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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) 64.5This question was addressed to me in semester exam.Query is from Average Power topic in section Power and Power Factor of Network Theory |
Answer» 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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