A capacitor of `1 muF` is placed in series with a resistor of 2 mega-ohm and a battery of emf 2 V. Calculate the time after which the charge will grow to 86.74% of its max value.

Explanation

Here, `C = 1 mu F = 10^(-6) F`,
`R = 2 mega-ohm = 2 xx 10^(6) Omega, V = 2`volt
`t = ? (q)/(q_(0)) = (86.47)/(100)`
`tau = RC = 2 xx 10^(6) xx 10^(-6) = 2 s`
From `q = q_(0) (1 - e^(-t//tau))`
`(q)/(q_(0)) = 1 - e^(-t//tau)`
or `e^(-t//tau) = 1 - (q)/(q_(0)) = 1 = (86.74)/(100) = (13.53)/(100)`
`e^(-t//2) = (13.53)/(100)` or `e^(t//2) = (100)/(13.53)`
`(t)/(2) = (log_(e) 100 - log_(e) 13.53)`
`= 2.303 (2 - 1.1310)`
`t = 2 xx 2.303 (0.8690) = 4.0 s`