An inductor of `2` henry and a resistance of `10` ohms are connected in series with a battery of `5` volts. The initial rate of change of current is

Options

A. `0.5 A s^(-1)`
B. `2.0 A s^(-1)`
C. ` 2.5 A s^(-1)`
D. ` 0.25 A s^(-1)`

Answer

C

Explanation

`I=I_o(1-e^(-t//tau))`
`(d tau)/(dt)=-I_0(-tau)e^(-t//tau)`
at `t=0,(dI)/(dt)=I_0//tau=v/R R/L`
`rarr=(d tau)/(dt)=r/L=5/2=2.5A//S`