A coil of inductance `300mh` and resistance `2Omega` is connected to a source of voltage `2V`. The current reaches half of its steady state value in

Options

A. `0.3 s`
B. `0.15 s`
C. `0.1 s`
D. `0.05 s`

Answer

C

Explanation

`L = 300 xx 10^(-3) H`
`R = 10 Omega`
`I = (I_(0))/(2)`
Using `I = I_(0)(1-e^(-Rt//L))`, we get
`(I_(0))/(2) = I_(0) (1-e^(-Rt//L))`
`rArr (1)/(2) = 1 - e^(-Rt//L) rArr e^(-Rt//L) = (1)/(2)`
`rArr e^(Rt//L) = 2 rArr (R )/(L)t = log_(e) 2 = 0.693`
`rArr t = (0693 xx 300 xx 10^(-3))/(2)s = 0.1 s`