Rankers Physics

Capacitors: Practice Problem & Solution

A capacitor of capacity $C_1$ charged upto $V$ volt and then connected to an uncharged capacitor of capacity $C_2$. Then final P.D. across each will be: (2002)
$\frac{C_2V}{C_1+C_2}$
$\frac{C_1V}{C_1+C_2}$
$\left(1+\frac{C_2}{C_1}\right)$
$\left(1-\frac{C_2}{C_1}\right)V$

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Capacitors. Understanding the underlying formula is key to arriving at the correct answer below:

The common potential $V'$ after connection is the total charge divided by the total capacitance. Initial total charge is $q = C_1V + 0$. Total capacitance is $C_1 + C_2$. Therefore, $V' = \frac{C_1V}{C_1 + C_2}$.

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