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2. Electric Potential and Capacitance
easy
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
$24\,m\,F$
Solution
(a) $U = \frac{1}{2}C{V^2}$
so $24 \times 60 \times 60 = \frac{1}{2}C{(1200)^2}$
$⇒ C=120 \,mF$
Standard 12
Physics