# Norton's theorem application problem

The first image shows the circuit we start with. The second image shows the circuit in which I had the load removed (as I know from Thevenin's theorem).

After that I'm getting confused because of the 2 voltage sources not knowing what to do. I tried finding videos on You-Tube but it's difficult to find a decent one that I can understand.

As you can see I'm in a rather terrible predicament and am clueless as to what I should do next that's why I would like to ask if there is someone to just tell me how to proceed.

If there is something missing I am willing to provide any additional details if needed.

Also it is important to note that I'm trying to find the current,power and voltage across the resistor R3.

Thank you!

Norton's theorem says that any voltage source in series with a resistor can be converted to a current source in parallel with a resistor:

simulate this circuit – Schematic created using CircuitLab

If you'd like to switch between the two circuits, the relationship is $V_{TH} = I_NR$; the resistor doesn't change.

In your case, the circuit simplifies to:

simulate this circuit

(I'll let you figure out the values of the components)

With this new circuit, it's pretty easy to tell that the total current split between the three resistors is $I_1$ + $I_2$. Using that total current, you can figure out the current (and then the voltage and power) by treating the circuit as a current divider.

As your title is "Norton's theorem", so let's use it to solve your problem.

First, you may need read this link or your book carefully. As you only concern about R3, the left of the R3 maybe transformed to a Norton equivalent circuit, then R3 is it's load resistor.

1. First, we find the Norton current $I_{No}$. Short R3, you can use "source transformation" to transform the two voltage source branches to current source branches, then because the current sources are both shorted, so all the current flow from the shorted path, no current flow in the resistors.

$$I_{NO}=I_{S1}+I_{S2}= V_{1}/R_{1}+V_{2}/R_{2}$$

1. Find the Norton resistance. Short the two independent voltage sources, it's two resistors paralleling.

$$R_{NO}= R_{1}||R_{2}$$

Now the circuit is a current source paralleled with two resistors. Then you can easily get the voltage drop on $R_{3}$.

Actually, there are many methods to solve this problem. But, as your title is "Norton's theorem", so here we use "Norton's theorem".

Note: "source transformation" also is an application of "Norton's theorem".

Simply you can see two below images are equivalent.

simulate this circuit – Schematic created using CircuitLab

Note that "Norton's Theorem" is useful when we have non linear element and want to solve circuit and is useful when we have a resistor network with one capacitor and want to find voltage of capacitor for this we can obtain Norton or Thevnin from two point of your capacitor and then solve problem easily.or when you write KCL you like to have current source instead of voltage source so you can transform it!In general Norton's Theorem is very useful. For obtain Thevenin or Norton circuit in general you can put a voltage source from two point that you have Norton's circuit with voltage V and suppose that's current is I now write KCL and KVL and find relation between V and I in linear circuit you obtain V=AI+B so A is your Thevenin's resistance and B is your Thevenin's voltage and you can easily convert it into Norton's circuit. Good luck