# Thevenin equivelant circuit

#### Erica

Joined May 18, 2011
15
See attached steps I used for obtaining an open-circuit voltage Vab. The voltage I got is zero.

Did I make any mistakes on this? #### Attachments

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#### The Electrician

Joined Oct 9, 2007
2,887
No. Have you got a calculator that can do complex arithmetic?

#### WBahn

Joined Mar 31, 2012
26,398
See attached steps I used for obtaining an open-circuit voltage Vab. The voltage I got is zero.

Did I make any mistakes on this? One of the beautiful things about circuit analysis is that there is almost always a way to check your work.

Once you got this result and question whether it could possibly be right (which is a reasonable reaction, so good for you), you can set it up a bit differently to directly answer that question. If the voltages at 'a' and 'b' are the same, that means that the two voltage dividers have to have the same ratio, so the question becomes:

$$\frac{j10 \Omega}{5 \Omega+j10 \Omega}\;=^?\;\frac{6 \Omega}{6 \Omega-j3 \Omega}$$

Notice how I track the units above. Get in the habit of doing that. I originally kept the units to the very end, but decided that it made things a bit clear if I let them cancel out. Note that they DID cancel, I didn't just decide to drop them.

Simplifying the fractions above

$$\frac{j2}{1+j2}\;=^?\;\frac{2}{2-j1}$$

Dividing both sides by 2

$$\frac{j1}{1+j2}\;=^?\;\frac{1}{2-j1}$$

Multiply the RHS by j/j

$$\frac{j1}{1+j2}\;=^?\;\frac{j1}{1+j2}$$

So, yes, the thevenin voltage will be identically zero.

This is actually a nice problem. I would not have expected it was possible from a casual glance. Only after seeing the result and looking at the problem with that in mind does it become obvious that this is, in fact, the case. There are a couple of ways of seeing this. If you were asked on a quiz to describe, in words, why this is possible, what would you say?

#### Erica

Joined May 18, 2011
15
Thanks for your help. This was a problem in a practice exam. The problem also asked to determine a load (Zload) to be connected between terminals a and b to get the maximum power output in Zload.

There would be no power output to Zload if the terminal voltage is zero. It appears this problem is questionable.

#### WBahn

Joined Mar 31, 2012
26,398
While you make a very valid point (and one that I recommend you point out in your write-up), the Zload that produces maximum power transfer to the load is dependent only on the effective impedance of the source and not on the thevenin voltage. So, for the purposes of finding that value of Zload, assume that the thevenin voltage is nonzero.