Two capacitors of \( 3 \mu \mathrm{F} \) and \( 6 \mu \mathrm{F} \) are connected in series and a potential difference of \( 900 \mathrm{~V} \) is applied across the combination. They are then disconnected and reconnected in parallel. The potential difference across the combination is

If \( Q \) is charge on capacitor and \( C \) is capacitance then, potential difference \( V=\frac{Q}{C} \)
When two capacitors are connected in series, then equivalent capacitance is
\(\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{1}{3 \mu F}+\frac{1}{6 \mu F} \)
\(\Rightarrow C=\frac{6 \times 3}{3+6}=2 \mu F \)
Now, \(\mathrm{Q}=C \times V \)
Given, \(\mathrm{V}=900 \mathrm{~V}\)
Therefore, \(Q=2 \times 900=1800 \mu C\)
When two capacitors are connected in parallel, then equivalent capacitance is
\(C=C_{1}+C_{2}=3 \mu F+6 \mu F=9 \mu F \)
Therefore \(V=\frac{Q}{C}=\frac{1800 \mu C}{9 \mu F}=200 \mathrm{~V}\)
Thus, potential difference across the combination is \( 200 \mathrm{~V} \)