A $1 \mu \mathrm{~F}$ capacitor is charged to 50 V and is then discharged through 10 mH inductor of…

A $1 \mu \mathrm{~F}$ capacitor is charged to 50 V and is then discharged through 10 mH inductor of negligible resistance. The maximum current in the inductor is
  1. 0.5 A
  2. 1.5 A
  3. 1 A
  4. 0.15 A

Solution

Energy stored in capacitor $=\frac{1}{2} \mathrm{CV}^2$ For maximum current energy stored in inductor, $\frac{1}{2} \mathrm{CV}^2=\frac{1}{2} \mathrm{LI}_0^2$. $\mathrm{I}_0=$ Maximum current $\mathrm{I}_0^2=\frac{\mathrm{CV}^2}{\mathrm{~L}}=\frac{10^{-6} \times 50 \times 50}{10 \times 10^{-3}}$ $=25 \times 10^{-2}$ $\mathrm{I}_0=\sqrt{25 \times 10^{-2}}$ $\mathrm{I}_0=0.5 \mathrm{~A}$

Asked in: MHT CET 2024 (11 May Shift 2)

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