Q. Two capacitors each of capacity 2$\mu$F are connected in parallel. If they are connected to 100 V battery ,then energy stored in them is
A) 0.02 J
B) 0.04 J
C) 0.01 J
D) 200 J
Explanation
(a) The energy stored is given by
$E = \frac{1}{2}CV^{2}$
When capacitors are connected in parallel, resultant capacitance is
$C^{'} = \ C_{1} + C_{2}$
$= 2\mu F + 2\mu F = 4\mu F$
$V = 100$ volt
$\therefore = \frac{1}{2} \times 4 \times 10^{- 6} \times (100)^{2}\ $
$E = 0.02$J
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.