A 2 mu F capacitor is charged to 50 V by a ba | vedclass

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(B) The capacitor is charged to a potential difference of $V = 50 \ V$. When the battery is removed and the capacitor is connected to the inductor,the

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$5000$

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(B) The capacitor is charged to a potential difference of $V = 50 \ V$. When the battery is removed and the capacitor is connected to the inductor,the circuit forms an $LC$ oscillator.At $t=0$,the charge on the capacitor is maximum,so the potential difference across it is $V_{max} = 50 \ V$.The voltage across the inductor is given by $V_L = L \frac{dI}{dt}$.In an $LC$ circuit,the sum of potential differences is zero: $V_C + V_L = 0$,which implies $|V_L| = |V_C|$.At $t=0$,the current $I=0$,so the entire potential of the capacitor appears across the inductor.Thus,$L \left( \frac{dI}{dt} ight)_{max} = V_{max}$.Given $L = 10 \ mH = 10 	imes 10^{-3} \ H$ and $V_{max} = 50 \ V$.$\left( \frac{dI}{dt} ight)_{max} = \frac{50}{10 	imes 10^{-3}} = \frac{50}{0.01} = 5000 \ A s^{-1}$.

$A$ $2 \mu F$ capacitor is charged to $50 V$ by a battery. The battery is removed after the capacitor is fully charged. At time $t=0$,a $10 mH$ coil is connected in series with the capacitor. The maximum rate at which the current changes in the circuit is (in $A s^{-1}$)

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