We were given this question in our intro to EE course:

Find the Thevenin equivalent network between points a and b for the circuit shown in the image below:

AC Circuit with two dependent sources

For the short-circuit current, I redrew the circuit as follows:

Simplified Short-circuit current circuit

I'm a little confused about where to proceed from here, though. My prof said a cap across 0V potential has zero resistance and should be treated as a short-circuit. However, this creates a wire loop surrounding the dependent current source. I wonder if, since 0V = the potential across the cap, if the cap should be treated as an open circuit instead (since the cap's voltage is at the potential on either side of it, and this would remove the wire loop)?

Any ideas on how to proceed? Is there a mistake?

Bonus question: what would be the best way to approach finding the Open circuit voltage (Vx, in this case)? I've tried but get an ugly 3x3 complex matrix to solve.



2 Answers 2


Welcome to EE.SE. I actually find this a bit of a fascinating question. You're completely right in that this question is a bit tricky.

My prof said a cap across 0V potential has zero resistance and should be treated as a short-circuit.

While it shouldn't make a difference in most cases, it does complicate matters for this circuit. It is probably better to leave the capacitor in there.

\$I_x\$ will choose the path of least impedance. This means it will be absorbed by the controlled voltage source. The same applies to the current generated by the controlled current source: it will follow the path of the least impedance, in this case through the short-circuit. This leads to the simple equation

$$I_{sc} = -I_x = -\frac{25V}{100\Omega}$$

In order to find the open-loop voltage, you can probably best solve it using a system of 2 equations. The first one expresses the current \$I_x\$, while the second one is the KCL equation in the output node (\$V_x\$).

$$\begin{align} I_x &= \frac{25V - 4V_x}{100\Omega} \\ \frac{4V_x - V_x}{-j25\Omega} &= I_x + \frac{V_x}{10\Omega} \end{align}$$

\$V_x\$ in this case is the open-circuit voltage (the voltage across the \$10\Omega\$ resistor).


Let's assume you have to find Thevenin's equivalent circuit and not Norton's. It'd be nice to find difference between them and I'm leaving that on you, however you need find Isc if you have to find Nortan's eq ckt and not in case you want Thevenin's eq one.

In either case the procedure to find Rth (or Rn) and its value remains same. Now I have legit no idea why your professor wants you to treat capacitor as short circuit, it acts as one if frequency tends to infinity, but here everything looks DC to me and it has to act open. Rth is resistance as seen from AB terminals, and when dependent sources are there, you simply apply some arbitrary source at AB and calculate how much current network is drawing and then ratio of applied source voltage to drawn current is your Rth. Of course, you have to set independent sources to zero first. Other method is to calculate quotient of Voc to Isc.

Now to find Rth, apply Vab in such a way that your calculations will become easy, here lets say Vab = 25V, which is same as Vx hence Vx = 25V, this will give you 4x25 = 100V for dependent source. Remember to set independent sources to zero! Now, as capacitor acts as an open ckt, you have Ix = -(100/100) = -1A. Minus cause it's opposite to given direction. Apply KCL at node A and you'll have 1.5A coming out from 25V source you connected and hence Rth is 25/1.5 Ohm.

How will you find Voc now? Simple, capacitor acts as open, so you have two loops, two unknowns Vx and Ix with Voc= Vx and 2x2 matrix to solve! Cheers and welcome to ee.se!

  • \$\begingroup\$ Wait a minute, doesn't the voltage source being given in a phasor form suggest it being an AC source? \$\endgroup\$ Jul 24, 2018 at 13:54
  • \$\begingroup\$ Well then you need frequency to solve this further...since this is introductory course and as OP seems to be new to this topic, I strongly feel that this is DC one and maybe OP mistakenly wrote voltage in phasor form cuz he couldn't find the right symbol... (i said maybe..).. had this been power system problem then I'd have taken 50Hz,but most of othet guys would take 60Hz...see? We need frequency to solve it further if its AC source \$\endgroup\$
    – Deep
    Jul 24, 2018 at 14:53
  • \$\begingroup\$ This is an AC circuit. Our prof just prefers the DC notation, since the only real difference is the sine curve written inside the sources. I also assume that, because we are given the capacitance in phasor form that we are to find the Thevenin equivalent circuit (we have NOT learned Norton at all) in phasor form as well (since we are also not given frequency). \$\endgroup\$
    – Nalorin
    Jul 24, 2018 at 18:12
  • \$\begingroup\$ I do - but I didn't think/wasn't sure if supermeshes could be used around dependent current sources? Are there any additional considerations to be aware of when doing so? \$\endgroup\$
    – Nalorin
    Jul 24, 2018 at 19:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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