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.
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.
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:
The graphs are asymptotic (like the one for radioactive decay) , i.e. in theory the capacitor does not completely discharge but in practice, it does. The product RC (capacitance of the capacitor × resistance it is discharging through) in the formula is called the time constant. The units for the time constant are seconds.
That is the rate of voltage rise across the capacitor will be lesser with respect to time. That shows the charging time of the capacitor increase with the increase in the time constant RC. As the value of time ‘t’ increases, the term reduces and it means the voltage across the capacitor is nearly reaching its saturation value.
Capacitors oppose changes of voltage. If you have a positive voltage X across the plates, and apply voltage Y: the capacitor will charge if Y > X and discharge if X > Y. calculate a capacitance value to discharge with certain voltage and current values over a specific amount of time