric impulses that originate in these fibers gradually spread from cell to cell throughout the right and left atrial muscles, causing them to contract. The pulse that passes through the muscle cells is often called a depolarization wave because of its effect on individual cells. If an individual muscle cell were examined in its resting state, a double-layer electric charge distribution would be found on its surface, as shown in Figure 17.14a (page 584). The impulse generated by the SA node momentarily and locally allows positive charge on the outside of the cell to flow in and neutralize the negative charge on the inside layer. This effect changes the cell's charge distribution to that shown in Figure 17.14b. Once the depolarization wave has passed through an individual heart muscle cell, the cell recovers the resting-state charge distribution (positive out, negative in) shown in Figure 17.14a in about 250 ms. When the impulse reaches the atrioventricular (AV) node (Fig. 17.13), the muscles of the atria begin to relax, and the pulse is directed to the ventricular A P P L I C AT I O N Electrocardiograms
Sinoatrial (SA) node
Purkinje fibers
LA RA LV RV
Atrioventricular (AV) node Figure 17.13 The electrical conduction system of the human heart. (RA: right atrium; LA: left atrium; RV: right ventricle; LV: left ventricle.)
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Figure 17.14 (a) Charge distribution of a muscle cell in the atrium before a depolarization wave has passed through the cell. (b) Charge distribution as the wave passes.
Depolarization wave front (a) (b)
1.0 Voltage (mV) 0.5 P 0 Q
R
T
0.5 0 0.2
S 0.4 0.6 Time (s)
Figure 17.15 An EKG response for a normal heart.
A P P L I C AT I O N Cardiac Pacemakers
muscles by the AV node. The muscles of the ventricles contract as the depolarization wave spreads through the ventricles along a group of fibers called the Purkinje fibers. The ventricles then relax after the pulse has passed through. At this point, the SA node is again triggered and the cycle is repeated. A sketch of the electrical activity registered on an EKG for one beat of a normal heart is shown in Figure 17.15. The pulse indicated by P occurs just before the atria begin to contract. The QRS pulse occurs in the ventricles just before they contract, and the T pulse occurs when the cells in the ventricles begin to recover. EKGs for an abnormal heart are shown in Figure 17.16. The QRS portion of the pattern shown in Figure 17.16a is wider than normal, indicating that the patient may have an enlarged heart. (Why?) Figure 17.16b indicates that there is no constant relationship between the P pulse and the QRS pulse. This suggests a blockage in the electrical conduction path between the SA and AV nodes which results in the atria and ventricles beating independently and inefficient heart pumping. Finally, Figure 17.16c shows a situation in which there is no P pulse and an irregular spacing between the QRS pulses. This is symptomatic of irregular atrial contraction, which is called fibrillation. In this condition, the atrial and ventricular contractions are irregular. As noted previously, the sinoatrial node directs the heart to beat at the appropriate rate, usually about 72 beats per minute. However, disease or the aging process can damage the heart and slow its beating, and a medical assist may be necessary in the form of a cardiac pacemaker attached to the heart. This matchboxsized electrical device implanted under the skin has a lead that is connected to the wall of the right ventricle. Pulses from this lead stimulate the heart to maintain its proper rhythm. In general, a pacemaker is designed to produce pulses at a rate of about 60 per minute, slightly slower than the normal number of beats per minute,
R
R
R
P Q S (a)
T
P
P
T P
P
T P
Q (b)
Q
R
R
R
R
R
Figure 17.16
Abnormal EKGs.
(c)
17.9
Electrical Activity in The Heart
585
but sufficient to maintain life. The circuitry basically consists of a capacitor charging up to a certain voltage from a lithium battery and then discharging. The design of the circuit is such that, if the heart is beating normally, the capacitor is not allowed to charge completely and send pulses to the heart.
An Emergency Room in Your Chest
In June 2001, an operation on Vice President Dick Cheney focused attention on the progress in treating heart problems with tiny implanted electrical devices. Aptly termed "an emergency room in your chest" by Cheney's attending physician, devices called Implanted Cardioverter Defibrillators (ICD's) can monitor, record, and logically process heart signals and then supply different corrective signals to hearts beating too slowly, too rapidly, or irregularly. ICD's can even monitor and send signals to the atria and ventricles independently! Figure 17.17a shows a sketch of an ICD with conducting leads that are implanted in the heart. Figure 17.17b shows an actual titanium-encapsulated dual-chamber ICD. The latest ICD's are sophisticated devices capable of a number of functions: 1. monitoring both atrial and ventricular chambers to differentiate between atrial and potentially fatal ventricular arrhythmias, which require prompt regulation; 2. storing about a half hour of heart signals that can easily be read out by a physician; 3. being easily reprogrammed with an external magnetic wand; 4. performing complicated signal analysis and comparison; 5. supplying either 0.25- to 10-V repetitive pacing signals to speed up or slow down a malfunctioning heart, or a high-voltage pulse of about 800 V to halt the potentially fatal condition of ventricular fibrillation, in which the heart quivers rapidly rather than beats (people who have experienced such a high-voltage jolt say that it feels like a kick or a bomb going off in the chest); 6. automatically adjusting the number of pacing pulses per minute to match the patient's activity. ICD's are powered by lithium batteries and have implanted lifetimes of 4 6 years. Some basic properties of these adjustable ICD's are given in Table 17.3 (page 586). In the table, tachycardia means "rapid heartbeat" and bradycardia A P P L I C AT I O N Implanted Cardioverter Defibrillators
Dual-chamber ICD
Blowup of defibrillator/ monitor lead
Courtesy of Medtronic, Inc.
(a)
(b)
FIGURE 17.17 (a) A dual-chamber ICD with leads in the heart. One lead monitors and stimulates the right atrium, and the other monitors and stimulates the right ventricle. (b) Medtronic Dual Chamber ICD.
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TABLE 17.3
Properties of Implanted Cardioverter Defibrillatorsa
Physical Specifications Mass (g) Size (cm) Antitachycardia Pacing Number of Bursts Burst Cycle Length (ms) Number of Pulses per Burst Pulse Amplitude (V) Pulse Width (ms) High-Voltage Defibrillation Pulse energy ( J) Pulse Amplitude (V) Bradycardia Pacing 85 7.3 6.2 1.3 (about five stacked silver dollars) ICD delivers a burst of critically timed low-energy pulses 1 15 200 552 2 20 7.5 or 10 1.0 or 1.9 37 stored/33 delivered 801 A dual-chamber ICD can steadily deliver repetitive pulses to both the atrium and the ventricle 40 100 0.25 7.5 0.05, 0.1 1.5, 1.9
Base Frequency (beats/minute) Pulse Amplitude (V) Pulse Width (ms)
aFor
more information see www.photonicd.com/specs.html.
means "slow heartbeat." A key factor in developing tiny electrical implants that serve as defibrillators is the development of capacitors with relatively large capacitance (125 f ) and small physical size.
SUMMARY
Take a practice test by logging into PhysicsNow at www.cp7e.com and clicking on the Pre-Test link for this chapter. of the charges, and A is the cross-sectional area of the conductor.
17.4 Resistance and Ohm's Law 17.1 Electric Current
I Q t [17.1] The electric current I in a conductor is defined as The resistance R of a conductor is defined as the ratio of the potential difference across the conductor to the current in it: R V I [17.3]
where Q is the charge that passes through a cross section of the conductor in time t. The SI unit of current is the ampere (A); 1 A 1 C/s. By convention, the direction of current is the direction of flow of positive charge.
17.2 A Microscopic View: Current and Drift Speed
The current in a conductor is related to the motion of the charge carriers by I nqvd A [17.2]
The SI units of resistance are volts per ampere, or ohms ( ); 1 1 V/A. Ohm's law describes many conductors, for which the applied voltage is directly proportional to the current it causes. The proportionality constant is the resistance: V IR [17.4]
17.5 Resistivity
If a conductor has length l and cross-sectional area A, its resistance is R l A [17.5]
where n is the number of mobile charge carriers per unit volume, q is the charge on each carrier, vd is the drift speed
Problems
587
where , is an intrinsic property of the conductor called the electrical resistivity. The SI unit of resistivity is the ohmmeter ( m).
17.8 Electrical Energy and Power
If a potential difference V is maintained across an electrical device, the power, or rate at which energy is supplied to the device, is I V as V [17.8]
17.6
Temperature Variation of Resistance
Over a limited temperature range, the resistivity of a conductor varies with temperature according to the expression
0[1
Because the potential difference across a resistor is IR, the power delivered to a resistor can be expressed V2 R
(T
T0)]
[17.6]
where is the temperature coefficient of resistivity and 0 is the resistivity at some reference temperature T0 (usually taken to be 20C). The resistance of a conductor varies with temperature according to the expression R R 0[1 (T T0)] [17.7]
I 2R
[17.9]
A kilowatt-hour is the amount of energy converted or consumed in one hour by a device supplied with power at the rate of 1 kW. This is equivalent to 1 kWh 3.60 106 J [17.10]
CONCEPTUAL QUESTIONS
1. Car batteries are often rated in ampere-hours. Does this unit designate the amount of current, power, energy, or charge that can be drawn from the battery? 2. We have seen that an electric field must exist inside a conductor that carries a current. How is that possible in view of the fact that in electrostatics we concluded that the electric field must be zero inside a conductor? 3. Why don't the free electrons in a metal fall to the bottom of the metal due to gravity? And charges in a conductor are supposed to reside on the surface -- why don't the free electrons all go to the surface? 4. In an analogy between traffic flow and electrical current, what would correspond to the charge Q ? What would correspond to the current I ? 5. Newspaper articles often have statements such as "10 000 volts of electricity surged through the victim's body." What is wrong with this statement? 6. Two lightbulbs are each connected to a voltage of 120 V. One has a power of 25 W, the other 100 W. Which bulb has the higher resistance? Which bulb carries more current? 7. When the voltage across a certain conductor is doubled, the current is observed to triple. What can you conclude about the conductor? 8. There is an old admonition given to experimenters to "keep one hand in the pocket" when working around high voltages. Why is this warning a good idea? 9. What factors affect the resistance of a conductor? 10. Some homes have light dimmers that are operated by rotating a knob. What is being changed in the electric circuit when the knob is rotated? 11. Two wires A and B with circular cross section are made of the same metal and have equal lengths, but the resistance of wire A is three times greater than that of wire B. What is the ratio of their cross-sectional areas? How do the radii compare? 12. What single experimental requirement makes superconducting devices expensive to operate? In principle, can this limitation be overcome? 13. What could happen to the drift velocity of the electrons in a wire and to the current in the wire if the electrons could move through it freely without resistance? 14. Use the atomic theory of matter to explain why the resistance of a material should increase as its temperature increases. 15. When is more power delivered to a lightbulb, just after it is turned on and the glow of the filament is increasing or after it has been on for a few seconds and the glow is steady?
PROBLEMS
1, 2, 3 straightforward, intermediate, challenging = full solution available in Student Solutions Manual/Study Guide = coached problem with hints available at www.cp7e.com = biomedical application carries a current of 2.50 A. Find the drift speed of the electrons in the conductor. 3. A 1.00-V potential difference is maintained across a 10.0- resistor for a period of 20.0 s. What total charge passes through the wire in this time interval? 4. In a particular television picture tube, the measured beam current is 60.0 A. How many electrons strike the screen every second?
Section 17.1 Electric Current Section 17.2 A Microscopic View: Current and Drift Speed 1. If a current of 80.0 mA exists in a metal wire, how many electrons flow past a given cross section of the wire in 10.0 min? Sketch the direction of the current and the direction of the electrons' motion. 2. A certain conductor has 7.50 1028 free electrons per cubic meter, a cross-sectional area of 4.00 10 6 m2, and
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Current and Resistance
5. In the Bohr model of the hydrogen atom, an electron in the lowest energy state moves at a speed of 2.19 106 m/s in a circular path having a radius of 5.29 10 11 m. What is the effective current associated with this orbiting electron? 6. If 3.25 10 3 kg of gold is deposited on the negative electrode of an electrolytic cell in a period of 2.78 h, what is the current in the cell during that period? Assume that the gold ions carry one elementary unit of positive charge. 7. A 200-km-long high-voltage transmission line 2.0 cm in diameter carries a steady current of 1 000 A. If the conductor is copper with a free charge density of 8.5 1028 electrons per cubic meter, how many years does it take one electron to travel the full length of the cable? 8. An aluminum wire carrying a current of 5.0 A has a crosssectional area of 4.0 10 6 m2. Find the drift speed of the electrons in the wire. The density of aluminum is 2.7 g/cm3. (Assume that one electron is supplied by each atom.) 9. If the current carried by a conductor is doubled, what happens to (a) the charge carrier density? (b) the electron drift velocity? Section 17.4 Resistance and Ohm's Law Section 17.5 Resistivity 10. A lightbulb has a resistance of 240 when operating at a voltage of 120 V. What is the current in the bulb? 11. A person notices a mild shock if the current along a path through the thumb and index finger exceeds 80 A. Compare the maximum possible voltage without shock across the thumb and index finger with a dry-skin resistance of 4.0 105 and a wet-skin resistance of 2 000 . 12. Suppose that you wish to fabricate a uniform wire out of 1.00 g of copper. If the wire is to have a resistance R 0.500 , and if all of the copper is to be used, what will be (a) the length and (b) the diameter of the wire? 13. Calculate the diameter of a 2.0-cm length of tungsten filament in a small lightbulb if its resistance is 0.050 . 14. Eighteen-gauge wire has a diameter of 1.024 mm. Calculate the resistance of 15 m of 18-gauge copper wire at 20C. 15. A potential difference of 12 V is found to produce a current of 0.40 A in a 3.2-m length of wire with a uniform radius of 0.40 cm. What is (a) the resistance of the wire? (b) the resistivity of the wire? 16. Make an order-of-magnitude estimate of the cost of one person's routine use of a hair dryer for 1 yr. If you do not use a blow dryer yourself, observe or interview someone who does. State the quantities you estimate and their values. 17. A wire 50.0 m long and 2.00 mm in diameter is connected to a source with a potential difference of 9.11 V, and the current is found to be 36.0 A. Assume a temperature of 20C, and, using Table 17.1, identify the metal out of which the wire is made. 18. A rectangular block of copper has sides of length 10 cm, 20 cm, and 40 cm. If the block is connected to a 6.0-V source across two of its opposite faces, what are (a) the
maximum current and (b) the minimum current that the block can carry? 19. A wire of initial length L 0 and radius r 0 has a measured resistance of 1.0 . The wire is drawn under tensile stress to a new uniform radius of r 0.25r 0. What is the new resistance of the wire? Section 17.6 Temperature Variation of Resistance 20. A certain lightbulb has a tungsten filament with a resistance of 19 when cold and 140 when hot. Assume that Equation 17.7 can be used over the large temperature range involved here, and find the temperature of the filament when it is hot. Assume an initial temperature of 20C. 21. While taking photographs in Death Valley on a day when the temperature is 58.0C, Bill Hiker finds that a certain voltage applied to a copper wire produces a current of 1.000 A. Bill then travels to Antarctica and applies the same voltage to the same wire. What current does he register there if the temperature is 88.0C? Assume that no change occurs in the wire's shape and size. 22. A metal wire has a resistance of 10.00 at a temperature of 20C. If the same wire has a resistance of 10.55 at 90C, what is the resistance of the wire when its temperature is 20C? 23. At 20C, the carbon resistor in an electric circuit connected to a 5.0-V battery has a resistance of 200 . What is the current in the circuit when the temperature of the carbon rises to 80C? 24. A wire 3.00 m long and 0.450 mm2 in cross-sectional area has a resistance of 41.0 at 20C. If its resistance increases to 41.4 at 29.0C, what is the temperature coefficient of resistivity? 25. The copper wire used in a house has a cross-sectional area of 3.00 mm2. If 10.0 m of this wire is used to wire a circuit in the house at 20.0C, find the resistance of the wire at temperatures of (a) 30.0C and (b) 10.0C. 26. A 100-cm-long copper wire of radius 0.50 cm has a potential difference across it sufficient to produce a current of 3.0 A at 20C. (a) What is the potential difference? (b) If the temperature of the wire is increased to 200C, what potential difference is now required to produce a current of 3.0 A? (a) A 34.5-m length of copper wire at 27. 20.0C has a radius of 0.25 mm. If a potential difference of 9.0 V is applied across the length of the wire, determine the current in the wire. (b) If the wire is heated to 30.0C while the 9.0-V potential difference is maintained, what is the resulting current in the wire? 28. A toaster rated at 1 050 W operates on a 120-V household circuit and a 4.00-m length of nichrome wire as its heating element. The operating temperature of this element is 320C. What is the cross-sectional area of the wire? 29. In one form of plethysmograph (a device for measuring volume), a rubber capillary tube with an inside diameter of 1.00 mm is filled with mercury at 20C. The resistance of the mercury is measured with the aid of electrodes sealed into the ends of the tube. If 100.00 cm of the tube is wound in a spiral around a patient's upper arm, the blood flow during a heartbeat causes the arm to expand, stretching the tube to a length of 100.04 cm. From this observation, and
Problems
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assuming cylindrical symmetry, you can find the change in volume of the arm, which gives an indication of blood flow. (a) Calculate the resistance of the mercury. (b) Calculate the fractional change in resistance during the heartbeat. [Hint: The fraction by which the cross-sectional area of the mercury thread decreases is the fraction by which the length increases, since the volume of mercury is constant.] Take Hg 9.4 10 7 m. 30. A platinum resistance thermometer has resistances of 200.0 when placed in a 0C ice bath and 253.8 when immersed in a crucible containing melting potassium. What is the melting point of potassium? [Hint: First determine the resistance of the platinum resistance thermometer at room temperature, 20C.] Section 17.8 Electrical Energy and Power 31. A toaster is rated at 600 W when connected to a 120-V source. What current does the toaster carry, and what is its resistance? 32. If electrical energy costs 12 cents, or $0.12, per kilowatthour, how much does it cost to (a) burn a 100-W lightbulb for 24 h? (b) operate an electric oven for 5.0 h if it carries a current of 20.0 A at 220 V? 33. How many 100-W lightbulbs can you use in a 120-V circuit without tripping a 15-A circuit breaker? (The bulbs are connected in parallel, which means that the potential difference across each lightbulb is 120 V.) 34. A high-voltage transmission line with a resistance of 0.31 /km carries a current of 1 000 A. The line is at a potential of 700 kV at the power station and carries the current to a city located 160 km from the station. (a) What is the power loss due to resistance in the line? (b) What fraction of the transmitted power does this loss represent? 35. The heating element of a coffeemaker operates at 120 V and carries a current of 2.00 A. Assuming that the water absorbs all of the energy converted by the resistor, calculate how long it takes to heat 0.500 kg of water from room temperature (23.0C) to the boiling point. 36. The power supplied to a typical black-and-white television set is 90 W when the set is connected to 120 V. (a) How much electrical energy does this set consume in 1 hour? (b) A color television set draws about 2.5 A when connected to 120 V. How much time is required for it to consume the same energy as the black-and-white model consumes in 1 hour? 37. What is the required resistance of an immersion heater that will increase the temperature of 1.50 kg of water from 10.0C to 50.0C in 10.0 min while operating at 120 V? 38. A certain toaster has a heating element made of Nichrome resistance wire. When the toaster is first connected to a 120-V source of potential difference (and the wire is at a temperature of 20.0C), the initial current is 1.80 A. However, the current begins to decrease as the resistive element warms up. When the toaster reaches its final operating temperature, the current has dropped to 1.53 A. (a) Find the power the toaster converts when it is at its operating temperature. (b) What is the final temperature of the heating element? A copper cable is designed to carry a 39. current of 300 A with a power loss of 2.00 W/m. What is the required radius of this cable?
40. A small motor draws a current of 1.75 A from a 120-V line. The output power of the motor is 0.20 hp. (a) At a rate of $0.060/kWh, what is the cost of operating the motor for 4.0 h? (b) What is the efficiency of the motor? 41. It has been estimated that there are 270 million plug-in electric clocks in the United States, approximately one clock for each person. The clocks convert energy at the average rate of 2.50 W. To supply this energy, how many metric tons of coal are burned per hour in coal-fired electric generating plants that are, on average, 25.0% efficient? The heat of combustion for coal is 33.0 MJ/kg. 42. The cost of electricity varies widely throughout the United States; $0.120/kWh is a typical value. At this unit price, calculate the cost of (a) leaving a 40.0-W porch light on for 2 weeks while you are on vacation, (b) making a piece of dark toast in 3.00 min with a 970-W toaster, and (c) drying a load of clothes in 40.0 min in a 5 200-W dryer. 43. How much does it cost to watch a complete 21-hour-long World Series on a 180-W television set? Assume that electricity costs $0.070/kWh. 44. A house is heated by a 24.0-kW electric furnace that uses resistance heating. The rate for electrical energy is $0.080/kWh. If the heating bill for January is $200, how long must the furnace have been running on an average January day? 45. An 11-W energy-efficient fluorescent lamp is designed to produce the same illumination as a conventional 40-W lamp. How much does the energy-efficient lamp save during 100 hours of use? Assume a cost of $0.080/kWh for electrical energy. 46. An office worker uses an immersion heater to warm 250 g of water in a light, covered, insulated cup from 20C to 100C in 4.00 minutes. The heater is a Nichrome resistance wire connected to a 120-V power supply. Assume that the wire is at 100C throughout the 4.00-min time interval. Specify a diameter and a length that the wire can have. Can it be made from less than 0.5 cm3 of Nichrome? 47. The heating coil of a hot-water heater has a resistance of 20 and operates at 210 V. If electrical energy costs $0.080/kWh, what does it cost to raise the 200 kg of water in the tank from 15C to 80C? (See Chapter 11.) ADDITIONAL PROBLEMS 48. One lightbulb is marked "25 W 120 V," and another "100 W 120 V "; this means that each converts its respective power when plugged into a constant 120-V potential difference. (a) Find the resistance of each bulb. (b) How long does it take for 1.00 C to pass through the dim bulb? How is this charge different upon its exit from, versus its entry into, the bulb? (c) How long does it take for 1.00 J to pass through the dim bulb? How is this energy different upon its exit from, versus its entry into, the bulb? (d) Find the cost of running the dim bulb continuously for 30.0 days if the electric company sells its product at $0.070 0 per kWh. What physical quantity does the electric company sell? What is its price for one SI unit of this quantity? m and a 49. A particular wire has a resistivity of 3.0 10 8 cross-sectional area of 4.0 10 6 m2. A length of this wire is to be used as a resistor that will develop 48 W of power when connected across a 20-V battery. What length of wire is required?
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Chapter 17
Current and Resistance
50. A steam iron draws 6.0 A from a 120-V line. (a) How many joules of internal energy are produced in 20 min? (b) How much does it cost, at $0.080/kWh, to run the steam iron for 20 min? 51. An experiment is conducted to measure the electrical resistivity of Nichrome in the form of wires with different lengths and cross-sectional areas. For one set of measurements, a student uses 30-gauge wire, which has a crosssectional area of 7.30 10 8 m2. The student measures the potential difference across the wire and the current in the wire with a voltmeter and an ammeter, respectively. For each of the measurements given in the following table taken on wires of three different lengths, calculate the resistance of the wires and the corresponding value of the resistivity: L (m) 0.540 1.028 1.543 V (V) 5.22 5.82 5.94 I (A) 0.500 0.276 0.187 R( ) ( m)
56.
57.
58.
59. What is the average value of the resistivity, and how does this value compare with the value given in Table 17.1? 52. Birds resting on high-voltage power lines are a common sight. The copper wire on which a bird stands is 2.2 cm in diameter and carries a current of 50 A. If the bird's feet are 4.0 cm apart, calculate the potential difference across its body. 53. You are cooking breakfast for yourself and a friend using a 1 200-W waffle iron and a 500-W coffeepot. Usually, you operate these appliances from a 110-V outlet for 0.500 h each day. (a) At 12 cents per kWh, how much do you spend to cook breakfast during a 30.0 day period? (b) You find yourself addicted to waffles and would like to upgrade to a 2 400-W waffle iron that will enable you to cook twice as many waffles during a half-hour period, but you know that the circuit breaker in your kitchen is a 20-A breaker. Can you do the upgrade? 54. The current in a conductor varies in time as shown in Figure P17.54. (a) How many coulombs of charge pass through a cross section of the conductor in the interval from t 0 to t 5.0 s? (b) What constant current would transport the same total charge during the 5.0-s interval as does the actual current?
60.
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6 Current (A) 4 2 0
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0
1
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Time (s) Figure P17.54
64.
55.
An electric car is designed to run off a bank of 12.0-V batteries with a total energy storage of 2.00 107 J. (a) If the electric motor draws 8.00 kW, what
is the current delivered to the motor? (b) If the electric motor draws 8.00 kW as the car moves at a steady speed of 20.0 m/s, how far will the car travel before it is "out of juice"? (a) A 115-g mass of aluminum is formed into a right circular cylinder, shaped so that its diameter equals its height. Calculate the resistance between the top and bottom faces of the cylinder at 20C. (b) Calculate the resistance between opposite faces if the same mass of aluminum is formed into a cube. A length of metal wire has a radius of 5.00 10 3 m and a resistance of 0.100 . When the potential difference across the wire is 15.0 V, the electron drift speed is found to be 3.17 10 4 m/s. On the basis of these data, calculate the density of free electrons in the wire. A carbon wire and a Nichrome wire are connected one after the other. If the combination has a total resistance of 10.0 k at 20C, what is the resistance of each wire at 20C so that the resistance of the combination does not change with temperature? (a) Determine the resistance of a lightbulb marked 100 W @ 120 V. (b) Assuming that the filament is tungsten and has a cross-sectional area of 0.010 mm2, determine the length of the wire inside the bulb when the bulb is operating. (c) Why do you think the wire inside the bulb is tightly coiled? (d) If the temperature of the tungsten wire is 2 600C when the bulb is operating, what is the length of the wire after the bulb is turned off and has cooled to 20C? (See Chapter 10, and use 4.5 10 6/C as the coefficient of linear expansion for tungsten.) In a certain stereo system, each speaker has a resistance of 4.00 . The system is rated at 60.0 W in each channel. Each speaker circuit includes a fuse rated at a maximum current of 4.00 A. Is this system adequately protected against overload? A resistor is constructed by forming a material of resistivity 3.5 105 m into the shape of a hollow cylinder of length 4.0 cm and inner and outer radii 0.50 cm and 1.2 cm, respectively. In use, a potential difference is applied between the ends of the cylinder, producing a current parallel to the length of the cylinder. Find the resistance of the cylinder. The graph in Figure P17.62a shows the current I in a diode as a function of the potential difference V across the diode. Figure P17.62b shows the circuit used to make the measurements. The symbol represents the diode. (a) Using Equation 17.3, make a table of the resistance of the diode for different values of V in the range from 1.5 V to 1.0 V. (b) Based on your results, what amazing electrical property does a diode possess? An x-ray tube used for cancer therapy operates at 4.0 MV, with a beam current of 25 mA striking the metal target. Nearly all the power in the beam is transferred to a stream of water flowing through holes drilled in the target. What rate of flow, in kilograms per second, is needed if the rise in temperature ( T ) of the water is not to exceed 50C? A 50.0-g sample of a conducting material is all that is available. The resistivity of the material is measured to be 11 10 8 m, and the density is 7.86 g/cm3. The material is to be shaped into a solid cylindrical wire that has a total resistance of 1.5 . (a) What length of wire is required? (b) What must be the diameter of the wire?
Problems
591
+0.10 I (A) +0.08
Variable voltage source I
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I (a) Figure P17.62 (b)
65. (a) A sheet of copper ( 1.7 10 8 m) is 2.0 mm thick and has surface dimensions of 8.0 cm 24 cm. If the long edges are joined to form a tube 24 cm in length, what is the resistance between the ends? (b) What mass of copper is required to manufacture a 1 500-m-long spool of copper cable with a total resistance of 4.5 ? 66. When a straight wire is heated, its resistance changes according to the equation R R 0[1 (T T0)]
+
Touch objects with these wires
(Eq. 17.7), where is the temperature coefficient of resistivity. (a) Show that a more precise result, which includes the fact that the length and area of a wire change when it is heated, is R R 0[1 (T [1 T0)][1 (T 2 (T T0)] T0)]
Figure A17.1
where is the coefficient of linear expansion. (See Chapter 10.) (b) Compare the two results for a 2.00-m-long copper wire of radius 0.100 mm, starting at 20.0C and heated to 100.0C. 67. A man wishes to vacuum his car with a canister vacuum cleaner marked 535 W at 120 V. The car is parked far from the building, so he uses an extension cord 15.0 m long to plug the cleaner into a 120-V source. Assume that the cleaner has constant resistance. (a) If the resistance of each of the two conductors of the extension cord is 0.900 , what is the actual power delivered to the cleaner? (b) If, instead, the power is to be at least 525 W, what must be the diameter of each of two identical copper conductors in the cord the young man buys? (c) Repeat part (b) if the power is to be at least 532 W. [Suggestion: A symbolic solution can simplify the calculations.] ACTIVITIES 1. Connect one terminal of a D-cell battery to the base of a flashlight bulb with insulated wire, tape a second wire to the other battery terminal, and tape a third wire to the center conductor of the bulb, as in Figure A17.1. Make
sure to remove about 1 cm of insulation from the ends of all wires before making the connections. Now bridge the gap between the open wires with different objects, such as a plastic pen, an aluminum can, a penny, a rubber band, and a spoon. Which objects make the bulb light up? Explain your observations. 2. When the lightbulbs in your home are turned on, they are always connected across the same potential difference. Which do you believe has a filament with the highest resistance when cool, a 60-W bulb or a 100-W bulb? To check your prediction, ask your instructor to lend you a device called an ohmmeter and to instruct you in its use. A resistor must always be disconnected from a circuit when its resistance is measured with an ohmmeter. 3. Examine the labels on several appliances, such as a toaster, a television set, a lamp, a stereo system, an air conditioner, and a clock. From each label, determine the power rating of the device in watts. Check the billing statement from your electric utility company to find the cost of electrical energy per kilowatt-hour. (Prices usually range from about a nickel to 20 cents.) Calculate the cost of running each appliance for 1 h. Estimate how many hours per day each appliance is used. Then, on the basis of your daily estimate, calculate the monthly cost of using each appliance.