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Find the current in the 5 ohm resistance using Norton’s theorem.(a) 1A(b) 1.5A(c) 0.25A(d) 0.5AThe question was asked during a job interview.This interesting question is from Norton’s Theorem topic in chapter Network Theorems of Basic Electrical Engineering

Answer»

The correct option is (d) 0.5A

To explain I would say: SHORTING all voltage sources and opening all current sources we have:

RN=(3||6)+10 = 12 ohm.

Since the 5 ohm is the load resistance, we SHORT it and find the resistance through the short.

If we apply source transformation between the 6 ohm resistor and the 1A source, we get a 6V source in series with a 6 ohm resistor. Now we have two meshes. LET us consider I1 flowing in the first mesh and I2 flowing in the SECOND mesh.

The mesh equations are:

9I1-6I2=4

-6I1+16I2=6

On solving these equations simultaneously, we get I2=0.72A, which is the short circuit current.

Connecting the current source in PARALLEL to RN which is in turn connected in parallel to the load resistance=5ohm, we get Norton’s equivalent circuit.

Using current divider: I = 0.72*12/(12+5) = 0.5 A.


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