When three identical capacitors are connected in series, their equivalent capacitance is `2muF` . Now they are connected in parallel across a source of e.m.f. 200 V. The total energy stored is

Options

A. 0.36 J
B. 0.48 J
C. 0.16 J
D. 0.08 J

Answer

A

Explanation

`C_(1)=C_(2)=C_(3)=C,C_(s)=2muF,V=200V,E_(P)=?`
`C_(s)=C/n" "thereforeC=C_(s)*n=2xx3=6muF`
`E_(P)=1/2C_(P)V^(2)`
`=1/2[C_(1)+C_(2)+C_(3)]V^(2)`
`=1/2xx18xx10^(-6)xx4xx10^(4)`
`= 0.36J`