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

Question

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?

Answer 1

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.)

Answer 2

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

Answer 3

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!