A cell of `1.5V` is connected across an inductor of `2mH` in series with a `2 Omega` resistor. What is the rate of growth of current immediately after the cell is switched on.

Explanation

`E = L(dI)/(dt) + IR`, therefore, `(dI)/(dt) = (E- =IR)/(L)`
`E = 1.5 "Volt", R = 2 Omega, L = 2mH = 2 xx 10^(-3)H`
When the cell is switched on, `I = 0`
Hence `(dI)/(dt) = (E)/(L) = (1.5)/(2 xx 10^(-3))As^(-1) = 750 As^(-1)`