The steady state current waveform iLt in an inductor of 50 m

The steady state current waveform i_L(t) in an inductor of 50 mu H. Determine the period, T. Determine the frequency. Determine the rate of current rise-time in the inductor. Determine the rate of current fall-time in the inductor. Determine the maximum voltage of the inductor, V_L max Determine the minimum voltage of the inductor, V_L min Carefully sketch the inductor current i_L(t) and the inductor voltage v_L(t) waveforms using the same time axis.

Solution

a) i) the period T = time of one complete cycle of the signal.in this case it is 5us. ii)) frequecy= 1/T, therefore in this case it is 1/5us = 0.2 MHz

b) rate of the current rise time is given by the rising slop from time 0 to 3us. it is (4-3)/(3-0)us = 0.33 * 10⁶ A/sec

c) rate of current fall time is given by the decreasing slope from 3us to 5 us. = 4-3/(3-5)us = -0.5 * 10⁶ A/sec.

d) the voltage is the differentiation of current multiplied with imductance. so, VLmax = 50* 0.33*10⁶ volt