Physics 122C
Electricity and Magnetism
Lecture 14
DC Circuits and Power
April 27, 2007
Martin Savage
Lecture 14 Announcements
 Lecture HW Assignment #5 has been posted
on the Tycho system and is due Wednesday at
10 PM.
 Remember that Midterm 2 is next friday
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Physics 122B - Lecture 14
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Circuit Elements & Diagrams
These are some of the symbols we will use to represent objects in circuit
diagrams.
Variable elements are indicated by adding an arrow.
Other symbols: inductance, transformer, diode, transistor, ammeter, etc.
A
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A Circuit Diagram
Actual Circuit
Circuit Diagram
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Kirchhoff’s Laws
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Basic Circuit
Vloop   (V )i  Vbat  VR  0
i
Vbat  E ; VR  Vdownstream  Vupstream   IR
E  IR  0; I 
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E
R
6
Plumber’s Analogy*
V1
High pressure
I
P1
I
V2
I
The “plumber’s analogy” of this circuit is a
pump (=battery) pumping water in a closed loop
of pipe that includes a constriction (=resistor).
The pressure (=V1) is the upper part of the
loop is higher than in the lower part (=V2).
There is a pressure drop (=V1-V2) across the
constriction and a pressure rise across the
pump. The water flow I is the same at all
points around the loop.
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Physics 122B - Lecture 14
Pump
Constriction
I
P2
Low pressure
Pump = Battery
Constriction = Resistor
Pressure = Potential
Water Flow =Current
7
Example:
A Single-Resistor Circuit
I
E (1.5 V)

 0.10 A
R (15 )
Vbat  E  1.5 V; VR  E  1.5 V
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Example:
A More Complex Circuit (1)
Analyze the circuit shown, to:
(a) Find the current in and potential
difference across each resistor.
(b) Graph the potentials around the
circuit
Vclosed loop   (V )i  Vb1  VR1  Vb 2  VR 2  0
i
Vb1  E1 ; VR1   IR1 ; Vb 2  E2 ; VR2   IR2
 (V )
i
I
i
 E1  IR1  E2  IR2  E1  E2  I ( R1  R2 )  0
E1  E2
 0.50 A
R1  R2
VR1   IR1  (0.50 A)(4 )  2.0 V
VR2   IR2  (0.50 A)(2 )  1.0 V
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Example:
A More Complex Circuit (2)
I
E1  E2
 0.50 A
R1  R2
VR1   IR1  (0.50 A)(4 )  2.0 V
VR2   IR2  (0.50 A)(2 )  1.0 V
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Question
What is the V across the indicated circuit element in
the direction of the current, I?
(a) -4 V;
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(b) -2 V; (c) 0 V; (d) +2 V; (e) +4 V;
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Applying Kirchhoff’s
Loop Law to Many Loops*
1. Define a minimum set of current
loops. Label all elements.
R1=
2. Write a loop equation for each
loop. (Battery or 0 = SV).
3. Solve equations for currents
4. Calculate other variables of
interest.
i1 – i3
R2=
i1
Vbat=
R4=
R3=
Loop equations:
R5=
i2
Loop 1: Vbat  R1i1  R2 (i1  i3 )  R3 (i1  i2 )
Loop 2: 0  R3 (i2  i1 )  R5 (i2  i3 )
Loop 3: 0  R2 (i3  i1 )  R4i3  R5 (i3  i2 )
i3
i1 – i2
Loop L:
 (i
L
 ii ,other ) Ri  0
i
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Solving for the loop currents
18 -8
-6
-8 32 -24
-6 -24 54
I1
I2
I3
V
= 0
0
18 -8
-1 4
-1 -4
I1
I2
I3
V
= 0
0
-6
-3
9
Gaussian Elimination
0
-1
0
0
0
2
36
3
-3
I1
I2
I3
V
= 0
0
I1 = 2 A , I2 = 1 A , I3 = 2/3 A
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Applying Kirchhoff’s Junction
Law to Many Junctions*
1. Define a minimum set of junction
potentials. You can select one
ground point. Label all elements.
2. Write a junction equation for
each unknown junction .
J1 R5=
R1=
Vbat
V1
R3=
R2=
3. Solve these equations for the
unknown junction potentials.
4. Calculate the other variables of
interest.
R4=
Junction equations:
V  V 0  V1 V2  V1 V2  V1
J1: 0  bat 1 


R1
R2
R3
R5
J2: 0 
0  V2 V1  V2 V1  V2


R4
R3
R5
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J2 V2
V=0
Junction J:

i
Physics 122B - Lecture 14
Vi ,away  VJ
Ri
0
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Energy and Power (1)
P  IE = I 2 R  E 2 / R
Example: A 90  load
resistance is connected
across a 120 V battery.
How much power is delivered
by the battery?
P
dU
 power  rate of energy transfer
dt
U  qVbat  qE
Pbat 
E (120 V)

 1.33 A
R (90 )
P  IE  (1.33 A)(120 V)  160 W
dU bat d
dq
 qVbat 
Vbat  I Vbat
dt
dt
dt
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I
P  E 2 / R  (120 V)2 /(90 )  160 W
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Energy and Power (2)
Echem  U  K  Eth
W  F s  (qE )d  K  Eper collision
Eth  qEL  qVR
dEth dq
PR 

VR  I VR
dt
dt
VR  E so PR  Pbat
The energy delivered by the battery is dissipated by the resistor.
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Example: The Power of Light
How much current is “drawn” by a 100 W
light bulb connected to a constant 120 V
outlet?
I
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P
(100 W)

 0.833 A
VR (120 V)
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Example: The Power of Sound
Most loudspeakers are designed to have a
resistance of 8 . If such a loudspeaker is
connected to a stereo amplifier with a rating of
100 W, what is the maximum possible current in
the loudspeaker?
8 
Pmax  I max Vmax  ( I max )2 R
I max 
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Pmax
(100 W)

 3.5 A
R
(8 )
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Example: A Dim Bulb
How much power is used by a 60 W light bulb
rated at 120 V if it is operated using a dimmer
switch at 100 V? (Assume the resistance is
constant.)
P
P
(V )2 (120 V)2
V  
so R 

 240 
I V / R
P
(60 W)
P  (V )2 / R  (100 V) 2 /(240 )  42 W
(Actually, the resistance at this lower operating potential will be less
than the R-value calculated above because the temperature will be
lower, so the actual power used will be somewhat higher.)
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End of Lecture 14
 Before the next lecture, read Knight, sections
31.5 through 31.8.
 Lecture HW Assignment #5 has been posted on
the Tycho system and is due next Wednesday at 10
PM.
Remember that next Friday we will have Exam 2.
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Physics 122B - Lecture 14
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