Understanding Thevenin's Theorem

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Emspak
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Homework Statement


You have a circuit (attached pic) and want to find the Thevenin equivalent resistance and voltage.

Homework Equations



See below

The Attempt at a Solution


OK< here's what confuses me. If I simply look at the total resistance of the circuit from A and B, it seems to me that I should be adding up R1 with the parallel sum of R2 and R3. From there I would get the total resistance of the circuit and if I hook up A and B to a voltmeter, I should take the current in the circuit and multiply that by R1, giving me a voltage drop. I take that drop, subtracting from V1, and run that across the parallel sum of R2 and R3. Current remains the same, so the total voltage there (call it V3) should be V1 - I*R1 - I*R2.

But the weird thing to me is that the actual answer is treating R2 and R1 as though they are in parallel. I get that with the A and B connections that's how it would look. OK then, but if that's the Thevenin resistance then should not Vth = I * Rth?

And if that's the case, total I (current) is going to be V(source) / total resistance, which is (R1 + (R2R3/ R2 + R3)). Bt that isn't what I am told it is, Vth = R2*V / R1+ R2. This is not making sense to me.

So any help in parsing this would be appreciated.
 

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There are a couple of ways to go about determining the Thévenin equivalent.

Choose any two of the following tasks:

-▸ place yourself at AB and look into the network once all voltage sources have been short-circuited; what you see is the Thévenin resistance. Here, you'll see the 3 resistors are all in parallel.

-▸ use circuit theory, and calculate how much current will flow into short-circuited terminals AB.

-▸ use circuit theory to determine open-circuit voltage, VAB.

Finally, put any two pieces of this data together and form the Thévenin equivalent representation.
Good luck! http://thumbnails112.imagebam.com/37333/0363e9373324851.jpg
 
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NascentOxygen said:
There are a couple of ways to go about determining the Thévenin equivalent.

Choose any two of the following tasks:

-▸ place yourself at AB and look into the network once all voltage sources have been short-circuited; what you see is the Thévenin resistance. Here, you'll see the 3 resistors are all in parallel.

-▸ use circuit theory, and calculate how much current will flow into short-circuited terminals AB.

-▸ use circuit theory to determine open-circuit voltage, VAB.

Finally, put any two pieces of this data together and form the Thévenin equivalent representation.
Good luck! http://thumbnails112.imagebam.com/37333/0363e9373324851.jpg
OK, when you say "ue circuit theory" which one on particular do you mean? KVL? KCL?
 
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Emspak said:
OK, when you say "ue circuit theory" which one on particular do you mean? KVL? KCL?
KVL, KCL, nodal analysis, mesh analysis, source replacement (successive Thevenin/Norton applications with network reduction), parallel/series reductions,... anything you want, really, it's your choice. There are many ways to tackle any given circuit.
 
OK, when you say "ue circuit theory" which one on particular do you mean? KVL? KCL?
Anything that works.

If I simply look at the total resistance of the circuit from A and B, it seems to me that I should be adding up R1 with the parallel sum of R2 and R3. From there I would get the total resistance of the circuit and if I hook up A and B to a voltmeter, I should take the current in the circuit and multiply that by R1, giving me a voltage drop. I take that drop, subtracting from V1, and run that across the parallel sum of R2 and R3. Current remains the same, so the total voltage there (call it V3) should be V1 - I*R1 - I*R2.
Total R = R1 + R2.R3/(R2+R3)
∴ current in R1 = V/(Total R)

Open circuit VAB = current x (R2.R3/(R2+R3)

= ## {V}\over {1 + \frac{R_1(R_2 + R_3)}{(R_2 \cdot R_3)}}##

You'll get this result whatever analytical approach you use.
 
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