Answer
(a) $I(t) = 4 -4e^{-5t}$
(b) The current after 0.1 seconds is 1.57 A
Work Step by Step
(a) $L\frac{dI}{dt}+RI = E(t)$
$\frac{dI}{dt}+\frac{R}{L}I = \frac{E(t)}{L}$
$\frac{dI}{dt}+\frac{10}{2}I = \frac{40}{2}$
$\frac{dI}{dt}+5I = 20$
We can find the integrating factor:
$I(x) = e^{\int 5~dt} = e^{5t}$
We can multiply by the integrating factor:
$(e^{5t})~\frac{dI}{dt}+5e^{5t}~I = 20e^{5t}$
$\frac{d}{dt}(e^{5t}~I) = 20e^{5t}$
We can integrate both sides of the equation:
$e^{5t}~I = \int 20e^{5t}~dt$
$e^{5t}~I = 4e^{5t} + C$
$I = 4 + Ce^{-5t}$
We can use the initial condition to find $C$:
$I(0) = 0$
$4 + Ce^{-5(0)} = 0$
$4+C = 0$
$C = -4$
We can write the solution:
$I(t) = 4 -4e^{-5t}$
(b) We can find the current after 0.1 seconds:
$I(t) = 4 -4e^{-5t}$
$I(0.1) = 4 -4e^{-5(0.1)}$
$I(0.1) = 4 -4e^{-0.5}$
$I(0.1) = 1.57$
The current after 0.1 seconds is 1.57 A
