Alternating Current

JEE Main level, numerical answer, medium

87 of 190

Question 87 of 190

A capacitor of capacitance 500μF500\mu F is charged completely using a de supply of 100V100V. It is now connected to an inductor of inductance 50mH50mH to form an LC circuit. The maximum current in LC circuit will be _____ AA.

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Answer: 10

Energy stored in capacitor
E=12CV2=12500×10−6×104E=\dfrac{1}{2}C{V}^{2}=\dfrac{1}{2}500\times {10}^{-6}\times {10}^{4}
⇒E=52J\Rightarrow E=\dfrac{5}{2}J
Current will be maximum when whole energy of capacitor becomes energy of inductor.
⇒12LI2=52\Rightarrow \dfrac{1}{2}L{I}^{2}=\dfrac{5}{2}
⇒I=5L=550×10−3=10A\Rightarrow I=\sqrt{\dfrac{5}{L}}=\sqrt{\dfrac{5}{50\times {10}^{-3}}}=10A

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