# [Circuits] Calculating the Thevenin Equivalent #1

1. Jan 21, 2014

### ainster31

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

2. Relevant equations

3. The attempt at a solution

I got $R_{ th }=4\quad Ω$ easily but I am having trouble finding $V_{th}$. Assume the negative terminal of the 32V voltage source is ground. Here is the nodal analysis at node 1:

$$\frac { 32-V_{ 1 } }{ 4 } +2=\frac { V_{ 1 } }{ 12 } +V_{ 1 }-V_{ 2 }$$

But I only have one equation and there are two variables.

Last edited: Jan 21, 2014
2. Jan 21, 2014

### Staff: Mentor

When you remove the load RL, and after assigning a common reference node, there's only one essential node remaining. So you should have one equation in one unknown.

Note that with the load removed there will be no current flow through the 1 Ω resistor...

3. Jan 21, 2014

### ainster31

Opps, I should've added that I'm trying to find the voltage between (a) and (b).

$V_1$ is the voltage on the node in between the 4 ohm and 12 ohm resistor and $V_2$ is the voltage of (a). (b) is ground.

If there is no current flow through the 1 ohm resistor, does that mean that the voltage of $V_1$ and $V_2$ is the same?

4. Jan 21, 2014

### Staff: Mentor

Yup. No current means no potential drop.