In the circuit above, The potential across each capacitor is 100V. The total energy stored in the two capacitors is
Attached Figure / Diagram
A
3.0 x 104J
B
3.0 x 102J
C
2.5 x 10-2J
D
6.0 x 10-3J
Verified Correct Answer: C.
Physical Principles & Numerical Solution
Energy stored in a capacitor is given by
E = \(\frac{1}{2}\)CV\(^2\)
But the effective capacitance for the parallel arrangement is given by
C = C\(_1\) + C\(_2\)
= 2 + 3
= 5\muF = 5 x 10\(^{-6}\) x (100)\(^2\)
∴E = \(\frac{1}{2}\)CV\(^2\)
= \(\frac{1}{2}\) x 5 x 10\(^{-6}\) x 10\(^4\)
2.9 x 10\(^{-2}\)J
Source: Myschool Past Questions Archive (Record #11272)