# Finding a voltage in a circuit with only resistors and current sources

1. Feb 9, 2014

### Deluxe489

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

Find V0 in the circuit.

2. Relevant equations
Kirchhoff's Current Law: All currents entering a node must sum to zero.
Kirchhoff's Voltage Law: All voltages around a loop must sum to zero.

3. The attempt at a solution

So far, I have taken the top of the center branch to be a node with five currents entering/leaving it. Using Kirchhoff's Current Law, I have done:

0 = 2Vy + (3Vx / 3Ω) + 6A - (2Vy / 4Ω) - 4Vx

I assumed that the voltage across the 2Ω resistor in the second branch from the right was also Vy, and I assumed that the voltage across the 2Ω resistor in the second branch from the left was 2Vx. I know this because I know that the current is constant within each branch.

Solving the equation for Vx, I got:

Vx = (1/2)Vy + 2A

At this point, though, I'm confused about what to do. I've thought of using Kirchhoff's Voltage Law using the loop formed by the two resistor branches (so basically ignoring all the branches that current sources in them), but even if I do that and make the substitution I derived from Kirchhoff's Current Law, I don't know how I'm supposed to get V0 from doing that. Am I supposed to find the current in those two branches and then just add up the voltage drops across the four resistors? If so, I'm not sure how to find the currents. Am I supposed to find the voltages across the current sources?

I would really appreciate any help.

Thank you.

2. Feb 9, 2014

### Staff: Mentor

It would, perhaps, be easier to first write the node equation without reference to Vx or Vy except for their controlled source contributions. Then separately write expressions for Vx and Vy in terms of Vo and substitute them into the node equation later. As it stands, I don't understand your terms for the resistor branches. Surely the current in the 1Ω + 2Ω branch would be =Vx/1Ω ?

Anyways. Try writing the equation without referencing Vx and Vy for the resistive branches. Afterwards substitute expressions for Vx and Vy for the controlled source terms.