thevenin's theorem

Thread Starter

exidez

Joined Aug 22, 2008
26
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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Ratch

Joined Mar 20, 2007
1,070
exidez,

Can someone push me in the right direction?
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
 
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Thread Starter

exidez

Joined Aug 22, 2008
26
exidez,



The neatest way to do this problem, avoiding mesh, branch, and node analysis, is by superposition and potientiometer 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
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
 

Thread Starter

exidez

Joined Aug 22, 2008
26
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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hgmjr

Joined Jan 28, 2005
9,029
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
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
 
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