Answer
a) $I(t)$ = $\frac{V}{R}(1-e^{-(\frac{R}{L})t})$
b) $0.632\frac{V}{R}$
c) $0.0184$ seconds
Work Step by Step
a)
$\frac{dI}{dt}+\frac{R}{L}I$ = $\frac{1}{L}V$
$A(t)$ = $\frac{R}{L}$
$B(t)$ = $\frac{1}{L}V(t)$
$α(x)$ = $e^{(\frac{R}{L})t}$
$[e^{(\frac{R}{L})t}I]'$ = $e^{(\frac{R}{L})t}(\frac{V}{L})$
$e^{(\frac{R}{L})t}I$ = $\frac{V}{R}e^{(\frac{R}{L})t}+C$
$I(t)$ = $\frac{V}{R}+Ce^{-(\frac{R}{L})t}$
$I(0)$ = $0$
$0$ = $\frac{V}{R}+C$
$C$ = $-\frac{V}{R}$
$I(t)$ = $\frac{V}{R}(1-e^{-(\frac{R}{L})t})$
b)
$t \to \infty$, $e^{-(\frac{R}{L})t}) \to 0$ so
$I(t)$ = $\frac{V}{R}$
$t$ = $\frac{L}{R}$
$I(\frac{L}{R})$ = $\frac{V}{R}(1-e^{-(\frac{R}{L})(\frac{L}{R})})$ $\approx$ $0.632\frac{V}{R}$
c)
$\frac{9}{10}$ = $1-e^{-(\frac{R}{L})t}$
$t$ = $\frac{L}{R}\ln{10}$
$L$ = $4$
$R$ = $500$
so
$t$ $\approx$ $0.0184$ seconds
