Superposition Problem with Four Voltage and Current Sources


Determine  V_x and  I_x using the superposition method.
A circuit with four voltage and current sources to be solved by the superposition method
Solution
I. Contribution of the  -5V voltage source:

To find the contribution of the  -5V voltage source, other three sources should be turned off. The  3V voltage source should be replaced by short circuit. The current source should be replaced with open circuits, as shown below.

Superposition - Finding the contribution of the -5V voltage source

It is trivial that  I_{x1}= \frac{-5 V}{2 \Omega}=-2.5 A. The current of the  3\Omega resistor is zero. Using KVL, -(-5V)+V_{3\Omega}-V_{x1}=0 \to V_{x1}=-(-5V)=5V.

II. Contribution of the  3V voltage source:
Similarly, the  -5V voltage source becomes a short circuit and the current source should be replaced with open circuits:
Superposition - Contribution of the 3V voltage source

The current of the  2\Omega resistor is zero because of being short circuited. It is trivial that  I_{x2}= 0 A (current of an open circuit). The current of the  3\Omega resistor is also zero. Using KVL, -(3V)+V_{2\Omega}+V_{x2}+V_{3\Omega}=0 \to V_{x2}=3V.

III. Contribution of the  -1A current source:
The voltage sources should be replaced by short circuits and the  2A current source becomes with open circuit:

Superposition - Contribution of the -1A current source
Again, the  2\Omega resistor is short circuited and its current is zero. it is clear that  I_{x3}= 1 A . The current of the  3\Omega resistor is equal to  -1A . Using KVL, V_{x3}+V_{3\Omega}=0 \to V_{x3}+(-1A)\times (3\Omega)=0 \to V_{x3}=3V.

IV. Contribution of the  2A current source:
Likewise, the voltage sources should be replaced by short circuits and the  -1A current source becomes with open circuit:
Superposition - Contribution of the 2 A current source

Again, the  2\Omega resistor is short circuited and its current is zero. it is also trivial that  I_{x4}= 0A . The current of the  3\Omega resistor is  2A . Using KVL, V_{x4}+V_{3\Omega}=0 \to V_{x4}+(2A)\times (3\Omega)=0 \to V_{x4}=-6V.

V. Adding up the individual contributions algebraically:

V_x=V_{x1}+V_{x2}+V_{x3}+V_{x4}=5V+3V+3V-6V\to V_x=5V
I_x=I_{x1}+I_{x2}+I_{x3}+I_{x4}=-2.5A+1A+0A-0A\to I_x=-1.5A

17 thoughts on “Superposition Problem with Four Voltage and Current Sources

  1. Sir this all numericals are very helpful but I have some small questions arrise in my mind which I want to know xo sir plz tell me how can I contact u plz

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