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

Options

A. 2s
B. 0.693s
C. 0.5s
D. 1.0s

Answer

(b)

Explanation

Charging , `q=q_(0)( 1 - e^(-1/CR))`
`rArr e^(-1/CR) = 1 - q/q_(0) = 1 - 0.5`
`e^(-1/CR) = 0.5 `
`1/CR = 2.3.26log_(10)^(2) = 0.693 S`