Text sections 28.1 - 28.3
• Emf, internal resistance
• Series and parallel rules for resistors
Practice:
Chapter 28, problems 7, 9, 15, 17
“Direct Current”: current always flows in one
direction.
(or, often: current is constant in time)
1
E ≡ external work per unit charge
Units: J/C = volts (not actually a force)
e.g.: Battery
(chemical energy  electrical energy)
Generator
(mechanical energy  electrical energy)
Solar cell, etc.
-
Example: battery.
Eg.
12 V
I=2A
R
+
What does the energy balance
look like in this circuit?
What is the resistance R?
2
-
Example: battery
+
I=2A
Eg.
R
12 V
Battery: supplies power,
P = IE = 24W
(Chemical energy  electrical energy)
Resistor: Electrical energy  heat,
P = 24W=I2R ⇒ R = 6Ω
I
A
r
C •
I
I
r = “internal resistance”
RL (external “load”)
B
VA – VB = “terminal voltage”  measured
VC – VB = E (but “C” is not accessible for measurement)
And VA – VC = - I r
“Terminal voltage”
3
For this model of a battery:
If the load resistance is decreased, then
A)  the battery terminal voltage increases
B)  the battery terminal voltage decreases
C)  the battery terminal voltage is unchanged
For this model of a battery:
If the load resistance is decreased, then
A)  the battery internal resistance increases
B)  the battery internal resistance decreases
C)  the battery internal resistance is unchanged
4
At terminals
Find:
Automobile battery:
12.8 V (with 20 A current)
9.2 V (with 200 A current)
E and rinternal of battery
What is the maximum current the battery can provide?
What is the maximum current we can extract
from the battery in the previous example?
5
Notes:
i)  As I 0, Vload E
ii)  Imax = E /r (when RL = 0, a short circuit)
Question: How would you select a load (e.g., lightbulb)
to get the maximum power output, for a given battery?
A famous theorem: Maximum power transfer is
achieved when the load resistance matches the
source resistance.
 Maximum load power when RL= r
Proof:
Set
and solve for RL (exercise)
6
VL = E – rIL
Max. power point
VLoad
ILoad
Pmax
PLoad
r
R1
R2
RLoad
R3
•  Same current through all resistors
•  Voltages add:
ΔV = ΔV1 + ΔV2 + ΔV3 + …
 IReff = IR1 + IR2 + IR3 + …
So,
7
I
I1
R1
I2
R2
I3
R3
I
•  same voltage across each resistor
•  currents add:
I = I1 + I2 + I3 + …
so,
R1
11 V
R4
R3
R2
Find:
R5
R6
All resistors = 1 Ω
a)  Battery current and power
b)  Current in R5, R2
Homework: Calculate the power dissipated in each resistor, and
check that it adds up to the power supplied by the battery.
8
Solution:
a) Resistor network:
1Ω
3Ω
1 Ω || 3 Ω 
Total,
R1
V1
V
R2
Show that
V2
and
9
Resistor A and resistor B are connected in parallel
across 120 volts; and resistor A dissipates twice as
much power as resistor B.
If instead they are connected together in series
across 120 volts, then:
A) resistor A will dissipate more power than B
B) resistor B will dissipate more power than A
C) they will dissipate equal powers
A regular “40 watt” bulb and a “60 watt” bulb are
connected in SERIES across 120 V.
What is the total power dissipated by both bulbs?
(Assume resistances don’t change with temperature
—these are special bulbs.)
A) less than 40W
B) between 40W and 60W
C) between 60W and 100W
D) more than 100W
10
A regular “40 watt” bulb and a “60 watt” bulb are
connected in SERIES across 120 V.
What power does each bulb give?
(Assume resistances don’t change with temperature
—these are special bulbs.)
Junction Rule: total current in = total current out
at each junction (from conservation
of charge).
Loop Rule: Sum of emfs and potential differences
around any closed loop is zero (from
conservation of energy).
11
Junction Rule: conservation of charge.
I2
I1
I 1 = I2 + I 3
I3
or
I 1’
I2’ (= -I2)
I 1’ + I 2 ’ + I 3 ’ = 0
I3’ (= -I3)
Loop Rule: conservation of energy.
Follow a test charge q around a loop:
around any loop in
circuit.

R
I
R
I
-
+
-Q
+Q
changes going from left to right
ΔV = -IR
ΔV = +IR
ΔV =
ΔV = Q/C
C
12
18 Ω
9V
b
6Ω
I2
I1
c
3Ω
3V
a
I3 = - (I1 + I2)
d
18 Ω
9V
b
3V
a
c
6Ω
I2
I1
Find the current
through each battery.
3Ω
Find the current
through each battery.
I3 = - (I1 + I2)
d
Answers: I1 = 400 mA, I2 = 200 mA
13
18 Ω
9V
b
I2
I1
c
6Ω
3V
a
Junction b:
I 1 + I2 + I3 = 0
so I3 = -(I1 + I2)
3Ω
I3 = - (I1 + I2)
d
Solution:
Loop abcda:
..... 1
Loop dbcd:
..... 2
 I2 = 200 mA
7× 2 - 1 : 60 I2 = 12A
3 × 1 - 2 : 60 I1 = 24A 
10 Ω
6V
b
20 Ω
I2
I1
a
c
3V
40 Ω
I1 = 400 mA
Find the current
through each battery.
I3 = - (I1 + I2)
d
14
10 Ω
6V
b
20 Ω
I2
I1
40 Ω
3V
a
Find the current
through each battery.
c
I3 = - (I1 + I2)
d
Answers: I1 = 171 mA, I2 = -64.3 mA
10 Ω
6V
b
I2
I1
20 Ω
3V
a
Solution:
Loop abcda:
c
40 Ω
I3 = - (I1 + I2)
Junction b:
I 1 + I2 + I3 = 0
so I3 = -(I1 + I2)
d
Loop dbcd:
4x 1 - 5x 2 : -140 I2 = 9
..... 1
..... 2
 I2 = -64 mA
3x 1 - 2x 2 : 70 I1 = 12  I1 = 171 mA
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c
12 V
200 Ω
200Ω
a
b
600 Ω
300Ω
d
What is Vab (i.e., Va - Vb ) when the switch is open?
Exercise for fun: Find the current through the
switch when it is closed.
c
2Ω
1Ω
Reff = ?
a
1Ω
2Ω
b
1Ω
d
“Series and parallel“ rules don’t help in this case. You
have to go back to the fundamentals—Kirchhoff’s
Circuit Rules.
(Answer: Reff = 1.4Ω )
16
Reff = VTOTAL/ITOTAL
ITOTAL
c
I1
VTOTAL
1Ω
a
I4
I3
2Ω
b
1Ω
2Ω
d
Show that:
I2
I5
1Ω
I1 = I 5 = 3I 3
I 2 = I 4 = 2I 3
ITOTAL= 5I3
VTOTAL= (7Ω) I3
1)  Use Kirchhoff’s rules to write everything in terms
of, one variable, e.g., I3.
2)  Divide VTOTAL/ITOTAL .
Reff = VTOTAL/ITOTAL
ITOTAL
c
I1
VTOTAL
1Ω
a
I4
I3
2Ω
b
1Ω
2Ω
d
Show that:
I2
1Ω
I5
I1 = I 5 = 3I 3
I 2 = I 4 = 2I 3
ITOTAL= 5I3
VTOTAL= (7Ω) I3
1)  Use Kirchhoff’s rules to write everything in terms
of, e.g., I3. (Results: I1 = I5 = 3I3, I2 = I4 = 2I3; ITOTAL= 5I3;
VTOTAL= (7Ω) I3.)
2)  Divide VTOTAL/ITOTAL .
17
Solution to part A:
c
250
12 V
I1
a
I1 + I 3
Solution to part b:
I2
100
I3
b
Loop adba:
400
250
d
I1 - I3
1
Loop abca:
2
Then:
 Solve…
18
R1
V1
V
R2
V2
and
(from
; and
etc)
c
3Ω
1Ω
Reff = ?
a
5Ω
2Ω
b
4Ω
d
Notice that “series and parallel“ don’t help in this
case. You have to go back to Kirchhoff’s Rules.
(Answer: Reff = ( 155/74 ) Ω )
19