# Thévenin's Theorem - What to do when ending up with a voltage source in paralell with one resistor?

I want to find the Thévenin and Norton equivalents for the circuit below.

To find $$\R_{Th}\$$ we first kill all the sources. Then we have 2 paralell resistors, in series with one resistor and all of those merged are in paralell with one more resistor. $$(\frac{1}{3} + \frac{1}{6})^{-1} + 2 = 4 \Omega$$ $$(\frac{1}{4} + \frac{1}{4})^{-1}= 2 \Omega$$ Since $$\R_{Th} = R_{N}\$$ we get: $$\R_{Th} = R_{N} = 2 \Omega\$$

If I did that correctly then I get this:

And now I am lost. When using Thévenin's Theorem we should end up with a voltage source in series with a resistor. With Norton we should end up with a current source in paralell with a resistor. I usually use source transformation but the only methods I have seen used needs a voltage source in series with a resistor or a current source in paralell with a resistor. But I have a voltage source in paralell with a resistor. What should I do to continue from here?

• Try again, more carefully this time. If you short circuit the original circuit, the 3 ohm resistor will limit the current in some way or other. Your circuit reduction will deliver infinite current if short circuited. => you have done it incorrectly. Commented Sep 23, 2023 at 11:24
• But the Vab = Vth is not equal to 20V. You need to calculate this voltage first. Take a look at this example it may help you electronics.stackexchange.com/questions/377467/… Rth is indeed equal to 2 ohms.
– G36
Commented Sep 23, 2023 at 11:54