# Circuit Analysis Toughy! finding Thevenin Equivelent

Discussion in 'Homework Help' started by damm1371, Feb 22, 2012.

1. ### damm1371 Thread Starter New Member

Feb 22, 2012
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So the object is to find the thev equivelant circuit for AB . The techniques I have available include, thev/norton, lakeville, KVL,KCL, ohms, and superposition.

Where I have started so far is trying lakville theorum on V1, but that only gets me so far, and Im not sure it is even correct. If you have an idea of how to work this please let me know step by step. Thank you.

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Feb 22, 2012
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3. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Superposition is your friend in this situation.

Since voltage source V1 sets the voltage across current source I1 you can ignore I1 altogether.

You can readily find Rth by setting all the current sources as open circuit and the voltage source as a short circuit and then evaluating the resistance seen 'looking into' terminals A-B.

Then by superposition calculate the contribution to VAB from both the voltage source V1 and the current source I2 with each considered in isolation.

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4. ### damm1371 Thread Starter New Member

Feb 22, 2012
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Ok thanks very helpful. Using lakeville I was getting rid of I1 as well. Anyways, so I have attempted construction of what you have said. Superposition about V1 and I1. I end up with two thev equiv circuits, which is great. How would I go about combining them into one thev equiv circuit? Just add the voltages and the two resistors. I have pictured my work. Please point out any and all mistakes. Thanks.

the first pic is superpostion about V1 and the second is for I2.

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5. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Not sure what you mean by lakeville.

You seem to be over-complicating the solution. It's not really that difficult a problem.

I've already stated how to find Rth.

Superposition approach to find the Thevenin voltage ...

For the V1 voltage source:

V1, R1 & R2 form a voltage divider with the voltage drop across R2 being the V1 part contribution to Vab. Call this Vab1

For the I2 source:

I2 flowing into the parallel combination of R1 & R2 will create the I2 part contribution to Vab. Call this Vab2.

Add the two values together to give the actual Thevenin equivalent at terminals A-B

i.e. Vab=Vab1+Vab2

Keep in mind that the polarities of Vab1 & Vab2 may be different.

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