The circuit shown contains an inductor L,a ca | vedclass

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(C) Initially,the capacitor has a charge $q_0 = CE$ and the current in the inductor is $i = 0$.When the switch is closed,the circuit forms an $LC$ cir

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$3CE$

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(C) Initially,the capacitor has a charge $q_0 = CE$ and the current in the inductor is $i = 0$.When the switch is closed,the circuit forms an $LC$ circuit connected to a battery of $EMF$ $2E$.Using the conservation of energy,the total energy at any time $t$ is the sum of the energy in the inductor,the energy in the capacitor,and the work done by the battery.Let the charge on the capacitor be $q$. The energy stored in the capacitor is $\frac{q^2}{2C}$ and in the inductor is $\frac{1}{2}Li^2$.The work done by the battery is $W = \int (2E) i \, dt = 2E \int dq = 2E(q - q_0)$.Applying the energy conservation principle: $\frac{q_0^2}{2C} + W = \frac{q^2}{2C} + \frac{1}{2}Li^2$.At maximum charge $q_{max}$,the current $i = 0$. Thus,$\frac{q_0^2}{2C} + 2E(q_{max} - q_0) = \frac{q_{max}^2}{2C}$.Substituting $q_0 = CE$: $\frac{(CE)^2}{2C} + 2E(q_{max} - CE) = \frac{q_{max}^2}{2C}$.$\frac{C^2E^2}{2C} + 2Eq_{max} - 2CE^2 = \frac{q_{max}^2}{2C} \Rightarrow \frac{CE^2}{2} + 2Eq_{max} - 2CE^2 = \frac{q_{max}^2}{2C}$.Multiplying by $2C$: $C^2E^2 + 4CEq_{max} - 4C^2E^2 = q_{max}^2$.$q_{max}^2 - 4CEq_{max} + 3C^2E^2 = 0$.Solving the quadratic equation: $q_{max} = \frac{4CE \pm \sqrt{16C^2E^2 - 12C^2E^2}}{2} = \frac{4CE \pm 2CE}{2}$.$q_{max} = 3CE$ or $q_{max} = CE$. Since the charge increases,the maximum charge is $3CE$.

The circuit shown contains an inductor $L$,a capacitor $C$ with an initial charge $CE$,and a battery of $EMF$ $2E$. Find the maximum charge on the capacitor after the switch is closed at $t = 0$.

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