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CURRENT AND RESISTANCE
22
Q22.1. Reason: You can use a compass placed beside the wire and observe the compass needle change direction
when the battery is disconnected and reconnected in the opposite direction. So something must be happening in the
wire. You can feel the heat and see the light emanating from the light bulb. You can use charged glass and plastic
rods to find that charge does not accumulate at the light bulb. Energy is transferred somehow from the battery to the
bulb, so we conclude that something carries that energy.
Assess: The compass indicates that the space around the wire is somehow changed when the circuit is connected,
and we also know that the light bulb is giving off energy that must come from somewhere. We conclude that charges
flow through the wires.
Q22.2. Reason: In order for a bulb to light, charged particles must move through the bulb and give up energy in
the process. We can look at the battery and determine the terminal with the higher electric potential (the positive end)
and hence the direction of the electric field through the external circuit. Knowing the direction of the electric field we
can establish which way positive and/or negative charges will travel. An ammeter will assure us that indeed charges
are moving through the circuit. But the ammeter will not tell us which direction the charges are moving, hence we
cannot distinguish their charge.
Assess: There are no simple observations and/or measurements you can make on this circuit to distinguish a current
composed of positive charge carriers from a current composed of negative charge carriers.
Q22.3. Reason: In order for electrons to move through a wire as a current, the wire must be connected to a source
of electric potential difference. This source of electric potential difference will establish an electric field in the circuit
and it is this electric field that causes electrons to travel through the circuit.
Assess: Frequently, we use misleading language when describing the current in a circuit. In fact, you may have used
the misleading statement “flow of current.” Current is defined as the motion of charges. It is the charges that flow,
not the current. Current is the flow of charges.
Q22.4. Reason: (a) Because there are no junctions or branching of the wires, they must carry the same current.
This is because of conservation of charge. Charges can’t appear or disappear. The current is the same everywhere in a
single-loop circuit, including in the filament itself.
(b) Because of the answer to part (a) the brightness of the bulb will stay the same. The current will be the same if the
wires are switched.
Assess: This question is asked because many students have difficulty with this concept. Concentrate on this idea
until it makes complete sense.
Q22.5. Reason: The circuit has no place to store electrons and no way to generate electrons. Also, for every
electron that enters the circuit, another electron leaves the circuit. Based on these two statements, we can conclude
that I1 is equal to I 4 . As the circuit shows, the current I1 splits into two currents ( I 2 and I 3 ) which then combine to
give I 4 . This allows us to conclude that both I1 and I 4 are greater than I 2 and I 3 . Since the wires that carry the
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22-1
22-2
Chapter 22
current I 2 and I 3 are identical (they each have the same resistance), the current I1 will divide in half to give I 2 and
I 3 . This allows us to make the following statements:
I 2 = I 3 < I1 = I 4 , and ( I 2 = I 3 ) = (1/ 2)( I1 = I 4 )
Assess: We cannot create or destroy charges in the wire, neither can we store them in the junction. The rate at which
electrons flow into one or many wires must be exactly balanced by the rate at which they flow out of the other wires.
We are conserving the amount of charge and the flow of that charge (current).
Q22.6. Reason: Equation 22.1 gives the definition of current: I = ΔQ / Δt.
The initial current is 4 A: I1 = Δ Q1/ Δt1 = 4 A.
Now we double the charge and halve the time interval:
I2 =
Δ Q2 2Δ Q1
= 1
= 4( I1 ) = 4(4 A) = 16 A
Δ t2
2 Δ t1
Assess: Watch the dependencies. Had it been the other way, half the charge in twice the time, the final current would
be 1 A.
Q22.7. Reason: If negative charges move opposite the field, we still say it is a positive current in the direction of
the field. So both the positive ions and the negative ions contribute to a current in the direction of the field.
Assess: The two ion currents do not cancel out, even if the ions were moving at the same speed.
Q22.8. Reason: A device that can actively separate charge, leading to a potential difference, is a source of emf.
The ion pumps do exactly that, making them a source of an emf.
Assess: A key aspect of an emf, as opposed to just a potential difference, is that a source of emf is also a source of
energy: It gives electric potential energy to each charge that passes through it.
Q22.9. Reason: (a) The electric potential difference (3 V as produced by the battery) establishes an electric field
from the point of higher electric potential (+ terminal of the battery at point 2) to the point of lower electric potential
(− terminal of the battery at point 3). The electrons travel in a direction opposite the direction of the electric field,
hence they will travel counterclockwise through the wire.
(b) The electron’s electric potential energy decreases as it moves through the wire. It must use some of its energy to
“work” its way through the wire and overcome the “atomic-level friction.”
(c) As the electrons move through the wire they continually collide with the positively charged ions of the metal.
These collisions cause the ions to move randomly; we saw in Chapter 10 that randomly moving atoms in a material
have energy in the form of thermal energy. So as the electrons move through the wire they transfer their electric
potential energy into thermal energy of the wire—it gets warmer.
(d) We established in part (a) that the electrons travel through the external circuit from the negative terminal to the
positive terminal. As a result, in order to get back to the negative terminal (in order to make another trip), they must
go through the battery from the positive terminal to the negative terminal.
(e) The electron’s electric potential energy increases as it moves through the battery. The chemical reactions
occurring in the battery provide the energy needed to move the electron internally from the positive to the negative
terminal. Since work has been done on the electron (at the expense of chemical energy), its electric potential energy
has increased.
(f) The increase in the electron’s electric potential energy comes at the expense of chemical energy.
Assess: It is important to understand not only each concept dealt with in this question, but also the connections
between the concepts.
Q22.10. Reason: The resistance of a metal wire depends on the resistivity, length, and cross sectional area of the
wire according to R = ρ L / A. If the temperature of metal wire changes, the resistivity, length, and cross section area
of the wire change. If the temperature of a metal increases, all three quantities of interest ( ρ , L and A) increase and
it the temperature decreases, all three of these quantities decrease. Without knowing rates of changes for length and
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Current and Resistance
22-3
resistivity, it would be difficult to say exactly what will happen; however, in general, as the temperature of a metal
wire increases, the resistance increases.
Assess: This course will help develop your ability to look at an expression and answer “what if” questions.
Q22.11. Reason: The resistance of an object is given by Equation 22.7: R = ρ L /A.
For objects with a circular cross section, A = π r 2. Since all five wires are made of the same material, ρ is the same in
each equation. Now we must compute the resistance of all five wires.
ρL
π r2
ρL
1 ρL 1
=
= R1
R2 =
2
π (2r )
4 π r2 4
ρ (2 L) 2 ρ L 1
=
= R1
R3 =
π (2r ) 2 4 π r 2 2
ρ (2 L)
ρL
= 2 2 = 2 R1
R4 =
π r2
πr
ρ (4 L) 4 ρ L
=
= R1
R5 =
π (2r ) 2 4 π r 2
R1 =
Therefore
R4 > R1 = R5 > R3 > R2
Assess: This question helps cement the concept that longer wires have higher resistance while fatter wires have
lower resistance.
Q22.12. Reason: For every electron that enters each circuit there is an electron that leaves the circuit. As a result,
the current is the same everywhere in each circuit. This allows us to conclude that I1 = I 2 and I 3 = I 4 . The wire in
the circuit to the right is longer hence has more resistance and consequently less current for the same size battery.
This allows us to conclude that ( I1 = I 2 ) > ( I 3 = I 4 ).
Assess: The circuit with more resistance will have a smaller current.
Q22.13. Reason: Conservation of charge requires I1 = I 2 = I 3 and I 4 = I 7 . Symmetry requires I 5 = I 6.
Conservation of charge requires I 4 = I 5 + I 6, so I 4 > I 5 .
So far we know I 4 = I 7 > I 5 = I 6. Now we compare the two circuits. The potential difference from 1 to 3 to 2 is the
same as the potential difference from 4 to 6 to 7 and the resistance of the wires around each path is the same, so
I = Δ V / R is the same for both of these paths. So I 3 = I 6.
Putting this all together, I 4 = I 7 > I1 = I 2 = I 3 = I 5 = I 6.
Assess: The stated assumptions (identical batteries, wires of equal radius) are important. Conservation of charge,
symmetry, and Ohm’s law are all helpful here.
Q22.14. Reason: (a) The source of all the energy used in a circuit—thermal energy dissipated by resistors, but also
energy used to run motors or create light—ultimately comes from the chemical energy of the battery that is the
source of emf for the circuit. This statement is true.
(b) A battery maintains a constant potential difference, equal to its emf, across its terminals, at least until its chemical
energy is depleted. So this statement is true.
(c) Electrons moving through the circuit are the source of the current. The electric potential difference of the battery
establishes an electric field in the external circuit, which causes the electrons to move. As the chemicals in the battery
are depleted, the terminal electric potential difference established by the battery decreases, the electric field
established in the wire decreases and the flow of electrons and hence current decreases. Statement (c) is false.
Assess: This question illustrates why we need to be very careful and precise with our language when dealing with
electricity. A statement that on the surface sounds correct may be misleading.
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22-4
Chapter 22
Q22.15. Reason: Equation 22.8 gives the current in an ohmic wire: I = ΔV /R.
We now compute the current in each wire.
2V
= 1A
2Ω
1V 1
I2 =
= A
2Ω 2
2V
=2A
I3 =
1Ω
1V
=1A
I4 =
1Ω
I1 =
Therefore
I 3 > I1 = I 4 > I 2
Assess: This exercise helps one see that potential difference causes the current while resistance, well, resists it.
Q22.16. Reason: A plot of the electric potential versus position in the circuit is shown in the following figure.
Notice that each battery increases the electric potential, the connection between the batteries, and the wire
(considered to be ideal) connecting the battery, and the resistor causes no decrease in electric potential. Finally, the
electric potential decrease at the resistor is the same as the increase by all the batteries.
Assess: If the electric potential decrease at the resistor was not equal to the increase at the batteries, you would have
a point in the circuit with two different electric potentials—this does not happen.
Q22.17. Reason: After the lightning strike, you are standing in the electric field along the ground. There is an
electric potential at every point in an electric field. If you stand with your feet close together, the electric potential
difference between your feet is decreased. Since your body has a fixed resistance (from one foot to the other) the
smaller the electric potential difference across your feet, the smaller the current you will experience.
Assess: This question requires us to combine our knowledge of electric fields and Ohm’s law.
Q22.18. Reason: If the corrosion eats away at the wire and reduces its cross-sectional area, the electrical resistance
will be affected. Reducing A increases R.
Assess: If impurities are introduced into the wire material as a result of the corrosion that would further increase the
resistance.
Q22.19. Reason: The brightness of the bulb is a function of the current through the bulb. As the filament
deteriorates, its resistance increases causing a decrease in current and hence brightness.
Assess: It seems reasonable that an old bulb will not be as bright as a new bulb.
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Current and Resistance
22-5
Q22.20. Reason: The power dissipated by a resistor may be determined by P = ΔV 2/ R. This expression allows us
to determine the power dissipated by each resistor and then rank the values.
P1 =
ΔV 2
(ΔV /2) 2 1 ⎛ ΔV 2 ⎞
, P2 =
= ⎜
⎟
2R
8⎝ R ⎠
R
P3 =
⎛ ΔV 2 ⎞
(2Δ V ) 2
= 8⎜
⎟
( R/2)
⎝ R ⎠
P4 =
⎛ ΔV 2 ⎞
(2Δ V ) 2
= 2⎜
⎟
2R
⎝ R ⎠
Comparing these values, we see that P3 > P4 > P1 > P2 .
Assess: The comparison was made possible by identifying a common factor (Δ V 2/ R ) in each expression.
Q22.21. Reason: The brightness is determined by the power dissipated in each bulb. When they both operate at
(ΔV )2
to see that the 100 W bulb has a lower resistance than the 60 W bulb.
R
But when they both have the same current through them we use the other version of the power equation P = I 2 R to
show that the bulb with the lower resistance (the 100 W bulb) dissipates less power and is dimmer than the 60 W
bulb.
Assess: The wattage label on a regular household bulb assumes it will be operated at 120 V; the power dissipated
will be different if the potential difference is not 120 V.
120 V we use the power equation P =
Q22.22. Reason: The current through each light bulb may be determined by I = P/ΔV . This expression allows us
to determine the current in each light bulb and then rank the values.
IA =
PA
0.8 W
=
= 0.53 A
Δ VA 1.5 V
IB =
PB
6W
=
=2A
Δ VB 3 V
IC =
PC
4W
=
= 0.89 A
ΔVC 4.5 V
ID =
PD
8W
=
= 1.3 A
ΔVD 6 V
Comparing these values, we see that: I B > I D > I C > I A.
The correct choice is B.
Assess: We have just applied the basic relationship between power, current, and electric potential difference to
obtain these values.
Q22.23. Reason: We know the power dissipated by a resistor is a product of the voltage (potential difference) and
the current, PR = I ΔVR. For objects that obey Ohm’s law, this can also be written as Equation 22.13. While not
strictly true, we will assume the light bulb filaments in this question obey Ohm’s law so we can apply this relation:
PR = (ΔVR ) 2 / R.
Solve for R and apply to each case.
A.
R=
(ΔVR )2 (1.5 V)2
=
= 2.8 Ω
PR
0.8 W
B.
R=
(Δ VR ) 2 (3 V) 2
=
= 1.5 Ω
PR
6W
C.
R=
(Δ VR ) 2 (4 . 5 V) 2
=
= 5.1 Ω
PR
4W
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22-6
Chapter 22
D.
R=
(Δ VR ) 2 (6 V) 2
=
= 4.5 Ω
PR
8W
The correct choice is C.
Assess: The fact that ΔVR is squared carries the day for C.
Q22.24. Reason: The resistance of a wire may be determined by R = ρ L /A. In this case the resistivity is a
constant. The resistivity is a function of the material used to manufacture the wire and that will not change as
the wire is stretched. At this point the problem is complicated by the fact that as the length of the wire increases, the
cross-sectional area decreases. The problem can be simplified by realizing that the cross sectional area can be related
to the length. This can be done by realizing that the mass and volume of the wire remain constant as it is stretched.
We aren’t creating or destroying any wire, we are just changing its shape. We could use the mass as the quantity that
is constant, but that would result in our using the mass density and volume of the wire—so let’s just agree to use the
volume. Recall that the volume of the wire is determined by V = LA. We can use this expression for the volume to
eliminate either the length L or the area A; let’s agree to eliminate the area. In that case the expression for the
resistance becomes R = ρ L / A = ρ L /(V / L) = ρ L2 / V .
Since the resistivity and the volume are constant, we see that as the wire is stretched the resistance will increase.
Furthermore we are able to say how much it will increase. For example, if the length of the wire doubles, the
resistance will increase by a factor of four.
Based on this information, as the wire is stretched, the resistivity remains the same and the resistance increases. The
correct choice is C.
Assess: The question was simplified by reducing the expression for the resistance to just one variable. We used
length as the variable. You should repeat the solution using the area and see if you obtain the same result.
Q22.25. Reason: If the potential difference is increased then the current will also increase, and so will the power
dissipated ( PR = I ΔVR ). The length is fixed, so the electric field will also increase ( E = ΔV /d ).
Resistance is a function of the wire itself (the material it is made of and the temperature), but not directly a function
of the potential difference.
The correct choice is C.
Assess: Indirectly, since the increased potential difference increases the current and power dissipation, the
temperature may rise and cause the resistance to change too, but we don’t know whether the temperature coefficient
is positive or negative (the resistance might go up or down).
Q22.26. Reason: The power output for the speaker may be determined by P = ΔV 2/R. Since the resistance of the
speaker is a constant, the power output depends on the square of the electric potential difference. If you increase the
potential difference across the speaker by a factor of 2 you will double the power output of the speaker. This new
potential difference may be calculated by Δ Vnew = ( 2 )Δ Vold = 1.41(5.0 V) = 7.1 V.
The correct choice is A.
Assess: We have obtained this value by inspecting the basic relationship between power, resistance, and electric
potential difference.
(ΔV )2
Q22.27. Reason: Use P =
for the power dissipated. Since the resistor is the same, R doesn’t change, and
R
we see that the power scales as the square of the potential difference. Doubling the potential difference (from 3.0 V
to 6.0 V) changes the power dissipated by a factor of 4. (4)(1.0 W) = 4.0 W. The answer is D.
Assess: This means the current increases also.
Q22.28. Reason: Reading about energy and time in the question makes us think of power = energy/time. The
other information in the question gives us voltage and current, and we also know that P = I Δ V . So we tie these ideas
together.
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Current and Resistance
time =
22-7
energy energy
5000 J
=
=
= 2777 s ≈ 46 min
power
I ΔV
(1 .2 A)(1 . 5 V)
The correct choice is D.
Assess: Rechargeable batteries are often rated in mA ⋅ h, which, assuming the standard voltage 1 . 2 V for most
rechargeables, allows us to calculate the energy stored. For example, for a 2000 mA ⋅ h battery (a typical value for a
NiMH AA battery)
energy = PΔt = I Δ V × Δt = ( I Δt ) × Δ V = (2000 mA ⋅ h)(1 .2 V)(3600 s/h) = 8 . 6 kJ
This result is comparable to the one in the question, with the times (46 min versus 1 h), currents (1.2 A versus 2 A), and
voltages (1.5 V versus 1.2 V), a little different but in the same general range.
Q22.29. Reason: For every charge that enters the wire at one end, another charge will leave the wire at the other
end. This indicates the same amount of charge is moving through each section of the wire at each unit of time. Since
our concept of current is that current is a description of the rate at which a charge moves through a conductor, we
conclude that I1 = I 2 = I 3 . The correct choice is C.
Assess: Charge is not created or destroyed; whatever charge flows into one end of the wire each unit of time must
also flow out of the other end of the wire.
Q22.30. Reason: Resistance is related to resistivity, length and cross sectional area by R = ρ L / A. A section of the
body may be modeled as two resistors (fat and muscle) in parallel. In this case the muscle branch is not affected,
however as the person gains weight (adds girth) the area for the fat branch in increases. This lowers the resistance of
the fat branch and hence for the entire segment of the body. As a person gains weight, the electrical resistance of the
limbs will decrease. The correct response is C.
Assess: If the resistance of one branch of two parallel resistors decreases, the resistance of the parallel combination
of resistors will decrease.
Problems
P22.1. Prepare: We will find the total charge that flows through the hair dryer and then divide it by the electron
charge to find the number of electrons.
Solve: Equation 22.2 is Q = I Δt . The amount of charge delivered is
⎛
60 s ⎞
Q = (10.0 A) ⎜ 5.0 min ×
⎟ = 3000 C
1
min ⎠
⎝
The number of electrons that flow through the hair dryer is
N=
Q
3000 C
=
= 1.9 × 1022
e 1.60 × 10−19 C
Assess: This is an enormous amount of charge and is typical of such devices.
P22.2. Prepare: Equation 22.2 defines current as the ratio of total charge and time during which charge flows.
Solve: From Equation 22.2,
I=
Q Ne (2.0 × 1013 )(1.60 × 10−19 C)
=
=
= 0.0032 C/s = 0.0032 A = 3.2 mA
Δt Δt
1.0 × 10−3 s
Assess: This is a typical current for a transistor and is reasonable.
P22.3. Prepare: By conservation of charge, the sum of the currents into a junction must equal the sum of the
currents leaving the junction.
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22-8
Chapter 22
Solve: (a) Therefore, we know that the current in wire 3 must be 0.65 A − 0.40 A = 0.25 A. This is 0.25 C per
second.
⎞
0.25 C ⎛
1e
18
⎜
⎟ = 1.6 × 10 electrons/s
1 s ⎝ 1.6 × 10−19 C ⎠
(b) Because more current is coming out of the junction in wire 2 than is going in in wire 1, the 0.25 A in wire 3 must
be going into the junction.
Assess: Electrons can’t disappear, so each one that comes into the junction must leave it.
P22.4. Reason: The relationship between current, charge, and time is I = Δ Q /Δt .
Solve: The current is I = Δ Q/ Δt = 9.0 × 10−12 C /(5.0 × 10−4 s) = 1.8 × 10−8 A
Assess: We expect a small current due to very small flow of charge.
P22.5. Prepare: Electric current, charge, and time are related by I = Δ Q /Δt . The amount of charge due to
N electrons is Δ q = Ne.
Solve: The number of electrons passing a given point in 1.0 s may be determined by
N = ΔQ/ e = I Δt /e = (15 × 10−6 A)(1.0 s)/(1.6 × 10−19 C) = 9.4 × 1013 electrons
Assess: Since each electron has such a small charge, we expect a large number of electrons.
P22.6. Prepare: Electric current, charge, and time are related by I = Δ Q/ Δt.
Solve: The current during the lightning strike is I = ΔQ/Δt = 2.5 C/(2.0 × 10−4 s) = 1.3 × 104 A.
Assess: As expected, this is a large amount of current.
P22.7. Prepare: Current is defined in terms of the amount of charge ΔQ passing through a cross section of the
wire in a time interval Δ t : I = Δ Q/ Δt.
We are given that Δ t = 40 μs = 40 × 10−6 s and Δ Q = (1 − 0 .13)(6 . 0 × 10−4 C) = (0 . 87)(6 . 0 × 10−4 C) = 5 . 22 × 10−4 C.
Solve:
I=
ΔQ 5 . 22 × 10−4 C
=
= 13 A
Δt
40 × 10−6 s
Assess: 13 A is a fairly large current, but that is because capacitors discharge very quickly when the plates are
connected with a wire.
In the next chapter you will learn more specifically how the current decays in a discharging capacitor; it isn’t at a
constant rate or in a linear fashion. But since all we’re asked for in this problem is the average current, we are already
equipped to solve the problem.
P22.8. Prepare: The direction of the current in a material is opposite to the direction of motion of the negative
charges and is the same as the direction of motion of positive charges. That is, I = I + − I − = q+ /t − q− /t = q+ − q− /t
Solve: The charge due to positive ions moving to the right per second is
q+ = N + (2e) = (5.0 × 1015 )(2 × 1.60 × 10−19 C) = 1.60 × 10−3 C
The charge due to negative ions moving to the left per second is
q− = N (−e) = (6.0 × 1015 s)( −1.60 × 10−19 C) = − 0.96 × 10−3 C
Thus, the current in the solution is
I=
q+ − q− 1.60 × 10−3 C − (− 0.96 × 10−3 C)
=
= 2.56 × 10−3 A = 2.6 mA
t
1s
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Current and Resistance
22-9
The direction of the current in both cases is to the right.
Assess: This is a typical current for ionic solutions.
P22.9. Prepare: The starter motor draws a current of 150 A and the motor runs for 0.80 s until the car engine starts.
The charge that flows through the starter motor can be obtained from I = Q / Δ t.
Solve: The charge delivered in time Δ t is
Q = I Δ t = (150 A)(0.80 s) = 120 C
Assess: Because the motor draws a lot of current, the charge delivered was expected to be large.
P22.10. Prepare: We will use Equation 22.2 to find the charge that leaves the battery.
Solve: The total charge in the battery is
Q = I Δt = (90 A)(3600 s) = 3.2 × 105 C
Assess: As expected, this is a large amount of charge.
P22.11. Prepare: We believe in conservation of current at a junction, expressed in Equation 22.3: Σ I in = Σ I out . We
apply this first at the left junction (because there is only one unknown current there), and then at the right.
Solve: At the left junction:
Σ I in = Σ I out
7A + I C = 5A
So I C = 5A − 7 A = −2A, where the negative sign indicates the current goes in the opposite direction from the arrow
(or, to the right).
At the right junction:
Σ I in = Σ I out
3A = I B + I C
3A = I B + ( −2 A)
So I B = 3 A + 2 A = 5 A.
Assess: On the diagram, replace the unknown current labels with their newly known values. Use the negative sign
on I C to change the direction of the arrow. Verify that current is conserved at each junction and everything makes
sense.
P22.12. Prepare: For a junction, the law of conservation of charge requires Σ I in = Σ I out. This says that the total
current into a junction must equal the total current out of that junction.
Solve: First let’s redraw the figure, label the junctions, and give the unknown currents a name.
Now write a junction equation for each junction:
Junction 1: 5A = 3A + i1 or i1 = 2A
Junction 2: i1 = iB + 1 A or iB = ii − 1 A = 1 A
Junction 3: 3 A = i2 + 2 A or i2 = 1 A
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22-10
Chapter 22
Junction 4: 1 A + i2 = iC or iC = 1 A + i2 = 2 A
In summary, I B = 1 A to the right, I C = 2 A to the right.
Assess: Notice that the total current going into this arrangement and the total current going out is 5 A.
P22.13. Prepare: The change in energy is the charge multiplied by the change in potential.
Solve:
ΔU = qΔV = q (Vf − Vi ) = (1.0 × 10−6 C)(1.5 V) = 1.5 × 10−6 J
Assess: The work done by the escalator on the charge is stored as electric potential energy of the charge.
P22.14. Prepare: The emf is defined as the work done per unit charge by the charge escalator or the battery.
Solve:
ε=
Wchem
0.60 J
=
= 12 V
q
0.050 C
Assess: An emf of 12 V is reasonable for a battery.
P22.15. Prepare: Charge, electric current, and time are related by I = Δ Q/ Δt . Work, electric potential difference,
and charge are related by W = Δ qΔ V.
Solve: (a) The charge transferred is Δ q = I Δ t = (2.5 × 10−3 A)(5.0 hr)(3.6 × 103 s/hr) = 45 C
(b) The change in potential energy of these charges by the battery is Δ U = Δ qΔ V = (45 C)(9.0 V) = 4.1 × 102 J.
Assess: These values are reasonable for this case.
P22.16. Prepare: Review Conceptual Example 22.4 to see that the emfs of batteries add up if the batteries are in
series. We need to put enough 0.75 V fuel cells in series to add up to 24 V to run the motor.
Solve: We need to solve the following equation for n, where n is the number of fuel cells.
n(0 . 75 V) = 24 V ⇒ n =
24 V
= 32
0 .75 V
Assess: The voltage given is indeed the voltage produced by a hydrogen–oxygen fuel cell, and it is also true that
they are often combined in series and parallel in a “fuel cell stack” of more than 45 cells depending on the design
criteria (so 32 would not be an outlandish number).
P22.17. Reason: The total potential difference is related to the potential difference for one cell by Vtotal = NVcell .
Solve: The number of cells needed to generate a total potential difference of 350 V is
N = Vtotal / Vcell = 350 V /(0.110 V) = 3200 cells.
Assess: Since the potential difference per cell is small, we expect a large number of cells.
P22.18. Prepare: The potential difference between the terminals of a battery is the battery’s emf. For an ideal
battery, Δ Vbat = Δ Vwire = ε = 6.0 V. Using Equation 22.5, I = Δ Vwire / R.
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Current and Resistance
22-11
Solve:
(a) Δ Vwire = 6.0 V, I = Δ Vwire / R = 6.0 V/(1.0 Ω) = 6.0 A.
(b) Δ Vwire = 6.0 V, I = Δ Vwire / R = 6.0 V/(2.0 Ω) = 3.0 A.
(c) Δ Vwire = 6.0 V, I = Δ Vwire / R = 6.0 V/(3.0 Ω) = 2.0 A.
Assess: The current decreases with an increase in resistance, as we would expect.
P22.19. Prepare: Resistance and resistivity are related through Equation 22.7. Resistivity depends only on the type
of material and not on the geometry of the wire. Equation 22.7 can also be written as R = ρ L / A = ρ L /(π r 2 ),, where
the wire has length L and radius r.
Solve: (a) Wires 1 and 2 are made of the same material, so ρ 2 = ρ1 and thus ρ 2 / ρ1 = 1.0.
(b) Because the two wires have the same resistivity,
2
2
R2 ρ L2 /(π r 22 ) ⎛ r1 ⎞ L2 ⎛ 1 ⎞ 2 1
=
=⎜ ⎟
=
= = 0.50
R1 ρ L1/(π r 12 ) ⎝ r2 ⎠ L1 ⎜⎝ 2 ⎟⎠ 1 2
Assess: A thicker wire has smaller resistance and a larger wire has higher resistance.
P22.20. Prepare: The resistance of a wire may be determined by R = ρ L / A and the volume of the wire may be
determined by V = LA. We can use the expression for the volume to eliminate the area A in the expression for the
resistivity and obtain R = ρ L / A = ρ L /(V / L) = ρ L2 / V .
This expression informs us that the resistance of a wire varies as the square of the length if the volume remains
constant.
Solve: Since the length has doubled and the resistance varies as the length squared, the resistance of the stretched
wire will be four times that of the unstretched wire.
Rstretched = 4 Runstretched = 4(0.010 Ω) = 0.040 Ω
Assess: This is a reasonable resistance for wire.
P22.21. Prepare: Resistivity is related to resistance, length, and cross-sectional area by R = ρ L /A, and the area is
related to the width and thickness of the leaf by A = WT .
Solve: Combining these two expressions and solving for the resistivity, we obtain
ρ = RA/ L = RWT / L = (2.0 × 106 Ω)(2.5 × 10−2 m)(2.0 × 10−4 m) /(0.20 m) = 50 Ω ⋅ m
Assess: This value for the resistivity is the same order of magnitude as other organic materials listed in Table 22.1.
P22.22. Prepare: From Table 22.1, copper and iron have resistivities of 1.7 × 10−8 Ω ⋅ m and 9.7 × 10−8 Ω ⋅ m. We
will use Equation 22.7, which connects resistance with resistivity.
Solve: (a) The resistance is
R=
ρL
A
=
ρ L (1.7 × 10−8 Ω ⋅ m)(1 m)
=
= 0.087 Ω
π r2
π (2.5 × 10−4 m) 2
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22-12
Chapter 22
(b) The resistance is
R=
ρL
A
=
ρL
d2
=
(9.7 × 10−8 Ω ⋅ m)(0.1 m)
= 0.0097 Ω
(0.001 m) 2
Assess: Small wires or pieces of metal such as copper and iron have low resistance as obtained previously.
P22.23. Prepare: Resistance, current, and electric potential difference are related by Ohm’s law R = V / I .
Resistance, resistivity, length, and cross-sectional area are related by R = ρ L / A.
Solve: Combining these two expressions and solving for the length of the wire obtain:
L=
AV π r 2V π (12.5 × 10−5 m) 2 (12 V)
=
=
= 8.7 m
Iρ
Iρ
(4.0A)(1.7 × 10−8 Ω ⋅ m)
Assess: This is a reasonable length and can be sewn into a vest.
P22.24. Prepare: The resistance of a “resistor” (the blood-filled artery in this case) is given by Equation 22.7:
R = ρ L / A.
We are asked for the resistances of the blood itself, so we ignore the artery walls. We are given L = 0. 20 m and
A = π r 2 = π (d /2) 2 = π (0 .0050 m) 2 = 7 .85 × 10−5 m 2. We also look up the resistivity of blood in Table 22.1:
ρ blood = 1 .6 Ω ⋅ m.
Solve:
R=
ρL
A
=
(1 . 6 Ω ⋅ m)(0 . 20 m)
= 4100 Ω
7 . 85 × 10−5 m 2
Assess: 4100 Ω is a reasonable value of resistance, neither extremely large nor extremely small. The value of the
resistivity of blood, as given in Table 22.1, is also neither extremely large nor extremely small, so we are satisfied
with our answer.
P22.25. Prepare: The current I in a wire when a potential difference is applied to the ends of the wire can be
obtained from Equation 22.5, I = Δ V / R, and Equation 22.7, R = ρ L / A. The resistivity of nichrome from Table 22.1
is 1.5 × 10−6 Ω ⋅ m.
Solve:
⎛ A ⎞ (3.0 V)π (0.40 × 10−3 m) 2
I = ΔV ⎜
= 2.0 A
⎟=
−6
⎝ ρ L ⎠ (1.5 × 10 Ω ⋅ m)(0.50 m)
Assess: The resistivity of nichrome is small, so a current of 2.0 A is reasonable.
P22.26. Prepare: The potential difference between the ends of a copper wire that carries a current can be obtained
from Equation 22.5, I = Δ V / R, and Equation 22.7, R = ρ L / A. The resistivity of copper from Table 22.1 is
1.7 × 10−8 Ω ⋅ m.
Solve:
Δ V = IR = I ρ
L (3.0 A)(1.7 × 10−8 Ω ⋅ m)(0.20 m)
=
= 13 mV
π (0.5 × 10−3 m) 2
A
Assess: Because copper’s resistivity is small, a potential difference of 13 mV across a 1.0 mm diameter and 20-cm
long wire is reasonable.
P22.27. Prepare: Resistance is related to resistivity, length, and cross-sectional area by R = ρ L / A. An electrical
worker could have palms that are 10 cm by 12 cm or 1.2 × 10−2 m 2 .
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Current and Resistance
22-13
Solve: The resistance is R = ρ L / A = (1 × 106 Ω ⋅ m)(1.5 × 10−3 m) /(1.2 × 10−2 m 2 ) = 1 × 105 Ω
Assess: This built-in safety factor is a good thing and a reasonable value.
P22.28. Prepare: The resistivity of aluminum is 2.65 × 10−8 Ω ⋅ m.
Solve:
L
(37 km)
= (1100 A)(2.65 × 10−8 Ω ⋅ m)
= 1.9 kV
2
A
⎛ 2.7 cm ⎞
π⎜
⎟
⎝ 2 ⎠
Assess: Since the voltage drop is proportional to the current it is better to use very high voltages and low currents for
long-distance transmission lines.
Δ V = IR = I ρ
P22.29. Prepare: The slope of the I versus ΔV graph, according to Ohm’s law, is the reciprocal of the resistance
R. The resistance is directly proportional to the length of the resistor according to Equation 22.7.
Solve: (a) From the graph in the figure, the reciprocal of the slope and hence the resistance of the resistor is
ΔV
1
10 V
=R=
=
= 2.0 Ω
I
Slope 5.0 A
(b) Doubling the length of the wire will double the resistance, which means the slope will decrease. The currentversus-potential-difference graph is shown.
Assess: Since the resistance is the reciprocal of the slope, a smaller slope means a higher resistance.
P22.30. Prepare: The slope of the I versus ΔV graph, according to Ohm’s law, is the reciprocal of the resistance R.
Solve: From the graph in the figure, the resistance of the cylinder is
R=
1
1
Δ V 100 V
=
=
=
= 50 Ω
slope ( I/ ΔV )
I
2A
Assess: 1 Ω = 1(V/A). A 50 Ω resistance is reasonable.
P22.31. Prepare: We’ll assume the filament is ohmic so we can use Ohm’s law. Equation 22.7 will help us find
the resistance, and we need to recall E = ΔV /d .
Solve: (a) E = Δ V / d = 120 V/(0 . 60 m) = 200 V/m.
(b) All else being equal, the new field strength would be E = Δ V / d ′ = 120 V/(1. 20 m) = 100 V/m.
(c) Assuming the filament is ohmic, the original current would be
I=
ΔV 120 V
=
= 0 . 50 A
R
240 Ω
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22-14
Chapter 22
We are reminded that the current is proportional to the electric field, so halving the electric field (from 200 V/m to
100 V/m) means we also halve the current from 0.50 A to 0.25 A.
(d) We can use Ohm’s law with the new current to calculate the new resistance.
R′ =
ΔV 120 V
=
= 480 Ω
I′
0 .25 A
Assess: Another way to see the new resistance: Since R = ρ L /A, then doubling the length doubles the resistance,
and R′ = 2 × 240 Ω = 480 Ω. This method is consistent with the other one.
The power dissipated (the brightness of the bulb) will be reduced correspondingly because P = (ΔV ) 2 / R. ΔV is still
120 V but R has doubled.
P22.32. Prepare: Assume the battery and the wires are ideal. Connecting the wire to the battery leads to an electric
field inside the wire that is given by Ewire =
Δ Vwir e
L
.
Solve: From this information, Δ Vwire = LEwire = (0.30 m)(0.010 V/m) = 3.0 mV.
Assess: We expected a small electric potential difference because the electric field is very small.
P22.33. Prepare: Assume the copper wire is a resistor. We’ll need Ohm’s law ΔV = IR.
Solve: The text says the field in a current-carrying resistor is
E=
ΔV IR I ( ρ LA ) I ρ (20 A)(1.7 × 10−8 Ω ⋅ m)
=
=
=
=
= 0.43 V/m
L
L
L
A
π (0.50 mm) 2
Assess: This is not a large field strength.
P22.34. Prepare: The solution may be modeled after Example 22.10 in the text. The zero point of the potential is
the negative terminal of the battery. Charges are raised to a higher potential by the battery, experience no change in
potential as they move through ideal connecting wires and lose half the potential gained at the battery as they move
through each bulb (since the bulbs are identical), and finally return to the negative terminal of the battery with no
potential.
Solve: (a)
(b) Notice that the charges gain potential as they move through the battery, experience no change in potential as they
move through the ideal wires, and lose potential as they travel through the bulbs.
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Current and Resistance
22-15
Assess: A graphic representation such as the one developed in this problem can be a tremendous aid to your
conceptualization of a number of related concepts.
P22.35. Prepare: We use the fact that ΔU = qΔV .
Solve:
Δ U = qΔV = (2000 C)(1.5 V) = 3000 J
Assess: These are typical values for a battery.
P22.36. Prepare: The 1500 W rating is for operating at 120 V. That is, the hair dryer dissipates 1500 W at
Δ VR = 120 V. We will use Equation 22.13 and Ohm’s law.
Solve: (a) The hair dryer’s resistance using Equation 22.13 is
R=
(Δ VR ) 2 (120 V) 2
=
= 9.6 Ω
PR
1500 W
(b) The current in the hair dryer when it is used is given by Ohm’s law:
I=
ΔVR 120 V
=
= 12.5 A ≈ 13 A
R
9.6 Ω
Assess: Devices such as a hair dryer that operate at 120 V usually have relatively small resistance but large current.
P22.37.
Prepare: The text explains that the potential energy gained is Δ U = (Δq )ε .
Solve:
ε = ΔU
Δq
=
7.2 J
= 6.0 V
1.2 C
Assess: This is a typical emf for a battery.
P22.38. Prepare: Electric power, potential, and resistance are related by P = (ΔV ) 2 /R. Electric potential, current,
and resistance are related by Ohm’s Law R = Δ V/ I .
Solve: (a) The resistance of the wire in the blanket is R = ( ΔV ) 2 / P = 4.6 Ω (b) The current in the electric blanket is
I = Δ V/ R = 3.9 A
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22-16
Chapter 22
Assess: These are reasonable values for an electric blanket.
P22.39. Prepare: The resistance, potential difference, and current in a wire are related by ΔV = IR. The power
dissipated by the wire may be determined by any of the following: P = I Δ V = I 2 R = Δ V 2 / R. Based on these
expressions, we can approach the problem three different ways. Let’s do two methods and if the answers agree, we
can feel confident that we are correct.
Solve: (a) The simplest approach is to determine the current by
I = P / Δ V = 45 W/(120 V) = 0.38 A
(b) The resistance may be determined by
R = ΔV / I = 120 V/(0.375 A) = 320 Ω
Also
R = P/ I 2 = 45 W/(0.375 A) 2 = 320 Ω
Assess: Since we have determined the resistance by two methods and obtained the same result, we can be confident
that our answer is correct.
P22.40. Prepare: The relationships needed for this problem are P = IV , P = ΔU / Δt and Q = I Δt.
Solve: (a) The power is P = IV = 360 W.
(b) The total energy is Δ U = PΔ t = 0.36 J.
(c) The total charge that flows is Q = I Δt = 8.0 × 10−4 C.
Assess: Given the electric potential, current, and time pulse, these are reasonable values for the power, total energy,
and total charge.
P22.41. Prepare: The dimensions of current times time (the rating printed on the battery) are not yet the
dimensions of energy until we multiply by the dimensions of voltage. That is, in SI units, A ⋅ s = C, not J. However,
if we multiply both sides by V then the result is an energy unit: A ⋅ s ⋅ V = J.
The upshot of this is that the battery rating mA ⋅ h doesn’t tell us the energy capacity unless we also know the
voltage of the battery. But we do know the voltage of the battery, so we proceed.
Solve:
⎛ 1 A ⎞⎛ 3600 s ⎞
(450 mA ⋅ h)(9 V) ⎜
⎟⎜
⎟ = 14 580 A ⋅ s ⋅ V = 14 580 J ≈ 15 000 J
⎝ 1000 mA ⎠⎝ 1 h ⎠
Assess: This is a reasonable amount of energy. Fatter batteries hold more chemicals and therefore more chemical
potential energy, so a D-cell would have a higher mA ⋅ h rating than a AA-cell (both at 1.5 V).
The 9 V battery likely has six 1.5 V cells connected in series inside.
P22.42. Prepare: The electric potential difference, resistance, and current are related by Δ V = IR. The current is
related to the charge and time by I = Δ Q/ Δt. We can combine these two expressions to determine the amount of
charge that moves through the bulb in the specified time.
Solve: The amount of charge that moves through the bulb in a specified amount of time may be determined by
ΔQ = I Δt = (ΔV/ R)Δt = (3.0 V)(6.0 × 102 s)/(6.0 Ω) = 300 C
Assess: This is a reasonable amount of charge to move through a flashlight bulb in 10 min.
P22.43.
Prepare: We are given the power and the voltage of the toaster, as well as the time interval.
Solve:
P = I ΔV =
ΔQ
PΔ t (900 W)(60 s)
ΔV ⇒ ΔQ =
=
= 450 C
120 V
Δt
ΔV
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Current and Resistance
22-17
Assess: These are typical values for a toaster.
P22.44. Prepare: Ohm’s law may be used to determine the resistance of the wire ( R = V/ I ). The total amount of
heat
energy
needed
to
warm
the
ice
to
0° C
and
then
melt
it
may
be
determined
by
Q = mcΔT + mL f = m(cΔT + L f ). The power of the heater may be determined by P = IV . Finally, the time needed
for this heater to warm and melt the ice may be determined by P = Q /Δ t .
Solve: (a) The resistance of the heating wire is R = V / I = 120 V/5.0 A = 24 Ω (b) The amount of heat needed to
warm and melt the ice is Q = m(cΔT + L f ) = (0.720 kg)[(2090 J/(kg ⋅ K))(10 K) + 33.3 × 104 J/kg] = 2.55 × 105 J
The power of the heater is P = IV = (5.0 A)(120 V) = 600 W.
The time needed to warm and melt the ice is Δt = Q/P = 2.55 × 105 J / 600 W = 425 s = 7.1 min
Assess: The value for the resistance is reasonable and 7.1 minutes is a reasonable time needed for a 600 W heater to
warm and melt 720 g of ice.
P22.45. Prepare: Knowing the area of the plate and the thickness of the desired plating, we can determine the
volume of zinc to be plated (don’t forget we have to plate both sides). Knowing the volume of zinc needed and the
density of zinc, we can determine the mass of the zinc needed. Knowing the molecular mass of zinc and Avogadro’s
number, we can determine the mass of each zinc ion and then the number of ions needed to obtain the desired
platting mass. Knowing the number of ions needed and the charge on each ion, we can determine the amount of
charge to be transferred. Finally, knowing the amount of charge to be transferred and the current, we can determine
the amount of time needed to transfer this charge.
Solve: The volume of zinc to be plated is
V = 2( Δ x) A = 2(1.00 × 10−7 m)(4.0 × 10−4 m 2 ) = 8.0 × 10−11 m3
The mass of zinc needed is
Μ = ρV = (7.14 × 103 kg/m3 )(8.0 × 10−11 m3 ) = 5.71 × 10−7 kg
The mass of one zinc ion is
m = M / N A = (65.4 × 10−3 kg/mole)/(6.02 × 1023 /mole) = 1.09 × 10−25 kg
The number of zinc ions needed is
N = Μ / m = 5.71 × 10−7 kg/(1.09 × 10−25 kg) = 5.24 × 1018
The amount of charge carried by this number of zinc ions is
Δq = Nqion = N (2e) = (5.24 × 1018 )(2)(1.6 × 10−19 C) = 1.68 C
The amount of time needed to transport this amount of charge with the specified current is
Δ t = Δ q / I = 1.68 C/(1.0 × 10−3 A) = 1.68 × 103 s = 28 min
Assess: This is a reasonable plating time. If you hang out with physics faculty long enough, you will eventually hear
one mutter “sweet.” When they do that, they have just thought up a problem similar to this one. Notice the large
number of concepts involved and the scope and range of these concepts. This type problem is excellent for
developing your logic skills. It also is excellent final exam material.
P22.46. Prepare: This is an integrated problem and we will need to use information from previous chapters such
as Equation 12.23, Q = McΔT , to get the energy Q, and the equation for power, P = Q /Δt , along with the equation
for electrical power, P = ( ΔV ) 2 / R (assuming the hot dogs are ohmic).
We’ll solve the middle equation for Δ t , insert P from the last equation and Q (which is energy, not charge) from
the first.
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22-18
Chapter 22
Known
Δ V = 120 V
R = 150 Ω
M = 0 . 060 kg
Δ T = 80 °C − 20 °C = 60 °C = 60 K
c = 2500 J/(kg ⋅ K) (from Table 12.4)
Find
Δt
Solve:
Δt =
Q
McΔT
(0 . 060 kg)(2500 J/(kg ⋅ K))(60 K)
=
=
= 94 s
2
(120 V) 2 /(150 Ω)
P (ΔV ) / R
Assess: The answer is just over one and a half minutes. This is not as fast as a modern microwave, but it was
considered pretty quick before microwave ovens came along. Hot dogs cooked this way do not have a nice grilled
look or flavor; they don’t smell as good as grilled hot dogs either. But kids will eat anything.
It is especially in the integrated problems that we are likely to run into different uses of the same symbol. Recently
we’ve been using Q as charge, but in this problem it is energy.
The units deserve some attention. The kg and K cancel in the numerator. We’re left with J ⋅ Ω/ V 2. Further, Ω = V/A
and V = J/C.
J⋅Ω
J V J V
J C C
= 2 =
=
= =s
V2
V A A V2 A J A
P22.47 Prepare: In order to work this problem, we will need to combine a number of concepts. A list of the
concepts we must be familiar with are as follows:
(a) The capacitance of the capacitor is determined by C = ε o A / d .
(b) Knowing the capacitance, the initial charge on the capacitor is determined by Q = CV .
(c) The resistance of the air between the plates of the capacitor is determined by R = ρ d /A.
(d) Knowing the resistance, the current flowing between the plates of the capacitor may be determined by I = V /R.
(e) Knowing the current, we can determine the charge that flows by Δ Q = I Δt .
(f) Finally, knowing the initial charge Q and the charge that flows Δ Q, the percent of charge that flows is
determined by percent = (Δ Q /Q )(100).
Solve: The capacitance of the capacitor is
C = ε o A/ d = (8.85 × 10−12 F/m)(10−2 m 2 ) /(1.2 × 10−3 m) = 7.38 × 10−11 F
Knowing the capacitance of the capacitor, we can determine the initial charge on the capacitor as
Q = CV = (7.38 × 10−11 F)(250 V) = 18.5 × 10−9 C
The resistance of the air between the plates of the capacitor is determined by
R = ρ d/ A = (3×1013 Ω ⋅ m)(1.2 × 10−3 m) /(10−2 m 2 ) = 3.6 × 1012 Ω
Knowing the resistance, the current flowing between the plates of the capacitor may be determined by
I = V / R = 250 V/(3.6 × 1012 Ω) = 69.4 × 10−12 A
Knowing the current and the time the current flows, we can determine the charge that flows by
Δ Q = I Δt = (69.4 × 10−12 A)(60 s) = 4.16 × 10−9 C
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Current and Resistance
22-19
Finally, knowing the initial charge (Q) and the charge that flows (Δ Q ), the percent of charge that flows is
determined by
Percent = ( ΔQ / Q )(100) = (4.16 × 10−9 C /(18.5 × 10−9 C))(100) = 20 %
Assess: This is a reasonable percent of the charge to flow through the capacitor in 60 s. After Chapter 23, we will
have a more sophisticated method of working this problem. You should note with a sense of accomplishment the
number of different concepts that had to be mastered in order to solve this problem.
P22.48. Prepare: The current is the rate at which the charge of the ions moves through the ion channel. A current
of 1.8 pA for the potassium ions means that a charge of 1.8 pC flows through the potassium ion channel per second.
Potassium ions are monovalent and each ion carries a charge of 1.6 × 10−19 C.
Solve: (a) The number of potassium ions that pass through the ion channel per second is
1.8 × 10−12 C/s
= 1.125 × 107 s −1
1.6 × 10−19 C
Since the channel opens only for 1.0 ms, the total number of potassium ions that pass through the channel is
(1.125 × 107 s −1 )(1.0 × 10−3 s) = 1.1 × 104 ions.
Assess: A small current over a small time means that a small number of ions pass through the channel.
P22.49. Prepare: The resistance is given by Equation 22.7: R = ρ L / A.
We are given L = 5. 0 × 10−9 m, A = π r 2 = π (d /2) 2 = π (0. 30 nm/2) 2 = 7 . 07 × 10−20 m 2, and ρ = 0 . 60 Ω ⋅ m.
Solve:
R=
ρL
A
=
(0. 60 Ω ⋅ m)(5 .0 × 10−9 m)
= 4 . 2 × 1010 Ω = 42 G Ω
7 . 07 × 10−20 m 2
Assess: This is a tremendously large resistance (due to the small cross-sectional area); in many electronic circuit
applications it would be simply approximated as infinite. So not many charge carriers (ions) will get through the ion
channel (the current will only be a few pA, depending on the potential difference); but that is precisely the point—we
want the cell to control the ion flow very closely.
P22.50. Prepare: Current, charge, and time are related by I = Δ Q/ Δt. Power, current, and electric potential
difference are related by P = I ΔV .
Solve: (a) The current in the channel is
I = Δq/Δt = (1.0 × 107 ions/s)(1e/ion)(1.6 × 10−19 C/e) = 1.6 × 10−12 A
(b) The power dissipated in the channel is
P = I Δ V = (1.6 × 10−12 A)(7 × 10−2 V) = 1.1 × 10−13 W
Assess: On a biological molecular scale we should expect very small values.
P22.51. Prepare: The electric potential difference, current, and resistance are related by Ohm’s law I = ΔV /R.
The amount of charge, the current, and the time it takes the charge to flow are related by Δ Q = I Δt .
Solve: The current is obtained by I = Δ V / R = 1.2 V/22 Ω = 54.5 mA. The amount of time the battery can deliver this
current is Δ t = Δ Q / I = 1800 mA ⋅ hr / 54.5 mA = 33 h.
Assess: Since the current is very small (54.5 mA) we expect the battery to able to deliver this current for a long time.
P22.52. Prepare: The electric power is P = I Δ V . The amount of charge, the current, and the time it takes the
charge to flow are related by Δ Q = I Δ t.
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22-20
Chapter 22
Solve:
Δt =
ΔQ
ΔQ
ΔQ
15 A ⋅ h
=
=
ΔV =
(1.5 V ) = 15 h
1.5 W
I
P/ΔV
P
Assess: This seems about right.
P22.53. Prepare: The electric potential difference across a resistor is related to the current through the resistor and
the resistance of the resistor by Δ V = IR. The resistance of a wire is related to its resistivity, length, and crosssectional area by R = ρ L / A.
Solve. (a) The resistance of the hot resistor is R = Δ V / I = 120 V/8.0 A = 15 Ω.
(b) Using this value for the resistance we determine the resistivity of the wire to be
ρ = RA / L = Rπ r 2 / L = (15.0 Ω)π (3.0 × 10−4 m) 2 /(2.0 m) = 2.1 × 10−6 Ω ⋅ m
The value listed in Table 22.1 for the resistivity of nichrome is ρ nichrome = 1.5 × 10−6 Ω ⋅ m. The calculated value is
greater due to the fact it was determined from a value for the resistance when the wire was hot and the resistance is a
function of the temperature. For nichrome wire, as the wire becomes hotter, its resistance increases.
Assess: This problem demonstrates that it is important to know the conditions under which data are obtained to
determine a physical quantity.
P22.54. Prepare: We assume that Ohm’s law applies to the situation.
I=
ΔV
R
We also use Equation 22.7 which gives R in terms of ρ , L, and A.
R=
ρL
A
We are given that Δ V = 9 . 0 V, L = 0 . 050 m, I = 230 μ A, and A = π r 2 = π (d /2) 2 = π (1 . 5 mm/2) 2 = 1 . 77 × 10−6 m 2.
Solve: Combine the two previous equations.
ρ=
RA ΔV A (9 . 0 V)(1 . 77 × 10−6 m 2 )
=
=
= 1.4 Ω ⋅ m
L
I L (230 × 10−6 A)(0. 050 m)
Assess: This resistivity is reasonably close to the value for blood (1.6 Ω ⋅ m) given in Table 22.1. One of the m’s
cancels and V/A = Ω.
P22.55. Prepare: Assume the bulb is to be used in a 120 V circuit. The resistivity for tungsten is given in
Table 22.1 for both temperatures of interest. The resistance of the filament may be determined by R = V 2 / P and the
length of the filament may be determine by R = ρ L / A. The cross-sectional area of the filament is
A = π r 2 = 1.26 × 10−9 m 2 .
Solve: (a) The resistance of the filament is R = V 2 / P = (120 V) 2 /40 W = 360 Ω. The length of the filament is
L = RA/ ρ = (360 Ω)(1.26 × 10−9 m 2 ) /(5.0 × 10−7 Ω ⋅ m) = 0.90 m.
(b) The resistance of the cold filament is R = ρ L / A = (5.6 × 10−8 Ω ⋅ m)(0.90 m) /(1.26 × 10−9 m 2 ) = 40 Ω
Assess: These are reasonable values for the length and cold resistance of a tungsten filament.
P22.56. Prepare: We assume the extension cord is ohmic so we can use Δ V = IR. We also use R = ρ L / A.
We are given L = 50 ft = 15 . 2 m, I = 10 A, and A = π r 2 = π (d /2) 2 = π (1. 3 mm/2) 2 = 1. 33 × 10−6 m 2. We also look
up the resistivity of copper in Table 22.1: ρ Cu = 1 . 7 × 10−8 Ω ⋅ m.
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Current and Resistance
22-21
Solve: Combine the two equations.
(1 . 7 × 10−8 Ω ⋅ m)(15 . 2 m)
⎛ ρL ⎞
Δ V = IR = I ⎜
= (10 A)
= 1.9 V
⎟
1 . 33 × 10−6 m 2
⎝ A ⎠
Assess: 1.9 V is less than 2 percent of 120 V, but in some situations it can be important. The longer the extension
cord the greater the resistances (if the diameter stays the same); the greater the voltage drop across the cord, the more
power is dissipated as thermal energy in the cord. If your extension cord gets hot to the touch, then you are
dissipating too much power in the cord. Get a shorter or fatter extension cord. If you buy a long extension cord, make
sure it has a low-gauge wire (large diameter) so the resistance will be low.
P22.57. Prepare: The voltage drop along the wire is related to the resistance of the wire and the current in the wire
by Δ V = IR. The resistance of a wire is related to its resistivity, length, and cross-sectional area by R = ρ L / A. The
area of a wire is related to its diameter by A = π d 2 /4.
Solve: The resistance of wire is
R = ΔV / I = 0.50 V/(3 × 102 A) = 1.67 × 10−3 Ω
The cross-sectional area of the wire is
A = ρ L / R = (1.7 × 10−8 Ω ⋅ m)(1.0 m)/(1.67 × 10−3 Ω) = 1.02 × 10−5 m 2
The diameter of the wire is
d = 4 A / π = 4(1.02 × 10−5 m 2 )/π = 3.6 × 10−3 m = 3.6 mm
Assess: The next time you are working on your car, take a look at the cable between the battery and the starter. You
will find that this is a good value.
P22.58. Prepare: The moving electrons are a current, even though they’re not confined inside a wire. The current
is the rate at which the charge Q = N e e = I Δ t moves. The number of moving electrons per unit time is thus
N e / Δt = I/ e.
Solve: (a). The electron current is
Ne I
50 × 10−6 A
= =
= 3.12 × 1014 s −1
Δ t e 1.60 × 10−19 C
This means during the time interval Δ t = 1 s, 3.1 × 1014 electrons strike the screen.
(b) The acceleration can be found from kinematics:
v12 = (4.0 × 107 m/s) 2 = v 02 + 2aΔx = 2aΔ x ⇒ a =
(4.0 × 107 m/s) 2
= 1.60 × 1017 m/s 2
2(5.0 × 10−3 m)
But the acceleration is a = F / m = eE/ m. Consequently, the electric field must be
E=
ma (9.11 × 10−31 kg)(1.60 × 1017 m/s 2 )
=
= 9.1 × 105 N/C
1.6 × 10−19 C
e
(c) When they strike the screen, each electron has a kinetic energy
K=
1 2 1
mv1 = (9.11 × 10−31 kg)(4.0 × 107 m/s) = 7.288 × 10−16 J
2
2
Power is the rate at which the screen absorbs this energy. The power of the beam is
P=
ΔE
N
= K e = (7.288 × 10−16 J)(3.12 × 1014 s −1 ) = 0.227 J/s = 0.23 W
Δt
Δt
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22-22
Chapter 22
Assess: Power delivered to the screen by the electron beam is reasonable because the screen over time becomes a
little warm.
P22.59. Prepare: current is conserved, the currents in the two segments of the wire are the same. The currents in
the two segments of the wire are related by I1 = I 2 = I . Δ V1 = I1R1 and Δ V2 = I 2 R2 . Because from Equation 22.7
R = ρ L / A so R1 = ρ1L1/ A1 and R2 = ρ 2 L2 / A2 . The wire’s diameter is constant, so A1 = A2 . Also L1 = L2 .
Solve: From this information,
ΔV1 I1R1 ⎛ I1 ⎞⎛ ρ1L1 ⎞⎛ A2
=
= ⎜ ⎟⎜
⎟⎜
Δ V2 I 2 R2 ⎝ I 2 ⎠⎝ A1 ⎠⎝ ρ 2 L2
⎞ ρ1 1
=
⎟=
⎠ ρ2 2
Assess: Notice the beauty of being asked to find a ratio of two quantities—most other quantities cancel.
P22.60. Prepare: Recall that P = I Δ V ⇒ Δ V =
P
. We are given the power dissipated per meter, or P /L = 6.0 W.
I
Solve: The electric field strength is
E=
ΔV P / I P / L 6.0 W/m
=
=
=
= 7.5 V/m
0.80 A
L
L
I
Assess: This is a reasonable value for the electric field strength.
P22.61. Prepare: Since the problem asks us to identify the likely material by its electrical properties, we probably
want to find the resistivity of the material. The equation that contains resistivity is R = ρ L / A.
Fortunately, we are also given L and A (indirectly), and we can compute R from Ohm’s law, R = Δ V / I . We’ll
compute ρ and then look up the value in Table 22.1 to see if we can identify the material.
We are given L = 2. 3 m, Δ V = 1 . 2 V, I = 0 . 61 A, and A = π r 2 = π (d /2) 2 = π (0 . 38 mm/2) 2 = 1 . 13 × 10−7 m 2.
Solve: Solve the first equation for ρ and insert R from Ohm’s law.
ρ=
RA ΔV A (1 .2 V)(1 . 13 × 10−7 m 2 )
=
=
= 9 . 7 × 10−8 Ω ⋅ m
(0 .61 A)(2 . 3 m)
L
I L
This value matches the value given in the table for iron, so we assert the wire is made of iron.
Assess: We are gratified that the answer matched one of the values in the table. The units work too, as V/A = Ω.
P22.62. Prepare: Since we are asked for L and given ρ and A (indirectly), we realize we want to solve the
following for L: R = ρ L / A.
We will need to solve for R in P = ( ΔV ) 2 / R and insert it into the equation for L.
A = π r 2 = π (d /2) 2 = π (0 .035 mm/2) 2 = 9 . 62 × 10−10 m 2.
Known
Δ V = 120 V
P = 100 W
ρ = 5. 0 × 10−7 Ω ⋅ m
A = 9. 62 × 10−10 m 2
Find
L
Solve:
L=
RA
ρ
=
(Δ V ) 2 A (120 V) 2 9 . 62 × 10−10 m 2
=
= 0 . 28 m = 28 cm
P ρ
100 W 5 .0 × 10−7 Ω ⋅ m
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Current and Resistance
22-23
Assess: The data and result match reasonably closely those in Example 22.5, and this comforts us. The wattages of
the bulbs aren’t the same, and the given diameters are slightly different, so we don’t expect a perfect match, but our
answer is about within a factor of two of the answer in the example for a similar situation.
The units deserve some attention. One of the m’s cancels. Split the V 2 apart, and for one of them use V = J/C. Also
apply W = J/s. and A = C/s. At the end we will use V = A ⋅ Ω.
V V m V CJ m V s m V m
=
=
=
=m
W Ω J/s Ω
C Ω AΩ
P22.63. Prepare: The resistivity of copper is 1.7 × 10−8 Ω ⋅ m and we want the wire’s resistance to be 1.0 Ω. We will
use Equation 22.7 to find the wire’s length in terms of its area of cross section or the wire’s radius r. Because the mass
of the wire’s material is given, we can use the definition of mass density to connect the wire’s length and radius r.
Solve: Using the formula for the resistance of a wire,
R=ρ
L
L
⇒ 1.0 Ω = (1.7 × 10−8 Ω ⋅ m) 2 ⇒ L = (1.848 × 108 m −1 )r 2
A
πr
We need another relation connecting L and r. Making use of the mass density of copper,
1.0 × 10−3 kg
= 8900 kg/m3 ⇒ r 2 L = 3.577 × 10−8 m3
π r 2L
Using the value of L obtained previously,
r 2 (1.848 × 108 m −1 )r 2 = 3.577 × 10−8 m3 ⇒ r 4 = 1.936 × 10−16 m 4 ⇒ r = 1.18 × 10−4 m = 0.118 mm
Thus, the diameter of the wire is 0.24 mm and the length is
L = (1.848 × 108 m −1 )(1.18 × 10−4 m) 2 = 2.6 m
Assess: It is reasonable to make a 2.6 m long wire with a diameter of 0.24 mm from a copper block of 1.0 g.
P22.64. Prepare: We’ll assume the resistors are ohmic, so we can use both expressions for the power dissipation:
(ΔV )2
= I 2 R.
R
Solve: When the resistors are connected to identical batteries (same ΔV ) we can find the ratio of the resistances.
P=
P1 = 3P2 ⇒
(ΔV )2
(ΔV )2
=3
⇒ R2 = 3R1
R1
R2
Now we use the other version of the power equation when the current is the same for both resistors.
P1 I 2 R1
R
1
= 2 = 1 =
P2 I R2 3R1 3
Assess: Now resistor 2 dissipates more power.
P22.65. Prepare: The power dissipated by the heater is related to the potential difference across the heater and its
resistance by P = ΔV 2 / R. The amount of energy needed to raise a mass m of water by an amount Δ T is Q = mcΔ T.
The power dissipated is related to the energy dissipated and the time to dissipate that energy by P = W /Δ t .
Solve: (a) The resistance of the heater is
R = ΔV 2 / P = (120 V) 2 /(300 W) = 48 Ω
(b) The amount of energy that must be dissipated by the heater in order to heat the water from 18oC to 100oC is
Q = mcΔ T = (0.40 kg)(4.186 × 103 J/(kg ⋅ °C))(82 °C) = 1.37 × 105 J
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22-24
Chapter 22
The time required to dissipate this much energy is
Δt = Q/ P = 1.37 × 105 J/(3 × 102 W) = 460 s = 7.6 min
Assess: This is not an unreasonable amount of time to wait for a good cup of tea.
P22.66. Prepare: Rewrite I = ΔQ / Δ t as Δ Q = I Δt for a constant current that is straightforward. If I varies, as in
our problem, ΔQ would be the area under an I vs. t graph, which we can figure out in sections.
In part (b) we’ll use P = I 2 R, and, since energy = power × time, PΔ t = I 2 RΔt , in sections.
We are given that R = 1. 0 Ω.
Solve: (a) Between t = 0 s and t = 2 s there is no area under the graph.
Between t = 2 s and t = 5 s the area is (2. 5 A)(3. 0 s) = 7. 5 A ⋅ s = 7. 5 C.
Between t = 5 s and t = 10 s the area is (1 .25 A)(5 .0 s) = 6 . 25 A ⋅ s = 6 . 25 C.
The total area under the graph is 7 . 5 C + 6 .25 C = 13 . 75 C ≈ 14 C.
(b) Between t = 0 s and t = 2 s there is no current so I 2 RΔt = 0 J.
Between t = 2 s and t = 5 s, I 2RΔt = (2. 5 A) 2 (1 .0 Ω)(3. 0 s) = 18 .75 J.
Between t = 5 s and t = 10 s, I 2RΔt = (1 .25 A) 2 (1. 0 Ω)(5. 0 s) = 7 . 81 J.
The energy dissipated is 0 J + 18 . 75 J + 7 . 81 J = 26 . 6 J ≈ 27 J.
Assess: These results both seem reasonable, not obviously out of the ballpark. Calculus would allow one to find the
total Δ Q for a smoothly varying current, as it allows one to find the area under a curve.
P22.67. Prepare: We will use Integrated Example 22.16 (b) as a model for our solution. Consider the muscle and
fat tissue to be resistors in parallel with the total current equal to the sum of the current through the fat tissue and the
muscle ( I T = I F + I M ). The fraction of the fat ( f F ) plus the fraction of the muscle ( f M = 1 − f F ) constitutes the leg
segment. Give the potential difference, we can use Ohm’s law to determine the current through the fat and muscle, if
we can determine the resistance of each. The resistance is related to the resistivity, length, and cross-sectional area by
R = ρ L / A.
Solve: (a) Determine the cross-sectional area of the leg by A = π r 2 = π (6.0 × 10−2 m) 2 = 1.13 × 10−2 m 2 .
Determine an expression for the resistance of the fat tissue
RF =
ρF L
AF
=
ρF L
fF A
=
(25 Ω ⋅ m)(0.40 m) 8.85 × 102 Ω
=
f F (1.13 × 10−2 m 2 )
fF
Determine an expression for the resistance of the muscle.
RM =
ρM L
AM
=
ρM L
fM A
=
ρM L
(1 − f F ) A
=
(13 Ω ⋅ m)(0.40 m)
4.60 × 102 Ω
=
(1 − f F )(1.13 × 10−2 m 2 )
(1 − f F )
Determine an expression for the current in the fat tissue.
IF =
(0.75 V) f F
V
=
= 8.48 × 10−4 f F A
RF (8.85 × 102 Ω)
Determine an expression for the current in the muscle.
IM =
V
(0.75 V)(1 − f F )
=
= 16.3 × 10−4 (1 − f F ) A
RM
(4.60 × 102 Ω)
Since we are modeling the fat tissue and muscle as two resistors in parallel, the total current is the sum of these two
currents.
I T = I F + I M ⇒ 16 × 10−4 A = 8.48 × 10−4 f F A + 16.3 × 10−4 (1 − f F ) A
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Current and Resistance
22-25
This expression may be solved for the fraction of the fat tissue f F = 3.8 % = 1/ 26. The fraction of the muscle is then
f M = 96.2 % = 25/ 26.
(b) The fraction of fat is 1/26.
Assess: This would be reasonable values for an athletic person.
P22.68. Prepare: We’ll use Ohm’s law to calculate the current in each case.
Solve: (a)
I=
ΔV
120 V
=
= 0 . 24 mA
R
5 × 105 Ω
This does not exceed the maximum safe current of 5 mA, so it is not dangerous.
(b)
I=
ΔV
120 V
=
= 120 mA
R 1000 Ω
This exceeds the maximum safe current and is dangerous.
Assess: The warnings against using electrical equipment while wet (in the shower, for example) make a lot of sense.
Your resistance from one fingertip to another on the other hand is much larger than the resistance between fingertips
on the same hand, so you might get a good shock touching the terminals with the same hand. The good thing is that
the current wouldn’t go through your heart in this case.
P22.69. Prepare: The resistance of a wire is related to its resistivity, length, and cross-sectional area by
R = ρ L / A. The area of a wire is related to its diameter by A = π d 2 /4. The potential drop across a resistor is related to
the resistance of the resistor and the current through the resistor by Δ V = IR. The power dissipated is related to
potential difference, current, and resistance by P = VI = I 2 R = V 2 / R.
Solve: (a) The resistance between the hands is
R = ρ L / A = ρ L(π d /4) = 4 ρ /(π d 2 ) = 4(5.0 Ω ⋅ m)(1.6 m)/(π (0.10 m) 2 ) = 1.0 × 103 Ω
(b) The potential difference across a resistor of this size when there is a current of 100 mA is
ΔV = IR = (0.10 A)(1.02 × 103 Ω)=100 V
Assess: This result tells us that a potential difference of 100 V can be lethal. This should give you new respect for
the potential difference at all of your electrical outlets.
P22.70. Prepare: The problem says we can approximate the resistance as R ≈ Δ V /I (which isn’t exactly right for a
curve such as this one).
Solve:
R=
6 .0 V
ΔV
=
= 37 .5 Ω
0 . 160 A
I
The correct choice is D.
Assess: The units work out. We see the slope getting shallower as the potential difference increases; this means the
1
resistance is increasing since R = slope
.
P22.71. Prepare: We have been informed that the filament gets progressively thinner as the bulb ages and we
know that the resistance of the filament may be determined by R = ρ L / A.
Solve: As the filament ages it gets thinner, that is, the cross-sectional area gets smaller. If this is the case the
resistance will get larger. The correct choice is A.
Assess: Answering what-if questions by examining expressions relating physical quantities is a valuable skill.
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22-26
Chapter 22
P22.72. Prepare: The temperature monotonically increases to a stable value and so the resistance does the same.
Solve: As the resistance rises and levels off, the current decreases to a stable value. This eliminates choice B. We
expect a graph of I versus t to be monotonic, so we eliminate C.
The correct choice is A.
Assess: The current does not need time to build up; it is there as soon as the potential difference is applied, so we
eliminate the two choices that have zero current at zero time. It does take a short time to heat the filament to glowing
temperature.
P22.73. Prepare: Since the power dissipated is P = I 2 R, the current is the critical factor and the filament melts
due to a current surge.
Solve: In order to increase the lifetime of the bulb, we must find a way to prevent a current surge. Limiting the
maximum current through the bulb will do this. The correct response is B.
Assess: Limiting the maximum current will prevent a current surge.
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