# Thevenin Equivalent

Discussion in 'Homework Help' started by jstrike21, Oct 1, 2009.

1. ### jstrike21 Thread Starter Member

Sep 24, 2009
104
0
Here is a problem on thevenin equivalent. What i have done is take out the ten amp source and make the branch with the 33 ohm resistor have the resistor and a 330v source. After this im not sure what to do. I don't think there is any way to simplify the circuit more and when I applied mesh current analysis to the circuit i got no answer

Help!

Edit: I used mesh current to solve for the short circuit current and got 37.48 A but that doesn't seem right to me

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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You might show what work you have done so far - it then helps to make suggestions about how to solve the problem from your perspective.

3. ### jstrike21 Thread Starter Member

Sep 24, 2009
104
0
k i took out the 10 amp source and made the branch below that have a 330v source and a 33 ohm resistor. After that I used mesh current to try to solve for the short circuit current.

Mesh A: 500= 16(Ia-Ib)+11(Ia-Ic)
Mesh B: 330 = 33Ib+10(Ib-Ic)+16(Ib-Ia)
Mesh C: 0= 11(Ic-Ia) + 10(Ic-Ib)

solved this and got Ia=48.69 Ib=25.15 Ic=37.48
solved for Isc and got 37.48A

I think this is wrong and I don't know how to get the next answers

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Well you've obtained the correct answer for Isc - so you're off to a good start.

To find Voc do another mesh analysis. In this case there are only two equations required since Ic is zero. Solve for Ia and Ib with this condition. Then find the (IxR) voltage drops across Rc and Rd - then you're almost there!

Hint: Keep in mind that Voc = VRc + VRd

Also remember that Vth = Voc and Rth=Voc/Isc