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Q.

A capacitor of capacitance 160 μF is charged to a potential difference of 200 V and then connected across a discharge tube which conducts until the potential difference across it has fallen to 100 V. The energy dissipated in the tube is

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a

6.4 J

b

4.8 J

c

3.2 J

d

2.4 J

answer is D.

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Detailed Solution

E=12CV2−V′2=12×160×10−6(200)2−(100)2=12×160×10−6×3×104=2.4J
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