The energy stored in the above capacitor is
-
2.25J
- 6.25J
- 18.25J
- 24.25J
The correct answer is: A
Explanation
The energy stored in the capacitor = \(\frac{1}{2}\)\(\frac{q^2}{C}\)
Where C = 2F, q = 3C
= \(\frac{1}{2}\)\(\frac{3^2}{2}\) = \(\frac{9}{4}\) = 2.25J
There is an explanation video available .