Question 20.7: A pulse is applied to the RL integrator in Figure 20–36. Det...
A pulse is applied to the RL integrator in Figure 20–36. Determine the complete waveshapes and the values for I, V_{R} , and V_{L}.
The circuit time constant is
\tau=\frac{L}{R}=\frac{5.0 mH }{1.5 k \Omega}=3.33 \mu sSince 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 = 0 A
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 \text { in } 16.7 \mu s
v_{R} increases exponentially to 10 V in 16.7 μs
v_{L} decreases exponentially to zero in 16.7 μs
At the falling edge of the pulse,
i = 6.67 mA
v_{R}=10 V
v_{L}=-10 V
After the pulse,
i decreases exponentially to zero in 16.7 μs
v_{R} decreases exponentially to zero in 16.7 μs
v_{L} increases exponentially to zero in 16.7 μs
The waveforms are shown in Figure 20–37.
