# Thevenin Equivalent Circuits

Discussion in 'Homework Help' started by dWilksEE, Aug 31, 2014.

1. ### dWilksEE Thread Starter New Member

Aug 31, 2014
2
0
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.

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2. ### mayilsamy New Member

Mar 26, 2013
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0
Rth is 7.5 ohm,
Vth is 200.25 V

3. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,500
511
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...

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
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

5. ### dWilksEE Thread Starter New Member

Aug 31, 2014
2
0
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!

6. ### mayilsamy New Member

Mar 26, 2013
5
0
Yes that is correct. I made mistake writing Isc 56.667 as 26.667.