# Circuit nodal analysis

1. May 27, 2013

### xlu2

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

Find Vx.

2. Relevant equations
KCL
V=IR

3. The attempt at a solution
First since the undrawn voltage source is in parallel with the 2 ohms resistor. Therefore, I concluded that the voltage source is also -3V

For node labeled V1: Iin-(-3V/2ohms)+I from V3 to V1=0. (The first question is how can I get I in and I from V3 to V1?)

For node labeled V3:7A=I from V3 to V1 + I entering the parallel part

For the parallel part: V5-V6=Vx; Vx/6ohms+Is = I from V5 to V6.

For node labeled V7: I from V5 to V6 + (-4Vx)-(-2V/2ohms)=0
==> (V5-V6)/6ohms+Is-4(V5-V6)+1=0 (How do I find Is?)

Would someone please give me some insight on this circuit?

2. May 27, 2013

### Staff: Mentor

You may be over-thinking this exercise. Suppose you take the bottom rail (node) as the ground reference. Can you determine by inspection the potentials for the nodes at the tops of the 2Ω resistors? What then is the potential difference across the 6Ω resistor?

3. May 27, 2013

### xlu2

I see the 2 ohms resistor on the left side has a potential of -3V.

Since the 7A is in parallel with that resistor, the voltage across it is also -3V.

For the 2 ohms resistor on the right side, the voltage across is given to be -2V.

So the potential difference between Vx = V3 and V7 = (-3)-(-2)=-1 V

4. May 27, 2013

### Staff: Mentor

Yes, that's right.

5. May 27, 2013

### xlu2

Thank you so much!