First slide

L - C Oscillations

Question

A capacitor of capacitance $2 μF$ 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

Moderate

A

0.32 V
B

0.16 V
C

79.5 V
D

159 V

Solution

Reactance of capacitor or capacitive reactance is given by
$X_{C} = \frac{1}{ωC} = \frac{1}{2 πfC}$
where f is frequency of the alternating current
Given, $f = 1 \times 10^{3} Hz, C = 2 \times 10^{- 6} F$
$∴$ $X_{C} = \frac{1}{2 \times 3 .14 \times 10^{3} \times 2 \times 10^{- 6}}$
$= \frac{10^{3}}{4 \times 3.14}$
Hence, voltage across the capacitor is
$V = {iX}_{C} = 2 \times 10^{- 3} \times \frac{10^{3}}{4 \times 3 .14} = 0 .16 V$

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