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Electromagnetic Induction Physics Ncert Solutions Class 12th Physics Class 12th

A capacitor of capacitance 500 $μ F$ is charged completely using a dc supply of 100 V. It is now connected to an inductor of inductance 50 mH to form an LC circuit. Then maximum current in LC circuit will be_______ A.

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Payal Gupta | Contributor-Level 10

2 months ago

C = 500 $μ F,$ V = 100 v, L = 50 mH

In this LC – oscillation

q = q₀ cos $ω t$

$i = \frac{- d q}{d t} = q_{0} ω s i n ω t ω = \frac{1}{\sqrt[]{2 c}} = \frac{1}{\sqrt[]{50 \times 1 0^{- 3} \times 5 \times 1 0^{- 4}}}$

= $\frac{1000}{5} = 200$

So, i_max = $q_{0} ω = 500 \times 1 0^{6} \times 100 \times 200$

= 10A

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A capacitor of capacitance 500 $μ F$ is charged completely using a dc supply of 100 V. It is now connected to an inductor of inductance 50 mH to form an LC circuit. Then maximum current in LC circuit will be_______ A.

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A capacitor of capacitance 500 μF is charged completely using a dc supply of 100 V. It is now connected to an inductor of inductance 50 mH to form an LC circuit. Then maximum current in LC circuit will be_______ A.

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A capacitor of capacitance 500 $μ F$ is charged completely using a dc supply of 100 V. It is now connected to an inductor of inductance 50 mH to form an LC circuit. Then maximum current in LC circuit will be_______ A.