I asked this question before in relation to calculating power before but it seems I might have not been clear enough. I'll draw the original circuit and then the circuits for calculating Thevenin and Norton:



simulate this circuit – Schematic created using CircuitLab

Here's how I would go about writing the circuit to find I short circuit or I Norton (EDIT: just noticed, is my arrow for Isc backward given V0? BTW, this isn't my main trouble for this problem):


simulate this circuit

Here's what I would do for V open circuit or V Thevenin:


simulate this circuit

I could calculate the open circuit voltage and the short circuit current easy if I had more independent sources but the dependent sources make these calculations harder.

I'm not sure how or where to use source transformation, mesh or nodal analysis, etc on this circuit. Each time I try one I end up in a spot that I can't figure out how to use it or I get equations that don't give me values.

For example, If I try to find I short circuit by using mesh analysis, I do not have any resistors that will give me a current for I short circuit. I can't solve the system of equations for I short circuit after getting the other meshes. I could try and find the other currents and then apply KCL at one of the nodes that I short is going into or out of, but I can't ever find the correct equations for the other currents in the other meshes.

For V open circuit, I just don't know where to start with that one.

This circuit seems impossible to solve and I have been looking at it for hours now. If someone could please give me some guidance/solution you will save my sanity. Thank you.

  • \$\begingroup\$ You can apply the superposition principle to solve this, as detailed in electronics.stackexchange.com/questions/107435/…. Some people might tell you that you can't use superposition to solve for dependent sources. This is incorrect. For more information, read eprints.soton.ac.uk/271202/1/superposition.pdf \$\endgroup\$
    – BeB00
    Oct 6, 2018 at 7:29
  • \$\begingroup\$ Where is I_zero? Why don't you give us the original question instead of a re-drawn, re-worded version? \$\endgroup\$
    – Chu
    Oct 6, 2018 at 16:39
  • \$\begingroup\$ There @Chu, I added where I_zero is. Thank you for taking a look at it. \$\endgroup\$
    – JustHeavy
    Oct 6, 2018 at 19:09

2 Answers 2


First, Try redrawing the circuit, and label all the nodes, giving letters to the unknown ones and values to the known ones. Choose a 0V node:


simulate this circuit – Schematic created using CircuitLab

From this, you can tell that a is always 5V.

You can then set up a system of equations for each case. First, for the open circuit case, we can say that:

  • Vo=Va-Vb=5-Vb
  • Io=0
  • V2=2*0=0
  • Vc=5+0=5V
  • Ir1 = Vc/1=5/1=5A

We now need to figure out the current I1. we can say directly that:

  • Vb = 10 - Ir3*2

The current through R3 is the sum of the current through R2 and the current through I1. Both of these currents directly depend on Vb.

  • Ir3 = Ir2 + I1
  • I1 = 2*(Vo) = 2*(Va-Vb)=2*(5-Vb)
  • Ir2 = Vb/4

We now can solve for Vb:

  • Vb = 10 - Ir3*2 = 10 - ((Vb/4) + 2*(5-Vb))*2
  • Vb = 10 - Vb/2 - (20 - 4Vb) = 3.5Vb - 10
  • 0 = 2.5Vb - 10
  • Vb = 4V

Now lets check that this works. If Vb = 4V, Vo = 1V, so I1 = 2A, so Ir3 = 2 + 4/4 = 3A. 3 amps through a 2 ohm resistor (R3) drops 6V, and 10-6 = 4V, meaning that we were correct.

You can do a similar procedure for the short circuit current.

It's probably faster to use superposition, but it's a bit more mathsy and slightly less obvious.


Taking the bottom rail as \$\small 0 \:V\$ reference, nodal analysis at node b gives:

$$\small \frac{V_b-10}{2}+\frac{V_b}{4}+2(V_a-V_b)=0 $$

and \$\small V_a=5 \:V\$, therefore:

$$\small V_b=4 \:V $$

hence: $$\small V_{TH} =V_a-V_b=1 \:V$$

\$\small R_{TH}\$ is easily found by turning all sources to zero, thus:

$$\small R_{TH} = 2\:\Omega \:||\:4\Omega=\frac{4}{3}\:\Omega$$

  • \$\begingroup\$ Are you sure it is alright to turn off the dependent sources? I thought you had to leave those alone, but since it is just resistance then I guess it might not matter. \$\endgroup\$
    – JustHeavy
    Oct 7, 2018 at 6:01

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