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Determine the average power delivered to the circuit consisting of an impedance Z = 5+j8 when the current flowing through the circuit is I = 5∠30⁰. (a) 61.5 (b) 62.5 (c) 63.5 (d) 64.5 This question was addressed to me in semester exam. Query is from Average Power topic in section Power and Power Factor 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)

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Correct answer is (b) 62.5 Explanation: The expression of the average power delivered to the circuit is Pavg = Im^2 R/2. Given Im = 5, R = 5. So the average power delivered to the circuit = 5^2×5/2 = 62.5W.

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