Calculate the energy stored in a 20\(\mu F\) capacitor. If the p.d between the plates is 40V
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Correct Answer: Option B
Explanation:
W = \(\frac{cv^2}{2} = \frac{20 \times 10^{-6} \times 40^{2}}{2}\)
= 1.6 x 10-2J
W = \(\frac{cv^2}{2} = \frac{20 \times 10^{-6} \times 40^{2}}{2}\)
= 1.6 x 10-2J