Hi Again,

Regarding the following circuit:

I am trying to find current through the 5.1Ω resistor. I have looked through some old threads but all the circuits are a bit simpler than what i have.

Do i need to break it down to individual loops or can i find it as a total? I understand i need to remove the resistor and treat that loop as open but that is about it.

The Eoc is the same throughout each loop as far as i understand due to all components being in parallel.

Any advice would be much appreciated.

 

Have you learned superposition theorem, yet?

If so, that would be a good way to analyse this circuit.

 

I'll have a look at it. I was absent on the day we did this stuff. Thanks again.

 

Thanks for the tip, that makes much sense to me than thevenins. The question does specifically ask me to 'verify thevenins theorum' though.

I have got the answer for the current through the resisitor by this method but is superposition part of Thevenins or a seperate entity?

 

Superposition is observed in all linear time-invariant systems (and some instances of periodic analysis). It is separate from Thevenin's theroem.
http://www.allaboutcircuits.com/vol_1/chpt_10/7.html

 

To do this in thevenins theorem, you need to follow these steps.

First remove the resistor in question. This case R3.

Then using any method solve for the voltage, that will be present between the terminals that R3 were hooked up to.

That voltage will be your thevenin voltage. (Vth)

Then solve for the resistance that is present between those two terminals as well.
To solve for resistance, you need to DEactivate the voltage sources, then solve for the resistance.

This resistance will then be the thevenin resistance. (Rth)

Then draw a circuit with your Thevenin voltage source, in series with your thevenin resistance, then connect the R3 resistor in series with this, to solve for the current value through R3.

On a side note:
The thevenin voltage found at those terminals, where R3 is hooked up to, is only that value with R3 out of the branch.
When you insert the R3 back into that branch, the voltage between those two terminals, will now become
(Vth. x R3) / (R3 + Rth).

 

And the penny drops... Thanks. So the fact that there's 3 sources is irrelevent, just short them all. Much appreciated...

 

Being somewhat relevant to Thenenin's Theorem, I would the Norton's one for every source-resistor pair, making all the resistors and the current sources parallel. Then, it's a walk in the park.