# Thevenins Theorem

Can someone please help, I'm struggling with this circuit. I don't quite understand what I need to do, mainly with whats series and parallel. My tutor is not helpfull in the slightest! He' only given me some very very basic examples to look at and when I ask him any questions he mutters a few things and walks off! I've searched online to find a circuit thats similar but can't find anything.

The way i understand it is you remove the load, remove the voltage source, then calculate the resistance looking from the left

Any help to steer me in the right direction would be very much appreciated • If a resistor shares the same node with another resistor, then those two resistors are in parallel. For solving the Thevenin resistance, start from the right side of this circuit.
– user103380
Mar 17 '19 at 16:28
• Thanks KingDuken,So would that calculate as R7+R6 // R5+R4 // R3+R2 // R1+R8 Mar 17 '19 at 16:34
• @samflatt: Use brackets please: you equation is open for interpretation. As KingDuken says start at the right hand side so start with R9. (R9...) Mar 17 '19 at 16:38
• Is R9 supposed to be the load, or part of the circuit you want the equivalent of? Mar 17 '19 at 17:01
• There might be some significance to the fact that R9 has its nodes labelled A and B. Mar 17 '19 at 17:03

I'll try to point you in the right direction.

You say that R9 is the load, so you remove it. The voltage source (an independent source) becomes a short (0V). With those changes, your circuit reduces to: simulate this circuit – Schematic created using CircuitLab

With that, it should not be that difficult to find the equivalent Thevenin resistor looking into the circuit from A and B. Notice that in the previous circuit R1 and R2 are in parallel. You can solve that first and apply a similar procedure as needed. In the end, you will end up with an equivalent resistance for the combined R1, R2, R3, R4, R5, and R6 and that equivalent will need to be added with R7 and R8.

Hope it points you in the right direction.

Thanks @Big6. I think/hope I understand!  You say that R9 is the load, so you remove it

Yes. For finding the Thevenin's equivalent resistance, you have to short circuit independent voltage sources and open circuit independent current sources. If there were to be a dependent voltage/current source, you can evaluate the Thevenin's resistance by adding a test voltage/current (say 1V or 1A) across the terminals of the load.

Okay, moving on. You can observe that R1 and R2 are in parallel and you can simplify it. Let this be Req1. This is in series with R3. In a similar manner, you can simplify your circuit to find the equivalent resistance between A and B.

For determining the Thevenin's voltage, you may perform KCL or KVL.