A \( 300\,\mu\mathrm{F} \) capacitor is charged by 90 volt. Once it is charged battery is removed. Now another uncharged capacitor of capacitance \( 600\,\mu\mathrm{F} \) is connected across it (in parallel). The value of common potential is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 30 volt
The battery is removed before the second capacitor is connected, so the charge is conserved: the charge on the first capacitor is shared until both are at the same common potential.
\( \displaystyle Q = C_1 V_1 = 300\,\mu\mathrm{F} \times 90\,\mathrm{V} = 27000\,\mu\mathrm{C} \)
In parallel the total capacitance is \( C_1 + C_2 \):
\( \displaystyle V = \frac{C_1 V_1}{C_1 + C_2} = \frac{27000\,\mu\mathrm{C}}{(300 + 600)\,\mu\mathrm{F}} = 30\,\mathrm{V} \)