# How do I use Norton's theorem to determine the values of an equivalent current source on the right?

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We transform the circuit on the left into a Norton equivalent circuit on the left (see attached picture below) .

The equivalent Norton current source `I_(No)` is the current that flows between points A and B if they are short-circuited.

`I_(No) = V/R_1 = 15/15 =1 A`

The equivalent Norton resistance `R_(No)` is the resistance between terminals A and B if all the voltage sources are short-circuited.

`R_(No) = R1 "parallel" R2 = (R1*R2)/(R1+R2) =(15*30)/(15+30) = 10 Omega`

Therefore the circuit on the left can be transformed into a current source `I_(No)=1 A` in parallel with a resistor `R_(No) =10 Omega` .

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