An alternating voltage v = 300 2 sin (60t) (in volts) is connected across a 4 μF capacitor through an a.c. ammeter. The reading of ammeter will be

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402 mA

302 mA

72 mA

40 mA

answer is C.

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Detailed Solution

V0=3002 ∴ Vrms=V02=300 VoltI=VrmsXC=vrms×ωc=300×60×4×10−6I=72×10−3A;I=72 mA

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