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Q.

A 6 μF capacitor is charged from 10 volts to 20 volts. Increase in energy will be

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a

18×10−4 J

b

9×10−4 J

c

4.5×10−4 J

d

9×10−6 J

answer is B.

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Detailed Solution

ΔE=EFinal −EInifial =12CVFinal 2−VInitial 2=12×6×202−102×10−6=3×(400−100)×10−6=3×300×10−6=9×10−4 J
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A 6 μF capacitor is charged from 10 volts to 20 volts. Increase in energy will be