Deterimning Thevenin's circuit

1. Apr 30, 2014

johnsy1312

I have to determine the Thevenin's equivalent circuit for the circuit attached.
I am stuck on the first and essential part, finding the thevenin's resistance.

Answer should be 10ohms, my attempt:

$R_2||4 = \frac{5*16}{5+16} = 3.81ohms$

$R_1,3,5 = 20 + 12 + 2 = 34ohms$

2. Apr 30, 2014

johnsy1312

Ok, so R2 AND R4 are not in parallel. Are R2 and R1 in parallel?

3. Apr 30, 2014

phinds

Be a lot easier to figure out what you are talking about if you actually showed the circuit.

4. Apr 30, 2014

Staff: Mentor

Yes, when the source voltage E is suppressed then R1 and R2 will be in parallel.

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5. Apr 30, 2014

johnsy1312

Sorry, i did but it wouldn't allow me since i already posted it in another post.

So:

$R1||2=\frac{20*5}{20+5}=4ohms$

6. Apr 30, 2014

johnsy1312

Are R3 an R4 parallel?

7. Apr 30, 2014

Staff: Mentor

No. Once you've combined R1 and R2, redraw your circuit including the "new" resistance (maybe call it R12) and re-evaluate the layout.

8. Apr 30, 2014

johnsy1312

Is that new resistance positioned where R2 is or where R1 is?

9. Apr 30, 2014

Staff: Mentor

Either. It's replacing both, which were in parallel, so they make the same node connections in the circuit. Note that the orientation (vertical, horizontal, diagonal,...) on a drawing makes no difference to the electrical behavior of the circuit. The only thing that matters is the connections.

10. Apr 30, 2014

johnsy1312

so...

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11. Apr 30, 2014

Staff: Mentor

Sure. Keep on going...

I assume that you're using the voltage source in your image as a placeholder for the eventual Thevenin voltage source. If you're clever you can give it a value as you work your way through the circuit reductions; you can successively replace the portions of the circuit you reduce with Thevenin equivalents (Voltage and resistance) and eventually end up with the final overall Thevenin equivalent.

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