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Chapter 25
Electric Currents and
Resistance
Copyright © 2009 Pearson Education, Inc.
25-4 Resistivity
Example 25-5: Speaker wires.
Suppose you want to connect
your stereo to remote
speakers. (a) If each wire must
be 20 m long, what diameter
copper wire should you use to
keep the resistance less than
0.10 Ω per wire? (b) If the
current to each speaker is 4.0
A, what is the potential
difference, or voltage drop,
across each wire?
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25-4 Resistivity
For any given material, the resistivity
increases with temperature:
  T   0 1    T  T0  
 R  T   R0 1    T  T0  
Semiconductors are complex materials, and
may have resistivities that decrease with
temperature.
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25-4 Resistivity
Example 25-7: Resistance thermometer.
The variation in electrical resistance with
temperature can be used to make precise
temperature measurements. Platinum is
commonly used since it is relatively free from
corrosive effects and has a high melting point.
Suppose at 20.0°C the resistance of a platinum
resistance thermometer is 164.2 Ω. When
placed in a particular solution, the resistance is
187.4 Ω. What is the temperature of this
solution?   Pt   0.003927   m 
Copyright © 2009 Pearson Education, Inc.
ConcepTest 25.3a
Wires I
Two wires, A and B, are made of the
1) dA = 4dB
same metal and have equal length,
2) dA = 2dB
but the resistance of wire A is four
times the resistance of wire B. How
do their diameters compare?
3) dA = dB
4) dA = 1/2dB
5) dA = 1/4dB
ConcepTest 25.3a
Wires I
Two wires, A and B, are made of the
1) dA = 4dB
same metal and have equal length,
2) dA = 2dB
but the resistance of wire A is four
3) dA = dB
times the resistance of wire B. How
4) dA = 1/2dB
do their diameters compare?
5) dA = 1/4dB
The resistance of wire A is greater because its area is less than
wire B. Since area is related to radius (or diameter) squared, the
diameter of A must be two times less than the diameter of B.
R ρ
A
25-5 Electric Power
Power, as in kinematics, is the energy
transformed by a device per unit time:
or
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25-5 Electric Power
The unit of power is the watt, W.
For ohmic devices, we can make the
substitutions:
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25-5 Electric Power
Example 25-8: Headlights.
Calculate the resistance of a 40-W
automobile headlight designed for 12 V.
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25-5 Electric Power
What you pay for on your electric bill is
not power, but energy – the power
consumption multiplied by the time.
We have been measuring energy in
joules, but the electric company
measures it in kilowatt-hours, kWh:
1 kWh = (1000 W)(3600 s) = 3.60 x 106 J.
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25-5 Electric Power
Example 25-9: Electric heater.
An electric heater draws a steady 15.0
A on a 120-V line. How much power
does it require and how much does it
cost per month (30 days) if it operates
3.0 h per day and the electric company
charges 9.2 cents per kWh?
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25-7 Alternating Current
Current from a battery
flows steadily in one
direction (direct current,
DC). Current from a
power plant varies
sinusoidally (alternating
current, AC).
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25-7 Alternating Current
The voltage varies sinusoidally with time:
,,
as does the current:
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25-7 Alternating Current
Multiplying the current and the voltage gives
the power:
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25-7 Alternating Current
Usually we are interested in the average power:
.
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25-7 Alternating Current
The current and voltage both have average
values of zero, so we square them, take the
average, then take the square root, yielding the
root-mean-square (rms) value:
Copyright © 2009 Pearson Education, Inc.
25-7 Alternating Current
Example 25-13: Hair dryer.
(a) Calculate the resistance and the peak current
in a 1000-W hair dryer connected to a 120-V line.
(b) What happens if it is connected to a 240-V line
in Britain?
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25-8 Microscopic View of Electric
Current: Current Density and Drift
Velocity
Electrons in a conductor have large, random
speeds just due to their temperature. When a
potential difference is applied, the electrons
also acquire an average drift velocity, which is
generally considerably smaller than the
thermal velocity.
Copyright © 2009 Pearson Education, Inc.
25-8 Microscopic View of Electric
Current: Current Density and Drift
Velocity
We define the current density (current per
unit area) – this is a convenient concept
for relating the microscopic motions of
electrons to the macroscopic current:
If the current is not uniform:
.
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(Remember the
water in the pipe)
25-8 Microscopic View of Electric
Current: Current Density and Drift
Velocity
This drift speed is related to the current in the
wire, and also to the number of electrons per unit
volume:
and
Copyright © 2009 Pearson Education, Inc.
25-8 Microscopic View of Electric
Current: Current Density and Drift
Velocity
Example 25-14: Electron speeds in a wire.
A copper wire 3.2 mm in diameter carries a 5.0A current. Determine (a) the current density in
the wire, and (b) the drift velocity of the free
electrons. (c) Estimate the rms speed of
electrons assuming they behave like an ideal
gas at 20°C. Assume that one electron per Cu
atom is free to move (the others remain bound
to the atom).
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25-8 Microscopic View of Electric
Current: Current Density and Drift
Velocity
The electric field inside a current-carrying
wire can be found from the relationship
between the current, voltage, and resistance.
Writing R = ρ l/A, I = jA, and V = El , and
substituting in Ohm’s law gives:
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25-8 Microscopic View of Electric
Current: Current Density and Drift
Velocity
Example 25-15: Electric field inside a wire.
What is the electric field inside the wire of
the earlier example? (The current density
was found to be 6.2 x 105 A/m2.)
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25-9 Superconductivity
In general, resistivity
decreases as
temperature decreases.
Some materials,
however, have
resistivity that falls
abruptly to zero at a
very low temperature,
called the critical
temperature, TC.
Purely quantum mechanical; CANNOT be
explained using classical physics.
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25-9 Superconductivity
Experiments have shown that currents, once
started, can flow through these materials for
years without decreasing even without a
potential difference.
Critical temperatures are low; for many years no
material was found to be superconducting above
23 K.
Since 1987, new materials have been found that
are superconducting below 90 K, and work on
higher temperature superconductors is
continuing.
Copyright © 2009 Pearson Education, Inc.
Summary of Chapter 25
• A battery is a source of constant potential
difference.
• Electric current is the rate of flow of electric
charge.
• Conventional current is in the direction that
positive charge would flow.
• Resistance is the ratio of voltage to current:
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Summary of Chapter 25
• Ohmic materials have constant resistance,
independent of voltage.
• Resistance is determined by shape and
material:
• ρ is the resistivity.
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Summary of Chapter 25
• Power in an electric circuit:
• Direct current is constant.
• Alternating current varies sinusoidally:
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Summary of Chapter 25
• The average (rms) current and voltage:
• Relation between drift speed and current:
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Chapter 26
DC Circuits
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26-1 EMF and Terminal Voltage
Electric circuit needs battery or generator to
produce current – these are called sources of
emf.
Battery is a nearly constant voltage source, but
does have a small internal resistance, which
reduces the actual voltage from the ideal emf:
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26-1 EMF and Terminal Voltage
This resistance behaves as though it were in
series with the emf.
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26-1 EMF and Terminal Voltage
Example 26-1: Battery with internal resistance.
A 65.0-Ω resistor is
connected to the
terminals of a battery
whose emf is 12.0 V and
whose internal
resistance is 0.5 Ω.
Calculate (a) the current
in the circuit, (b) the
terminal voltage of the
battery, Vab, and (c) the
power dissipated in the
resistor R and in the
battery’s internal resistance r.
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26-2 Resistors in Series and in
Parallel
A series connection has a single path from
the battery, through each circuit element in
turn, then back to the battery.
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26-2 Resistors in Series
The current through each resistor is the
same; the voltage depends on the
resistance. The sum of the voltage
drops across the resistors equals the
battery voltage:
V  V1  V2  V3  IR1  IR2  IR3
 I  R1  R2  R3   IReq
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 Series
26-2 Resistors in Series
From this we get the equivalent resistance (that
single resistance that gives the same current in
the circuit):
Unless an internal
resistance r is
specified assume V
constant.
Copyright © 2009 Pearson Education, Inc.
ConcepTest 26.1a
Series Resistors I
1) 12 V
Assume that the voltage of the battery
is 9 V and that the three resistors are
identical. What is the potential
difference across each resistor?
2) zero
3) 3 V
4) 4 V
5) you need to know the
actual value of R
9V
ConcepTest 26.1a
Series Resistors I
1) 12 V
Assume that the voltage of the battery
is 9 V and that the three resistors are
identical. What is the potential
difference across each resistor?
2) zero
3) 3 V
4) 4 V
5) you need to know the
actual value of R
Since the resistors are all equal,
the voltage will drop evenly
across the 3 resistors, with 1/3 of
9 V across each one. So we get a
3 V drop across each.
9V
ConcepTest 26.1b
Series Resistors II
1) 12 V
In the circuit below, what is the
2) zero
voltage across R1?
3) 6 V
4) 8 V
5) 4 V
R1 = 4 
R2 = 2 
12 V
ConcepTest 26.1b
Series Resistors II
1) 12 V
In the circuit below, what is the
2) zero
voltage across R1?
3) 6 V
4) 8 V
5) 4 V
The voltage drop across R1 has
to be twice as big as the drop
across R2. This means that V1 =
R1 = 4 
R2 = 2 
8 V and V2 = 4 V. Or else you
could find the current I = V/R =
(12 V)/(6  = 2 A, and then use
12 V
Ohm’s law to get voltages.
Follow-up: What happens if the voltage is 24 V?
26-2 Resistors in Parallel
A parallel connection splits the current; the
voltage across each resistor is the same:
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26-2 Resistors in Parallel
The total current is the sum of the currents
across each resistor:
,
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26-2 Resistors in Parallel
This gives the reciprocal of the equivalent
resistance:
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26-2 Resistors in Parallel
An analogy using water
may be helpful in
visualizing parallel
circuits. The water
(current) splits into two
streams; each falls the
same height, and the total
current is the sum of the
two currents. With two
pipes open, the resistance
to water flow is half what
it is with one pipe open.
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26-2 Resistors in Series and in
Parallel
Conceptual Example 26-2: Series or parallel?
(a) The lightbulbs in the figure are identical.
Which configuration produces more light? (b)
Which way do you think the headlights of a car
are wired? Ignore change of filament resistance R
with current.
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ConcepTest 26.2a
Parallel Resistors I
1) 10 A
In the circuit below, what is the
2) zero
current through R1?
3) 5 A
4) 2 A
5) 7 A
R2 = 2 
R1 = 5 
10 V
ConcepTest 26.2a
Parallel Resistors I
1) 10 A
In the circuit below, what is the
2) zero
current through R1?
3) 5 A
4) 2 A
5) 7 A
The voltage is the same (10 V) across each
R2 = 2 
resistor because they are in parallel. Thus,
we can use Ohm’s law, V1 = I1R1 to find the
R1 = 5 
current I1 = 2 A.
10 V
Follow-up: What is the total current through the battery?
ConcepTest 26.2b
Points P and Q are connected to a
Parallel Resistors II
1) increases
battery of fixed voltage. As more
2) remains the same
resistors R are added to the parallel
3) decreases
circuit, what happens to the total
4) drops to zero
current in the circuit?
ConcepTest 26.2b
Parallel Resistors II
Points P and Q are connected to a
1) increases
battery of fixed voltage. As more
2) remains the same
resistors R are added to the parallel
3) decreases
circuit, what happens to the total
4) drops to zero
current in the circuit?
As we add parallel resistors, the overall
resistance of the circuit drops. Since V =
IR, and V is held constant by the battery,
when resistance decreases, the current
must increase.
Follow-up: What happens to the current through each resistor?
26-2 Resistors in Series and in
Parallel
Conceptual Example 26-3: An illuminating surprise.
A 100-W, 120-V lightbulb and a 60-W, 120-V lightbulb
are connected in two different ways as shown. In each
case, which bulb glows more brightly? Ignore change
of filament resistance with current (and temperature).
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26-2 Resistors in Series and in
Parallel
Example: Current in one branch.
What is the current through the 500-Ω resistor
shown?
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26-2 Resistors in Series and in
Parallel
Example 26-8:
Analyzing a circuit.
A 9.0-V battery whose
internal resistance r is
0.50 Ω is connected in
the circuit shown. (a)
How much current is
drawn from the
battery? (b) What is
the terminal voltage of
the battery?
Note: slight error in figure and text
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