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(i1i2)+12(i1)=0 > 20*i1  8*i2 = 500
8(i2i1)+30(i210)+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.
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(i1i2)+12(i1)=0 > 20*i1  8*i2 = 500
8(i2i1)+30(i210)+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.
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