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Chapter 26
DC Circuits
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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.
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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 W
R2 = 2 W
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 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 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
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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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 W
R1 = 5 W
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 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
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 26-2: Series or parallel?
(a) The light bulbs 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.
Note: brightness is
proportional to power
Copyright © 2009 Pearson Education, Inc.
26-2 Resistors in Series and in
Parallel
Conceptual Example 26-2: Series or parallel?
(a) The light bulbs 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.
Series:
V
V2
2
I 
 P1  I 1R 
2R
4R
Parallel:
V
I1 
R
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V2
 P1  I R 
R
2
1
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
Conceptual Example 26-3: An illuminating surprise.
120V 
120V 
V2


P
 R100 
 144W & R60 
 240W
R
100W
60W
V
120V
Series: I 

 0.3125 A
R60  R100 144W  240W
2
2
 P60  I 2R60   0.3125  240  23.4W
2
& P100  I 2R100   0.3125  144  14.1W
2
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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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
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.
I3  I1  I2
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26-3 Kirchhoff’s Rules
Loop rule: The sum of
the changes in
potential around a
closed loop is zero.
0  12V  I  400W   I  290W 
12V
I 
 0.017 A
 400  290  W
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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. Each new junction or
loop must include a new element.
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.
Copyright © 2009 Pearson Education, Inc.
ConcepTest 26.10
Kirchhoff’s Rules
The lightbulbs in the
1) both bulbs go out
circuit are identical. When
2) intensity of both bulbs increases
the switch is closed, what
3) intensity of both bulbs decreases
happens?
4) A gets brighter and B gets dimmer
5) nothing changes
ConcepTest 26.10
Kirchhoff’s Rules
The lightbulbs in the
1) both bulbs go out
circuit are identical. When
2) intensity of both bulbs increases
the switch is closed, what
3) intensity of both bulbs decreases
happens?
4) A gets brighter and B gets dimmer
5) 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
26-3 Kirchhoff’s Rules
Example: Using Kirchhoff’s rules.
Calculate the currents I1, I2, and I3 in the three
branches of the circuit in the figure.
I1 
I2 
3.9 W  9.8 W  13.7 W
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 I3
Solution:
Junction A: I2  I1  I3
Left loop:
0  12  13.7I1  1.2I2
12
1.2
 I1 

I2  0.876  0.088I2
13.7 13.7
Right loop: 0  9  6.7I3  1.2I2
9
1.2
 I3 

I2  1.343  0.179I2
6.7 6.7
 I2   0.876  0.088I2    1.343  0.179I2 
2.219
 1.751 A
1.267
 I1  0.876  0.088  1.751  0.722 A
 2.219  0.267I2
 I2 
 I3  1.343  0.179  1.751  1.029 A
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ConcepTest 26.12
More Kirchhoff’s Rules
1) 2 – I1 – 2I2 = 0
Which of the equations is valid
2) 2 – 2I1 – 2I2 – 4I3 = 0
for the circuit below?
3) 2 – I1 – 4 – 2I2 = 0
4) I3 – 4 – 2I2 + 6 = 0
5) 2 – I1 – 3I3 – 6 = 0
1W
I2
2W
6V
22 VV
4V
I1
1W
I3
3W
ConcepTest 26.12
More Kirchhoff’s Rules
1) 2 – I1 – 2I2 = 0
Which of the equations is valid
2) 2 – 2I1 – 2I2 – 4I3 = 0
for the circuit below?
3) 2 – I1 – 4 – 2I2 = 0
4) I3 – 4 – 2I2 + 6 = 0
5) 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
26-4 Series and Parallel EMFs
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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26-5 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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I
26-5 RC Circuits
To find the voltage as a function of time, we
write the equation for the voltage changes
around the loop:
Q t 
0 
 I t  R
C
Since Q = dI/dt, we can integrate to find the
charge as a function of time:
Q  t   C  1  e  t /RC 
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26-5 RC Circuits
The voltage across the capacitor is VC = Q/C:
VC  t     1  e
 t / RC

The quantity RC that appears in the exponent
is called the time constant of the circuit:
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26-5 RC Circuits
The current at any time t can be found by
differentiating the charge:
dQ  t    t /RC
I t  
 e
dt
R
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26-5 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 RC Circuits
If an isolated charged
capacitor is
connected across a
resistor, it discharges:
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26-5 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 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 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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26-6 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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Summary of Chapter 26
• A source of emf transforms energy from
some other form to electrical energy.
• A battery is a source of emf in parallel with an
internal resistance.
• Resistors in series:
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Summary of Chapter 26
• Resistors in parallel:
• Kirchhoff’s rules:
1. Sum of currents entering a junction
equals sum of currents leaving it.
2. Total potential difference around closed
loop is zero.
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Summary of Chapter 26
• RC circuit has a characteristic time constant:
  RC
V0  t /
Charging: I  t   e
& VC  t   V0  1  e  t / 
R
V0  t /
Discharging: I  t   e & VC  t   V0e  t /
R
• To avoid shocks, don’t allow your body to
become part of a complete circuit.
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