Chapter 9, Problem 38P

An RL circuit includes a voltage source v_s, a resistor R = 1.8O, and an inductor L = 0.4 H, as shown in the figure. The differential equation that describes the response of the circuit is

$\frac{L}{R}\frac{d{i}_L}{dt}+i_L+\frac{v_s}{R}$

where i_Lis the current in the inductor. Initially i_L= 0, and then at t = 0 the voltage source is changed. Determine the response of the circuit for the following three cases:

(a) $v_s=10\sin (30\pi t)$ V for t $t\ge 0$ .  

(b) $v_s=10e^{-t/0.06}\sin (30\pi t)$ V for $t\ge 0$ .

Each case corresponds to a different differential equation. The solution is the current in the inductor as a function of time. Solve each case for $0\le t\le 0.4$ s. For each case plot v_sand i_Lversus time (make two separate plots on the same page).