JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Electromagnetic Induction

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. The maximum current in LC circuit will be ______ A.

Answer: 10

Solution

The energy stored in the fully charged capacitor is:

U_{C} = \frac{1}{2} C V^{2} = \frac{1}{2} \times 500 \times 1 0^{- 6} \times (100)^{2} = 2.5 J

In an LC circuit, energy oscillates between the capacitor and inductor. Maximum current occurs when all capacitor energy is transferred to the inductor:

\frac{1}{2} L I_{ma x}^{2} = U_{C}

I_{ma x} = \frac{2 U _{C}}{L} = \frac{2 \times 2.5}{50 \times 1 0 ^{- 3}} = \frac{5}{0.05} = 100 = 10 A

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Integer answer

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