Question 193254
ln i/I=Rt/L

FIrst of all i didnt get this point (the current to reach 0.600 of the initial value, but still im considering it to be 0.6 of initial value)

Calculate in seconds it take the current to reach 0.600 of the initial value if I=0.584A, R=7.44 ohms and L=1.41H.

ANS) i here is the current.
I is the current at time t= 0.
we need to calculate the time in "seconds" 
         when????
         when current reaches 0.600 i.e. i = 0.6 of the initial value.
        
            what is the initial value given I = 0.584A

             hence i = 0.6 of I = 0.6*0.584 = 0.3054A

therefore

ln(i/I) = R*t/L   ==>    ln(0.6*I/I) = 7.44*t/1.41   

      ==>   ln(0.6)* 0.1895 = t (in seconds)