# Need help with this mesh/nodal analysis problem

Discussion in 'Homework Help' started by mesaman000, Jun 29, 2010.

1. ### mesaman000 Thread Starter New Member

Jun 29, 2010
16
0
So guys. I'm having trouble with this problem..

I try using nodal analysis but don't know how to treat the voltage sources going into the nodes. So i decided to use mesh analysis and found all Ia, Ib, Ic, and Id.. I got Vo= .3 V but I'm not sure if this is right.. And also I have no idea how you're supposed to approach it using nodal analysis either.

Firstly can someone verify that the answer is .3 V? Cuz I got the current Ic going through that 1k resistor on the right (which is Vo) is .3 mA , therefore .3*1= .3 V..

And secondly can someone help me on how I can start this problem using nodal analysis?

I would REALLY appreciate it. Thanks a lot!

2. ### tskaggs New Member

Jun 17, 2010
26
3
There are three unknowns in this circuit: $V_{y}$, $I_{x}$, and $V_{0}$.

From examination of the center node you can see that $I_{x}$ and $V_{y}$ are related by the equation

$I_{x}=\frac{V_{y}-2V}{1k}$

So now you only need two equations to solve the rest of the problem.

Perform nodal analysis with the supernode (i.e., the node that includes $V_{y}$ and the source $2V_{y}$) and the node on the center/right-side ($3V_{y}-V_{o}$).

Do you need help writing the equations from this point?

3. ### mesaman000 Thread Starter New Member

Jun 29, 2010
16
0
ermm..well for the supernode i'd have to include the current for the bottom left (we don't know the current going through that 2 v voltage source)

4. ### tskaggs New Member

Jun 17, 2010
26
3
Actually the 5 currents leaving the supernode correspond to the five resistors. So the supernode equation would be:

$0=\frac{V_{y}}{1k}+\frac{V_{y}-2}{1k}+\frac{3V_{y}-2}{1k}+\frac{3V_{y}-(3V_{y}-V_{0})}{1k}+\frac{V_{y}-(3V_{y}-V_{o})}{1k}$