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

Answer

C

Explanation

`X_C = (1)/(omegaC)= (1)/(2pifC)`
`=(1)/(2xx3.14xx110^3xx2xx10^(-6))=(10^3)/(4xx3.14)`
voltage, `V=ixxX_C=2xx10^(-3)xx(10^(-3)/(4xx3.14)=0.16V.`