Question

Determine the voltage across (2+j5) Ω impedance considering 20∠30⁰ voltage source. (a) 45.69∠-110.72⁰ (b) 45.69∠110.72⁰ (c) 46.69∠-110.72⁰ (d) 46.69∠110.72⁰ I got this question in an international level competition. The query is from Superposition Theorem topic in section Steady State AC Analysis of Network Theory Select the correct answer from above options network theory questions and answers, network theory questions pdf, network theory question bank, network theory gate questions and answers pdf, mcq on network theory pdf, gate network theory questions and solutions, network theory mcq Test , ies network theory questions, Network theory Questions for GATE EC Exam, Network Theory MCQ (Multiple Choice Questions)

Answer 1

Correct answer is (d) 46.69∠110.72⁰ The best I can explain: The voltage across (2+j5) Ω impedance considering 20∠30⁰ voltage source is V2 = 8.68∠42.53^o (2+j5) = 46.69∠110.72⁰V.