Thevenin equivalent with two current source

Chalielogan

Joined Apr 11, 2022
8
I am bit confused with this question. I tried to analyse and tried some questions but this is bit hard for me. So solution 1 I thought was mesh analysis. But it seems its not a good idea to use in this question. Also, with current source, should I have to find Nortan analysis?

WBahn

Joined Mar 31, 2012
29,827
You need to show your best attempt at a solution. It doesn't have to be correct or taken all the way to the end. Set things up as best you can and show your work as far as you can. That will let us see what your thinking is and where you are going right and where you are going wrong.

You diagram appear so say that the CCVS is -Rm*I, but I don't see where the current "I" is defined. Also, is that a negative sign, or just a stray mark?

Chalielogan

Joined Apr 11, 2022
8
You need to show your best attempt at a solution. It doesn't have to be correct or taken all the way to the end. Set things up as best you can and show your work as far as you can. That will let us see what your thinking is and where you are going right and where you are going wrong.

You diagram appear so say that the CCVS is -Rm*I, but I don't see where the current "I" is defined. Also, is that a negative sign, or just a stray mark?
Hi. Thank you and I tried to figure it out but I was not sure with the way I go through it. The problem is I have to find I. And it is -2A.

WBahn

Joined Mar 31, 2012
29,827
Hi. Thank you and I tried to figure it out but I was not sure with the way I go through it. The problem is I have to find I. And it is -2A.
We are not mind readers and our crystal balls have repeatedly been proven to be useless. We can't figure out what you did wrong unless you show us what you did.

You have a current-controlled voltage source in which you are stating that the output voltage is equal to the value of Rm multiplied by the value of the current labeled "I" (possibly with a minus sign, as well). But you don't have a current labeled I anywhere on your schematic.