# Thévenin equivalent with multiple sources circuit

I am asked to find out the value of the voltage between points A and B, aka Uab, and I am supposed to use the Thévenin equivalent.

I found 12.4 Ω for the Thévenin resistance, but for that I replaced every source by the equivalent, and that short-circuited the 4 Ω resistance on the left, because of the replaced 20 V voltage source.

When it comes to the Thévenin voltage, I don't know what to do; the two resistances above are bothering me. Do I need to use the Kennely theorem?

• Analyze what is happening at the node in the middle, and ask your self you can simplify the circuit any. Sep 23, 2019 at 21:00
• Hi, thanks but am I allowed to do the Norton and Thevenin equivalents on any source? I mean converting the only current source by a Thevenin one would simplify a lot, right? Thanks in advance! Sep 23, 2019 at 21:05
• Yes - find the Thevenin equivalent source resistance of the current source by setting all the other sources to zero. It should be obvious what I'm hinting at then. Sep 23, 2019 at 21:12
• That's a cool circuit. Look, when load resistance (R_AB) is removed, you know exactly voltage in the middle node; and you know exactly the current in another 10 ohm resistor. Now it's very easy to find Uab using KVL or Ohm's law. Feb 20, 2021 at 14:25

I think your first step should be to simplify this circuit with some very trivial replacements of combinations of resistors, and removal of redundant resistors:

• The top 1Ω and 3Ω resistors are in series, and can be replaced by a single resistance.

• The single resistance from the previous step is in parallel with the horizontal 12Ω resistor, and these two can also be replaced by a single resistance.

• The vertical 12Ω resistor in parallel with the 20V source does not affect the potentials anywhere in the circuit, since those are fixed by the 20V source. Consequently, it has no effect on any currents anywhere, and can be removed without altering the circuit's behaviour in any way.

What you are left with is:

simulate this circuit – Schematic created using CircuitLab

It's not obvious, but for the same reason we were able to remove the 12Ω resistance across V2, we are also able to disregard V1, R1 and R2! They cannot possibly influence the potentials at either end of voltage source V2. You could still perform a nodal analysis at this stage, to prove this to yourself, but perhaps you can see that any current emerging from the right end of R1 must be the same, by KCL, as current entering the bottom of V1. That equality means this current effectively cancels itself out from KCL equations for the right-hand loop.

You also have current source I1 in parallel with R4, which can be replaced by a Thevenin equivalent:

simulate this circuit

No Δ-Y transformations necessary (I assume that's what you mean by Kennely theorem).

Finally I've got something of interesting. I was told to use the Thevenin theorem between points C and D instead. C is located on the middle node and D under the 10 ohms resistance. I've summed up everything on that picture, the Rth is 0 ohm because the the circuited voltage source gives a wire with 0 ohms along it, and it is in parallel with the two other resistance, right? And for Eth, I knew the voltage between two branches is supposed to be the same, thats why Eth equals 20 volts. Thanks you Phil!

A easy way to calculate Thevenin equivalent resistance is:

1. Connect 1A current source between A and B.
2. Ground terminal B.
3. Open circuit current sources and short circuit voltage sources.
4. Keep dependent sources as as they are.
5. Then use nodal analysis to calculate $$\V_A\$$ which would be Thevenin resistance between A and B.

To calculate Thevenin voltage and resistance with LTspice, follow these steps:

1. Use LT spice to do a DC sweep by connecting a 1A DC source in downward direction towards B with the terminal B connected to ground.It is important to connect the current source this way.
2. Do a DC sweep for this current source between -1A to 10A with 0.1 increment.
3. When you look at the DC sweep output, the voltage on left you obtain for 0A would be Thevenin voltage $$\V_{th}\$$.
4. Now, see the current at zero voltage on left. This would be the short circuit current $$\I_{sc}\$$.
5. If you want to calculate thevenin resistance, it is $$\\frac{V_{th}}{I_{sc}}\$$