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Question

# An alternating voltage (in volts) given by V=200√2sin(100)t is connected to 1μF capacitor through an ideal ac ammeter in series. The reading of the ammeter and the average power consumed in the circuit shall be

A
20 mA,4 W
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B
20 mA,0
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C
202mA,8W
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D
202mA,42W
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Solution

## The correct option is B 20 mA,0On comparing V=200√2sin(100t) withV=V0sinωt, we get V0=200√2V,ω=100rads−1∴Vrms=V0√2=200√2V√2=200VThe capacitive reactance is XC=1ωC=1100×1×10−6=104Ωac ammeter reads the arms value of current . Therefore, the reading of the ammeter is Irms=VrmsXC=200V104Ω=20×10−3A=20mAThe average power consumed in the circuit,P=IrmsVrmscosϕIn an pure capacitive circuit , the phase difference between current and voltage is π2∴cosϕ=0∴P=0.

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