Question

A 10 ohm resistor and 10 henry inductor are attached in series to a 10 volt DC supply. If the current i(0) = 2.5 amperes, what is the current after 1 second. (round in the tenths place)

Answer 1

\[U=u_L+u_R\]\[U=L\frac{di}{dt}+Ri\]\[\frac{U}{L}=\frac{di}{dt}+\frac{R}{L}i=\frac{di}{dt}+\frac{1}{\tau}i\quad,\quad\tau=\frac{L}{R}=1s\]\[i(t)=C\exp{(-t/\tau)}+\frac{U}{R}=i(t)=\left(I_0-\frac{U}{R}\right)\exp{(-t/\tau)}+\frac{U}{R}\]\[i(t=1s)=\left(I_0-\frac{U}{R}\right)\exp{(-1)}+\frac{U}{R}=(2.5-1)\exp{(-1)}A+1A\approx 1.55A\]