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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 :

11381_c8c0645114eb453080836dfb344c5854.png


A
22×106J
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B
11×106J
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C
(323)×106J
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D
(163)×106J
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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 $$=\frac {1}{2}QV$$
$$=\frac {1}{2}\times 11\times 10^{-6}\times 2J$$
$$=11\times 10^{-6}J$$
64246_11381_ans_a3250b2f08054416a46068b8db519b3a.png

Physics
NCERT
Standard XII

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