Three capacitors $C_1,\,C_2$ and $C_3$ are connected as shown in the figure to a battery of $V\,volt$. If the capacitor $C_3$ breaks down electrically the change in total charge on the combination of capacitors is
AIIMS 2010, Diffcult
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Equivalent capacitance of circuit,
$\frac{1}{C_{e q}}=\frac{1}{C_{3}}+\frac{1}{C_{1}+C_{2}}$
(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 $=q^{\prime}-q$
$=\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right) \mathrm{V}-\frac{\mathrm{C}_{3}\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right) \mathrm{V}}{\mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}}$
$=\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right) \mathrm{V}\left[1-\frac{\mathrm{C}_{3}}{\mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}}\right]$
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