Question

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

• A. $0.16\, V$
• B. $0.32\, V$
• C. $79.5 \, V$
• D. $159 \, V$

Answer 1

Answer: (A)

Reactance of capacitor or capacitive reactance is given by
$X_C=\frac{1}{\omega C}=\frac{1}{2 \pi f C}$
where $f$ is frequency of the alternating current.
Given, $f=1 \times 10^3 Hz , C=2 \times 10^{-6} F$
$\Rightarrow 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
$V=i X_C=2 \times 10^{-3} \times \frac{10^3}{4 \times 3.14}=0.16 \,V$