00:01
So in this question we are given there is an rc circuit basically.
00:06
So this is r, this is c and charging is happening basically here.
00:15
So voltage v.
00:16
So when you close this, what will happen? the charge on this capacitor will slowly rise and hence the potential will also rise.
00:24
So the variation of charge with the passage of time is given as q is equal to cb in to 1 minus e -rest to the power minus t upon rc so this is the variation of charge which is a very standard formula this is known to us and now what we need to do is we need to find the potential on the capacitor also right so charge in a capacitor is known to us we know that q is equal to cb right so now the potential on the capacitor would be equal to q by c so potential on the i'm calling this v -dash as a potential on the capacitor which would be equal to q by c so the value of q is here so we can write it c v into 1 minus series to power minus t by rc this is q divided by c so v -dash is coming out to be v -dash is the voltage across the capacitor as a function of time so this value is coming out to be v which is battery potential into 1.
01:32
1 minus a to power minus t by rc.
01:37
So now we are also given certain data.
01:40
The data which is given to us, i'm writing it here.
01:43
We are given the value of resistance, which is 280 .0.
01:48
We are given the value of capacitor, which is one microperid, which means 10 to power minus of 6 ferret.
01:55
Okay, we are given the voltage, which we are applying this voltage battery.
02:01
It is given as 6 bolt and we need to find the time at which the time at which the potential v prime will become equal to 4 .1 so now so from here we are putting here b prime is equal to 4 .1 which is equal to v battery potential which is given to us 6 volt 6 into 1 minus e to the power minus t and then divided by rc.
02:33
The value of r is 2 ,800...