## Fundamentals of Electrical Engineering

$i = -13.667 mA$
To start this problem, turn the graph into a piecewise function with $(7*10^5 )t$ defined from 0 to 5 us and -1.9 defined from 5 to infinity. To find the current through a 0.75mH inductor, integrate the equation $v = L*\frac{di}{dt}$ to get $i(t) = 1/L * \int^{t}_{t_0} {v(t')*dt' + i(0)}$ Apply this equation to the piecewise defined function at $t=15*10^{-6} s$ $i = 1333.333*(3.5*10^5 * t^2|^{5*10^{-6}}_{0} - (1.9*t)|^{15*10^{-6}}_{5*10^{-6}})$ $i = -13.667 mA$ Be sure to convert from micro seconds to seconds, and from mH to H before integrating to avoid any wrong units.