Q131 / 180 · AIIMS 2003 · Physics
A capacitor of capacitance $2 \mu \mathrm{F}$ is connected in the tank circuit of an oscillator oscillating with a frequency of $1 \mathrm{kHz}$. If the current flowing in the circuit is $2 \mathrm{~mA}$, the voltage across the capacitor will be
- A$0.16 \mathrm{~V}$
- B$0.32 \mathrm{~V}$
- Correct answerC$79.5 \mathrm{~V}$
- D$159 \mathrm{~V}$.
Solution
50.
$$
\begin{aligned}
& \text { } X_C=\frac{1}{\omega C}=\frac{1}{2 \pi f C} \\
& =\frac{1}{2 \times 3.14 \times 1 \times 10^3 \times 2 \times 10^{-6}}=\frac{10^3}{4 \times 3.14}
\end{aligned}
$$
Voltage, $V=i \times X_C=2 \times 10^{-3} \times \frac{10^{-3}}{4 \times 3.14}=0.16 \mathrm{~V}$
Hence, the correct answer is option: 'C'.