Question 20.2: Determine how much the capacitor in Figure 20–10 will charge...

Calculate the time constant

Because the pulse width is 5 ms, the capacitor charges for 2.27 time constants (5 ms/2.2 ms = 2.27). Use the exponential formula from Chapters 12 (Eq. 12–19) to find the voltage to which the capacitor will charge. With V_{F} = 25  V and t = 5 ms, the calculation is as follows:

= (25  V )\left(1- e ^{-5  ms / 2.2  ms }\right)

= (25  V )(1-0.103)=(25  V )(0.897)=22.4  V

These calculations show that the capacitor charges to 22.4 V during the 5 ms duration of the input pulse. It will discharge back to zero in five time constants when the pulse goes back to zero.