00:01
In the given problem, if the maximum charge stored over a capacitor initially is q0, then the charge remaining over its plate after a given time is given as q is equal to q0 into e ratio bar minus t by tau.
00:27
Where this tau is the time constant and here it becomes the equation of a discharging capacitor.
00:40
Now, if we use the relation for the charge in terms of capacitance and potential across the plates of the capacitor, here this q is given as cv.
00:53
So for q0, it will be c into v0.
00:56
The maximum voltage is stored over the plates of the capacitor.
01:00
And here this v is the inverse.
01:01
Instantaneous voltage and c is the capacitance which remains fixed into e ratio bar minus t by tau.
01:10
So canceling this c we get an equation for the potential remaining over the discharging capacitor.
01:19
Here it is given that v is equal to 0 .10 % of initial voltage means we can say here this is 0 .10 divided by 100 .0 .00, canceling this 0...