Two capacitors of capacitance $4 \mu F$ and $12 \mu F$ respectively are connected in series and the combination is connected momentarily across a $200\; V$ battery. The charged capacitors are now isolated and connected in parallel, similar charged plates being connected together. Calculate the common potential.

Given,
$C_1 = 4 \; \mu F$
$C_2 = 12 \; \mu F$
Potential difference $(V) = 200\;V$

When the capacitors are connected in series, then the equivalent capacitance $(C)$ is obtained as,

$C = $ $ \frac{C_1 \; * \; C_2}{C_1 \;+\;C_2} = \frac{4 \; * \; 10^{-6} * \; 12\; * \; 10^{-6}}{4 \;  * \; 10^{-6} \; + \; 12 \; * \; 10^{-6}}$ $ = 3\; * \; 10^{-6}\;F$

In series combination, charge on each capacitor;

$q = q_1 = q_2 = C\;V = 3\;*10^{-6}\; * \; 200 = 6\; * \; 10^{-4}\;C$

Let V be the common potential difference after parallel combination of capacitors, then

Total charge in parallel combination

$ = (C_1 \; + \; C_2)\;V = (4 \; + \; 12)\;*\;10^{-6}\;V = 16 \; * \; 10^{-6}\;V$

Since total charge is conserved,
Total charge in series = Total charge in parallel

or, $q_1 \; + \; q_2 = 16\; * \; 10^{-6}\; V$

⇒ $V = $ $\frac{12 \; * \; 10^{-4}}{16 \; * \; * 10^{-6}}$

∴ $V = 75\;V$

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