Question

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∠-42.53⁰

I 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 1

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.