Norton Current is Determined as

• Answer: B. 244 mA

 

Solution:

Norton’s theorem states that a linear two-terminal circuit can be replaced by an equivalent circuit which consists of a Norton equivalent current source I_N in parallel with a Norton equivalent resistor of resistance R_N.

Here I_N is the short circuit current through the terminals, RN is the equivalent resistance seen from the terminals when independent sources are turned off.

For the circuit given in the problem, we need to construct Norton equivalent circuit at terminals A and B.

Finding I_N:

As we have discussed, I_N is the short circuit current through terminals A and B. We need to short circuit terminals A and B.

First, we need to find node Voltage V, which is shown in the above figure.

Applying KCL at node gives,

(V-75)/68+V/68+V/120=0

(V-75)120+120V+68V=0

308×V=75×120

V=(75×120)/308

Then I_N=V/120 =[(75×120)/308×120]=243.506 mAmps≈244 mAmps.

Norton’s current I_N at terminals A, and B for the given circuit is 244 mA.

☛ Related Questions:

• Ohm’s law is true for
• Ohm’s law states that
• Parallel resonant circuit