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I have the following exercise:

I need to use thevenin's theorem to get the equivalent voltage source and resistor enter image description here

I have redrawn it in LTSpice for better understanding: enter image description here

I knew how to get the equivalent resistance : (R1*R2)/(R1+R2)

But how to get the equivalent voltage source? When i use the superposition theorem it's V = V1 + V2 * R2/(R1+R2) but that is wrong. How can i solve this ? And what have i done wrong using the superposition theorem ?

Reference for the picture : https://stefan-schenke.de/get/challenges/Training_Ersatzquellen.pdf

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    \$\begingroup\$ Please provide a reference for the image, as required by the rules here. Does the asymmetry in your result suggest anything? Also you do need to indicate a reference node or your sign might be wrong. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Nov 26 at 8:28
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    \$\begingroup\$ Are those current sources or voltage sources? \$\endgroup\$
    – Ste Kulov
    Commented Nov 26 at 8:34
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    \$\begingroup\$ Both are voltage sources \$\endgroup\$
    – Tim Graf
    Commented Nov 26 at 8:37
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    \$\begingroup\$ You short out one supply at a time, which I guess you have. Then you need to be aware of where the voltage is measured. How do you come to the conclusion that it is V1 + something? \$\endgroup\$
    – Tyassin
    Commented Nov 26 at 8:44
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    \$\begingroup\$ @TimGraf - Hi, I have removed the new question from your self-answer. Although there are a couple of different approaches you could use if you want to ask a new, related question, the easiest (and least confusing for future readers) is to post a new self-contained question containing what you now want to ask, with a link to this one for context, if relevant. || The thing to avoid is changing the question, or asking a new one, inside an "answer" since then it isn't an answer any more. TY \$\endgroup\$
    – SamGibson ♦
    Commented Nov 26 at 10:01

1 Answer 1

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Thanks to Tyassin I now know what I have done wrong in the superposition theorem. It's V = V1 * R2/(R1+R2) + V2 * R1/(R1+R2) = (V1*R2+V2*R1) / (R1+R2).

To calculate with mesh currents, get the total current which is (V1-V2)/(R1+R2) then get the voltage drop on R2 which is I * R2 and then add those two up to get the voltage on OUT. So the voltage on out is : (V1-V2)/(R1+R2) * R2 + V2

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