It's actually a maximum power question, but I'm suspecting that my answer is wrong...it's a past exam question and no solution was given.

For V open circuit, using Va as the top mode, Vb as above 4Ohm resister, Vc as top right, and bottom as reference node, I got

Va = 60 + 3Vb (At Va)
11Vb - 6Vc = 0 (At Vb)
2Vc - Va - Vb = 0 (At Vc)

giving me -1380V, -480V, -930V, and the Vopen would be -930V (I'm thinking that this is wrong...of course)

For I short circuit, using bottom left as Ia, Bottom right as Ib, and top as I3

-10Ia + 4Ib + 6Ic = -100
4Ia - 6Ib + 2Ic = -20
18Ia - 10Ib - 10Ic = 0

givin me 86.38A, 84.54A, 70.91A, and the Ishort would be 84.54A

(Rth = -11Ohm, Max power = -78.6k using my values...)

 

Your values look fine. Given the result includes a negative source resistance I'm not sure about the maximum power transfer condition. If you make R_Load=11Ω then you would have infinite current giving infinite load power.

 

Hmm...got and used a circuit solver app, and apparently my current is correct, but the voltage should be 116.25V. Trying to figure out what I did wrong.

 

If you are saying the circuit solver gave 116.25V for the Vth value than that's clearly a problem with the simulation. The values you found are correct for the schematic as given. Or is 116.25V the answer given in the question source book?

 

your values are perfectly fine... negative thevinin's resistance is possible in case of circuits with dependent sources..
refer page 3 of
https://docs.google.com/viewer?url=.../Syllabi/ES250/Thev/ThevVCVS_HOsoln.pdf&pli=1

 

Thanks for the replies...but this is very weird.

I went and bugged the prof about it today, and the voltage is actually 116.25V, solved with loop analysis. I looked over the numbers again, but couldn't find anything wrong either.

For the simulation, it doesn't show Vth, but I simply broke the circuit, and found the voltage difference between the two nodes.

 

Well you are either looking at the solution for another problem or your professor needs to re-check the work. The answer of 116.25V is incorrect.

I'll post a loop analysis solution shortly.

 

Here 'tis ....

 

Well you are either looking at the solution for another problem or your professor needs to re-check the work. The answer of 116.25V is incorrect.

I'll post a loop analysis solution shortly.

I solved the network with both nodal analysis and loop analysis, and I get Vth = -930 volts.

The maximum power theorem would normally say that max power is developed in the load when the load equals Rth. So in this case the load should be -11 ohms for maximum power???

 

Hi Electrician,

Interesting question about the maximum power transfer condition. Not sure where one would obtain the -11Ω resistor. Maybe a canny electronic parts collector has one in their resistor bin.

 

I went to the prof again, and found that he simply used the same simulator I was using, and didn't actually solve it himself. It turned out that the answer was right.

I tried loop as well, and got the same number.

Now everything is cleared up, thanks!

 

Your prof needs a reality check.

 

Very strange because my simulation shows Vth = -930V
What simulation program have you use ?

 

My simulation agrees with the calculations ... until I substitute a 11 ohm resistor in the load.

Then the simulaiton goes awry.

I did a dc analysis, sweeping the load from 9 to 13 ohms and the results look way out of wack. That result of course was much different than the DC results by just "energizing" the simulator in the DC mode with an 11 ohm load.

I've seen this type of error before, not with electronic simulations but with a simulator that computed the coverage are of a Loran-C baseline. At a point close to the baseline extension, you would have thought it was a Department of Justice gerrymandered election district. I will send that circuit to my simulator's programmer and let them come up with a reasonable answer.

Anyway, here are two graphics illustrating how the TINA simulation reacts.