The diagram above shows two capacitors P and Q of capacitance 2μF and 4μF respectively connected to a d.c source. The ratio of energy stored in P to Q is
Attached Figure / Diagram
A
4:1
B
2:1
C
1:4
D
1:2
Verified Correct Answer: D.
Physical Principles & Numerical Solution
Energy = 1/2CV2
∴Energy of P = 1/2 x 2 x V2 = V2
Energy of Q = 1/2 x 4 x V2 = 2V2
∴ P/Q = V2/2V2 = 1/2 => P:Q = 1:2
Source: Myschool Past Questions Archive (Record #11838)