In the circuit above, The potential across each capacitor is 100V. The total energy stored in the two capacitors is
-
A.
3.0 x 104J -
B.
3.0 x 102J -
C.
2.5 x 10-2J -
D.
6.0 x 10-3J
Correct Answer: Option C
Explanation
Energy stored in a capacitor is given by
E = 1/2CV2
But effective capacitance for the parallel arrangement is given by
C = C1 + C2
= 2 + 3
= 5μF = 5 x 10-6 x (100)2
∴E = 1/2CV2
= 1/2 x 5 x 10-6 x 104
2.9 x 10-2J