Here's a circuit for which I was asked to find the Thevenin Resistance:enter image description here

My professor said that there's a shorter way to do this. Apparently one could short-circuit voltage sources(in this case the 72 V source) and replace current sources by an open circuits. After that one could find the resistance across the said two points and this would be our Thevenin Resistance. I wonder why this trick works in the first place. I've been trying to deduce why, but keep making the circuit more complex.

  • 1
    \$\begingroup\$ Which trick don't you understand? The short-circuit part, or the rearranging the circuit part? \$\endgroup\$
    – gbarry
    Commented Dec 22, 2021 at 4:35
  • \$\begingroup\$ @gbarry The short-circuit part. How does that help? Won't short-circuiting change things? \$\endgroup\$ Commented Dec 22, 2021 at 4:43
  • 4
    \$\begingroup\$ @chocolateMousse I guess the better way to think about why an ideal voltage source is like a short circuit is that it has no internal resistance to it. It really is a short circuit. Yes, it jacks up the voltage difference between two nodes. But it does this perfectly (without any impedance to it.) Another way to see it is to imagine what happens if you try to impose a "signal" across a voltage source. It would immediately be "shorted out" and there would be no observable signal. All ideal voltage sources are also ideal short circuits. Goes hand in hand. \$\endgroup\$
    – jonk
    Commented Dec 22, 2021 at 5:02
  • 1
    \$\begingroup\$ You have a good answer for the why. About not seeing how to simplify the circuit: if you replace the 72V source with a wire, the 12Ω resistance goes directly between A and B, and the 5Ω in parallel to the 20Ω (remember: parallel -> the are connected to the same two nodes). Now the latter parallel is in series with the 8Ω (same current flowing -> series), and: 5//20=4, 8+4=12, and 12//12=6. \$\endgroup\$
    – Rmano
    Commented Dec 23, 2021 at 11:11

4 Answers 4


This is a classic example of superposition as applied to linear systems - if you consider your circuit as a function mapping excitations (the voltages and currents of independent voltage and current sources, respectively) to node voltages/currents (i.e. the solved circuit), it's linear if the circuit consists only of linear elements (linear dependent sources, R/L/C).

As a result, you can compute the superposition of two different sets of excitations - one representing the excitations of your original circuit (i.e. the 72 V source in your example, but in general including all independent voltage and independent current sources), and one representing the external excitation connected to the port.

In order to compute the response of the circuit to the external excitation, you must first null all of the internal sources in your circuit to leave only the external excitation.

The primary insight here is that an independent voltage source becomes a zero-volt voltage source. A zero-volt source is nothing more than a short between the two nodes - the two nodes necessarily have the same voltage because the voltage between them is zero, and there's no restriction or mathematical statement regarding the current flowing in that branch.

Likewise, you can null an independent current source by setting its current to zero. A branch with no current across it is simply a branch that doesn't exist and was open-circuited, and there's no restriction on the voltage between the two nodes.

Note that dependent current sources stay the same - they have linear effects in a linear circuit, but they themselves are not excitations.


You're not really short-circuiting the voltage sources - the voltage sources are already presenting a zero impedance to the circuit. You're just setting them to zero voltage so that the output voltage doesn't cloud your thinking.

The same thing with the current sources. They already present an infinite impedance to the circuit. Just set their output to zero current so that it doesn't get in the way of the resistance calculation.


Consider this: -

  • Would you say that the Thevenin resistance looking into the A and B terminals is any different if the voltage source was 36 volts or 1,000,000 volts?
  • Would you say that the Thevenin resistance looking into the A and B terminals is any different if the voltage source was 72 volts RMS AC?

If you can answer "no" and "no" then, it's the tiniest step to take to imagine that at any point on the AC voltage waveform, the Thevenin resistance is constant. And the last step is to consider what the Thevenin resistance is as the waveform passes through 0 volts and, of course, 0 volts is exactly the same as a dead short circuit.


The short circuit is because an ideal voltage source has zero output impedance.

Thus if you only look at impedances, and not voltages, the impedance of an ideal voltage source can be replaced by the impedance which it has so it would be a zero ohm resistor, which can be directly drawn as a short circuit.


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.