\(\frac{1}{C_{e q}}=\frac{1}{C_{3}}+\frac{1}{C_{1}+C_{2}}\)

(since \(\mathrm{C}_{1}\) and \(\mathrm{C}_{2}\) are in parallel and which is in series with \({\mathrm{C}}_{3}\) ).

ie, \(\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{\mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}}{\mathrm{C}_{3}\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right)}\)

\(\therefore C_{e q}=\frac{C_{3}\left(C_{1}+C_{2}\right)}{C_{1}+C_{2}+C_{3}}\)

since \(V\) is the voltage of battery, charge, \(\mathrm{q}=\mathrm{C}_{\mathrm{eq}} \mathrm{V}\)

\(=\frac{\mathrm{C}_{3}\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right) \mathrm{V}}{\mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}}\)

If the capacitor \(\mathrm{C}_{3}\) breaks down, then effective capacitance, \(C_{e q}^{\prime}=C_{1}+C_{2}\)

New charge \(q'=\) \(\mathrm{C}_{\mathrm{eq}}^{\prime} \mathrm{V}=\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right) \mathrm{V}\)

Change in total charge \(=q^{\prime}-q\)

\(=\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right) \mathrm{V}-\frac{\mathrm{C}_{3}\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right) \mathrm{V}}{\mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}}\)

\(=\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right) \mathrm{V}\left[1-\frac{\mathrm{C}_{3}}{\mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}}\right]\)