# Finding Voltage Thevenin via short circuit

I found this question online where $$\ R_1 = R_2 = R_3 = R_4 = 15 \Omega \$$ and $$\V_s = 15 V\$$ Finding $$\R_{th}\$$ first, we short circuit $$\V_s\$$,

We know $$\R_1\$$ and $$\R_2\$$ are in parallel which is then in series with $$\R_3\$$ which is then in parallel with $$\R_4\$$.

Substituting 15 ohms for all R values we get $$\R_{th} = 9 \Omega\$$

Now we want to find $$\V_th\$$ I attempt to do this by using the $$\i_{sc}\$$ short circuit method. By connecting the two terminals, I think we are short circuiting the $$\R_4\$$ wire So, applying ohm's law: $$\V_{s} = R_{1,2,3} * i_{sc}\$$

where $$\R_{1,2,3}\$$ is the equivalent resistance of $$\R_1\$$, $$\R_2\$$, $$\R_3\$$

Finding $$\R_{1,2,3}\$$, $$\R_{1,2,3} = 22.5 \Omega\$$

Solving for $$\I_{sc}\$$

$$V_S = R_{1,2,3} * i_{sc}$$

$$15V = 22.5\Omega * i_{sc}$$

$$i_{sc} = \frac {2}{3}$$

So, $$\V_{th}\$$ should just be $$\R_{Th}\$$ times $$\i_{sc}\$$

$$V = R_{Th} * i_{sc}$$

$$V = 9\Omega * \frac {2}{3}$$

$$V = 6V$$

However, my answer is incorrect(as given by the answer checker. Can someone please tell me where I went wrong?

• FYI, EE uses \$ instead of $ to start and end inline math. Oct 4 '19 at 4:28
• Isn't Rth = R1+(R2^-1+(R3+R4)^-1)^-1 = 25?, It has been some years since I have done this :) Oct 4 '19 at 4:33
• @Sorenp, to get Rth you need to look at it from the output port, not from the voltage source. OP has calculated Rth correctly: ((R1 || R2) + R3) || R4. Oct 4 '19 at 5:16
• @The Photon time to dust of the old notes, thanks :) Oct 4 '19 at 5:56

Your $$\I_{sc}\$$ is the current produced by the source $$\V_S\$$ when the output is shorted.

It isn't the current that goes through the short.

• @WilsonGuo, there's no "current through the whole circuit". There's current through different branches of the circuit. You calculated the current through the branch with the voltage source. What you want is the current through the short (which will be the same as the current through R3). Oct 4 '19 at 4:54
• Re-draw the circuit with the short applied across the outputs. What makes you think there's no current through R2? Oct 4 '19 at 5:11
• Re draw the circuit with the short. What makes you think there's no current through R2? You should actually see that R2 and R3 are in parallel. So if there's current through R3 there must also be current through R2. Oct 4 '19 at 5:12
• Again, I don't like the term "total current", but you already solved for it. That's what you got from your original calculation. Oct 4 '19 at 5:23
• You mean 0.333... A, right? Oct 4 '19 at 5:29

Well, first of all we are trying to analyze the following circuit: simulate this circuit – Schematic created using CircuitLab

Finding first $$\\text{R}_\text{th}\$$ we need to look at: simulate this circuit

Now, we see that:

$$\text{R}_\text{th}=\frac{\text{R}_4\cdot\left(\frac{\text{R}_1\text{R}_2}{\text{R}_1+\text{R}_2}+\text{R}_3\right)}{\text{R}_4+\frac{\text{R}_1\text{R}_2}{\text{R}_1+\text{R}_2}+\text{R}_3}\tag1$$

Finding secondly $$\\text{I}_\text{th}\$$ we need to look at: simulate this circuit

$$\text{I}_\text{th}=\frac{\text{R}_2}{\text{R}_2+\text{R}_3}\cdot\frac{\text{V}_\text{in}\left(t\right)}{\text{R}_1+\frac{\text{R}_2\text{R}_3}{\text{R}_2+\text{R}_3}}\tag2$$

So:

$$\text{V}_\text{th}=\text{R}_\text{th}\cdot\text{I}_\text{th}=\frac{\text{R}_4\cdot\left(\frac{\text{R}_1\text{R}_2}{\text{R}_1+\text{R}_2}+\text{R}_3\right)}{\text{R}_4+\frac{\text{R}_1\text{R}_2}{\text{R}_1+\text{R}_2}+\text{R}_3}\cdot\frac{\text{R}_2}{\text{R}_2+\text{R}_3}\cdot\frac{\text{V}_\text{in}\left(t\right)}{\text{R}_1+\frac{\text{R}_2\text{R}_3}{\text{R}_2+\text{R}_3}}\tag3$$