BEEE102: Basic Electrical and Electronics
Engineering
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Superposition Theorem
I The superposition principle states that the voltage across (or current
through) an element in a linear circuit is the algebraic sum of the
voltages across (or currents through) that element due to each
independent source acting alone
I It helps us to analyze a linear circuit with more than one independent
source by calculating the contribution of each independent source
I Turn off (voltage source short circuit and current source open circuit) all
independent sources except one source. Find the output (voltage or
current) due to that active source using KCL or KVL
I Repeat step 1 for each of the other independent sources
I Find the total contribution by adding algebraically all the contributions
due to the independent sources
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Superposition Theorem......
For the circuit shown apply superposition theorem and determine Io
8Ω
+
6V + 4Ω v 3A
−
−
Since there are two sources, let
v = v1 + v2 (1)
where v1 and v2 are the contributions due to the 6-V voltage source and the
3-A current source, respectively
To obtain v1 turn-off the current source (open circuit)
8Ω
+
6V + i1 4Ω v1
−
−
Applying KVL gives
8i1 + 4i1 = 6 ⇒ i1 = 0.5 A (2)
v1 = 4i1 = 2 V (3) 3/9
Superposition Theorem......
To get v2 , turn-off voltage source (short circuit)
8Ω i2
i3
+
4Ω v2 3A
−
Applying current divide rule gives
3×8
i3 = =2A (4)
4+8
v2 = 4i3 = 8 V (5)
The voltage v is
v = v1 + v2 = 10 V (6)
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Superposition Theorem......
Find the current I in the circuit shown below using the superposition
principle
2Ω
6Ω I 8Ω
16 V + + 12 V
− 4A −
Ans: 0.75 A
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Superposition Theorem.....
Find the voltage vo in the circuit shown below using the superposition
principle
6Ω
4A
4Ω 2Ω
+
40 V + 2A vo 3Ω
−
−
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Superposition Theorem.....
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Reference
Charles K Alexander, Mathew N O Sadiku, ’Fundamentals of Electric
Circuits’, Tata McGraw Hill.
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THANK YOU
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