Hi,

after calculations I end up in the circuit below¨ (Thevenin equivalent)



So now I have to say Rth=RL and I will find Rx ?


The initial circuit is this:


Thanks

 

So now I have to say Rth=RL and I will find Rx?

Yes, but it will be impossible to have Rth = RL = 20Ω in this case. Since any Rx resistor in parallel with 14Ω resistor will give you a lower equivalent resistance. Or maybe a negative resistance is allowed?

So, do you know the answer?

 

 

Look sensible.

 

But if Rth is 20 ohms, then an RL of 14 ohms doesn't give maximum power transfer. Show us how you obtained a value of 20 ohms for Rth.

 

 

The last schematic shown near the bottom of your calculations is confused. You have a node labeled A, and VA is the same as Vo. What is the voltage at the output node where you inject IN? Maybe you should call that VB; for sure don't call it Vo. You have made the same mistake you did in the other thread. You haven't got the correct voltage across the 10 ohm resistor. Fix the expression for the voltage across the 10 ohm resistor and do your calculations again.

By the way, if you are able to get a correct answer as a result of help you were given by forum members, it's good etiquette to show your correct result and say thank you. Those of us who help like to feel good about our efforts.

 

and Pmax=24.52 W

 

I made a mistake when checking your Rth; it is indeed 20 ohms. As you say, the practical answer is for Rx to be open circuit, although as Jony130 suggested a negative resistance (-140/3 ohms ) would give the maximum power transfer from the source.