1
\$\begingroup\$

I have a certain circuit in mind and I want to find its Thevenin equivalent resistor:

enter image description here

I see that there is an dependent voltage source so I add a voltage source of 1V between a and b and make the other voltage source equal to 0 (the independent 9v). when I use nodal analysis on the new circuit, I am not sure how to calculate the current i3. I thought about using Kirchhoff voltage law on the middle loop but for now I am confused. If I have three voltage sources, do I just take a wild guess about which way the current goes when I apply KVL on the middle loop? because if they didn't note the direction of the current on the Vx resistor, I would have said that Vx+2i3-2Vx+1=0, or would that be wrong because I assumed i3's direction to go to the right?

tl;dr

  • how would I do a KVL equation on the middle loop
  • is there an easier way to solve it using nodal analysis (specifically, to find i3)?

Sorry if the question seems a bit confused, because I'm confused.

\$\endgroup\$
  • 2
    \$\begingroup\$ Well it seems to me that it is a reasonable question. He discussed what he tried to do and what is still confusing to him. \$\endgroup\$ – WalyKu Jun 1 '14 at 12:34
0
\$\begingroup\$

This is how i think i would solve the problem:

I don't see it necessary to add a 1V excitation source since there already a 9v in the circuit. The trick when analyzing circuit with dependent sources is to avoid the Short circuiting you would normally do when having a circuit with independent sources alone:

_Things to note that may help

Focus on node A:

define a current from the 9v source direction to node A and also define a current from 6ohm to A

_not Vx =Va

9 - Vx = I1

Vx/6 = I2

and i reckon Vab = 2Vx..

Hope this helps you!

\$\endgroup\$
2
\$\begingroup\$

If your objective is to find the Thevenin resistance, then I'm not sure why you're approaching it this way.

If you can find the Thevenin resistance \$R'_{th}\$ of the circuit to the left of the \$1\Omega\$ resistor, the total equivalent resistance is just

$$R_{th} = R'_{th}|| 1\Omega $$

And, I recommend using a current test source which will make finding \$R'_{th}\$ almost trivial.

\$\endgroup\$
0
\$\begingroup\$

For this particular question, i think the source conversion technique would be the best.

You convert the independent voltage source and 3 ohm resistor to a current source of 3A with 3 ohm res in parallel.

Next, this 3 ohm res is in parallel with 6 ohm res. Reduce it to a single res of 2 ohm. So now you have a current source of 3A in parallel with 2 ohm res.

At this point again use source conversion. You get a voltage source of 6V in series with a 2 ohm res, combined in series with 2 ohm res and dependent source. The remainder of the network can be easily solved.

A major benefit of this technique is that you have reduced one loop completely.

Hope i am correct.

\$\endgroup\$

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.