Hello!

I need help finding the voltage in node A. I know that I have to do supernode-analysis and I understand how to do the KCL, the problem is how I should deal with the dependent voltage source, should I do supernode there to? The circuit is attached.

Thanks in advance!

 

V=2*V(A) does not compute.

 

You can not use supernode on B1 because it does not lie between two essential nodes.

 

Hello!

I need help finding the voltage in node A. I know that I have to do supernode-analysis and I understand how to do the KCL, the problem is how I should deal with the dependent voltage source, should I do supernode there to? The circuit is attached.

Thanks in advance!

Until it starts becoming obvious how to deal with it, take it in two steps.

Call the node that connects R5 and B1 Node C.

That gives you four unknowns.

Put a supernode around nodes A and B.

Develop your node equations for the supernode and Node D. Don't do one for Node C.

That gives you two equations. Now you need two more equations which you can get from the constraints.

Can you write an equation for the voltage at Node B in terms of the other node voltages?

Can you write an equation for the voltage at Node C in terms of the other node voltages?

Now you have your four equations.

With a bit of practice, you'll be able to streamline this a bit and write down just three equations in three unknown (you won't have Vc) by inspection. But it takes practice to get there. In the meantime, got step-by-step in however small of steps you need.

 

Just verify that which one is applicable... Simple nodal or Supernode circuit analysis....
if you cant decide, then here is a detail post for you Supernode circuit analysis
if you still face a problem, then please send me your assignment... i will forward the solved example.... Thanks

 

if you still face a problem, then please send me your assignment... i will forward the solved example.... Thanks

Gee, why don't you make it even more convenient and just submit the solution directly to the instructor?

 

Gee, why don't you make it even more convenient and just submit the solution directly to the instructor?

And there wasn't any fee involved. Truly magnanimous!

 

And there wasn't any fee involved. Truly magnanimous!

Isn't that how it starts. Someone does some 'favor' for you without wanting anything in return -- but then, at some point, it comes time to 'pay back' the favor....