A charging capacitor has charge deposited onto its plates and as the capacitor gets more charged it becomes increasingly difficult for further charge to build up on it (because of the increasing electrostatic charge). Therefore the current flow decreases over time. Since , we can write;

(2)

Since, by definition ;

(3)

Substituting equations and into gives;

Rearranging to give ;

In order to integrate and remove the ‘rate of change of time’, so exact values for charge at a particular time frame can be given;

Dividing both sides by gives;

Putting the equation into this form tells us to integrate the equation. If the capacitor is to be charged from a start time of , where it has no charge (), to another point in time where it will contain a charge we can write;

This equation becomes;

Using log rules whereby , this becomes;

Taking the exponential of both sides gives;

Rearranging for q gives;

Dividing by C gives;

Since

where, is the p.d. at time

where;
is the initial voltage supplied by the cell when the capacitor had no charge across it.
is the resistance supplied by the circuit in which the capacitor and cell are part of
is the capacitance of the capacitor

Thinking back to the original step where;

When charging the capacitor, and

Therefore and so

Since ,

Since the current at time can now be written as ;

where
is the initial current, at the start of the process of charging the capacitor.