Two Uncharged Capacitors

Between $t = 0$ and $t = t_0$ :

$\large \displaystyle E = \frac{1}{C} \int_0^{t} i(\tau) d\tau + L \frac{di}{dt}$

Apllying Laplace transform :

$\large \displaystyle \frac{E}{s} = \frac{1}{Cs} I(s) + Ls I(s)$

$\large \displaystyle I(s) = \frac{EC}{LCs^2+1}$

$\large \displaystyle I(s) = \frac{CE}{\sqrt{LC}} \cdot \frac{\frac{1}{\sqrt{LC}}}{s^2+ \frac{1}{LC}}$

So:

$\color{#20A900} \boxed{ \large \displaystyle i(t) = \frac{CE}{\sqrt{LC}} \sin \left ( \frac{1}{\sqrt{LC}} t \right ) }$

Likewise the voltage on capacitor $A$ is:

$\large \displaystyle V_A(s) = \frac{1}{Cs} I(s) = \frac{E}{s(LCs^2+1)}$

$\large \displaystyle V_A(s)= \frac{E}{s} - \frac{Es}{s^2+\frac{1}{LC}}$

$\color{#20A900} \boxed{ \large \displaystyle v_A(t) = E \left [ 1 - \cos \left ( \frac{1}{\sqrt{LC}} t \right ) \right ] }$

So:

$\color{#3D99F6} \large \displaystyle i(t_0) = \frac{CE}{\sqrt{LC}} \rightarrow \boxed{ \large \displaystyle \beta= 1}$

$\color{#3D99F6} \large \displaystyle v_A(t_0) = E \rightarrow q_A(t_0) = CE \rightarrow \boxed{ \large \displaystyle \alpha = 1}$

After $t_0$ :

On capacitor $A$ :

$\color{#D61F06} (i) \ \color{#333333} \large \displaystyle i(t) = C\frac{dv_A(t)}{dt}, v_A(t_0) = E$

On capacitor $B$ :

$\color{#D61F06} (ii) \ \color{#333333} \large \displaystyle i(t) = C\frac{dv_B(t)}{dt}, v_B(t_0) = 0$

On inductor:

$\color{#D61F06} (iii) \ \color{#333333} \large \displaystyle L \frac{di(t)}{dt} = v_L(t), i(t_0) = \frac{CE}{\sqrt{LC}}$

Applying laplace again on three equations:

$\color{#D61F06} (i) \ \color{#333333} \large \displaystyle I(s) = C(sV_A(s) - E)$

$\color{#D61F06} (ii) \ \color{#333333} \large \displaystyle I(s) = CsV_B(s)$

$\color{#D61F06} (iii) \ \color{#333333} \large \displaystyle L \left ( sI(s) - \frac{CE}{\sqrt{LC}} \right ) = V_L(s)$

Multuplying $(iii)$ by $Cs$ and rearranging:

$\color{#D61F06} (i) \ \color{#333333} \large \displaystyle I(s) = CsV_A(s) - CE$

$\color{#D61F06} (ii) \ \color{#333333} \large \displaystyle I(s) = CsV_B(s)$

$\color{#D61F06} (iii) \ \color{#333333} \large \displaystyle LCs^2 I(s) = Cs V_L(s) + CE \sqrt{LC}s$

Adding up the three equations and remembering that $V_A+V_B+V_L= 0$ , since we have a closed loop:

$\large \displaystyle (LCs^2+2) I(s) = CE \sqrt{LC}s - CE$

$\large \displaystyle I(s) = CE \left ( \frac{\sqrt{LC}s}{LCs^2+2} - \frac{1}{LCs^2+2} \right )$

$\large \displaystyle I(s) = CE \left ( \frac{1}{\sqrt{LC}} \cdot \frac{s} {s^2+\frac{2}{LC}} - \frac{1}{\sqrt{2LC}} \cdot \frac{ \sqrt{\frac{2}{LC}}} {s^2+\frac{2}{LC}} \right )$

$\large \displaystyle I(s) = \frac{CE}{\sqrt{2LC}} \left ( \sqrt{2} \cdot \frac{s} {s^2+\frac{2}{LC}} - \frac{ \sqrt{\frac{2}{LC}}} {s^2+\frac{2}{LC}} \right )$

So:

$\large \displaystyle i(t) = \frac{CE}{\sqrt{2LC}} \left [ \sqrt{2} \cos \left ( \sqrt{\frac{2}{LC}} t \right ) - \sin \left ( \sqrt{\frac{2}{LC}} t \right ) \right ]$

Multiplying above and below by $\sqrt{3}$ :

$\large \displaystyle i(t) = E \sqrt{\frac{3C}{2L}} \left [ \sqrt{\frac23} \cos \left ( \sqrt{\frac{2}{LC}} t \right ) - \sqrt{\frac13} \sin \left ( \sqrt{\frac{2}{LC}} t \right ) \right ]$

Let $\omega = \sqrt{\frac{2}{LC}}$ and $\phi$ such that $\cos(\phi) = \sqrt{\frac23}$ , we will also have $\sin(\phi) = \sqrt{\frac13}$ . So:

$\large \displaystyle i(t) = E \sqrt{\frac{3C}{2L}} \left [ \cos(\phi) \cos \left ( \omega t \right ) - \sin(\phi) \sin \left ( \omega t \right ) \right ]$

$\color{#20A900} \boxed{\large \displaystyle i(t) = E \sqrt{\frac{3C}{2L}} \cos(\omega t + \phi ) }$

So:

$\color{#3D99F6} \large \displaystyle I_{max} = E \sqrt{\frac{3C}{2L}} \rightarrow \boxed{ \large \displaystyle \gamma = 3, \delta = 2 }$

Thus:

$\color{#3D99F6} \boxed{ \large \displaystyle \alpha + \beta + \gamma + \delta = 7}$

Several intermediate calculation and simplification steps in the solution below are left out.

When $S_1$ is closed, capacitor B is disconnected and the circuit equation reads:

$-E + \frac{Q_A}{C}+L\dot{I}=0$ $I = \dot{Q}_A$

$Q_A(0)=0$ $I(0) = 0$

$\implies \ddot{Q}_A + \frac{Q_A}{LC}=\frac{E}{L}$

The solution to the above equation after considering initial conditions comes out to be:

$Q_A = CE\left(1-\cos\left(\frac{t}{\sqrt{LC}}\right)\right)$ $I = \dot{Q}_A = \frac{CE}{\sqrt{LC}}\cos\left(\frac{t}{\sqrt{LC}}\right)$

It follows immediately that at $t = \frac{\pi \sqrt{LC}}{2}$ :

$Q_A = CE$ $I = \frac{CE}{\sqrt{LC}}$

$\implies \boxed{\alpha = 1 \ ; \ \beta =1}$

At this instant, the switch $S_1$ is opened and $S_2$ is closed. For this scenario let the charge on capacitor A be $Q_A$ and that on capacitor $B$ be $Q_B$ . Treating this instant as time $t=0$ , the convention used here is that at a general time $t$ the current through the inductor flows upwards (as viewed in the given diagram) and capacitor A is discharging while B is charging. Keeping in mind, the circuit equation reads:

$L\dot{I} = \frac{Q_A}{C}-\frac{Q_B}{C} \ \dots (1)$ $I = -\dot{Q}_A = \dot{Q}_B$ $Q_A(0) = CE$ $Q_B(0) = 0$ $I(0) = -\frac{CE}{\sqrt{LC}}$

From (1): $\dot{I}(0) = \frac{E}{L}$

Re-arranging to obtain and ODE governing the evolution of $I$ leads to:

$\ddot{I}+\left(\frac{2}{LC}\right)I=0$

Let:

$\omega^2 = \frac{2}{LC}$

Solving the above equation and applying initial conditions gives:

$I = \frac{CE}{\sqrt{2LC}} \sin\left(\omega t\right) - \frac{CE}{\sqrt{LC}} \cos\left(\omega t\right)$

Using a bit of trigonometry, the current amplitude can be computed. The value is:

$I_{max} = \sqrt{\frac{3C}{2L}}E$

$\implies \boxed{\gamma = 3 \ ; \ \delta =2}$

The required answer is: $\boxed{7}$ .