I'm trying to replace a part of a circuit with the Thevenin equivalent. I managed to find \$R_{\text{th}}\$, but I'm not sure if my solutions for \$E_{\text{th}}\$ are correct. I have 3 sources, so the logical method for finding \$E_{\text{th}}\$ was the superposition theorem. Here are my results:

Here's the complete circuit:

enter image description here

1) 9V only

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So, the voltage between A and B should be: \$50 + 200 = 250 \Omega\$ and \$350 + 150 = 500 \Omega\$, then \$250 + 500 = 750 \Omega\$. This gives us a voltage of: $$U_{AB} = \frac{9\text{ V} \times 750 \Omega}{500 \Omega + 750 \Omega}$$

which is 5.4V?

2) 5V only

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$$(350 + 150) \parallel 500 + 200 = 450 \Omega$$ Then $$U_{AB} = \frac{5V \times 50}{450 + 50} = 0.5\text{ V?}$$

3) 12mA only -- This one is a real problem for me

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I solved the resistors the same as in 2), which gives: $$U_{AB} = 50 \times \frac{450 \times 12\text{ mA}}{50 + 50} = 0.54V$$

\$E_{\text{th}} = 6.44\text{ V}\$???

Please help! I'm not sure if this is correct.. Can someone explain how to solve these circuits the right way?

  • \$\begingroup\$ If you want to check your solution, you can always simulate it.. \$\endgroup\$
    – Eugene Sh.
    Jan 22, 2016 at 18:39
  • \$\begingroup\$ In your step 1, you didn't consider the 200 + 50 in parallel with the 500, which are then in series with the 350 and 150. \$\endgroup\$
    – Tyler
    Jan 22, 2016 at 18:44

1 Answer 1


There are times you ought to apply source transformation when solving with superposition, instead of attempting these shortcut you've shown above. Source transformations allows you to move resistors in series or parallel with a source, opening space for simpler and less prone to error mathematics.

For example, in the first instance of removing all but 9v source. You tried a shortcut but for some odd reason neglected one 500 ohm resistor, and thought you could add the 500 with the 250. Also you can't add 50 with 200, with respect to (A) + and (B) -, because they are parallel with respect to that thevenin.

You should try to convert transform the voltage source to current source, after adding 150 and 350 ohms in series. So you can have 500 ohms in parallel with the other 500 ohms, to truly calculate the next resistance in series/parallel. Then again transform the source from current source to voltage to continue. etc...

Here's the basic idea of transforming between current and voltage source enter image description here

Here's a demonstration, without giving the final answer of one of the instances.

enter image description here

  • \$\begingroup\$ Wow! I can't thank you enough for taking the time to explain this! In my class we never applied source transformation, but it obviously significantly simplifies the circuit. I solved the 9V case by your example, and I got 0.45V, I think it's correct? :) \$\endgroup\$
    – Quant
    Jan 22, 2016 at 22:36
  • \$\begingroup\$ P.S. Can I transform the current source in the 12mA case? Does it matter that the 350 resistor is up there? \$\endgroup\$
    – Quant
    Jan 22, 2016 at 22:44
  • \$\begingroup\$ yes, i got 0.45v as well. i also edited my question to provide the generic transformation equivalence example. no you cant transform the current source yet, because it's not in parallel with either resistor yet. i suggest you convert a set of three resistors from Y to Delta form. another fancy trick. see here en.wikipedia.org/wiki/Y-%CE%94_transform \$\endgroup\$
    – Iancovici
    Jan 22, 2016 at 23:10

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