At What Rate Is The Current Increasing After 1.2 ms After 45 V Potential Difference Is Applied To A Coil?

Question

A 45 V potential difference is suddenly applied to a coil with L= 50 mH and R= 180 Ohms. At what rate is the current increasing after 1.2 ms?

I don't know where to start on this problem... all I got is ΔV = (ΔU)/q

Answer 1

The general solution is Ldi/dt + Ri=45,so Li'+Ri=45,i'+R/L X i =V/L

then,(e^(R/L)t X i)'=V/LXe^(R/L)t then i=V/R +K X e^(-R/L)t.if i(0)=0,then

K=-V/R so i=45/180 {1-e^(-R/L)t}=0.25{1-e^(-3600)t}.Now t=1.2msec

then i=0.25{1-e^-3600x1.2/1000)=.25x(1-.0133)... amp =246 ma