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Chapter 26
DC Circuits
Copyright © 2009 Pearson Education, Inc.
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 and in
Parallel
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:
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26-2 Resistors in Series and in
Parallel
From this we get the equivalent resistance (that
single resistance that gives the same current in
the circuit):
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26-2 Resistors in Series and in
Parallel
A parallel connection splits the current; the
voltage across each resistor is the same:
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26-2 Resistors in Series and in
Parallel
The total current is the sum of the currents
across each resistor:
,
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26-2 Resistors in Series and in
Parallel
This gives the reciprocal of the equivalent
resistance:
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26-2 Resistors in Series and 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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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 26-4: Circuit with series and
parallel resistors.
How much current is drawn from the
battery shown?
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26-2 Resistors in Series and in
Parallel
Example 26-5: Current in one branch.
What is the current through the 500-Ω resistor
shown? (Note: This is the same circuit as in the
previous problem.) The total current in the circuit
was found to be 17 mA.
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26-2 Resistors in Series and in
Parallel
Conceptual Example 26-6:
Bulb brightness in a circuit.
The circuit shown has
three identical lightbulbs,
each of resistance R.
(a) When switch S is
closed, how will the
brightness of bulbs
A and B compare with
that of bulb C? (b) What
happens when switch S is
opened? Use a minimum of
mathematics in your answers.
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26-2 Resistors in Series and in
Parallel
Example 26-7: A two-speed fan.
One way a multiple-speed ventilation fan for a
car can be designed is to put resistors in
series with the fan motor. The resistors
reduce the current through the motor and
make it run more slowly. Suppose the current
in the motor is 5.0 A when it is connected
directly across a 12-V battery. (a) What series
resistor should be used to reduce the current
to 2.0 A for low-speed operation? (b) What
power rating should the resistor have?
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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? (c) What
is the current in the
6.0-Ω resistor?
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26-3 Kirchhoff’s Rules
Some circuits cannot be broken down into
series and parallel connections. For these
circuits we use Kirchhoff’s rules.
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26-3 Kirchhoff’s Rules
Junction rule: The sum of currents entering a
junction equals the sum of the currents
leaving it.
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26-3 Kirchhoff’s Rules
Loop rule: The sum of
the changes in
potential around a
closed loop is zero.
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26-3 Kirchhoff’s Rules
Problem Solving: Kirchhoff’s Rules
1. Label each current, including its direction.
2. Identify unknowns.
3. Apply junction and loop rules; you will
need as many independent equations as
there are unknowns.
4. Solve the equations, being careful with
signs. If the solution for a current is
negative, that current is in the opposite
direction from the one you have chosen.
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26-3 Kirchhoff’s Rules
Example 26-9: Using Kirchhoff’s rules.
Calculate the currents I1, I2, and I3 in the three
branches of the circuit in the figure.
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26-4 Series and Parallel EMFs;
Battery Charging
EMFs in series in the same direction: total
voltage is the sum of the separate voltages.
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26-4 Series and Parallel EMFs;
Battery Charging
EMFs in series, opposite direction: total
voltage is the difference, but the lowervoltage battery is charged.
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26-4 Series and Parallel EMFs;
Battery Charging
EMFs in parallel only make sense if the
voltages are the same; this arrangement can
produce more current than a single emf.
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26-4 Series and Parallel EMFs;
Battery Charging
Example 26-10: Jump
starting a car.
A good car battery is being used to jump
start a car with a weak battery. The good
battery has an emf of 12.5 V and internal
resistance 0.020 Ω. Suppose the weak
battery has an emf of 10.1 V and internal
resistance 0.10 Ω. Each copper jumper
cable is 3.0 m long and 0.50 cm in
diameter, and can be attached as shown.
Assume the starter motor can be
represented as a resistor Rs = 0.15 Ω.
Determine the current through the
starter motor (a) if only the weak battery
is connected to it, and (b) if the good
battery is also connected.
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26-5 Circuits Containing Resistor
and Capacitor (RC Circuits)
When the switch is
closed, the
capacitor will begin
to charge. As it
does, the voltage
across it increases,
and the current
through the resistor
decreases.
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26-5 Circuits Containing Resistor
and Capacitor (RC Circuits)
To find the voltage as a function of time, we
write the equation for the voltage changes
around the loop:
Since Q = dI/dt, we can integrate to find the
charge as a function of time:
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26-5 Circuits Containing Resistor
and Capacitor (RC Circuits)
The voltage across the capacitor is VC = Q/C:
The quantity RC that appears in the exponent
is called the time constant of the circuit:
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26-5 Circuits Containing Resistor
and Capacitor (RC Circuits)
The current at any time t can be found by
differentiating the charge:
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26-5 Circuits Containing Resistor
and Capacitor (RC Circuits)
Example 26-11: RC circuit,
with emf.
The capacitance in the circuit shown
is C = 0.30 μF, the total resistance is
20 kΩ, and the battery emf is 12 V.
Determine (a) the time constant, (b)
the maximum charge the capacitor
could acquire, (c) the time it takes
for the charge to reach 99% of this
value, (d) the current I when the
charge Q is half its maximum value,
(e) the maximum current, and (f) the
charge Q when the current I is 0.20
its maximum value.
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26-5 Circuits Containing Resistor
and Capacitor (RC Circuits)
If an isolated charged
capacitor is
connected across a
resistor, it discharges:
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26-5 Circuits Containing Resistor
and Capacitor (RC Circuits)
Once again, the voltage and current as a
function of time can be found from the
charge:
and
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26-5 Circuits Containing Resistor
and Capacitor (RC Circuits)
Example 26-12: Discharging RC circuit.
In the RC circuit shown, the battery has fully charged
the capacitor, so Q0 = CE. Then at t = 0 the switch is
thrown from position a to b. The battery emf is 20.0 V,
and the capacitance C = 1.02 μF. The current I is
observed to decrease to 0.50 of its initial value in 40
μs. (a) What is the value of Q, the charge on the
capacitor, at t = 0? (b) What is the value of R? (c) What
is Q at t = 60 μs?
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26-5 Circuits Containing Resistor
and Capacitor (RC Circuits)
Conceptual Example 26-13: Bulb in RC circuit.
In the circuit shown, the capacitor is originally
uncharged. Describe the behavior of the lightbulb
from the instant switch S is closed until a long time
later.
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26-5 Circuits Containing Resistor
and Capacitor (RC Circuits)
Example 26-14: Resistor
in a turn signal.
Estimate the order of
magnitude of the
resistor in a turn-signal
circuit.
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26-6 Electric Hazards
Most people can “feel” a current of 1 mA; a
few mA of current begins to be painful.
Currents above 10 mA may cause
uncontrollable muscle contractions, making
rescue difficult. Currents around 100 mA
passing through the torso can cause death by
ventricular fibrillation.
Higher currents may not cause fibrillation, but
can cause severe burns.
Household voltage can be lethal if you are wet
and in good contact with the ground. Be
careful!
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26-6 Electric Hazards
A person receiving a
shock has become part
of a complete circuit.
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26-6 Electric Hazards
Faulty wiring and improper grounding can be
hazardous. Make sure electrical work is done by
a professional.
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26-6 Electric Hazards
The safest plugs are those with three prongs;
they have a separate ground line.
Here is an example of household wiring – colors
can vary, though! Be sure you know which is the
hot wire before you do anything.
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26-7 Ammeters and Voltmeters
An ammeter measures current; a voltmeter
measures voltage. Both are based on
galvanometers, unless they are digital.
The current in a circuit passes through the
ammeter; the ammeter should have low
resistance so as not to affect the current.
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26-7 Ammeters and Voltmeters
Example 26-15: Ammeter design.
Design an ammeter to read 1.0 A at
full scale using a galvanometer with
a full-scale sensitivity of 50 μA and a
resistance r = 30 Ω. Check if the
scale is linear.
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26-7 Ammeters and Voltmeters
A voltmeter should not affect the voltage across
the circuit element it is measuring; therefore its
resistance should be very large.
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26-7 Ammeters and Voltmeters
Example 26-16: Voltmeter design.
Using a galvanometer with internal
resistance 30 Ω and full-scale
current sensitivity of 50 μA, design a
voltmeter that reads from 0 to 15 V. Is
the scale linear?
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26-7 Ammeters and Voltmeters
An ohmmeter measures
resistance; it requires a
battery to provide a
current.
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26-7 Ammeters and Voltmeters
Summary: An
ammeter must be in
series with the
current it is to
measure; a voltmeter
must be in parallel
with the voltage it is
to measure.
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26-7 Ammeters and Voltmeters
Example 26-17: Voltage
reading vs. true voltage.
Suppose you are testing an
electronic circuit which has two
resistors, R1 and R2, each 15 kΩ,
connected in series as shown in
part (a) of the figure. The battery
maintains 8.0 V across them and
has negligible internal resistance.
A voltmeter whose sensitivity is
10,000 Ω/V is put on the 5.0-V scale.
What voltage does the meter read
when connected across R1, part (b)
of the figure, and what error is
caused by the finite resistance of
the meter?
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Review Questions
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ConcepTest 26.1a
Series Resistors I
A) 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?
B) zero
C) 3 V
D) 4 V
E) you need to know the
actual value of R
9V
ConcepTest 26.1a
Series Resistors I
A) 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?
B) zero
C) 3 V
D) 4 V
E) 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
Follow-up: What would be the potential difference if
R = 1 W, 2 W, 3 W ?
ConcepTest 26.1b
Series Resistors II
A) 12 V
In the circuit below, what is the
B) zero
voltage across R1?
C) 6 V
D) 8 V
E) 4 V
R1 = 4 W
R2 = 2 W
12 V
ConcepTest 26.1b
Series Resistors II
A) 12 V
In the circuit below, what is the
B) zero
voltage across R1?
C) 6 V
D) 8 V
E) 4 V
The voltage drop across R1 has
to be twice as big as the drop
across R2. This means that V1 =
R1 = 4 W
R2 = 2 W
8 V and V2 = 4 V. Or else you
could find the current I = V/R =
(12 V)/(6 W) = 2 A, and then use
12 V
Ohm’s law to get voltages.
Follow-up: What happens if the voltage is doubled?
ConcepTest 26.2a
Parallel Resistors I
A) 10 A
In the circuit below, what is the
B) zero
current through R1?
C) 5 A
D) 2 A
E) 7 A
R2 = 2 W
R1 = 5 W
10 V
ConcepTest 26.2a
Parallel Resistors I
A) 10 A
In the circuit below, what is the
B) zero
current through R1?
C) 5 A
D) 2 A
E) 7 A
The voltage is the same (10 V) across each
R2 = 2 W
resistor because they are in parallel. Thus,
we can use Ohm’s law, V1 = I1R1 to find the
R1 = 5 W
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
A) increases
battery of fixed voltage. As more
B) remains the same
resistors R are added to the parallel
C) decreases
circuit, what happens to the total
D) drops to zero
current in the circuit?
ConcepTest 26.2b
Parallel Resistors II
Points P and Q are connected to a
A) increases
battery of fixed voltage. As more
B) remains the same
resistors R are added to the parallel
C) decreases
circuit, what happens to the total
D) 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?
ConcepTest 26.3a
Current flows through a
Short Circuit I
A) all the current continues to flow
through the bulb
connected across the
B) half the current flows through the wire,
the other half continues through the
bulb
bulb, what happens?
C) all the current flows through the wire
lightbulb. If a wire is now
D) none of the above
ConcepTest 26.3a
Current flows through a
Short Circuit I
A) all the current continues to flow
through the bulb
connected across the
B) half the current flows through the wire,
the other half continues through the
bulb
bulb, what happens?
C) all the current flows through the wire
lightbulb. If a wire is now
D) none of the above
The current divides based on the
ratio of the resistances. If one of the
resistances is zero, then ALL of the
current will flow through that path.
Follow-up: Doesn’t the wire have SOME resistance?
ConcepTest 26.3b
Two lightbulbs A and B are
connected in series to a
constant voltage source.
When a wire is connected
across B, bulb A will:
Short Circuit II
A) glow brighter than before
B) glow just the same as before
C) glow dimmer than before
D) go out completely
E) explode
ConcepTest 26.3b
Two lightbulbs A and B are
connected in series to a
constant voltage source.
When a wire is connected
across B, bulb A will:
Short Circuit II
A) glow brighter than before
B) glow just the same as before
C) glow dimmer than before
D) go out completely
E) explode
Since bulb B is bypassed by the wire,
the total resistance of the circuit
decreases. This means that the current
through bulb A increases.
Follow-up: What happens to bulb B?
ConcepTest 26.4a
Circuits I
The lightbulbs in the circuits below
A) circuit I
are identical with the same
B) circuit II
resistance R. Which circuit
produces more light? (brightness
 power)
C) both the same
D) it depends on R
ConcepTest 26.4a
Circuits I
The lightbulbs in the circuits below
A) circuit I
are identical with the same
B) circuit II
resistance R. Which circuit
produces more light? (brightness
 power)
In circuit I, the bulbs are in
parallel, lowering the total
resistance of the circuit. Thus,
circuit I will draw a higher current,
which leads to more light, because
P = IV.
C) both the same
D) it depends on R
ConcepTest 26.4b
The three lightbulbs in the circuit all have
Circuits II
A) twice as much
the same resistance of 1 W . By how
B) the same
much is the brightness of bulb B greater
C) 1/2 as much
or smaller than the brightness of bulb A?
(brightness  power)
D) 1/4 as much
E) 4 times as much
10 V
ConcepTest 26.4b
The three lightbulbs in the circuit all have
Circuits II
A) twice as much
the same resistance of 1 W . By how
B) the same
much is the brightness of bulb B greater
C) 1/2 as much
or smaller than the brightness of bulb A?
(brightness  power)
D) 1/4 as much
E) 4 times as much
We can use P = V2/R to compare the power:
PA = (VA)2/RA = (10 V)2/1 W = 100 W
PB = (VB)2/RB = (5 V)2/1 W = 25 W
Follow-up: What is the total current in the circuit?
10 V
ConcepTest 26.5a
More Circuits I
What happens to the voltage
A) increase
across the resistor R1 when the
B) decrease
switch is closed? The voltage will:
C) stay the same
R1
S
R3
V
R2
ConcepTest 26.5a
More Circuits I
What happens to the voltage
A) increase
across the resistor R1 when the
B) decrease
switch is closed? The voltage will:
C) stay the same
R1
With the switch closed, the addition of
R2 to R3 decreases the equivalent
S
resistance, so the current from the
battery increases. This will cause an
R3
V
increase in the voltage across R1 .
Follow-up: What happens to the current through R3?
R2
ConcepTest 26.5b
More Circuits II
A) increases
What happens to the voltage
across the resistor R4 when the
B) decreases
switch is closed?
C) stays the same
R1
S
R3
V
R2
R4
ConcepTest 26.5b
More Circuits II
A) increases
What happens to the voltage
across the resistor R4 when the
B) decreases
switch is closed?
C) stays the same
We just saw that closing the switch
causes an increase in the voltage
across R1 (which is VAB). The
voltage of the battery is constant,
so if VAB increases, then VBC must
A
R1
B
S
R3
V
R2
decrease!
Follow-up: What happens to the current through R4?
C
R4
ConcepTest 26.6
Even More Circuits
A) R1
Which resistor has the
B) both R1 and R2 equally
greatest current going
through it? Assume that all
C) R3 and R4
the resistors are equal.
D) R5
E) all the same
V
ConcepTest 26.6
Even More Circuits
A) R1
Which resistor has the
B) both R1 and R2 equally
greatest current going
through it? Assume that all
C) R3 and R4
the resistors are equal.
D) R5
E) all the same
The same current must flow
through the left and right
combinations of resistors.
On the LEFT, the current
splits equally, so I1 = I2. On
the RIGHT, more current will
go through R5 than R3 + R4,
since the branch containing
R5 has less resistance.
V
Follow-up: Which one has the
smallest voltage drop?
ConcepTest 26.7
Dimmer
A) the power
When you rotate the knob of a
B) the current
light dimmer, what is being
C) the voltage
changed in the electric circuit?
D) both (A) and (B)
E) both (B) and (C)
ConcepTest 26.7
Dimmer
A) the power
When you rotate the knob of a
B) the current
light dimmer, what is being
C) the voltage
changed in the electric circuit?
D) both (A) and (B)
E) both (B) and (C)
The voltage is provided at 120 V from the
outside. The light dimmer increases the
resistance and therefore decreases the current
that flows through the lightbulb.
Follow-up: Why does the voltage not change?
ConcepTest 26.8a
Lightbulbs
Two lightbulbs operate at 120 V, but
A) the 25 W bulb
one has a power rating of 25 W while
B) the 100 W bulb
the other has a power rating of 100 W.
C) both have the same
Which one has the greater
D) this has nothing to do
with resistance
resistance?
ConcepTest 26.8a
Lightbulbs
Two lightbulbs operate at 120 V, but
A) the 25 W bulb
one has a power rating of 25 W while
B) the 100 W bulb
the other has a power rating of 100 W.
C) both have the same
Which one has the greater
D) this has nothing to do
with resistance
resistance?
Since P = V2 / R , the bulb with the lower
power rating has to have the higher
resistance.
Follow-up: Which one carries the greater current?
ConcepTest 26.8b
Two space heaters in your living
room are operated at 120 V.
Space Heaters I
A) heater 1
Heater 1 has twice the resistance
B) heater 2
of heater 2. Which one will give
C) both equally
off more heat?
ConcepTest 26.8b
Two space heaters in your living
room are operated at 120 V.
Space Heaters I
A) heater 1
Heater 1 has twice the resistance
B) heater 2
of heater 2. Which one will give
C) both equally
off more heat?
Using P = V2 / R, the heater with the smaller resistance
will have the larger power output. Thus, heater 2 will
give off more heat.
Follow-up: Which one carries the greater current?
ConcepTest 26.9
Junction Rule
A) 2 A
What is the current in branch P?
B) 3 A
C) 5 A
D) 6 A
E) 10 A
5A
P
8A
2A
ConcepTest 26.9
Junction Rule
A) 2 A
B) 3 A
What is the current in branch P?
C) 5 A
D) 6 A
E) 10 A
The current entering the junction
in red is 8 A, so the current
leaving must also be 8 A. One
exiting branch has 2 A, so the
other branch (at P) must have 6 A.
S
5A
P
8A
Junction
2A
6A
ConcepTest 26.10
Kirchhoff’s Rules
The lightbulbs in the
A) both bulbs go out
circuit are identical. When
B) intensity of both bulbs increases
the switch is closed, what
C) intensity of both bulbs decreases
happens?
D) A gets brighter and B gets dimmer
E) nothing changes
ConcepTest 26.10
Kirchhoff’s Rules
The lightbulbs in the
A) both bulbs go out
circuit are identical. When
B) intensity of both bulbs increases
the switch is closed, what
C) intensity of both bulbs decreases
happens?
D) A gets brighter and B gets dimmer
E) nothing changes
When the switch is open, the point
between the bulbs is at 12 V. But so is
the point between the batteries. If
there is no potential difference, then
no current will flow once the switch is
closed!! Thus, nothing changes.
Follow-up: What happens if the bottom
battery is replaced by a 24 V battery?
24 V
ConcepTest 26.11
Wheatstone Bridge
An ammeter A is connected
A) l
between points a and b in the
B) l/2
circuit below, in which the four
C) l/3
resistors are identical. The current
D) l/4
through the ammeter is:
E) zero
I
V
ConcepTest 26.11
Wheatstone Bridge
An ammeter A is connected
A) l
between points a and b in the
B) l/2
circuit below, in which the four
C) l/3
resistors are identical. The current
D) l/4
through the ammeter is:
E) zero
Since all resistors are identical,
the voltage drops are the same
across the upper branch and the
lower branch. Thus, the
I
potentials at points a and b are
also the same. Therefore, no
current flows.
V
ConcepTest 26.12
More Kirchhoff’s Rules
A) 2 – I1 – 2I2 = 0
Which of the equations is valid
B) 2 – 2I1 – 2I2 – 4I3 = 0
for the circuit below?
C) 2 – I1 – 4 – 2I2 = 0
D) I3 – 4 – 2I2 + 6 = 0
E) 2 – I1 – 3I3 – 6 = 0
1W
I2
2W
6V
22 VV
4V
I1
1W
I3
3W
ConcepTest 26.12
More Kirchhoff’s Rules
A) 2 – I1 – 2I2 = 0
Which of the equations is valid
B) 2 – 2I1 – 2I2 – 4I3 = 0
for the circuit below?
C) 2 – I1 – 4 – 2I2 = 0
D) I3 – 4 – 2I2 + 6 = 0
E) 2 – I1 – 3I3 – 6 = 0
Eq. 3 is valid for the left loop:
The left battery gives +2 V, then
there is a drop through a 1 W
resistor with current I1 flowing.
Then we go through the middle
battery (but from + to – !), which
gives –4 V. Finally, there is a
drop through a 2 W resistor with
current I2.
1W
I2
2W
6V
22 VV
4V
I1
1W
I3
3W
Ch. 26 Homework
•
•
•
•
Finish reading Ch. 26.
Begin looking over Ch. 27.
Group problems #’s 6, 8, 18
Homework problems #’s 3, 7, 11, 19, 31,
33, 45
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