A`50` volt potential difference is suddenly applied to a coil with `L=5xx10^(-3)` henry and `R=180 ohm`. The rate of increase of current after `0.001` second is

Options

A. `27.3 amp//sec`
B. `27.8 amp//sec`
C. `2.73 amp//sec`
D. none of the above

Answer

D

Explanation

The rate of increase of current
`=(di)/(dt)=(d)/(dt)i_(0)(1-e^(-Rt//L))=(d)/(dt)i_(0)-(d)/(dt)i_(0)e^(-Rt//L)`
`=0-i_(0)e^(-Rt//L).(d)/(dt)(-(Rt)/(L))=i_(0)(R )/(L)e^(-Rt//L)`
`=(50)/(180)xx(180)/(5xx10^(-3))xxe^(-(180xx0.001)//(5xx10^(-3)))`