# Simple Nodal Analysis Problem

1. Mar 7, 2014

### Magnawolf

1. The problem statement, all variables and given/known data

2. Relevant equations

KVL and KCL

3. The attempt at a solution

Let V = V1 and the Voltage across the 2 ohm resistor be V2 just for reference.

Step 1: KCL @ Supernode: V1/3 + V2/2 + 21/4 + V3/6 = 0

I think my problem is here somewhere. Does V3 = V2 because they're in parallel? Also, am I wrong to say that 21/4 is a current going out of the supernode?

Step 2: V2-V1 = 9 (pretty sure this is right)

I can't get the correct answer. Any help is appreciated.

2. Mar 7, 2014

### Staff: Mentor

The 2 and 6 Ohm resistors are in parallel so they must share the same potential difference. If V1 is the potential at the top of the 3 Ohm resistor then the potential difference across the 4 Ohm resistor must be V1 - 21.

If you make a supernode of the nodes either side of the 9 V source, then really you need only one voltage variable to write the equation. Suppose you chose v to be that variable (after all it's already specified in the circuit diagram). Then on one side of the source the potential is v and on the other side it's v+9. No need to introduce V2 or V3 etc.

3. Mar 7, 2014

### Staff: Mentor

Hi. Your V3 is V2, it's the one and the same node. Keep it as just V2.

The current through the 4Ω is given by (voltage across the 4Ω resistor itself) / 4

So what is the voltage across that 4 ohm resistor? Hint: the expression will involve a subtraction

4. Mar 7, 2014

### Magnawolf

This. This is what I needed. Thanks so much!