A 300 μF capacitor is charged by 90 volt. Once it is charged battery is removed. Now another uncharged capacitor of capacitance 600 μF is connected across it (in parallel). The value of common potential is

30 volt

The correct answer is Option (1) → 30 volt

Common potential

$V=\frac{C_1 V_1+C_2 V_2}{C_1+V_2}$

$=\frac{300 \times 90+600 \times 0}{300+600}$

$=\frac{300 \times 90}{900}=30 V$