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Question: A \(6\mu F\) capacitor is charged from \(10\mspace{6mu} volts\) to\(20\mspace{6mu} volts\). Increase...

A 6μF6\mu F capacitor is charged from 10 6muvolts10\mspace{6mu} volts to20 6muvolts20\mspace{6mu} volts. Increase in energy will be

A

18×10−4J18 \times 10^{- 4}J

B

9×10−4J9 \times 10^{- 4}J

C

4.5×10−4J4.5 \times 10^{- 4}J

D

9×10−6J9 \times 10^{- 6}J

Answer

9×10−4J9 \times 10^{- 4}J

Explanation

Solution

ΔE=EFinal−EInitial=12C(VFinal2−VInitial2)\Delta E = E_{Final} - E_{Initial} = \frac{1}{2}C(V_{Final}^{2} - V_{Initial}^{2})

=12×6×(202−102)×10−6= \frac{1}{2} \times 6 \times (20^{2} - 10^{2}) \times 10^{- 6}

=3×(400−100)×10−6=3×300×10−6=9×10−4J= 3 \times (400 - 100) \times 10^{- 6} = 3 \times 300 \times 10^{- 6} = 9 \times 10^{- 4}J