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Here is my reasoning:

We have to sources so me must use superposition:

We first replace the current source by its resistance which is infinite for an ideal current source.

Now we need to measure the current and voltage across $R_4$. We consider the outer loop and using the voltage divider we get

$$V_{TH_1}=\frac{VR_4}{R_1+R_3+R_4}$$ $$R_{TH}= R_4//(R_3 + R_1//R_2)$$

Now we consider the current source and assume an ideal resistance of $0V$ through the voltage source. The current can take three paths so the current through $R_4$ is

$$I_4=\frac{(R_1 + R_2 + R_3 + R_4)I}{(R_3 + R_4) + (R_1 + R_2 + R_3 + R_4)}$$

So then we use superposition:

$$V_{TH} = V_{TH_1}+I_4 R_{TH}$$

And $$R_{TH} = R_4//(R_3 + R_1//R_2)$$

Does this make sense? What did I do wrong?



$$R_{TH} = R_4//(R_3 + R_1//R_2) + R_5$$

$$I_5=\frac{(R_1 + R_2 + 2R_3 + R_4 + R_5)I}{(R_3 + R_5) + (R_1 + R_2 + 2R_3 + R_4+ R_5)}$$

$$V_{TH} = V_{TH_1}+I_5 R_{TH}$$


1 Answer 1


Superposition is not necessary just put an open circuit where the current source is and a short circuit circuit where the voltage source is.
You then have a reasonably straightforward set of resistor for which to find the equivalent resistance and that it will be the Thevenin resistance.

You only have to worry about dependent current and voltage sources because you cannot remove them In this circuit there are none.

So it will be that $R_{\text{TH}}=(R_1||R_2+R_3)||R_4+R_5$

Your formula does not have $R_5$ in it which in fact must be there.

With superposition you need to short circuit the output and find the current through the short circuit. You then divide the Thevenin voltage by that current.
In your diagram the short circuit current will be the current through $R_5$

  • $\begingroup$ I followed your suggestions. I understand why $R_{TH}$ was wrong. I tried to fix $I_4$. Did I do it right? Do I need to count $R_3$ twice because of the two possible paths? And would then $V_{TH}$ be correct? $\endgroup$
    – 15yyyyy
    Feb 18, 2016 at 15:40
  • $\begingroup$ I now realize I needed to consider the current through $I_5$ to then divide it by $R_{TH}$ to get the voltage component. Did I do it right, this time? $\endgroup$
    – 15yyyyy
    Feb 18, 2016 at 15:49

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