A capacitor of capacitance $2 \mathrm{\mu F}$ 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

Reactance of capacitor or capacitive reactance is given by

${\mathrm{X}}_{\mathrm{C}}=\frac{1}{\mathrm{\omega C}}=\frac{1}{2\mathrm{\pi fC}}$

where f is frequency of the alternating current

Given, $\mathrm{f}=1 \times {10}^3 \mathrm{Hz}, \mathrm{C} = 2 \times {10}^{-6} \mathrm{F}$

$\therefore$    ${\mathrm{X}}_{\mathrm{C}}=\frac{1}{2\times 3.14\times {10}^3\times 2\times {10}^{-6}}$

$=\frac{{10}^3}{4\times 3.14}$

Hence, voltage across the capacitor is

$\mathrm{V}={\mathrm{iX}}_{\mathrm{C}}=2\times {10}^{-3}\times \frac{{10}^3}{4\times 3.14}=0.16\mathrm{V}$