00:01
Hi, in the given problem there is a combination of a capacitor here.
00:14
This is the capacitor which is joined in series with the help of a switch s with a resistor.
00:31
And then a source of emf.
00:36
This source of emf is having a value 12 .0 volt.
00:40
The capacitance is 0 .050 micro -farrate and the resistance is 40 .0 kilo -oom.
00:55
Here this terminal is marked as 1, this as 2 and this as 3.
01:04
The current passing through the capacitor is i -1 and that passing through the resistor.
01:12
Is marked as i2.
01:15
Now in the first part of the problem just after the switch is closed there is no charge stored over the capacitor means q is equal to zero so the potential drop taking place across its plates will be given by let it be vc and that will be given by q by c but as the charge is zero so this potential drop taking place across the plates of the capacitor will come out to be zero means we can say the potential whatever the potential will be at point 1 means v1 that will be equal to that taking place at point 2 means v2 and we know potential at point 1 is equal to the potential of battery means 12 .0 volt because there is no circuit element between the positive terminal of the battery and this point 1 so whatever the potential of this positive terminal of the battery is the same will be here at 0 .1 because it is given that the potential of terminal 3 is 0 so taking this as the base potential which is 0 the potential of the battery 12 volt will be equal to its potential at positive terminal.
03:29
That's why v1 is equal to v2.
03:31
And the current has both the elements, the capacitor and resistor are in series.
03:39
So the current passing through them will remain the same...