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If in an alternating current circuit, capacitance C is connected to a supply of 200V,50Hz. Current in the circuit is 1.89 A. Find the capacitance C.(a) 30 µF(b) 20 µF(c) 10 µF(d) 15 µFThe question was posed to me in an international level competition.My enquiry is from Resistance and Capacitance in Series in division Single Phase Series Circuits of Basic Electrical Engineering

Answer»

The correct CHOICE is (a) 30 µF

To explain I would say: XC= V/I = 200/1.89 = 106.1 Ω.XC=1/(2πfC)Substituting the VALUES we GET C = 30 µF.



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