A `mu F` capacitor is charged by a `400 V` supply through `0.1 M Omega` resistance. The time taken by the capacitor to develop a potential difference of `300 V` is : (Given `log_(10) 4 = 0.602`)

Options

A. `2.2 sec`
B. `1.1 sec`
C. `0.55 sec`
D. `0.48 "secs"`

Answer

D

Explanation

`tau = (L)/(R ) = RC`
`R = sqrt(L//C) = sqrt(8//2 xx 10^(-6)), R = 2 K Omega`