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A capacitor of capacitance 500 \mu \mathrm{F} is charged completely using a dc supply of 100 \mathrm{V}. It is now connected to an inductor of inductance 50 \mathrm{mH} to form an \mathrm{LC} circuit. The maximum current in \mathrm{LC} circuit will be__________\mathrm{A}.

Option: 1

10


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{C=500\mu F}

\mathrm{v=100v}

\mathrm{L=50mH}

The potential drop capacitor at steady state is \mathrm{100v}

\mathrm{t=0}

\mathrm{\left(U_{\text {max }}\right)_{\text {capacitor }}=\left(U_{\text {max }}\right)_{\text {Inductor }}}

\mathrm{\frac{1}{2} C V^2=\frac{1}{2} L I_0{ }^2}

\mathrm{I_0=\sqrt{\frac{C}{L}} \: V}

\mathrm{\text { Maximum\: current }=I_0 = \sqrt{\frac{500 \times 10^{-6}}{50 \times 10^{-3}}} \times 100}

\mathrm{I_{0}=10 \mathrm{~A}}

Posted by

Sanket Gandhi

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