The voltage drop across the capacitor alternates between V1 (1V) and discharging
down to zero according to the input voltage. Here in this example, the frequency
(and therefore the resulting time period, ƒ = 1/T = 1/100 Hz = 0.01s).
RC of the circuit is: 103 * 10-6 = 10-3 sec = 1ms.
This (10RC) time constant, 0.01s = 10ms, allows the capacitor to fully charge during the "ON" period (0-to-5RC) of the input waveform and then fully discharge during the "OFF" period (5-to-10RC) resulting in a perfectly matched RC waveform. If the time period of the input waveform is made longer (lower frequency, ƒ < 1/10RC) for example an "ON" half-period pulse width equivalent to say "8RC", the capacitor would then stay fully charged longer and also stay fully discharged longer producing an RC waveform as shown.
At time t = 0+ the capacitor is charging with the maximum current. The current through the capacitor develops a voltage drop across the resistor, since the current flow through the capacitor is at its peak the voltage drop across the resistor is also at its maximum. VR = IC x R. While capacitor is charging (red waveform) the current flow drops and the voltage drop across the resistor (yellow waveform) also drops.Download Multisim Blue Similation File Download docx File