0
\$\begingroup\$

enter image description here

For this problem, I am trying to find the Thevenin equivalent of the circuit. Is there any way to do repeated Thevenin to Norton transformations to solve this? Or is the only way to solve the open circuit voltage and closed circuit current.

If so, when trying to find the closed circuit current, does the wire from the short from A to B short out both the dependent current source and the resistor?

\$\endgroup\$
1
\$\begingroup\$

You can replace the 9*Ia dependent supply with a circuit that produces the same current, that is 9 repeated branches in parallel consisting of a 20 V supply in series with a 50 k resistor and then simplify.

But that is a long, improductive way to do it.

A better way to find the equivalent resistance is using the method of test supplies between A and B points. Remember to turn off the independent supplies. This is more useful as sometimes finding the circuit can be difficult.

If you want to find the short circuit current between A and B, it must be the current going through the short branch only, so keep the dependent supply and the 5 k resistor in your analysis.

\$\endgroup\$
  • \$\begingroup\$ Thanks, could you elaborate a little more on how to find the short circuit current? I have not learned about test supplies yet so I will look that up, but wouldn't connecting a wire between A and B short out the resistor and the current source? Or what happens when you connect the wire in parallel with the current source? Thanks \$\endgroup\$ – David M Mar 26 '16 at 18:59
  • \$\begingroup\$ The test supply is a very known method, you can find a fully detailed explanation in a circuit analysis textbook. One thing is solving a circuit, in that case the short simplifies all things in parallel; a different thing is finding the current in a specific branch, if you simplify that branch with another, then you loose the branch you were looking at in the first place. \$\endgroup\$ – berto Mar 26 '16 at 19:03
  • \$\begingroup\$ My thinking was that connecting a short between AB would mean all the current would flow through the short, bypassing the other two branches, and so I thought the short circuit current would just be the 20V supply with the 50k resistor in series. Is this a wrong way of thinking about it? \$\endgroup\$ – David M Mar 26 '16 at 19:09
  • \$\begingroup\$ That would be correct if there were only resistors in parallel, in that case all current goes by the short as you say, and no current goes by the resistors because the voltage over them is zero. BUT in the case of the dependent current supply it is a different situation, as a current goes through it no matter the voltage it has, "stealing" current from the short circuit . If you want to look it in that way, you can say that is ok to eliminate the resistor but not the dependent supplies. \$\endgroup\$ – berto Mar 26 '16 at 19:22
  • \$\begingroup\$ That makes sense. Thank you very much for your help \$\endgroup\$ – David M Mar 26 '16 at 19:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.