# Thevenin Voltage and Resistance Calculations

In order to calculate R_th, you have to made all voltage and current sources equal to 0 (creating open circuits at current sources and shorts at voltage sources), so you will wind up with only resistors. That means that:

$$R_{th}=[(R_1+R_2)^{-1}+1/R_3+1/R_4]^{-1}$$

The dependent voltage source has a transresistance, so it adds on to the value of R_th. This is not being accepted as my answer though. I tried with and without the transresistance, even though it shouldn't be removed.

• Nodal analysis is easier than mesh analysis in this case. – Chu Mar 31 '18 at 9:36
• Why would node analysis be easier? When dealing with the top voltage source, you end up forming a super node, so you sum all of the currents leaving the nodes to the right and left of the voltage source. This leads with you dividing over 0 on the right-hand side because you have the right node-a over that resistor, which is 0. Or do I not make that calculation and assume that the current in that piece of the circuit is 0 because it is an open loop? – Albert Garcia Mar 31 '18 at 16:00

For $\small V_{TH}$, let $\small V$ be the 25/55/25v node, then: $\frac{V-4I_x}{29}+\frac{V}{55}+\frac{V-25}{20}=0$, and $I_x=\frac{V}{55}$. Hence $\small V_{TH}=V-25$.