# Homework Help: Simple Thevenin equivalent resistance problem

1. Oct 4, 2011

### DrOnline

Gentlemen.

1. The problem statement, all variables and given/known data

Calculate the Thevenin equivalent resistance.

2. Relevant equations

Rseries - Rtot = R1 + R2 + ... + Rn

Rparallel - Rtot = 1/(R1^-1 + R2^-1 .... + Rn^-1)

3. The attempt at a solution

As the picture shows, The second ohm-meter tells me 5.7 Ohms is the correct answer. I also have this answer given to me in the task.

But I must be inept at calculating resistances, because I read it this way:

Rth = (R1||R2) + R3 + (R4||R5)

Rth = 2.045 + 5 + 3.6

Rth = 10,645

Obviously wrong. Even if I disregard R3, I get 5.6045 != 5.7.

2. Oct 4, 2011

### Staff: Mentor

R1 is not in parallel with R2. R3 is in the way! In order to be in parallel they'd have to share exactly two nodes.

3. Oct 4, 2011

### phinds

I don't understand the left side of your diagram. What are you doing with a multi-meter set to measure ohms and hooked up to a live circuit?

4. Oct 5, 2011

### DrOnline

Thanks! I'm new to this, and unfortunately terrible at recognizing what's parallel and what's serial. I'll review it again today.

This picture is just something I threw up real quick for illustration. The left side is largely irrelevant. I just didn't know how to draw terminals, so I felt I had to hook it up to the ohmmeter to illustrate.

5. Oct 5, 2011

### Staff: Mentor

So what's your new plan of attack for finding the equivalent resistance?

6. Oct 5, 2011

### phinds

OK, understood, but I'd avise you not to ever do that in a real situation unless you don't care if you burn out the ohmmeter.

7. Oct 5, 2011

### DrOnline

I'm trying to find another way of look at the circuit, perhaps redrawing it to make it more intuitive. I'm also gonna try just brute forcing it and then from that make a new drawing and see if I can understand it.

8. Oct 5, 2011

### Staff: Mentor

You may not be able to "redraw away" the inherent difficulty of not having any series or parallel opportunities to exploit for simplification. When brute force falters, try a little finesse

Two suggestions:
1) Investigate what are called Y-Delta and Delta-Y transformations.
2) Place a voltage source V where your Ohm meter is and use circuit analysis methods to find the current that this voltage would push into the circuit. Then R = V/I.

9. Oct 5, 2011

### DrOnline

Alright, so this circuit redrawing is not equivalent to the original, then.

Yeah we covered Y-Delta and Delta-Y in lectures, I've just never done it myself. I'm looking into it.

Thanks a lot by the way.

10. Oct 11, 2011

### DrOnline

I left this work for a while to focus on some other papers I had to hand in.

But I picked it up today, and I worked it out, by R3, R4 and R5 from a PI into a T, and got the 5.7 Ohm correct.