Determine $V_x$ and $I_x$ using 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. 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: 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: 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: 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:

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

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## Join the Conversation

1. 2. 3. 4. 5. 6. 1. Sophie Kovalevsky says:

Dr, where can I contact u?
Do you have a email where i could contact u? I have some problem to public in the web.

1. Hi Sophie,
My email is yaz at solved-problems dot com. I promoted you account and you should be able to submit problems from here

Thanks,
Yaz

2. Sophie Kovalevsky says:

I'm so really excited teacher.
Thank you so much to active my account to submit problem.

3. abrar says:

i want super position problems with 3 sources

4. sham says:

i will appreciate u please include the more problems

5. sham says:

i will appreciate u please include the more problems 4 Responses to Superposition Problem with Four Voltage and Current Sources

1. sham says:

k thank u

1. The voltage across the resistor is 2V. Therefore, $latex I_{10\Omega}=\frac{2V}{10 \Omega}=0.2 A$.

6. rehan says:

sir can u tell me some more example of nodal and superposition analyses

1. For superposition solved problems, see these.
For nodal analysis prolems, check out the ebook and these.

7. alveena says:

sir can you plz guide about some source transformation problems as well....

8. anubhab91 says:

9. anubhab91 says:

10. Awesome says:

Thank you so much. I love your notes. Short but precise !

11. anil says:

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

12. ansari anas says:

nice

13. adarsh minz says:

one requested problem from my side is a circuit containing two independent current sources using superposition theorem