Help out a brother here.The graph below shows the voltage across a 560 mH inductor vs. time
for a given circuit, V/I R the place although V I r variables, R (resistance) is const. So whilst u draw graph of V vs. I u continually get a at once line (no longer which consists of non uniform circuits like those having diff. densities factors at diff. places)
Since the plot can be separated into piecewise functions, the easiest way is to integrate the functions at the points you need Remember the current across an inductor cannot change instantaneously First off: Voltage across an inductor V L di/dt Rearranging for current i 1/L * [integral from t0 to t chosen] v(t) dt from t0 to t7*10^-6, solving the integral 1/560*10^-3 * [-400t] from t0 to t7 [-400*7*10^-6]/0.56 -5mA Similarly, repeat from t7us to t9us and sum the answers from t7*10^-6 to t9*10^-6, solving the integral 1/560*10^-3 * [320t] from t7*10^-6 to t9*10^-6 [320*9*10^-6 - 320*7*10^-6] / 0.56 1.14 mA Answer: Current at 9us -5mA + 1.14 mA -3.86mA
