First slide

Capacitance

Question

A capacitor of capacitance C is charged by connecting it to a battery. The battery is now removed and this capacitor is connected to a second uncharged capacitor of capacitance C'. If the charge distributes equally on the two capacitors, the ratio of the total energy stored in the capacitors after connection to the total energy stored in them before connection is:

Easy

A

1
B

1/2
C

$1 / \sqrt{2}$
D

1/3

Solution

$E_{i} = \frac{Q^{2}}{2 C} . . . . (1) After connection, E_{f} = \frac{{(Q_{2}^{'})}^{2}}{2 C} + \frac{{(\frac{Q}{2})}^{2}}{2 C^{'}} ∵ C = C^{'} \Rightarrow \frac{E_{f}}{E_{i}} = \frac{1}{2}$

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