Q. A $16\ \mu F$ capacitor is charged to a 20 volt potential. The battery is then disconnected and pure $40\ mH$ coil is connected across the capacitor so that LC oscillations are setup. The maximum current in the coil is
A) $0.2\ A$
B) $40\ mA$
C) $2\ A$
D) $0.4\ A$
Explanation
(d) $q = Q_{0}\cos{\omega t}$
$I = \frac{dq}{dt}\text{=} - Q_{0}\omega.\sin{\omega t}$
$I_{\max} = C\omega V = V\ \sqrt{\frac{C}{L}} = 20\sqrt{\frac{16 \times 10^{- 6}}{40 \times 10^{- 3}}} = 0.4A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.