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Determine the voltage across (2+j5) Ω impedance considering 20∠30⁰ voltage source.

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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

1 Answer

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Best answer
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.

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