# Thevenin Theorem

#### silverwing57

Joined Mar 25, 2007
3
Hey guys,
I'm a new electronics student and I can say that I love that fact that they're are forums like this. Anyway I'm having trouble with the thevenin theorem in this question. I have been stewing over it for about 2 hours and I just cant get it right. Any help would be more that appreciated, sorry if the photo is a little hazy, I took it with my phone. Thanks guys.

#### Dave

Joined Nov 17, 2003
6,969
Can you post up your workings thus far? It is easier to guide you with your existing work.

Dave

#### silverwing57

Joined Mar 25, 2007
3
I can't post it as I don't have a scanner and my phone could not produce a clear enough image. So I'll just text it.

So far I have taken out the load on the far right of the screen. Next I have tried to calculate Vth, I figured since the 4 ohm resistor above the load was incomplete, it did not count in the calculation. So after putting the other two 4 ohm resistors in series I gathered that the 2 ohm resistor dropped 6V and the two 4 ohm resistors also dropped the remaining 6 ohm each. So I figured that Vth = 6V. I also mixed around with a jumper, but not knowing the specifics of it I just omitted it and kept going.

Next I tampered with Vr. After reading a little on the net I came to the equation of (1/2) + (1/8), take the reciprocal = 1.6 + 4 = 5.6 ohms.

So I have technically "done" the question, problem is that I don't know how wrong I got it (and trust me it is), as I don't know how to check it. Any help would be greatly appreciated.

#### antseezee

Joined Sep 16, 2006
45
OK. I just did the whole problem and am going to go through it step-by-step. I have an upcoming test where we have to apply Thevenin's on Laplace circuits...so I consider it practice.

1. Remove load from A to B. Draw Thevenin Circuit with Eth, in series with Zth, and open node A, with open node B.

2. Find Zth. To do this, remove load between Node A & B of original circuit. Short out voltage sources with wire. Travel from point A to point B finding the total resistance between the 2 points:

Zth = 4 + (4 || 4 || 2)
Zth = 4 + 1
Zth = 5 ohms

3. Find Eth. To do this, replace sources in original circuits while leaving load open between points A & B. Find voltage from point A to B. This is where it gets tricky.

Look at the circuit, there's a 4 ohm resistor next to point A. But since there is an open from point A to B, there is no way that current can flow through this 4 ohm resistor. Therefore, there is no voltage drop across the initial 4 ohm resistor.

Now we have a 4 || 4, then a 12 V source in series with a 2 ohm resistor. Combine the 4 || 4 resistors to make them just 2 ohms. Now we have a completely series circuit left to calculate the voltage drops. Total of 4 ohms in series with the 12V source. Current = 12V / 4 ohms = 3A. Draw the current flow and mark the polarities of the voltage drops across the resistors. Now, the voltage from point A to B is -12V + 2 * (3A) = -6V = Eth. Remember that you are taking the voltages from A to B, so you must start right of the 12 V source (which has a negative polarity), and add the positive drop across the 2 ohm resistor equaling -6V.

4. Plug in values of Eth & Zth to Thevenin equivalent circuit. Plug in load at Thev. circuit. There ya go, that's all you need to do. The book problem doesn't say to find the RL voltage drop, and you theoretically cannot unless they give you a value. Calculating the potential voltage difference between the two points only is applicable in the Thevenin's equivalent circuit, and not the original one. You could calculate the voltage difference between both points for Eth, but you'll get the same answer as above (-6V).

It's a lot easier if I showed you the drawings. If you need more help, ask, and I'll post a quick sketch.

Ah, thank you so much, I think I love you (just kidding). I actually did this about 8 times, and 6V with 3A and 5 ohms was actually one of the answers I had, problem was that I just simplified the circuit wrong when I was checking the answer with a dummy value of 5 ohms for the load. But all is good now, I can move on with the paper, that was only the first question. 