So my first instinct on this problem was to ignore the sources and calculate a equivalent resistance. I got 7.5 Ohm, but I was wondering where I am going wrong for the long way to calculate the Thevenin resistance.

Rt= -Voc/Isc

i1 is the lower left loop, and i2 is the middle loop.
For Voc I got 425 V, from my two KVL equations:
-500 + 8(i1-i2)+12(i1)=0 -------> 20*i1 - 8*i2 = 500
8(i2-i1)+30(i2-10)+5.2(i2)=0 -------> -8*i1 + 43.2*i2 = 300
i1=30 A
i2=12.5 A
For Isc, I got 69.6656 A, from two KCL equations:
I just found the equivalent resistance with the sources in and then did a Norton to Thevenin on the top loop to get a circuit with just a voltage and a resistor.

583.798 V voltage source with a 8.3798827 Ohm resistor.

 

Rth is 7.5 ohm,
Vth is 200.25 V

 

For Isc is got 56.67 A. With Voc being 425 V, that gives Rth of 7.499 Ohm.

It looks like you setup your equations wrong for Isc. And I have 3 equations for it. You only have 2...

 

So my first instinct on this problem was to ignore the sources and calculate a equivalent resistance. I got 7.5 Ohm

Very good, Rth is indeed equal to 7.5 Ohm.

Rt= -Voc/Isc

Why "-" sign ?

i1 is the lower left loop, and i2 is the middle loop.
For Voc I got 425 V, from my two KVL equations:
-500 + 8(i1-i2)+12(i1)=0 -------> 20*i1 - 8*i2 = 500
8(i2-i1)+30(i2-10)+5.2(i2)=0 -------> -8*i1 + 43.2*i2 = 300
i1=30 A
i2=12.5 A

OK.

For Isc, I got 69.6656 A, from two KCL equations:
I just found the equivalent resistance with the sources in and then did a Norton to Thevenin on the top loop to get a circuit with just a voltage and a resistor.
583.798 V voltage source with a 8.3798827 Ohm resistor.

I don't see any KCL for Isc? And yore answer is wrong. The correct answer is Isc = 56.6667A

Isc = ( 500V/(8 +(5.2||12)) * 12/(5.2+12) ) + 500/30 + 10A = 30A + 16.666A + 10A = 56.6667

Rth is 7.5 ohm,
Vth is 200.25 V

Check you math. The correct answer is Vht = 425V

 

Thank you very much. I thought that it was probably Isc that I was messing up, but I couldn't find where I made the mistake. Thank you Jony130 and shteii01!

 

Very good, Rth is indeed equal to 7.5 Ohm.


Why "-" sign ?

OK.

I don't see any KCL for Isc? And yore answer is wrong. The correct answer is Isc = 56.6667A

Isc = ( 500V/(8 +(5.2||12)) * 12/(5.2+12) ) + 500/30 + 10A = 30A + 16.666A + 10A = 56.6667



Check you math. The correct answer is Vht = 425V

Yes that is correct. I made mistake writing Isc 56.667 as 26.667.