Here is my problem from Hayt Fig 5.7

I am trying to solve applying superposition principle, but I am not getting anywhere close to the correct answer.

As the first step I eliminated 2A current source. omit 2A

I tried to solve it using mesh analysis. In the first loop I introduced clockwise current called i_1, in the second loop I introduced clockwise current called i_2. Here are the equations that I wrote for it.

-3 + 7i_1+15i_1 + 5(i_1-i_2) = 0

i_2 = -4i

i_1-i_2 = i

I solved it and got i = -1/61 and v_2 = -5/61 which is far off from stated answer: -0.246V

What did I miss?

Then I short-circuited 3V voltage source:

Omit 3V

I solved it using node analysis. Nodes are labelled in the main picture. Here are my equations:

2= v_1/7 + (v_1 + v_2)/15

4i = (v_2 - v_1)/15 + v_2/5

v_2 = 5i

I solved it and got v_2 = -546, which is again far off from -1.148.

I double checked my calculations. Could somebody confirm that I got correct equations?


2 Answers 2


Please see the answer below.Do not kill dependent sources while applying SPST.

enter image description here

  • \$\begingroup\$ Thank you so much! I found my computational error. I did not kill dependent source. It is present in my system (i_2 = -4i). I like how you express current in terms of dependent current source. \$\endgroup\$ Dec 7, 2023 at 12:48
  • \$\begingroup\$ If it solves your problem, please accept the answer \$\endgroup\$
    – Hari
    Dec 7, 2023 at 13:24

In your mesh equations, you wrote i_1-i_2 = i. I think you meant i_2-i_1 = i. Your diagram didn't clearly show the directions of i_1 and i_2, so you made it hard for yourself to debug your own equations. I infer their directions from the two loop equations.

You appear to have declared the current from the dependent source to be 0 A. This is not valid unless current i is 0 A. When using superposition the procedure calls for killing independent sources, not dependent sources.

  • \$\begingroup\$ Thank you! I should have created better diagram. I should have added arrow indicating i_1 and i_2 direction instead of writing that they are clockwise. 0A dependent source is just artifact of multisim diagram, I had to put some value there. So I have chosen 0A, but I am not using this value anywhere. \$\endgroup\$ Dec 7, 2023 at 12:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.