A $10\,\mu F$ capacitor is charged to a potential difference of $50\;V$ and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes $20\;volt$. The capacitance of second capacitor is....$\mu F$
Medium
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(d) $V = \frac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}$ $==>$ $20 = \frac{{10 \times 50 + {C_2} \times 0}}{{10 + {C_2}}}$
$==>$ $200 + 20{C_2} = 500$ $==>$ ${C_2} = 15\,\mu F$
$==>$ $200 + 20{C_2} = 500$ $==>$ ${C_2} = 15\,\mu F$
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