Answer
(a) $I(t)=Ie^{\frac{-Rt}{L}}$
(b) $t=\frac{L}{R}ln2$
(c) $I(t) = I/e$
Work Step by Step
(a)
$L\frac{di}{dt}+RI=0$
$\frac{di}{dt}+\frac{R}{L}t=0$
It is a linear first order differential problem.
$I(t) = ce^-\int R/Ldt$
$I(t)=Ie^{\frac{-Rt}{L}}$
(b)
$(1/2)I = Ie^{\frac{-Rt}{L}}$
$ln(1/2) = -\frac{Rt}{L}$
$t = \frac{L\times ln2}{R}$
(c)
$I(t)=Ie^{\frac{-Rt}{L}}$
At t=L/R
$I=Ie^\frac{-RL}{LR}$
$I(t) = \frac{I}{e}$
