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Last Class
New Circuit
Resistors in series: R1
R3
Current through is same. Reffective = R1 + R2 + R3 + ...
Voltage drop across is IRi V1
Resistors in parallel: 1
Voltage drop across is same. Reffective
Current through is V/Ri =
R2
How Can We Solve This One?
1
1
1
+
+
+ ...
R1 R2 R3
R1
R3
Solved Circuits V1
R2
R1
V
V2
R3
R4
V
=
I1234
V2
R1234
V
=
R2
I1234
R12
THE ANSWER: Kirchhoff’s Rules Kirchhoff’s First Rule (“Loop Rule” or
“Kirchhoff’s Voltage Law”)
1
Rules
2!
I
R1
R2
The algebraic sum of the changes in potential encountered
in a complete traversal of any loop of circuit must be zero.
Voltage Gains enter with a + sign.
1
R1
"
"
2!
I
Voltage Drops enter with a - sign.
a
R2
- +
b
- +
b
a
Move
around
circuit:
b
a
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b
I
I
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a
Note:

always points from negative to positve
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1
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Loop Example
R1
b
a
I
c
R2
Internal Resistance of an emf Device
1
+
R4
f
+
-
I
d
e
R3
2!
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Resistors in Series
R3
R2
+
-
1!
R1
I
6
Another (intuitive) way
*When a potential difference, V, is
applied across resistances in series,
the resistances have identical
current.
I
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Consider two cylindrical resistors with
lengths L1 and L2
* Resistances in series can be replaced
with an equivalent resistance, REQ,
that has the same current I and the
same total potential difference V as
the actual resistances.
L1
A
L
R2 = ρ 2
A
R1 = ρ
I
R1
L1
V
L2
R2
Put them together, end to end to make a longer one...
Req = ρ
L1 + L2
= R1 + R2
A
R = R1 + R2
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8
2
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How to use Kirchhoff’s Law:
Kirchhoff’s Second Rule (Junction Rule or
“Kirchhoff’s Current Law”)
1!
R1
+
b
-
-
Analyze the circuit & identify all
circuit nodes:
c
+
R2
a
R3
The sum of the current entering any junction must be equal
to the sum of the currents leaving that junction.
2!
Identify all independent loops &
use Kirchhoff’s Voltage Law:
d
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Resistors in Parallel
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Another (intuitive) way
R2
R3
R1
L1
A
L
R2 = ρ 2
A
R1 = ρ
-
"
+
Consider two cylindrical resistors with crosssectional areas A1 and A2
*In parallel, the resistances all have the same potential differences.
* Resistances in parallel can be replaced with an equivalent resistance, REQ,
that has the same potential difference, V, and the same total current, I,
as the actual resistances.
V
A1
A2
R1
R2
Put them together, side by side … to make one “fatter”one,
Req =
ρL
( A1 + A2 )
⇒
1
A
A
1
1
= 1 + 2 =
+
Req ρ L ρ L R1 R2
1 1
1
=
+
R R1 R2
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Summary of Resistor & Capacitor Combinations
Resistors
Series R = n R
∑ i
eq
i =1
Capacitors
n
1
1
=∑
Ceq i =1 Ci
Ceq =
Parallel
n
1
1
=∑
Req i =1 Ri
Req =
Summary of Simple Circuits
• Resistors in series:
C1C2
for n = 2
C1 + C2
Current through is same;
n
Ceq = ∑ Ci
•  Resistors in parallel:
i =1
R1 R2
for n = 2
R1 + R2
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Req = R1 + R2 + R3 + ...
Voltage drop across is IRi
1
1
1
1
=
+
+
+ ...
Req R1 R2 R3
Voltage drop across is same;
13
Problem Solving Tips
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Current thru is V
/Ri
14
Example 25-16
•  When you are given a circuit, you must first carefully
analyze circuit topology.
–  find the nodes and distinct branches and pick
Linearly Independent subsets of each.
–  assign branch currents
•  Use Kirchhoff’s First Rule for all independent loops in
the circuit.
•  Use Kirchhoff’s Second Rule for all independent
nodes in circuit.
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Cramer’s Rule
If a1 x + b1 y = c1
a2 x + b2 y = c2
then c b
1
1
x=
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c2 b2
c b − c2b1
= 1 2
a1 b1 a1b2 − a2b1
a2 b2
y=
a1
c1
a2
a1
c2
a c − a2 c1
= 1 2
b1 a1b2 − a2b1
b2
a2
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