# Norton's theorem

#### KCHARROIS

Joined Jun 29, 2012
311
So there's a question in a book asking to solve a circuit using Norton's theorem and I've only been able to solve it using superposition. (Circuit attached)

So to solve using superposition here's what I did, I opened the current source and calculated 12V across R1 and 8V across R2, simple series circuit. Next I shorted the voltage source and calculated -6V across both resistors, simple parallel circuit. Then adding and subtracting the values I get 18V across R1 and 2V across R2 and this makes sense.

But how can I calculate using Norton's theorem, It may not be the best method but I'd like to learn.

Thanks

Edit: Forgot to mention that the 1mA source is actually an unknown resistor but has 1mA running through it.

#### daviddeakin

Joined Aug 6, 2009
207
You have a 20V source in series with R1. Convert that source into its Norton equivalent instead:
20/15k=0.1333mA

You can now re-draw the circuit as two Norton sources in parallel:
I4 in parallel with R2
and a 0.1333mA source in parallel with R1.

The total resistance becomes R1||R2 = 6k
And the total combined current is -1mA + 1.333mA= 0.333mA

Ohm does the rest.

• KCHARROIS