Question

Find the Norton’s current passing through ‘ab’ in the circuit shown below.

(a) 4.16∠126.8⁰

(b) 5.16∠126.8⁰

(c) 5.16∠-126.8⁰

(d) 4.16∠-126.8⁰

I had been asked this question by my school teacher while I was bunking the class.

I want to ask this question from Norton’s Theorem in portion Steady State AC Analysis of Network Theory

Answer 1

Right answer is (d) 4.16∠-126.8⁰

Easy explanation: The Norton’s current is equal to the current passing through the short circuit between the points a and b. The Norton’s current is I=(10∠0^o)/(3∠90^o)+(5∠90^o)/(2∠-90^o) =4.16∠-126.8^oA.