Thevenin Equivalent with Superposition Theorem

Question

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.

Answer 1

But if I deactivate the voltage source (short circuit) I don't know how to calculate the V2

I assume you mean that V2 is the voltage at the terminals when the voltage source is short circuit.

V2 is 4 amps multiplied by the equivalent resistance looking left into the 10 Ω resistor.

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).