JEE MAIN - Physics (2024 - 29th January Morning Shift - No. 5)
A capacitor of capacitance $$100 \mu \mathrm{F}$$ is charged to a potential of $$12 \mathrm{~V}$$ and connected to a $$6.4 \mathrm{~mH}$$ inductor to produce oscillations. The maximum current in the circuit would be :
2.0 A
3.2 A
1.5 A
1.2 A
Explanation
By energy conservation
$$\begin{aligned} & \frac{1}{2} \mathrm{CV}^2=\frac{1}{2} \mathrm{LI}_{\text {max }}^2 \\ & \mathrm{I}_{\max }=\sqrt{\frac{\mathrm{C}}{\mathrm{L}}} \mathrm{V} \\ & =\sqrt{\frac{100 \times 10^{-6}}{6.4 \times 10^{-3}}} \times 12 \\ & =\frac{12}{8}=\frac{3}{2}=1.5 \mathrm{~A} \end{aligned}$$
Comments (0)
