Question 20.7: A pulse is applied to the RL integrator in Figure 20-36. Det...

The circuit time constant is

Since 5\tau= 16.7 \ \mu s is less than t_{W}. the current will reach its maximum value and remain there until the end of the pulse. At the rising edge of the pulse,

i= 0A

v_{R}= 0 V

v_{L}= 10 V

The inductor initially appears as an open, so all of the input voltage appears across L During the pulse,

i increases exponentially to \frac{V_{p}}{R}= \frac{10 \ V}{1.5 \ k\Omega }= 6.67 \ mA  in 16.7 \ \mu s

v_{R} increases exponentially to 10 V in 16.7 \ \mu s

v_{L}decreases exponentially to zero in 16.7 \ \mu s

At the falling edge of the pulse,

i = 6.67 mA
v_{R} = 10V
v_{L} = -10 V

After the pulse,

i decreases exponentially to zero in 16.7 \ \mu s
v_{R} decreases exponentially to zero in 16.7 \ \mu s
v_{L} increases exponentially to zero in 16.7 \ \mu s

The waveforms are shown in Figure 20-37