Find the voltage across (2+j5) Ω impedance using Superposition theorem.(a) 40.85∠72.53⁰(b) 40.85∠-72.53⁰(c) 41.85∠72.53⁰(d) 41.85∠-72.53⁰The question was asked during an online exam.This intriguing question originated from Superposition Theorem in division Steady State AC Analysis of Network Theory

Find the voltage across (2+j5) Ω impedance using Superposition theor

1.	Find the voltage across (2+j5) Ω impedance using Superposition theorem.(a) 40.85∠72.53⁰(b) 40.85∠-72.53⁰(c) 41.85∠72.53⁰(d) 41.85∠-72.53⁰The question was asked during an online exam.This intriguing question originated from Superposition Theorem in division Steady State AC Analysis of Network Theory
Answer» The correct OPTION is (a) 40.85∠72.53⁰ Explanation: The voltage across (2+j5) Ω impedance USING Superposition THEOREM is the SUM of the voltages V1 and V2. V = V1+V2 = 29.16∠-9.28^o+46.69∠110.72^o=40.85∠72.53^o V.

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