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This Class
Kirchhoff’s First Rule (“Loop Rule” or
“Kirchhoff’s Voltage Law”)
Resistors in series: Current through is same. The algebraic sum of the changes in potential encountered
in a complete traversal of any loop of circuit must be zero.
Reffective = R1 + R2 + R3 + ...
Voltage drop across is IRi Resistors in parallel: 1
Voltage drop across is same. Reffective
Current through is V/Ri 1
1
1
1
+
+
+ ...
R1 R2 R3
=
I
V
R1
R2
Move
around
circuit:
Solve Circuits R2
R1
2!
R3
R4
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1
Rules
2!
I
R1
2
Loop Example
R2
b
a
I
Voltage Gains enter with a + sign.
Voltage Drops enter with a - sign.
a
- +
b
b
a
- +
b
a
R2
a
+
R4
f
+
-
I
d
e
R3
2!
b
I
I
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c
"
"
1
R1
Note:

always points from negative to positve
3
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1
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Internal Resistance of an emf Device
Resistors in Series
*When a potential difference, V, is applied
across resistances in series, the
resistances have identical current.
R1
R3
R2
+
-
1!
I
I
* 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.
I
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Another (intuitive) way
Consider two cylindrical resistors with
lengths L1 and L2
L1
A
L
R2 = ρ 2
A
R1 = ρ
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Kirchhoff’s Second Rule (Junction Rule or
“Kirchhoff’s Current Law”)
The sum of the current entering any junction must be equal
to the sum of the currents leaving that junction.
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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2
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How to use Kirchhoff’s Law:
-
c
+
R2
R3
R1
-
"
R2
b
-
+
+
Analyze the circuit & identify all
circuit nodes:
R3
a
Resistors in Parallel
2!
R1
1!
Identify all independent loops &
use Kirchhoff’s Voltage Law:
*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.
d
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Another (intuitive) way
L
R1 = ρ 1
A
L
R2 = ρ 2
A
Resistors
V
A1
Series R = n R
∑ i
eq
A2
R1
i =1
R2
ρL
Parallel
⇒
1
A
A
1
1
= 1 + 2 =
+
Req ρ L ρ L R1 R2
n
1
1
=∑
Ceq i =1 Ci
n
1
1
=∑
Req i =1 Ri
C1C2
for n = 2
C1 + C2
n
Ceq = ∑ Ci
i =1
RR
Req = 1 2 for n = 2
R1 + R2
1 1
1
=
+
R R1 R2
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Capacitors
Ceq =
Put them together, side by side … to make one “fatter”one,
( A1 + A2 )
10
Summary of Resistor & Capacitor Combinations
Consider two cylindrical resistors with crosssectional areas A1 and A2
Req =
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Summary of Simple Circuits
• Resistors in series:
Current through is same;
•  Resistors in parallel:
Problem Solving Tips
Req = R1 + R2 + R3 + ...
•  When you are given a circuit, you must first carefully
analyze circuit topology.
Voltage drop across is IRi
–  find the nodes and distinct branches and pick
Linearly Independent subsets of each.
1
1
1
1
=
+
+
+ ...
Req R1 R2 R3
Voltage drop across is same;
–  assign branch currents
•  Use Kirchhoff’s First Rule for all independent loops in
the circuit.
Current thru is V /Ri
•  Use Kirchhoff’s Second Rule for all independent
nodes in circuit.
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