How to Solve Circuit Nodal Analysis for Vx?

[xlu2]

Homework Statement

Find Vx.

Homework Equations

KCL
V=IR

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?

Many thanks in advance!

 

[gneill]

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?

 

[xlu2]

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?

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

 

[gneill]

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

Yes, that's right.

 

[xlu2]

Yes, that's right.

Thank you so much!