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

Options

A. 20 mA, 0
B. 20 mA, 4 W
C. `20sqrt2 mA, 8 W`
D. `20sqrt2 mA , 4sqrt2 W`

Answer

A

Explanation

Om comparing `V = 200sqrt2 sin (100t) ` with
` V = V_0 sin omega t , ` we get `V_0 = 200 sqrt2 V, omega = 100 rad//s`
`therefore V_(rms) = (V_0)/(sqrt2) = (200 sqrt2V)/(sqrt2) = 200V`
The capacitive reactance is
`X_C = (1)/(omega C) = (1)/(100 xx 1 xx 10^(-6)) = 10^4 Omega`
ac ammeter reads the rms value of current . Therefore , the reading of the ammeter is
`I_(rms) = (V_(rms))/(X_C) = (200V)/(10^4 Omega) = 20 xx 10^(-3) A =20 mA`
The average power consumed in the circuit ,
`P = I_(rms) V_(rms) cos phi`
In a pure capacitive circuit , the phase difference between current and voltage is `pi/2`
`therefore cos phi = 0`
` therefore P =0`