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Find the current through (2+j5) Ω impedance considering 20∠30⁰ voltage source.(a) 8.68∠-42.53⁰(b) 8.68∠42.53⁰(c) 7.68∠42.53⁰(d) 7.68∠-42.53⁰I got this question in a national level competition.The origin of the question is Superposition Theorem in portion Steady State AC Analysis of Network Theory

Answer»

Correct CHOICE is (b) 8.68∠42.53⁰

The explanation is: The CURRENT through (2+j5) Ω impedance considering 20∠30⁰ voltage SOURCE is I2 = 20∠30^o×j4/(2+j9) = 8.68∠42.53^o A.



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