# Thevenin Problem

1. Feb 11, 2013

### joe_cool2

1. The problem statement, all variables and given/known data
I give the entire question for context’s sake, but I am primarily concerned with A and B:

Suppose you have the network shown below. We want the Thevenin equivalent for the part of the circuit that surrounds R1.

A) Find Rth by zeroing sources and reducing the circuit.

B) Now find Ith by computing the short-circuit current with Rth and V0.

C) Using the values below, find Vth and Rth. Predict the voltage drop, and hence the current flow, across R1.

V0: 9V
R1: 100
R2: 330
R3: 220
R4: 2200
R5: 1000
R6: 100
R7: 3000
R8: 470

D) Build the circuit with these values and measure the actual value and see if it matches your prediction.

E) Build the Thevenin equivalent and connect it to R1. Is the voltage drop the same?

2. Relevant equations

N/A

3. The attempt at a solution
A) To find Rth, we have to look back through the circuit and determine which parts of the circuit are parallel or in series relative to the terminals.

$$\left (\frac{1}{R7}+\frac{1}{R3}\right)^{-1} + R5 + \left (\frac{1}{R8+R2}+\frac{1}{R6+R4}\right)^{-1}$$

This gets about 1.7 kOhms or so if I remember, and my multimeter confirms it.

B) This is where the examples I’ve seen regarding solving for Thevenin currents and voltages fail me. I have not yet seen an example of how to do one as complicated as this, and I have searched extensively. I do understand that in order to get the Thevenin equivalent circuit, we now need either the short circuit current or the open circuit voltage.

In the simpler examples I’ve seen, normally the open circuit voltage is found by looking for a parallel branch in the circuit, and finding the voltage drop across it. Perhaps I should be considering an equivalent branch made up of R7,R3, R5, R8, and R2?

The short circuit current often bypasses resistors on branches parallel to the short. I notice in this case, however, that there are resistors leading up to this short, so there is in fact a current division between the R8,R2 branch and the R5,R7,R3 branch. Therefore I cannot use this simplification.

Well, anyway, I gave it a shot. I thought maybe the Thevenin voltage could be found by the following:

$$V _{th}=V_{0}\frac{R_{th}}{R_{th}+R_{L}}$$
As the Thevenin voltage is connected in series with the load on the simplified diagram; a voltage divider. I am unsure of this result because it was not obtained by a method that resembles what I’ve seen before. Moreover, it relies on a load being there when we are in fact trying to find the open circuit voltage.

I understand that I might be completely off base here. I appreciate your help in advance should you choose to give it. I also understand that you do not just hand out answers to students on here. That's why I'm posting the topic here and not Yahoo answers. I wish to learn something here, not to regurgitate an answer. So if I'm way off, let's just start from square 1, whatever that might be.

2. Feb 11, 2013

### Staff: Mentor

One approach is to solve the problem by working your way from left to right through the circuit, making successive Thevenin equivalents as you go.

In this case you've already combined the obvious resistor combinations, and the resulting circuit looks something like this:

The resistor numbers refer to the original resistors that were combined into single resistances.

If you make a Thevenin equivalent for what's to the left of the leftmost red arrows, then it should be simple to tack on R357 and obtain the net Thevenin equivalent looking into AB.

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