# Why do we not include the current source when finding the Thévenin equivalent impedance?

The problem is from Basic Engineering Circuit Analysis, by J. David Irwin, Robert M. Nelms, chapter 9.3. When I tried to solve this problem, I tried to find the Thévenin equivalent circuit by finding the open circuit voltage, using this diagram: and then I tried to find the short circuit current using this diagram: and then calculating Zth using the formula Zth = Voc / Isc

However, this gave an incorrect answer. Apparently, the correct way to find the Thévenin equivalent is by cutting off both the load, and the current source to the right, like so: and to find the short circuit current, and then using the formula Zth = Voc / Isc to find the correct Zl which draws maximum power.

I don't understand why finding the Thévenin equivalent with the current source attached gives an incorrect answer. Any help is very much appreciated!

• According this site you choose the correct approach. Could you write out the equations, maybe there is an error in? – Huisman Jan 31 at 12:08
• The answer to the banner question is ... because an ideal current source has infinite impedance, hence looks like an open circuit. But if you're calculating the Thevenin impedance from Voc and Isc, as the question in the text implies, the impedance of the current source does not arise. – Chu Jan 31 at 15:49
• FYI, I am using this circuit as a learning exercise for CircuitLab. electronics.stackexchange.com/questions/478919/… – Mattman944 Feb 1 at 16:18

I had the same question when I first came across this and here's how I think of it. The whole point of Thevenin is to reduce a complicated circuit into a much simpler circuit consisting if a voltage (Vth) and a resistance (Rth). If you define Vth to be the O.C. voltage at the output of your complicated circuit then consider how this voltage might be formed by current through your complicated circuit flowing across some (as yet) unknown resistance.

To find this unknown resistance you might visualize the Vth generating current that flows back into your complicated circuit and think of all the paths it would take.

Now think of this Thevenin current attempting to make its way along the branch with the current source. An ideal current source will flow its current in any attached load and its model has an infinite resistance in parallel (in a real current source, this resistance is not infinite and will draw a varying current as the load resistance varies). You're only interested in the resistance of this path, so you switch off the current source but it's (infinite) parallel resistance remains. Since the resistance is infinite, no current will flow - it's regarded as an open circuit.

A voltage source is different. It's modelled as having a theoretically tiny (i.e., shorted) resistance in series. Current flowing through this voltage source encounters this tiny (i.e., non existant) resistance. And again, when you turn off the voltage source you're left with this infinitely tiny (i.e., short) resistance.

In reality, current flows around the circuit and out through your open circuit Vth, but it encounters the same resistances as if you had flowed the current into the circuit with the voltage and current sources off. It's you that defines Rth by following the established rules. But, just remember that what's happening is you're finding your back box resistance (Rth) that's establishing some voltage at the output (Vth). Hope that helps!

• "Now think of this Thevenin current trying to force its way against some internal current source." It seems you consider this current source outside the Thévenin equivalent circuit, but it is included. So, the Thévenin circuit will not try to push its current throug this current source. – Huisman Jan 31 at 10:49
• I probably could have worded that better. It's a device to show that traversing that path (the one with the current source) discovers an infinite resistance since the current source has an infinite resistance to change. – Buck8pe Jan 31 at 11:01
• Re-worded things to make it clearer (and more accurate - hopefully!) – Buck8pe Jan 31 at 11:25
• In the Thevenin equivalent circuit, there is no external current source (3 ∠10° A) anymore. So, there is no Thevenin current making its way along the branch with the current source. because that current source is incorporated within the Thévenin equivalent. – Huisman Jan 31 at 12:05
• @Huisman I understand that. The 3 ∠10° A source is "seen" by Voc (i.e., Vth). In other words, it may as well be inside the black box. Then what I said in my answer describes why the current source should be opened. – Buck8pe Jan 31 at 13:41

I don’t think you can remove the current source to get the Thevenin equivalent circuit. Here’s a simpler example. Voc = V1 + I1xR1 = 5 + 1x2 = 7 V

Ish = V1/R1 + I1 = 5/2 + 1 = 3.5 A

Rth = Voc/Ish = 2 Ω

The Thevenin equivalent circuit with Voc and Rth gets us this result, which is correct,

V_OUT = Voc/(Rth+RL) x RL = 7/(2+12) x 12 = 6 V

I_RL = V_OUT/RL = 6 / 12 = 0.5 A

If the current source, I1, is removed when figuring out the Thevenin equivalent circuit, we will get an obviously wrong answer, I_RL = V1/(R1+RL) = 5/14 A

So, you first method should be the right way to go.