capacitor is used to store $24\, watt\, hour$ of energy at $1200\, volt$. What should be the capacitance of the capacitor
A$120\,m\,F$
B$120\,\mu \,F$
C$24\,\mu \,F$
D$24\,m\,F$
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A$120\,m\,F$
a (a) $U = \frac{1}{2}C{V^2}$
so $24 \times 60 \times 60 = \frac{1}{2}C{(1200)^2}$
$⇒ C=120 \,mF$
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