00:02
Hi, in the given problem there is a circuit having a battery of 9 .0 volt, which is joined with a capacitor of 2 .0 micro ferret capacitance.
00:26
Then there is a switch.
00:29
This is terminal a, this is terminal b.
00:33
With the terminal b, there is a resistance having a value of 50 oom.
00:43
It is said that the switch was at position a for long.
00:50
So the capacitor will be fully charged initially.
00:56
The initial potential stored across drop the initial potential dropped across the plates of the capacitor will be equal to the potential of the battery.
01:08
Emf of the battery means 9 .0 .0 .0 .0 .m.
01:10
0 .0 volt.
01:12
The capacitance is having a value of 2 .0 micro ferret.
01:16
The resistance is having a value of 50 oom.
01:22
Now this switch is thrown to the position b.
01:26
So now what will happen? this capacitor will start discharging through this resistor.
01:34
So initially in part a we have to find the initial charge is stored over the of the capacitor and initial current passing through this resistor.
01:46
So as far as initial charge is stored is concerned this is simply given by the product of capacitance with the potential of the battery, emf of the battery.
01:58
So it is given as 2 .0 micro fared multiplied by 9 volt which comes out to be 18 microculum which is 1.
02:11
One of the answer of the given problem.
02:15
Now in the same part we have to find initial current also passing through the resistor.
02:22
So that is given by oms law.
02:25
I know t is equal to v0 by r.
02:27
So here this is just 9 by 50.
02:32
So it comes out to be 1 .8 into 10 dash to par minus 1 ampere or we can say this is 0 .18 ampere.
02:45
So these are the two answers for the first part of the problem.
02:50
Initial charge over the plates of the capacitor and initial current passing through the resistor.
02:56
Now we have to find the same values means the value of charge and the value of current passing through the resistance after a time t is equal to 50 microsecond...