1.

Find the load current in the circuit shown below.(a) 3.19∠166.61⁰(b) 3.19∠-166.61⁰(c) 4.19∠166.61⁰(d) 4.19∠-166.61⁰The question was posed to me in examination.My question is from Norton’s Theorem in chapter Steady State AC Analysis of Network Theory

Answer»

Right answer is (b) 3.19∠-166.61⁰

For explanation: The load CURRENT in the CIRCUIT is GIVEN by IL = I×(6∠-90^o)/(5+6∠-90^o) = 3.19∠-166.61^oA.



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