# How can I find the Thevenin's Equivalent on this problem using Superposition Theorem?

I am reading Foundations of Analog and Digital Electronic Circuits from Anant Agarwal.

In chapter 3 there is the exercise 3.11 at the end of the chapter that states :

Find the Norton equivalent at the terminals marked $$\xx\$$ in the following circuit.

Regarding $$\R_{th}\$$ 2 ohm resistor is in parallel with the 1 ohm resistor on the left side (left x) of the circuit and on the right side (right x) 4 ohm resistor is in parallel with the 2 ohm resistor.Those two parallels are in series.

$$\R_{th}= 2||1 + 4||2 = 2 \Omega\$$

But for $$\I_{sc}\$$ using superposition theorem I cannot understand how he finds I_{sc} = 1 A.

When the current source is turned off I have :

And when the voltage source is turned off I have :

But how he finds $$\I_{N} =1 A\$$.Any help?

It's easier (my view) to simply add a $$\0\:\text{V}\$$ source between the two x's and perform just a couple of simple steps:

Step one was to assign a ground. This gets me from top-left to top-right. The second step was to dump the $$\5\:\text{V}\$$ supply and replace it with a simple statement about the voltages at those wire points and to Theveninze the Nortion of $$\1\:\text{A}\$$ and $$\1\:\Omega\$$ and simply alter the voltage, as indicated.

Once that is done, the question of the current is quite simple. Convert the left side pair of voltages and resistances into a single Thevenin equivalent and then to the same thing for the right side pair of voltages and resistances. (I'll leave that to you.)

Then all you have is the difference between two voltages (just subtract one from the other, obviously) and a single resistance that is the sum of the two Thevenin resistances just calculated.

The current just falls out.

Also, since you know the voltage difference now and the current you can work out the Norton resistance, directly.

Added: Any two series resistors with voltages at either end of the pair (here that $$\V_1\$$ is at an end of $$\R_1\$$ and $$\V_2\$$ is at an end of $$\R_2\$$ and the two resistors are joined at their other ends) can be converted into a Thevenin equivalent of a single voltage and a single resistor this way: $$\V_{\text{TH}}=\frac{V_1\,\cdot\, R_2+V_2\,\cdot\, R_1}{R_1+R_2}\$$ and $$\R_{\text{TH}}=\frac{R_1\,\cdot\, R_2}{R_1+R_2}\$$. That is handing you a formula. (But you should be already able to create those two formulas on your own, with skills already at hand. Otherwise, you are in too deep right now.)

• It will help if you add the mathematics in the last step (subtracting the voltages).Thank you for your effort and your help. Commented Jul 12, 2023 at 16:06
• @HomerJ Before I do that, it worries me a little that you cannot perform a subtraction of two values. If, instead, you mean that you do not yet know how to calculate the Thevenin voltage and resistance from a pair of resistors and a pair of voltages, then that also worries me. But somewhat less. Still, you should not be engaging these problems before already having been well-trained (of having mastered) how to produce a single Thevenin voltage and resistance from a pair of voltages and series resistors. I'll add a general equation in the hope that it helps. Commented Jul 12, 2023 at 16:11
• which resistors are R1 and R2 in your addition ? Thanks for the advice.I am only 2 months newbie on electrical circuits.I will read again some notes. Commented Jul 12, 2023 at 16:35
• @HomerJaySimpson Either the 1 & 2 Ohm resistors on the left side or else the 2 & 4 Ohm resistors on the right side. You must do two conversions. Commented Jul 12, 2023 at 17:04
• @periblepsis What software did you use to draw those circuits? Commented Jul 13, 2023 at 6:23

Agree with Santi Ospina: solve Norton current through shorted XX nodes. It helps to re-draw the network, eliminating redundant parts. Identifying those parts that don't matter is a skill where you become more confident with practice.
Inspect the second circuit with care - you can solve In2 by inspection, rather than going through a lot of arithmetic.

simulate this circuit – Schematic created using CircuitLab

Use superposition to combine In1 with In2 to solve for short-circuit current for the Norton equivalent

• Regarding the first left circuit where the voltage source is short circuited I can find from current division on R2,$I_{sc1}=(1/3)*3=1 A$.But regarding the second one with the current source open circuited I don’t know how I can find 0 Amp.(It will be helpful for me if I can avoid the node or mesh analysis in this step,I find it more easier to use current-voltage division). Commented Jul 12, 2023 at 13:53
• 2nd circuit: you can further break it down by doing Thevenin equivalent of left arm (R3:1 ohm, R4:2 ohm) and then Thevenin equivalent of right arm (R6:2 ohm, R5:4 ohm). But then you discover that these two Thevenin voltages are the same value. That makes it very easy to solve for In2 Commented Jul 12, 2023 at 19:02
• @glen_gleek I think there's a mistake in your first circuit. When you shor circuit the xx nodes you would the 1Ω and 2Ω resitors in pararel as well as the 4 Ω and 2Ω leading to the circuit that you draw but with different resitances, that is, 2/3Ω and 4/3Ω. Commented Jul 13, 2023 at 6:26
• @SantiOspina Thank you. My slip only reinforces that these exercises should be practiced often (haven't done these for a long time). Commented Jul 13, 2023 at 13:05

To find the Norton equivalent circuit you should shorcircuit the xx nodes and then solve the circuit by superposition to find the norton equivalent current I_N.

First step = find equivalent Thevenin voltage between nodes xx by means of superposition

Second step = find equivalent norton current across the shorted path between nodes xx by means of superposition

Last step: Subsitute your circuit by the an equivalent current source with value equal to to value that you found with superposition and with a paralel resistance with value equal to V_th/I_N

Hope those hints helps!