In the question number 30, the net power absorbed by the circuit in on

1.	In the question number 30, the net power absorbed by the circuit in one complete cycle is
Answer» 5 W 10 W 15 W zero Solution :As `P = V_("rms")I_("rms")cos phi` In a PURE capacitance circuit, the phase difference between alternating VOLTAGE and CURRENT is `pi//2` . Hence `P = V_("rms")I_("rms")cos 90^(@) = 0`.

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In the question number 30, the net power absorbed by the circuit in one complete cycle is

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