A constant voltage is applied to a series `R-L` circuit by closing the switch. The voltage across inductor `(L=2H)` is `20V` at `t=0` and drops of `5V` and `20 ms`. The value of `R` in `Omega` is

Options

A. `100 ln 2Omega`
B. `100(1-ln2)Omega`
C. `100 ln 4Omega`
D. `100(1-ln4)`

Answer

C

Explanation

Value remains `1/4th` in `20ms` times. Hence, two half-lives equal to `20ms`. So, one half life is `10ms`.
`t_(1//2)=(ln 2)tau_C=(ln 2)L/R`
`:. R=((ln 2)L)/t_(1//2)`
`=((ln 2)(2))/(10xx10^-3)=(100 ln 4)Omega`