A capacitor of capacitance `2 mu F` is connected in the tank circuit of an oscillator oscillating with a frequency of 1 kHz. If the current flowing in the circuit is `2 mA`, the voltage across the capacitor will be

Options

A. `0.16 V`
B. `0.32 V`
C. `79.5 V`
D. `159 V`

Answer

A

Explanation

(a) Reactance of capacitor or capactive reactance is given by
`X_(C ) = (1)/(omega C) = (1)/(2 pi fC)`
where `f` is frequency of the alternating current
Given, `f = 1 xx 10^(3) Hz, C = 2 xx 10^(-6) F`
`implies X_(C ) = (1)/(2 xx 3.14 xx 10^(3) xx 2 xx 10^(-6))`
`= (10^(3))/(4 xx 3.14)`
Hence, voltage across the capacitor is `V = i X_(C ) = 2 xx 10^(-3) xx (10^(3))/(4 xx 3.14) = 0.16 V`