Engineering [Circuits] Calculating the Thevenin Equivalent #1

[ainster31]

Homework Statement

Homework Equations

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.

 

[gneill]

When you remove the load R_L, 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...

 

[ainster31]

When you remove the load R_L, 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...

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?

 

[gneill]

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?

Yup. No current means no potential drop.