Answer to Question #317101 in Differential Equations for bab

Question

Consider an electric circuit with an inductance of 5 henries

and a resistance of 12 ohms in series with an e.m.f 120sin200t

volts. Find the current

Answer 1

Due to Kirchhoff’s law:

$L\frac {di}{dt}+Ri=e$

$5\frac {di}{dt}+12i=120\sin{200t}$

The solutions to a nonhomogeneous equation are of the form

$i(t) = i_c(t) + i_p(t)$

where $i_c(t)$ is the general solution to the associated homogeneous equation and $i_p(t)$ is a particular solution.

The associated homogeneous equation:

$5\frac {di}{dt}+12i=0$

$i_c(t)=Ce^{-\frac {12}{5}t}$

$i_p(t)=A\sin{200t}+B\cos{200t}$

$i_p’(t)=A\cdot200\cos{200t}-B\cdot200\sin{200t}$

Now put these into the original differential equation to get:

$5 (A\cdot200\cos{200t}-B\cdot200\sin{200t} )+$ $12(A\sin{200t}+B\cos{200t})=120\sin{200t}$

$A=\frac{90}{62509}$ ; $B=-\frac{7500}{62509}$.

Answer: $i( t)=Ce^{-\frac {12}{5}t}+ \frac{90}{62509}\sin{200t}-\frac{7500}{62509}\cos{200t}$