A capacitor of capacitance $1 μ F$ is charged to a potential of $1 V$ . It is connected in parallel to an indicator of inductance $10^{- 3} H$ . The maximum current that will flow in the circuit has the value

\sqrt{1000} m A

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1 m A

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1 μ A

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1000 m A

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Solution

The correct option is A $\sqrt{1000} m A$
when capacitor is connected in parallel to inductor,
voltage across $C$ = voltage across $L$
$V_{C} = V_{L} \frac{1}{C} (Q_{0} - \int_{0}^{t} i (t) d t) = L \frac{d i}{d t}$
solving using Laplace transform,
$V_{C} = V_{L} \frac{1}{C} (Q_{0} - \int_{0}^{t} i (t) d t) = L \frac{d i}{d t} \frac{Q_{0}}{s} - \frac{I (s)}{s} = L C s I (s) I (s) = \frac{Q_{0}}{\sqrt{L C}} \frac{1 / \sqrt{L C}}{s^{2} + 1 / L C} i (t) = \frac{Q_{0}}{\sqrt{L C}} s i n (\frac{1}{\sqrt{L C}} t) Q_{0} = C V_{0}$
therefore maximum current = $\frac{C V_{0}}{\sqrt{L C}} = \sqrt{1000} m A$

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A capacitor of capacitance 1 μF is charged to a potential of 1 V. It is connected in parallel to an indicator of inductance 10−3 H. The maximum current that will flow in the circuit has the value

A capacitor of capacitance $1 μ F$ is charged to a potential of $1 V$ . It is connected in parallel to an indicator of inductance $10^{- 3} H$ . The maximum current that will flow in the circuit has the value