How Do You Find Thevenin's Resistance?

[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

 

[johnsy1312]

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

 

[phinds]

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

 

[gneill]

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

[You forgot to attach your image. I've "borrowed" it from your previous thread]

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

 

[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

 

[johnsy1312]

Are R3 an R4 parallel?

 

[gneill]

Are R3 an R4 parallel?

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

 

[johnsy1312]

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

 

[gneill]

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

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.

 

[johnsy1312]

so...

 

[gneill]

so...

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.