# How can I get the Thevenin equivalent of this circuit?

I need to do this circuit but I can't do it properly, I don't know if I done well the nodes analysis, the answers are Rth= 79.66 ohm and Vth= 5.29 V, thanks for your help.

• A very warm welcome to the site. Have corrected your question title. 'Thevenin' comes from Mr Thevenin. It's not 'Thevening', as if its been Thevened :-) Commented Aug 30 at 10:25
• Notice that V1 will short-out R5 and R1 resitors as well.
– G36
Commented Aug 30 at 11:05
• Elisa, Are you required to solve this in some particular way? Or are you allowed to use any approach that gets you the right answer? You mention nodal. Which is why I ask. Commented Aug 30 at 12:30
• @periblesis No it is not necessary do it with nodal. Commented Aug 30 at 12:44
• @Elisa Then I added a small note to help out. Otherwise, I think Andy has also done a lot to help you. Commented Aug 30 at 12:50

To find the Thevenin voltage at A and B you should try and simplify things.

The first thing you should note is this: -

This means that the circuit reduces to: -

And, the 9 volt source and the 5 volt source combine to be a single 14 volt source in series with R4.

You can also short out R7 because it is in series with a current source and plays no role: -

Can you take it from here?

• oops fixing thanks @periblepsis Commented Aug 30 at 12:31
• I will try do it by that way, thank for your answer Andy aka Commented Aug 30 at 12:50
• Added my +1. ;) Commented Aug 30 at 12:50

You missed just one detail in getting the right Thevenin resistance:

$$\R_1\$$, $$\R_2\$$, $$\I_1\$$, $$\V_2\$$, and $$\R_5\$$ can't alter anything as seen from A and B. What this subnet may alter is the current in $$\V_1\$$. But otherwise $$\V_1\$$ spans across that subnet and therefore bypasses it from the global perspective of terminals A and B.

Since you can use any method to get the Thevenin voltage, then I think Andy has set you up for that. You can either Nortonize the $$\14\:\text{V}\$$ source and the $$\100\:\Omega\$$ resistor, add that current source to $$\I_2\$$ and then Thevenize the resulting Norton. Or use some other approach.

The voltage at B will be more positive than the voltage at A, though. So be careful about specifying the resulting Thevenin voltage.

• Thank you for your answer periblepsis Commented Aug 30 at 12:54

The Thevenin voltage $$\V_{th}\$$ would be negative as the Norton current source generated with the 14V voltage source when added to 2A current source becomes a 0.6A current source towards ground.After this, convert the Norton current source back to a Thevenin voltage source.The Thevenin resistance is (82 || (2200+100+500)) which is 79.66Ω. Be slightly careful in your circuit analysis.