Finding Vth from Thevenin Equivalent of Circuit

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The discussion focuses on finding the Thevenin equivalent voltage (Vth) for a circuit after determining the Thevenin resistance (Rth). The user initially struggles with identifying the correct loops for their equations but learns that the two 1-ohm resistors are in parallel, simplifying the analysis. They are guided to understand that Vth can be calculated by summing potential changes along a path from point B to A, even without a closed loop. The conversation emphasizes the importance of recognizing parallel connections and the potential differences across components. Ultimately, the user gains clarity on how to approach finding Vth effectively.
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Homework Statement


Hello everyone!

I am attempting to find the thevenin equivalent for a circuit. I've got the Rth but am not struggling to find the Vth

Homework Equations



I have included a screen shot of the circuit

The Attempt at a Solution



I thought the best way to go about it was to divide the circuit up into three loops and make three equations.

I got
(1) 1 - 4I1 - 12(I1 - I3) =0
(2) I2 - 1(I2 - I3) = 0
(3) (I3 - I2) - 12(I3 -I1) - 5I3 = 0

If I solve these equations I get I1 = 1/16 A and I2 = 3/4 A so I am thinking that I can add up the two Is to give 13/16 A as the total current in the circuit. But I'm not sure how I can make a find loop which includes both AB and the voltage source to give me Vth.

Could anyone offer a hand?

Thanks very much.
 

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Vth is an open-circuit value, so terminals A-B are open. In that case I see only two loops.

However, there's a simpler way to find Vth here. What can you say about how the two 1 Ω resistors are connected?
 
gneill said:
Vth is an open-circuit value, so terminals A-B are open. In that case I see only two loops.

However, there's a simpler way to find Vth here. What can you say about how the two 1 Ω resistors are connected?

Oh, so are you saying that the path that goes through 3 resistors and AB isn't actually a loop as it isn't closed?

I'm pretty sure the two 1ohm resistors are in parallel, as they share the same nodes.
 
GBA13 said:
Oh, so are you saying that the path that goes through 3 resistors and AB isn't actually a loop as it isn't closed?
Right. A loop must be a closed path.
I'm pretty sure the two 1ohm resistors are in parallel, as they share the same nodes.
Correct. Does that suggest anything to you?
 
So that closes up the loops so there's only one loop so I can easily find the current flowing in the circuit using Kirchoff's voltage law. So the current I got was 1/16 Amps. So then I think I need to find a loop which passes through the voltage source and AB so I can rearrange to the find the voltage across AB, but if, like you said, there is only one loop and it doesn't include AB how can I do that?
Thanks for your help, you've been really useful so far.
 
You can sum potential changes between two points without there being a closed loop. So long as you know the potential change across each component on the path between two points in a circuit you can determine the potential difference between those points.

So, what's a path that will take you from B to A? Keep in mind that you've turned the two 1 Ω resistors into a single resistance.
 
Oh OK, that makes more sense.

So I would start at B and then through the 0.5 ohm resistor made by the two 1 ohm resistors. Then it gets a bit more complicated. I would assume that as the 12 ohm resistor is in parallel with the 1V supply and the 4 ohm resistor, it have the same voltage going across it as the supple and 4 omh resistor combined so doesn't matter which route I take? If that's true then I just go up either one of them and then through last resistor to A.
 
Right. Any path you take from B to A will yield the same result.
 
Great, thanks very much! I've got it from here.

Thanks very much for all your help! :)
 

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