Question

maximum current that will flow in the circuit.

Answer 1

$q_1=8C{V}_0$: $q_2=C{V}_0$
$q_1+q_2=9C{V}_0$
for $I_{max}$, all the energy stored in capacitors will be converted to energy in Inductor.

$1/2\times 2C\times (4V_o{)}^2+1/2\times C\times (V_o{)}^2=1/2\times L\times I_{max}^2$

$9/2C(V_o{)}^2=1/2L{I}_{max}^2$

$\therefore I_{max}=\sqrt{\frac{9C{V}_o^2}{L}}$