A capacitor of 2 `mu`F is charged as shown in the diagram . When the switch S is turned to position 2, the percentage of its stored energy dissipated is

Options

A. `20 %`
B. `75%`
C. `80%`
D. `0%`

Explanation

By connecting switch S from point 1 to point 2, constant charge on capacitor, q = `C_(i)V = 2`V · `:.` Energy stored in 2 `mu`F capacitor,
`:. U_(1)=(q^(2))/(2C_(i))=(q^(2))/(2xx2)=(q^(2))/(4)`
Now, by connecting S with point 2, both capacitors will be in parallel connection. So, equivalent capacitance, C = 2 + 8 = 10 `mu`F · `:.` Energy stored with this arrangement,
`U_(2) =(q^(2))/(2C) = (q^(2))/(2xx10)= (q^(2))/(20)`
`:.` Percentage of energy lost =`(U_(1)-U_(2))/(U_(1))xx100%`
`((q^(2))/(4)-(q^(2))/(20))/(q^(2)/(4))xx100%`
`=(q^(2)/(5))/((q^(2))/(4))xx100%`
= 80 %