Superposition Theorem With 2 Voltage Sources | Superposition Theorem Example
Analysis Of Superposition Theorem Example Which Circuit With 2 Voltage Sources And Superposition Theorem Steps And How We Apply Theorem In The Electrical Circuit
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" The Superposition Theorem states that the response in any element of a linear, bilateral network containing more than one independent sources can be obtained as the algebraic sum of the responses obtained by each independent source acting separately at a time and with all other independent source set equal to zero."
While evaluating the response from one independent source other source can not be removed bodily from the circuit.
Instead , you suppress all other Independent Source by setting their values to zero.
The suppressed source then behave like the following replacement.
- Ideal independent voltage source - short circuit
- Ideal independent current source - open circuit
Special care must be taken if the circuit also include the Controlled Sources.
In general a controlled source affect the individual contribution of each independent source, Consequently Controlled Source are not suppressed during analysis by super position.
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Example of Superposition Theorem With 2 Voltage Sources calculation:
Find The Current In 20 ohm Resistance In The Circuit Shown In Below Figer (Superposition Theorem Example Circuit - 1 ) By Using Superposition Theorem .
Answer :
Step : 1
Find Current I'¹ Due To 90 V Source Only :
From Figer [ Superposition Theorem Example Circuit - 1(B) ] We Get
R(t) = 20 + 4.8 = 24.8 ohm ;
I'¹ = 90/24.8 = 3.63 Amp ;
Step : 2
Find Current I''¹ Due To 60V Source Only :
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Example of Superposition Theorem With 2 Voltage Sources calculation:
Find The Current In 20 ohm Resistance In The Circuit Shown In Below Figer (Superposition Theorem Example Circuit - 1 ) By Using Superposition Theorem .
Superposition Theorem Example Circuit - 1 |
Answer :
Step : 1
Find Current I'¹ Due To 90 V Source Only :
Superposition Theorem Example Circuit - 1(A) |
Superposition Theorem Example Circuit - 1(B) |
From Figer [ Superposition Theorem Example Circuit - 1(B) ] We Get
R(t) = 20 + 4.8 = 24.8 ohm ;
I'¹ = 90/24.8 = 3.63 Amp ;
Step : 2
Find Current I''¹ Due To 60V Source Only :
Superposition Theorem Example Circuit - 1(C) |
Superposition Theorem Example Circuit - 1(D) |
From Figer [ Superposition Theorem Example Circuit - 1(D) ] We Get
R(t) = 12 + 5.71 = 17.71 ohm;
I(2) = 60 / 17.71 = 3.39 Amp;
I''¹ = ( 8/28 ) * ( 3.39 )
I''1 = 0.97 Amp ;
Step : 3
Find Current I'''1 Due To 2A Source Only :
Superposition Theorem Example Circuit - 1(E) |
Superposition Theorem Example Circuit - 1(F) |
From Figer [ Superposition Theorem Example Circuit - 1(F) ] We Get
I'''¹ = ( 4.8 * 2 ) / ( 20 + 4.8 ) = 0.39 Amp;
Step : 4
Calculate I(1) Current Through 20 ohm Resistance :
I(1) = I'¹ + I''¹ + I'''¹
= 3.63 - 0.97 - 0.39
= 2.27 Amp
So In Given Example The Current Flowing Through 20 ohm Resistance Is 2.27 Ampere.
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