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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

A

`2.5muF`

B

`15muF`

C

`1.5 muF`

D

`0.1muF`

Text Solution

Verified by Experts

The correct Answer is:
D
`C_(eq) = 2C+C/2 = (5C)/2`
`Q= C_(eq)xxV`
`implies 1.5 = (5C)/2 xx 6`
`implies C =1/10 muF =0.1 muF`
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