Electrical – How Do I Calculate The discharge Time of Capacitor with a resistor as a load

charge on a cap is a linear product of capacitance and voltage, Q=CV. If you plan to drop from 5V to 3V, the charge you remove is 5V*1F - 3V*1F = 2V*1F = 2 Coulombs of charge. One Amp is one Coulomb per second, so 2C can provide 0.01A for 2C / (0.01 C/sec) or 200 seconds. If you actually withdraw charge from the cap at a constant current, the voltage on the cap will decrease from 5V to 3V linearly with time, given by Vcap(t) = 5 - 2*(t/200).

Of course, this assumes you have a load that draws a constant 10mA even while the voltage supplied to it changes. Common simple loads tend to have relatively constant impedance, which means that the current they draw will decrease as the cap voltage decreases, leading to the usual non-linear, decaying exponential voltage on the cap. That equation has the form of V(t) = V0 * exp(-t/RC).

Ultimately how much any particular capacitor is charged depends on the total Coulombs it holds. The capacitance and the voltage level at which you consider it charged dictate the number of Coulombs you need to run thru it to "charge" it from the totally discharged state.

Charging a capacitor thru a resistor from a fixed voltage source is only one of many possible ways to dump the required Coulombs onto the capacitor. In that case, the voltage is a exponential approaching the voltage source value. In some cases the circuit may look more like a current source, in which case the capacitor voltage would rise linearly. In other cases the circuit could be non-linear, so all kinds of unusual voltage profiles are possible.