Four identical capacitors are connected as shown in diagram. When a battery of `6 V` is connected between `A` and `B`, the charges stored is found to be `1.5 muC`. The value of `C_(1)` is

Options

A. `2.5 muF`
B. `15 muF`
C. `1.5 muF`
D. `0.1 muF`

Answer

D

Explanation

The capacitance across `A` and `B`
`= (C_(1))/(2) + C_(1) + C_(1) = (5)/(2)C_(1)`
As `Q = CV`,
`1.5 muC = (5)/(2)C_(1) xx 6`
`rArrC_(1) = (1.5)/(15) xx 10^(-6) = 0.1 xx 10^(-6) F = 0.1muF`