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  • A capacitor is initially charged to 20 V and then shorted across a<img alt="\mathrm{6\mu H}" src="/latex-image/?%5Cma

A \mathrm{2\mu F}capacitor is initially charged to 20 V and then shorted across a\mathrm{6\mu H} inductor. What is the maximum value of the current?

Option: 1

5A


Option: 2

7.5A


Option: 3

9.5A


Option: 4

11.5A


Answers (1)

best_answer

The frequency of oscillation is independent of the initial charge and depends only on the values of the capacitance and inductance. The frequency is

\mathrm{\begin{aligned} & f=\frac{\omega}{2 \pi}=\frac{1}{2 \pi} \sqrt{\frac{1}{L C}}=\frac{1}{2 \pi \sqrt{L C}} \\ & =\frac{1}{2 \pi \sqrt{\left(6 \times 10^{-6}\right)\left(2 \times 10^{-6}\right)}}=4.59 \times 10^4 \mathrm{~Hz} \end{aligned}}

According to equation the maximum value of the current is related to the maximum value of the charge by

\mathrm{\mathrm{I}_{\mathrm{m}}=\omega \mathrm{Q}_0=\frac{\mathrm{Q}_0}{\sqrt{\mathrm{LC}}}}

The initial charge on the capacitor is

\mathrm{\mathrm{Q}_0=\mathrm{C} \mathrm{V}_0=(2 \mu \mathrm{F})(20 \mathrm{~V})=40 \mu \mathrm{C}}

Thus

\mathrm{ \mathrm{I}_{\mathrm{m}}=\frac{40 \mu \mathrm{C}}{\sqrt{(6 \mu \mathrm{H})(2 \mu \mathrm{F})}}=11.5 \mathrm{~A} }

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Rishi

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