# Thèvenin equivalent circuit

I calculated the Thèvenin equivalent circuit of a given circuit in two ways. I know that one is correct and the other is not (because I saw the result on my book), but I can't figure out why the other is wrong.

This is the starting circuit:

simulate this circuit – Schematic created using CircuitLab

First way :

$V_{eq}=V_1\cdot\frac{2R}{2R+2R}=\frac{V_1}{2}$. (this gives the right solution).

$R_{eq}=2R||2R=R$

simulate this circuit

Second way: I see the starting circuit as this:

simulate this circuit

That is with an infinite resistance in place of the open circuit. Now, I get:

$V_{eq}=V_1+V_1\cdot\frac{2R}{2R+2R}=...=V_1\cdot\frac{3}{2}$

$R_{eq}=(R+2R)||2R=...=\frac{6}{5}$

So,the equivalent Thèvenin circuit now is:

simulate this circuit

Where am I wrong with the second method?

EDIT:

When I calculate $V_{eq}$ with the 2nd method, I get this circuit:

simulate this circuit

And here it's true that:

$V_{eq}=V_1\cdot\frac{2R}{2R+2R}=\frac{V_1}{2}$

but it's also true that, with KVL:

$V_{eq}=V_1+V_1\cdot\frac{2R}{2R+2R}=...=V_1\cdot\frac{3}{2}$.

Where am I wrong with this last one?

• "Where am I wrong with this last one?" You got the polarity of the second term wrong: Veq = V1 - V1*2R/(2R+2R). May 12 '15 at 22:52

You're not combining series and parallel resistors correctly. To use the voltage divider equation with your second circuit, you first need to combine $R$ and $R_2$ in parallel with $2R$:

$$\frac{1}{R_{parallel}} = \frac 1 {2R} + \frac 1 {R + R_2}$$

$$\frac 1 {R_{parallel}} = \frac 1 {2R} + \frac 1 {R + \infty} = \frac 1 {2R} + 0 = \frac 1 {2R}$$

$$R_{parallel} = 2R$$

The voltage between the $2R$ resistors is then:

$$V_{mid} = V_1 \frac{2R}{2R + 2R} = \frac {V_1} 2$$

which is what you got for the first circuit.

I'm not sure why you didn't include the $R$ resistor in your Thevenin resistance. Normally in this kind of problem you want to end up with only one equivalent resistor.