Physics 122B
Electricity and Magnetism
Lecture 15 (Knight: 31.5 to 31.8)
Resistors in Circuits
April 30, 2007
Martin Savage
Lecture 15 Announcements
 Lecture HW Assignment #5 has been
posted on the Tycho system and is due at
10 PM on Wednesday.
The second midterm exam is this Friday.
It will cover everything up to the end of
Wednesdays lecture, emphasizing the
most recent material, but assumes
understanding of all material inclusive.
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Physics 122B - Lecture 15
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Units of Power
Power units
Energy units
watts  W  VA  J/s
1 kilowatt-hour  (1.0 103 W)(3600 s)
 3.6 106 J
At residential rates, Seattle City Light charges about 7.5¢ for a
kilowatt hour of electrical energy, so one million joules ( 1 MJ) of
electrical energy costs about 2¢. (Remarkably cheap!)
If you operate a 1500 W hair dryer for 10 minutes, you use 0.25
kilowatt hours or 0.9x106 J of energy, which adds about 1.8¢ to your
electric bill.
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Question
Which resistor dissipates the most power?
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Bulbs in Series
Question: How does the brightness
of bulb A compare with
that of bulbs B and C?
Answer: Bulb A is brighter than
bulb B and bulb C, which are of equal
brightness.
Reason: The potential drop across
bulb A is E, while the potential drop
across B and across C is E/2.
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Resistors in Series
Vab  V1  V2  IR1  IR2  I ( R1  R2 )
Vab I ( R1  R2 )
Rab 

 R1  R2
I
I
Req  R1  R2 
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 RN (series resistors)
Physics 122B - Lecture 15
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Example:
A Series Resistor Circuit
Req  15   4   8   27 
E
(9 V)
I

 0.333 A
Req (27 )
VR1   IR1  (15 )(0.333 A)  5.0 V
VR2   IR2  (4 )(0.333 A)  1.33 V
VR3   IR3  (8 )(0.333 A)  2.67 V
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Ammeters
x
x
I=?
Question: How do you measure
the current in a circuit?
Answer: You must break the
circuit and insert an ammeter
into the line of current flow.
Ideal Ammeter: To have a minimum
effect on the circuit being measured, the inserted ammeter must have
zero resistance, so that there is zero potential difference across the
ammeter. Electronic ammeters can give good approximations to this
condition, but electro-mechanical ammeters may not.
Note: “Clip on” ammeters that measure AC current without breaking the
circuit are commercially available. They use magnetic induction (see L20).
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Real Batteries (1)
An ideal battery provides a potential
difference that is a constant, independent of
current flow or duration of use.
But real batteries “sag” under load and
become “weak” or “dead” as their chemical
energy is used up. How can we include such
effects?
A reasonable approximation is to include an
internal resistance rint. The internal resistance
may increase as the battery ages and supplies
energy. The rule is that the larger and more
expensive the battery, the lower is rint.
A regulated electronic power supply provides
a very good approximation to a zero-resistance
constant-potential ideal battery.
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Real Batteries (2)
Vbat  E  I rint  E
I
E
E

Req R  rint
VR  IR 
Vbat
R
E
R  rint
Vbat  E  I rint  E 
rint
E
R  rint


r
R
=E 1  int  =
E
 R  rint  R  rint
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Question: How can you measure rint?
Answer: One (rather brutal) way is to vary
an external load resistance R until the
potential drop across R is ½E. Then R=rint
because each drops ½E.
Physics 122B - Lecture 15
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Example:
Lighting Up a Flashlight
A 6  flashlight bulb is powered by a 3 V battery having an internal
resistance of 1 . (Assume ideal wires.)
1/ What is the power dissipation of the bulb?
2/ What is the terminal voltage of the battery?
E
(3 V)
I

 0.43 A
R  r (6 )  (1 )
PR  I 2 R  (0.43 A) 2 (6 )  1.1 W
Vbat  E  I r  (3 V)  (0.43 A)(1 )  2.57 V
Note that in this situation, 14% of the available energy goes into
heating the battery rather than providing light from the bulb.
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A Short Circuit
What happens if you “short out” a
battery, i.e., connect an ideal
resistanceless wire across its
terminals, so that the potential
difference across the terminals
becomes zero?
Then, Ishort=E /r. In other words,
all of the battery’s potential is
dropped across its internal
resistance. Ishort is the maximum
possible current that a battery can
supply, and is a measure of the
internal resistance r of the battery
(r = E / Ishort).
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Example:
A Short-Circuited Car Battery
What is the short-circuit current of a 12 V car battery
with an internal resistance of 0.020 ?
What happens to the power supplied by the battery
when it is shorted?
I short
E
(12 V)
 
 600 A
r (0.020 )
P  I 2 r  (600 A)2 (0.020 )  7, 200 W
0.02 
12 V
All of this power would be dissipated internally in
the battery, making it likely to explode. Therefore,
do not short out your car’s battery!
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Bulbs in Parallel
Initially, bulbs A and B have the
same brightness and C is out.
What happens to the
brightness of the bulbs when
the switch is closed?
1. Bulb A gets brighter (because
the overall current increases);
2. Bulb B gets dimmer (because
current is diverted to Bulb C);
3. Bulb C glows with the same
brightness as B (because they
split the current equally).
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Resistors in Parallel
I  I1  I 2 
V1 V2

R1
R2
1 1
Vcd Vcd


 Vcd   
R1
R2
 R1 R2 
Rcd 
Vcd

I
Vcd
1
1
Vcd   
 R1 R2 
1
1
  
 R1 R2 
1
1
1 1
  
Req R1 R2
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
1
RN
(parallel resistors)
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Series vs. Parallel
V
R1
R2
Resistors in series divide the overall potential difference
between them in proportion to their resistances.
R1
I
R2
I
Resistors in parallel divide the overall current through
them in proportion to their inverse resistances. The
current splits and tends to favor “the path of least
resistance”.
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Example:
A Parallel Resistor Circuit
Three resistors are
connected across a 9 V
battery.
Find the current through
the battery.
Find the potential
differences across and
currents through each
resistor.
I
I1
I2
I3
Vbat
Vbat =V1 =V2 =V3  9.0 V
1
1
1 
 1
Req  


  2.26 
15

4

8



I1 
V1 (9.0 V)

 0.60 A
R1
(15 )
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I2 
I
E
(9 V)

 3.98 A
Req (2.26 )
V2 (9.0 V)

 2.25 A
R2
(4 )
Physics 122B - Lecture 15
I3 
V3 (9.0 V)

 1.13 A
R3
(8 )
17
Example:
A Combination of Resistors
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Voltmeters
Question: How do you measure the
potential difference between two points
in a circuit?
Answer: You can connect one lead of a
voltmeter to each point.
Ideal Voltmeter: To have a minimum
effect of the circuit being measured, the
connected voltmeter must have infinite
resistance, so that no current is diverted
through the voltmeter. Electronic
voltmeters can give good approximations
to this condition, but electro-mechanical
voltmeters may not.
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Voltmeters vs. Ammeters
An ideal voltmeter has infinite
internal resistance. It must be
connected between circuit elements
to measure the potential difference
between two points in the circuit.
An ideal ammeter has zero
internal resistance. It must be
inserted by breaking a circuit
connection to measure the current
flowing through that connection in
the circuit.
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Physics 122B - Lecture 15
V
A
I
X
I
20
Question
What is the correct order of bulb brightness (from brightest to dimmest)?
(The bulbs are identical)
(a) A=B=C=D; (b) A>B=C=D; (c) A=B>C=D; (d) A>B>C=D; (e) A>B>C>D
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Example:
Analyzing a Complex Circuit (1)
Four resistors are
connected to a 12 V battery
as shown.
Find the current through
the battery.
Find the potential
differences across and
currents through each
resistor.
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Example:
Analyzing a Complex Circuit (2)
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End of Lecture 15
 Before the next lecture, read Knight,
sections 31.9 through 31.10.
 Lecture HW Assignment #5 has been
posted on the Tycho system and is due at 10
PM Wednesday.
 Midterm 2 is Friday
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