The switch is in position A for a long time. | vedclass

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(A) When the switch is in position $A$ for a long time,the inductor acts as a short circuit (steady state). The current flowing through the inductor i

Vedclass · Accepted Answer

$(\sqrt{L C}) \frac{E}{R}$

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(A) When the switch is in position $A$ for a long time,the inductor acts as a short circuit (steady state). The current flowing through the inductor is $I_0 = \frac{E}{R}$.At $t=0$,the switch is moved to position $B$. The circuit now consists of an inductor $L$ and a capacitor $C$ in series,forming an $LC$ oscillator circuit. The initial current in the inductor is $I_0 = \frac{E}{R}$ and the initial charge on the capacitor is $q_0 = 0$.The energy stored in the inductor is converted into the energy stored in the capacitor. According to the conservation of energy:$\frac{1}{2} L I_0^2 = \frac{1}{2} \frac{q_{\max}^2}{C}$Substituting $I_0 = \frac{E}{R}$:$L \left(\frac{E}{R}\right)^2 = \frac{q_{\max}^2}{C}$$q_{\max}^2 = L C \frac{E^2}{R^2}$$q_{\max} = \sqrt{L C} \frac{E}{R}$

The switch is in position $A$ for a long time. At time $t=0$,it is shifted to position $B$. Find the maximum charge that will accumulate on the capacitor.

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