When a capacitor is discharged, the current will be highest at the start. This will gradually decrease until reaching 0, when the current reaches zero, the capacitor is fully discharged as there is no charge stored across it. The rate of decrease of the potential difference and the charge will again be proportional to the value of the current.
Capacitors will lose their charge over time, and especially aluminium electrolyts do have some leakage. Even a low-leakage type, like this one will lose 1V in just 20s (1000 μ μ F/25V). Nevertheless, YMMV, and you will see capacitors which can hold their charge for several months. It's wise to discharge them.
The time it takes for a capacitor to discharge 63% of its fully charged voltage is equal to one time constant. After 2 time constants, the capacitor discharges 86.3% of the supply voltage. After 3 time constants, the capacitor discharges 94.93% of the supply voltage. After 4 time constants, a capacitor discharges 98.12% of the supply voltage.
After 2 time constants, the capacitor discharges 86.3% of the supply voltage. After 3 time constants, the capacitor discharges 94.93% of the supply voltage. After 4 time constants, a capacitor discharges 98.12% of the supply voltage. After 5 time constants, the capacitor discharges 99.3% of the supply voltage.
C affects the discharging process in that the greater the capacitance, the more charge a capacitor can hold, thus, the longer it takes to discharge, which leads to a greater voltage, V C. Conversely, a smaller capacitance value leads to a quicker discharge, since the capacitor can't hold as much charge, and thus, the lower V C at the end.
For the equation of capacitor discharge, we put in the time constant, and then substitute x for Q, V or I: Where: is charge/pd/current at time t is charge/pd/current at start is capacitance and is the resistance When the time, t, is equal to the time constant the equation for charge becomes: