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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
2!
I
R1
Voltage Drops enter with a - sign.
"
"
a
R2
- +
b
- +
b
a
Move
around
circuit:
b
a
1
Loop Example
b
a
I
c
R2
+

always points from negative to positve
2
R4
f
+
Note:
Internal Resistance of an emf Device
1
R1
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b
I
I
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a
-
I
d
e
R3
2!
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4
1
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Resistors in Series
R1
R2
+
-
1!
I
R3
Another (intuitive) way
*When a potential difference, V, is
applied across resistances in series,
the resistances have identical
current.
I
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.
L
R1 = ρ 1
A
L
R2 = ρ 2
A
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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How to use Kirchhoff’s Law:
Kirchhoff’s Second Rule (Junction Rule or
“Kirchhoff’s Current Law”)
1!
a
R1
+
b
-
-
Analyze the circuit & identify all
circuit nodes:
c
+
R2
The sum of the current entering any junction must be equal
to the sum of the currents leaving that junction.
2!
R3
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Identify all independent loops &
use Kirchhoff’s Voltage Law:
d
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8
2
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Resistors in Parallel
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
1
A
A
1
1
= 1 + 2 =
+
Req ρ L ρ L R1 R2
⇒
( A1 + A2 )
1 1
1
=
+
R R1 R2
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Summary of Resistor & Capacitor Combinations
Resistors
Series R = n R
∑ i
eq
i =1
n
1
1
=∑
Req i =1 Ri
Summary of Simple Circuits
n
1
1
=∑
Ceq i =1 Ci
• Resistors in series:
C1C2
for n = 2
C1 + C2
Current thru is same;
n
Ceq = ∑ Ci
•  Resistors in parallel:
i =1
RR
Req = 1 2 for n = 2
R1 + R2
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Capacitors
Ceq =
Parallel
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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;
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Current thru is V
/Ri
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Problem Solving Tips
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
c b2
c b − c2b1
x= 2
= 1 2
a1 b1 a1b2 − a2b1
a2 b2
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a1
a2
y=
a1
a2
c1
c2
a c − a2 c1
= 1 2
b1 a1b2 − a2b1
b2
15
4