across the capacitor increases, the voltage across the resistor
decreases. At the end of two and a half time constants, the
capacitor voltage, as determined by the universal time constant
chart, is equal to approximately 92 percent of the applied voltage,
while the resistor voltage has decreased to approximately 8 percent
of the applied voltage. At this time, the applied voltage falls to
zero, and the capacitor starts to discharge through the resistor.
This causes a negative voltage to be developed across R equal to 92
percent of the maximum applied voltage. The discharge curve is
slightly more gradual than the charge curve for the first pulse. The
reason for this difference is that the capacitor charges from zero
toward E during the pulse duration time (t1), and discharges from 92
percent of E toward zero during the pulse rest time (t2). At the end
of five time constants Ec is equal to approximately 7 percent of the
maximum value of E, while ER is equal to approximately 7 percent of
E. At this time, another positive pulse is applied, and the process
is repeated.
(3) Now consider the output waveshapes when a square wave is
applied to an RC circuit with a comparatively shorttime constant
(Figure 25). In this case, the time constant is equal to 0.02 of a
frequency cycle. Thus, in 20 percent of a half cycle (duration
time), five time constants occur. As a result, the capacitor charges
very quickly to the maximum applied voltage. The early rise of Ec to
the full, applied voltage and its rapid decrease to zero when E falls
to zero cause the voltage waveform across the capacitor to resemble
the squarewave input. The short duration of the high charging and
The rapid rise and drop of ER cause the voltage waveform across the
resistor to be peaked at each squarewave changeover (change from
maximum voltage to zero or from zero to maximum). The amplitude of
each curve, at any instant of time, can be readily determined from
the universal time constant chart (Figure 20).
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