Question

# Two capacitors $$C_{1}=2\mu F$$ and $$C_{2}=6\mu F$$ in series, are connected in parallel to a third capacitor $$C_{3}=4\mu F$$ . This arrangement is then connected to a battery of e.m.f.=2 V, as shown in figure. The energy lost by the battery in charging the capacitors is :

A
22×106J
B
11×106J
C
(323)×106J
D
(163)×106J

Solution

## The correct option is B $$11\times 10^{-6}J$$total charge flown is $$Q=C\cdot V$$$$=\frac {11}{2}\times 120^{-6}\times 2$$$$Q=11\times 10^{-6}$$Now energy lost by battery in charging $$=\frac {1}{2}QV$$$$=\frac {1}{2}\times 11\times 10^{-6}\times 2J$$$$=11\times 10^{-6}J$$PhysicsNCERTStandard XII

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