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← JEE Main 2022 · 29 Jul · Shift 2
PhysicsInteger+4 / -1Question 60 of 90

JEE Main 2022 · 29 Jul · Shift 2, Physics question 60

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

Answer and worked solution

Official answer: 10

Energy 12CV2=12LI2\frac12CV^2=\frac12LI^2: I=VC/L=1005×10−45×10−2=10I=V\sqrt{C/L}=100\sqrt{\frac{5\times10^{-4}}{5\times10^{-2}}}=10 A.

Answer: 10