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A capacitor of capacity 0.1 muF connecte...

A capacitor of capacity `0.1 muF` connected in series to a resistor of `10 M Omega` is charged to a certain potential and then made to discharge through resistor. The time in which the potential will take to fall to half its original value is (Given, `log_(10) 2 = 0.3010`)

A

`2 s`

B

`0.693 s`

C

`0.5 s`

D

`1.0 s`

Text Solution

Verified by Experts

The correct Answer is:
B

By equation of charging,
`q = q_(0)(1 - e^(-t//CR))`
According to question,
`(q)/(q_(0)) = (1)/(2) = 0.50 :. 0.50 = 1 - e^(-t//CR)`
`e^(-t//CR) = 1 - 0.50 = 0.50 rArr e^(t//CR) = 2`
or (t)/(CR) = log_(e)2 or (t)/(CR) = 2.3026 log_(10)2`
or `t = CRxx2.3026log_(10)2`
or `t = 0.1xx10^(-6)xx10xx10^(6)xx2.3026 log_(10)2`
or `t = 2.3026xx0.3010 or t = 0.693 s`
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