If two capacitors `C_(1)` and `C_(2)` are connected in a parallel combination then the equivalent capacitance is `10 mu F `. If both the capacitors are connected across a l V battery, then energy stored in `C_(2)` is 4 times of that in `C_(1)`. The equivalent capacitance if they are connected in series is

Options

A. `16 mu F `
B. `1.6mu F `
C. `4 mu F `
D. `(1)/(4) mu F `

Answer

A

Explanation

For parallel connection
and from question
`4 ((1)/(2) C_(1)^(2))= (1)/(2) C_(2)V^(2)`
`:. 4 C_(1)= C_(2)`
By using equation (2) in equation (1),
`C_(1) + 4 C_(1) = 10 mu F `
` :. 5 C_(1) = 10 mu F`
`:. C_(1) = 2 mu F `
From equation (2)
`C_(2) = 4 C_(1)`
`= 4xx2`
`= 8 mu F `
Now, equivalent capacitance of series connection of `2mu`F and `8mu`F
`C = (C_(1)C_(2))/(C_(1)+C_(2))=(2xx8)/(2+8)= (16)/(10)`
`= 1.6 mu F `