Thevenin Equivalent

[evilsanta]

Have this circuit to use Thevenin, Nortons and Superposition on and its tripping me up because Im unsure as to how to do it.

Ive attempted the question using all the methods and I have 3 different answers which isnt good I imagine. Although two of the answers are close one is further out. I had the voltage going the wrong direction in the superposition example but have put the answer as a minus as I believe this is the correct answer. Any pointers would be great.

Thanks Alan

 

[Harald Kapp]

From IMG_0765.JPG:
Rtotal=292,631Ω is correct.
I=0.0752 A is correct, too.
Vdrop across R1, R2 = 6.947V is correct also (within accuracy of calculation).
I(Rl)=0.0316A is correct also. Note that you should use the expression "current through", not "across".

I do not understand the last line where you add 0.031A + 0.019A. What's the use of this line?

By the way, I encourage you to always keep units throughout your calculations. Instead of I=0.031+0.19=0.05A write I=0.031A+0.019A=0.05A. This may seem trivial in this case, but it greatly helps to discover computational errors if the units don't match (e.g. if at one point you come at R=1A+2V you can be 200% certain that this cannot be correct).

 

[evilsanta]

I added the total of the currents from the first loop and the second loop added to give the total current in RL, in Loop 1 I found the current in RL without E2 and then found the current in the second loop without E1 so added together, should I be subtracting them?

And I will add in the units of measurement for future equations.

How did the other equations go?

 

[evilsanta]

If going by the current direction from E1, then I1 would flow into RL from the top and the current from E2 would flow from the bottom so that would make the I1 - I2 for the superpostion thereom? Meaning 0.019A - 0.031A giving me an answer of 0.01A which is the same for the thevenin equivalent example. But the norton example seems to be wrong completely??

 

[Laplace]

The problem shows what amounts to two Thevenin equivalent circuits in parallel driving one load resistor. What if you converted each Thevenin into its Norton equivalent? Simple to do. Then you would have two Norton equivalent circuits in parallel driving one load resistor. Or two current sources in parallel driving three resistors in parallel. But two current sources in parallel will simplify as either the sum or difference into a single current source...

 

[evilsanta]

Thanks for the advice, redone all three theorems and got the same answer for all, once rounded up I got -0.0119A.