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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:

schematic

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\$

schematic

simulate this circuit

Second way: I see the starting circuit as this:

schematic

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:

schematic

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:

schematic

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?

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

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  • \$\begingroup\$ Thanks for the reply. I added an EDIT to the question. Could you please read it? \$\endgroup\$
    – sl34x
    May 12 '15 at 22:38
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    \$\begingroup\$ You got your signs backwards. As you go counterclockwise around the loop, you go from negative to positive through V1 (+) and positive to negative through the two resistors (-). Also, you need to include both resistors in your loop equation. \$\endgroup\$
    – Adam Haun
    May 12 '15 at 23:27

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