# thevenin's theorem

Discussion in 'Homework Help' started by exidez, Aug 22, 2008.

1. ### exidez Thread Starter Member

Aug 22, 2008
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I am just getting all too confused on this question.
I can work out the total resistance Rth but cannot do anything further from there. I want the voltage across CD as it is the same for AB but cannot work out the current according to the terms. I have tried mesh and nodal analysis but all return a vale of zero!!
Can someone push me in the right direction?

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2. ### hgmjr Moderator

Jan 28, 2005
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See if this material on Thevenin's Theorem in the AAC ebook helps you out.

hgmjr

3. ### exidez Thread Starter Member

Aug 22, 2008
26
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i came across that exact page earlier but didn't help much with this question

Last edited: Aug 22, 2008
4. ### Ratch New Member

Mar 20, 2007
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exidez,

The neatest way to do this problem, avoiding mesh, branch, and node analysis, is by superposition and potentiometer methods. You are right, you can ignore the R across A to C unless it is so large that it forms a voltage divider with your voltmeter.

Knowing that a perfect voltage source has zero ohms, we can short two voltage sources while we calculate the contribution of the remaining voltage source.

The eastern voltage source (VS) makes a contribution of 0.5*V at node C, the middle VS gives 0.25*V, and the western VS adds a contribution of 0.125*V. Adding it all up makes the voltage at node C equal to 0.875*V.

You should worry that you cannot analyze the circuit by any of the other methods. It is more work, but it is doable.

Ratch

Last edited: Aug 22, 2008

Oct 9, 2007
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6. ### exidez Thread Starter Member

Aug 22, 2008
26
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Thanks a lot for the help, i actually understand it now. I wish i read you post before i continued with mesh analysis! I ended up using maple (mathematics software) to rearrange my 3 eqns to find the voltage at node C in relation to the the current in loop 3. It got really really messy and ended up getting V at node C to be 14RI3.
I3 = current around loop 3
things would have been so easier if i used the superposition technique. I didnt even think of that.

Thanks again

7. ### exidez Thread Starter Member

Aug 22, 2008
26
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i understand this concept but i am trying to prove that the middle voltage supplies 0.25*V at node C. I know its right though. For example the attached circuit is the simplified circuit with the outter voltage sources short circuited. I also made R = 50 ohms

i know that node C (the point between the 50 and 100 ohm resister) is 2/3 of the voltage of at the point above the source. but how can i prove mathematically that point C is 0.25*V ??
the resister above the source stumps me

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8. ### hgmjr Moderator

Jan 28, 2005
9,030
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Observe that the 100 ohm resistor located on the left side of the circuit is actually in parallel with the combination of the 50 ohm and the 100 ohm resistor on the right side of the circuit. That means you can combine the resistors into one using the parallel resistor formula. That simplification should lead you to a final solution.

hgmjr

9. ### exidez Thread Starter Member

Aug 22, 2008
26
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i didnt see that, but now i do. Thanks again!

Jan 28, 2005
9,030
214

hgmjr

11. ### exidez Thread Starter Member

Aug 22, 2008
26
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yep, i got the final answer
now its just the other questions!

12. ### hgmjr Moderator

Jan 28, 2005
9,030
214
So what value did you get for the voltage at CD?

hgmjr

Last edited: Aug 24, 2008