# Norton's theorem

#### Ledwardz

Joined Oct 31, 2010
37
Hi, im trying to do the example using Norton's Theorem on the attached file named EEE1003. Ive also attached my attempt which is completely wrong. The voltage supply in the middle confuses me and i dont know if im doing any of it right. Any help appreciated. Thanks, Lee.

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#### Georacer

Joined Nov 25, 2009
5,182

#### Ledwardz

Joined Oct 31, 2010
37
Erm, thee initial question is in the attachment EEE1003 and the picture is just my rubbish attempt. It asks me to find the current in winding one but the attachment should make that all clear. Thanks for reply also.

#### Georacer

Joined Nov 25, 2009
5,182
You are right, I didn't read the pdf.

So, I take it you want to solve the problem with method 5, the Norton's theorem. The quantity for calclulation is the current through winding 1.

It is rather easy to apply Norton's theorem, but it requires a basic understanding of the method.

What I want you to do is to isolate the "battery" circuit and the "winding 1" circuit and draw them separately.
Then try to find their Norton equivalents. It should be easy, given that each circuit has only one resistor and a voltage source.
Finally connect the two equivalent circuits back together, along with the resistor of winding 2 and form the complete circuit.

You should now have one circuit with two parallel current sources and three parallel resistors. Merge the current sources two find their equivalent and use current division rule to calculate the current flowing through R1.

Post the steps of your work to point the points where you have trouble with.

#### Ledwardz

Joined Oct 31, 2010
37
when i do the equivalent circuit bit is it as easy as changing the 100v supply and 0.5 ohms resistor to 200A and leaving the 0.5 ohms resistor in or am i missing the point? Ill attach what ive done and you can have a good laugh at it.

#### Georacer

Joined Nov 25, 2009
5,182
Did you read the link I told you to? If you did you should notice that in the Norton equivalent circuit, the current source is placed paralelly to the equivalent resistance. Redraw your circuit to show that.

After you do so, use the current divider formula (http://www.allaboutcircuits.com/vol_1/chpt_6/3.html) to calculate the current through R1.