Circuit Theorems
Circuit Theorems Overview
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Dr. Mustafa Kemal Uyguroğlu
Linearity Property
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Linearity
Superpositions
Source Transformation
Thévenin and Norton Equivalents
Maximum Power Transfer
Linearity Property
A linear element or circuit satisfies the properties of
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Additivity: requires that the response to a sum of
inputs is the sum of the responses to each input
applied separately.
If v1 = i1R and v2 = i2R
then applying (i1 + i2)
v = (i1 + i2) R = i1R + i2R = v1 + v2
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Homogeneity:
If you multiply the input (i.e. current) by some
constant K, then the output response (voltage) is
scaled by the same constant.
If v1 = i1R
then K v1 =K i1R
Linearity Property
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Linearity Property - Example
A linear circuit is one whose output is linearly
related (or directly proportional) to its input.
i0
vs
+
Linear
Circuit
i
R
Solve for v0 and i0 as a function of Vs
Suppose vs = 10 V gives i = 2 A. According to
the linearity principle, vs = 5 V will give i = 1 A.
Linearity Property – Example
(continued)
Linearity Property - Example
Ladder Circuit
3A
5A
+6V-
2A
+3V-
2A
1A
+
+
+
14 V
8V
5V
-
-
-
This shows that assuming I0 = 1 A gives Is = 5 A; the actual source
current of 15 A will give I0 = 3 A as the actual value.
Superposition
Superposition
“In any linear circuit containing multiple
independent sources, the current or voltage
at any point in the circuit may be calculated
as the algebraic sum of the individual
contributions of each source acting alone.”
1kΩ
1kΩ
+
V1
+
–
Vout
1kΩ
+
–
V2
–
Superposition
1kΩ
How to Apply Superposition
1kΩ
1kΩ
+
V1
+
–
V’out
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1kΩ
+
1kΩ
+
–
V’out = V1/3
V’’out = V2/3
Vout = V’out + V’’out = V1/3 + V2/3
V’’out
1kΩ
+
–
V2
–
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To find the contribution due to an individual
independent source, zero out the other
independent sources in the circuit.
– Voltage source ⇒ short circuit.
– Current source ⇒ open circuit.
Solve the resulting circuit using your favorite
technique(s).
Superposition - Problem
2kΩ
2mA Source Contribution
12V
4mA
2kΩ
– +
2mA
1kΩ
2kΩ
2mA
1kΩ
I’0
I0
I’0 = -4/3 mA
4mA Source Contribution
2kΩ
2kΩ
12V Source Contribution
12V
2kΩ
4mA
– +
1kΩ
1kΩ
I’’0
I’’0 = 0
2kΩ
I’’’0
I’’’0 = -4 mA
2kΩ
Final Result
I’0 = -4/3 mA
I’’0 = 0
I’’’0 = -4 mA
I0 = I’0+ I’’0+ I’’’0 = -16/3 mA
Example
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find v using superposition
Superposition Procedure
1. For each independent voltage and current source (repeat the
following):
a) Replace the other independent voltage sources with a short
circuit (i.e., V = 0).
b) Replace the other independent current sources with an open
circuit (i.e., I = 0).
Note: Dependent sources are not changed!
c) Calculate the contribution of this particular voltage or current
source to the desired output parameter.
2. Algebraically sum the individual contributions (current and/or
voltage) from each independent source.
one independent source at a time,
dependent source remains
KCL: i = i1 + i2
Ohm's law: i = v1 / 1 = v1
KVL: 5 = i (1 + 1) + i2(2)
KVL: 5 = i(1 + 1) + i1(2) + 2v1
10 = i(4) + (i1+i2)(2) + 2v1
10 = v1(4) + v1(2) + 2v1
v1 = 10/8 V
Consider the other independent source
KCL: i = i1 + i2
KVL: i(1 + 1) + i2(2) + 5 = 0
i2(2) + 5 = i1(2) + 2v2
Ohm's law: i(1) = v2
v2(2) + i2(2) +5 = 0 => i2 = -(5+2v2)/2
i2(2) + 5 = i1(2) + 2v2
-2v2 = (i - i2)(2) + 2v2
-2v2 = [v2 + (5+2v2)/2](2) + 2v2
-4v2 = 2v2 + 5 +2v2
-8v2 = 5 => v2 = - 5/8 V
from superposition: v = -5/8 + 10/8
v = 5/8 V