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

Answer

B

Explanation

Reactance fo capacitor or capacitive reactance is given by
`X_(C)=1/(omegaC)=1/(2pifC)`
where `f` is frequency of the alternating current
Given, `f=1xx10^(3)Hz, C=2xx10^(6)F`
`:. X_(C)=1/(2xx3.14xx10^(3)xx2xx10^(-6))`
`=(10^(3))/(4xx3.14)`
Hence, voltage across the capacitor is
`V=iX_(C)=2xx10^(-30xx(10^(3))/(4xx3.14)=0.16 V`