# Simplifying circuit for Thevenin's theorem

I am supposed to find a Thevenin's equivalent circuit for this one:

I am struggling with simplifying it to calculate Rth. When I apply short circuit to voltage source I achieve a junction connected to R1, R3, R7+R6 and R4. How to handle it?

The first thing to do when you face a schematic that seems complicated is to redraw it. It's a skill, so you do have to work at developing it. But it's easy to learn and get better, with practice. So you should just start and practice.

In this case, let's try something like this:

simulate this circuit – Schematic created using CircuitLab

This is the same circuit. (You can check it out.) But I've removed the voltage source as a separate schematic item and replaced it with the $$\+5\:\text{V}\$$ rail symbols. This eliminates "busing power around with wires" and helps avoid some confusion over wire connections that aren't important for analysis purposes.

I think you can more easily see, now, that this circuit can be analyzed as two independent sub-circuits. Thevenize the left side and then the right side. Once you've done that, you will have two Thevenin voltage sources with their two Thevenin series resistances.

The difference between the two Thevenin voltages will be what appears across A and B, when open circuit. The current that flows via the wire connecting A and B, when shorted, will be the short-circuit current. This tells you all you need to know.

• Thank you for your answer. – feral witchchild May 18 '20 at 17:36

How to handle it?

There's nothing special about that situation.

Start looking for parallel and series combinations.

Or use KVL or KCL if the resistors are arranged in ways that can't be handled by parallel and series combinations.