Q125 / 150 · MHT CET 2022 (07 Aug Shift 1) · Physics
A battery connected to a capacity $20 \mu \mathrm{F}$ is charged to a potential of $35 \mathrm{~V}$. The battery is disconnected. A pure inductor coil of $200 \mathrm{mH}$ is connected across the capacitor so that LC oscillations are set up.The maximum current in the coil is
- A$0.025 \mathrm{~A}$
- B$0.25 \mathrm{~A}$
- C$0.035 \mathrm{~A}$
- Correct answerD0.35 A
Solution
Capacitance, $\mathrm{C}=20 \mu \mathrm{F}$
Voltage applied to capacitor, $\mathrm{V}=35$ volts
Inductance connected to the capacitor, $\mathrm{L}=200 \mathrm{mH}$
To find: The maximum current in the inducting coil.
The charge stored in the capacitor is given by:
$\mathrm{Q}_0=\mathrm{CV}$
The maximum current flowing in the series LC circuit is:
$I_{\text {max }}=Q_{\text {ow }}$
The frequency of a series LC circuit is given by:
$\omega=\frac{1}{\sqrt{\mathrm{LC}}}$
Substitute the values in equation of maximum current.
$\begin{aligned} & \Rightarrow I_{\max }=(\mathrm{CV}) \times \frac{1}{\sqrt{\mathrm{LC}}}=\sqrt{\frac{\mathrm{C}}{\mathrm{L}}} \mathrm{V} \\ & \therefore \mathrm{I}_{\max }=\sqrt{\frac{20 \times 10^{-6}}{200 \times 10^{-3}}} \times 35=0.35 \mathrm{~A}\end{aligned}$
Hence, the correct answer is option: 'D'.