Thevenin equivalent voltage

Thread Starter

chadchoud

Joined Dec 18, 2008
11
Hello

In this question, I have the voltage between A and B and I need to find Vin.
How do I do that?

I tried setting the ground at B, and I called V1 the node above A and V2 the node below A.

I did a node voltage:
V1/5 + (V1-VAB)/4 = 0
and same for V2

then I tried Vin = V1 - V2 but I did not reach the desired value.
Can somebody help me?
 

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hgmjr

Joined Jan 28, 2005
9,027
It would be helpful for you to show a bit more detail of your work in arriving at the answer you chose.


hgmjr
 

Thread Starter

chadchoud

Joined Dec 18, 2008
11
Sure :)

I attached an image showing where I chose V1 and V2. The ground is on B.

so:
V1/5 + (V1-VAB)/4 = 0
V1(1/5+1/4) = 10.46+6.04j
so V1 = 23.24+13.12j V

(V2 - VAB)/(-4j) + V2/(8.66j) = 0
V2(1/(-4j) + 1/(8.66j)) = -6.04+10.46j
so V2 = 77.73+44.88j

Now I want Vin which is equal to V1-V2 (right?) = -54.5-31.68j
= 63.03 < -149.83

Which is different from the correct answer.
 

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The Electrician

Joined Oct 9, 2007
2,970
V1/5 + (V1-VAB)/4 = 0
Your very first equation is wrong. Those two currents don't add up to zero. What about the current in the + line of the Vin source?

The rest of your calculations suffer from the same problem.

I would solve the problem by letting the - input be the reference and calculate two voltage dividers to nodes A and B, take the difference, and go from there.
 

Thread Starter

chadchoud

Joined Dec 18, 2008
11
Amazing! After a lot of trials, I reached the correct answer :)

Vin-VA = Vin*4 / (4-4j) ..... (1)
Vin-VB = Vin*5 / (5+8.66j) ..... (2)

Then do (2) - (1)
Isn't that what you meant?
 
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