Inductor and Capacitor

Calling the left capacitor as $A$ and the right one as $B$ :

$\large \displaystyle \color{#D61F06} (i) \ \color{#333333} i = C \frac{dv_A}{dt}, v_A(0) = V_0$

$\large \displaystyle \color{#D61F06} (ii) \ \color{#333333} i = 3C \frac{dv_B}{dt}, v_B(0) = 2V_0$

$\large \displaystyle \color{#D61F06} (iii) \ \color{#333333} L \frac{di}{dt} = v_L, i(0) = 0$

Applying Laplace transform :

$\large \displaystyle \color{#D61F06} (i) \ \color{#333333} I(s) = C(sV_A(s) - V_0)$

$\large \displaystyle \color{#D61F06} (ii) \ \color{#333333} I(s) = 3C(sV_B(s) - 2V_0)$

$\large \displaystyle \color{#D61F06} (iii) \ \color{#333333} LsI(s) = V_L(s)$

Multiplying $(i)$ by $3$ , $(iii)$ by $3Cs$ and rearranging:

$\large \displaystyle \color{#D61F06} (i) \ \color{#333333} 3I(s) = 3CsV_A(s) - 3CV_0$

$\large \displaystyle \color{#D61F06} (ii) \ \color{#333333} I(s) = 3CsV_B(s) - 6CV_0$

$\large \displaystyle \color{#D61F06} (iii) \ \color{#333333} 3LCs^2I(s) = 3CsV_L(s)$

Adding up the three equations, and having in mind that $V_A(s) + V_B(s) + V_L(s) = 0$ , since we have a closed loop:

$\large \displaystyle (3LCs^2+4)I(s) = -9CV_0$

$\large \displaystyle I(s) = -\frac{9CV_0}{3LCs^2+4}$

$\large \displaystyle I(s) = -V_0 \frac32 \sqrt{\frac{3C}{L}} \frac{ \frac{2}{\sqrt{3LC}} }{s^2+\frac{4}{3LC}}$

$\large \displaystyle \color{#20A900} \boxed{ i(t) = -V_0 \frac32 \sqrt{\frac{3C}{L}} \sin \left ( \frac{2}{\sqrt{3LC}} t \right ) }$

So, in magnitude:

$\large \displaystyle \color{#3D99F6} I_{max} = V_0 \frac32 \sqrt{\frac{3C}{L}} \rightarrow \boxed{ \large \displaystyle a = 3, b = 2, c = 3 }$

For capacitor $A$ , from $(i)$ :

$\large \displaystyle V_A(s) = \frac{I(s) + CV_0}{Cs}$

$\large \displaystyle V_A(s) = -\frac{9V_0}{s(3LCs^2+4)} + \frac{V_0}{s}$

$\large \displaystyle V_A(s) = -\frac{9V_0}{4} \left ( \frac{1}{s} - \frac{s}{s^2+\frac{4}{3LC}} \right ) + \frac{V_0}{s}$

$\large \displaystyle V_A(s) = \frac{9V_0}{4} \cdot \frac{s}{s^2+\frac{4}{3LC}} - \frac{5V_0}{4s}$

So:

$\large \displaystyle v_A(t) = \frac{9V_0}{4}\cos \left ( \frac{2}{\sqrt{3LC}}t \right ) - \frac{5V_0}{4}$

Notice that, as expected, the argument of the $\cos$ in the voltage is the same of the $\sin$ in the current. So, since we want the maximum current, the $\sin$ in the current was $1$ , so the $\cos$ in the voltage will be $0$ . So:

$\color{#3D99F6} \large \displaystyle v_A(t@I_{max}) = \left | - \frac{5V_0}{4} \right | = \frac{5V_0}{4} \rightarrow \boxed{\large \displaystyle d = 5, e = 4}$

Likewise for capacitor $B$ :

$\large \displaystyle V_B(s) = \frac{I(s) + 6CV_0}{3Cs}$

$\large \displaystyle V_B(s) = -\frac{9V_0}{4} \left ( \frac{1}{s} - \frac{s}{s^2+\frac{4}{3LC}} \right ) + \frac{2V_0}{s}$

$\large \displaystyle V_B(s) = \frac{3V_0}{4} \cdot \frac{s}{s^2+\frac{4}{3LC}} - \frac{5V_0}{4s}$

So:

$\large \displaystyle v_B(t) = \frac{3V_0}{4}\cos \left ( \frac{2}{\sqrt{3LC}}t \right ) - \frac{5V_0}{4}$

Likewise, since the $\cos$ will be $0$ :

$\color{#3D99F6} \large \displaystyle v_B(t@I_{max}) = \left | - \frac{5V_0}{4} \right | = \frac{5V_0}{4} \rightarrow \boxed{\large \displaystyle f = 5, g = 4}$

Thus:

$\color{#3D99F6} \boxed{\large \displaystyle a+b+c+d+e+f+g=26}$

Nice one.

Let the charge on the capacitor $C_1=C$ be $q_1$ and that on $C_2 = 3C$ be $q_2$ . Let the instantaneous current be $I$ . As the capacitors start discharging, the circuit equations read:

$\frac{q_1}{C_1} + \frac{q_2}{C_2} -L\dot{I}=0 \ \dots (1)$ $I = -\dot{q}_1 = -\dot{q}_2$

Simplifying the equations leads to a differential equation for circuit current as such: $\ddot{I} +\frac{4}{3LC}I = 0$ $I(0) = 0$

From (1):

$\dot{I}(0) = \frac{3V_o}{L}$

$\implies I = A \cos\left(\frac{2t}{\sqrt{3LC}}\right)+B \sin\left(\frac{2t}{\sqrt{3LC}}\right)$

Plugging in initial conditions and solving for $A$ and $B$ leads to the solution:

$\boxed{I = \frac{3V_o\sqrt{3C}}{2\sqrt{L}}\sin\left(\frac{2t}{\sqrt{3LC}}\right)}$

Now:

$\dot{q}_1 = -I$ $\dot{q}_2 = -I$

$q_1(0) = CV_o \ ; \ q_2(0)=6CV_o$

Solving for the charges on the capacitor gives:

$\boxed{q_1 = \frac{9CV_o}{4}\cos\left(\frac{2t}{\sqrt{3LC}}\right) - \frac{5CV_o}{4}}$ $\boxed{q_2 = \frac{9CV_o}{4}\cos\left(\frac{2t}{\sqrt{3LC}}\right) + \frac{15CV_o}{4}}$

The current is maximum when:

$\sin\left(\frac{2t}{\sqrt{3LC}}\right)=1$ $\implies \boxed{I_{max} =\frac{3V_o\sqrt{3C}}{2\sqrt{L}}}$

At the instant, the magnitude of potential difference across the capacitors are:

$\boxed{\frac{q_1}{C} = \frac{5V_o}{4}}$ $\boxed{\frac{q_2}{3C} = \frac{5V_o}{4}}$

Therefore:

$a = 3$ $b = 2$ $c=3$ $d = 5$ $e = 4$ $f = 5$ $g = 4$