# Thevenin Equivalent with Superposition Theorem

I have the following circuit that I found (here)

https://www.tutorialspoint.com/network_theory/network_theory_thevenins_theorem.htm

The author uses the node analysis in order to calculate the Thevenin equivalent. But I want to calculate $$\V_{th}\$$ with Superposition theorem. The $$\V_{th}\ =\frac{200}{3}\$$.

For the Superposition theorem I will have $$\V_{th}=V_{1}+V_{2}\$$.

If I deactivate the current source (open circuit) the 10 ohm resistor will play no role.So I am interested on voltage across the other 10 ohm resistor. Using the voltage divider I will have $$\V_{1}=\frac{10}{15}*20=\frac{200}{15}=\frac{40}{3}\$$.

But if I deactivate the voltage source (short circuit) I don't know how to calculate the $$\V_{2}\$$.

Side note: I do not want to use node analysis for this.

• Where is the voltage V2 ?
– LvW
Commented Jul 11, 2023 at 10:35
• Could you label the parts? "deactivate the current source (open circuit) the 10 ohm resistor" ... which 10 ohm resistor do you mean? Commented Jul 11, 2023 at 11:23

It looks like $$\5||10 + 10\$$ to me. That's 13.3333 Ω so, V2 will be 53.3333 volts.
And, if you work it out by adding V1 (13.3333 volts) then you get the same $$\V_{TH}\$$ answer as when solving it via source transformation (66.6667 volts).