Solution

Let q be the charge that flows from the positive plate of$C_0$ toward C. Then the charge on$C_0$ is $Q_0 - q$(with $Q_0 = C_0 V_0$) and the charge on C is q. The voltages are

$V_{C0} = \frac{Q_0 - q}{C_0}, \qquad V_C = \frac{q}{C}$

The circuit equation is obtained from the mesh:

$V_{C0} - V_C = L \frac{dI}{dt} = L \frac{d^2q}{dt^2}$

Substituting the above expressions,

$\frac{Q_0}{C_0} - \frac{q}{C_0} - \frac{q}{C} = L \ddot{q}$

Rearranging, we obtain the differential equation of the forced harmonic oscillator:

$L \ddot{q} + q\left(\frac{1}{C_0} + \frac{1}{C}\right) = V_0$

The natural angular frequency of the circuit is

$\omega = \frac{1}{\sqrt{L\,C_{\text{eq}}}} = \sqrt{\frac{C+C_0}{L\,C\,C_0}}$

where
$C_{\text{eq}} = \dfrac{C\,C_0}{C+C_0}$ is the equivalent series capacitance.

The general solution with the initial conditions $q(0)=0$(capacitor C starts uncharged) and$I(0)=\dot{q}(0)=0$ (the inductor prevents abrupt changes in current) is

$q(t) = C_{\text{eq}} V_0 \bigl(1 - \cos\omega t\bigr)$

Therefore, the voltage on capacitor C evolves as

$V_C(t) = \frac{q(t)}{C} = \frac{C_{\text{eq}}}{C}\,V_0 \bigl(1 - \cos\omega t\bigr) = \frac{V_0}{1 + \dfrac{C}{C_0}} \bigl(1 - \cos\omega t\bigr)$

The maximum value that V_C can reach is

$V_{C,\max} = \frac{2V_0}{1 + \dfrac{C}{C_0}}$

If the breakdown voltage V is greater than this maximum, capacitor C never breaks down.
Otherwise $(V < V_{C,\max})$ breakdown occurs at the instant $\tau$when $V_C(\tau) = V$

Solving for the cosine and then for time:

$\cos\omega\tau = 1 - \left(1 + \frac{C}{C_0}\right)\frac{V}{V_0}$

$\boxed{\tau = \frac{1}{\omega} \arccos\!\left[1 - \left(1 + \frac{C}{C_0}\right)\frac{V}{V_0}\right]}$