00:01
For the first part of this question, we are just looking at when the switch is closed for a very long time.
00:08
I want to find what is the charge on the capacitor.
00:11
To find the charge, we need to use the formula, q equals to c times v.
00:18
Now what is v? we know the capacitance, but we don't know what is the voltage across.
00:24
So when it is fully charged, that is after a very long time, we expect that there will be no current.
00:30
Right, no current flowing through or flowing past the capacitor.
00:38
And when there's no current, it means that there is no potential drop across the resistor 10 -on.
00:44
And the 10 -on resistor, no current means no potential drop.
00:54
It will mean that any potential across this entire length of this parallel circuit would be the across the capacitor.
01:06
Now, what is the potential drop across this? these two points? well, it is based on the 40 -ooms resistor.
01:16
Because there's no currents flowing in, over here, we can just treat the 6th circuit as just the 40 -oom and the 60 -oom resistor.
01:26
So what is the potential across the 40 -oom resistor? well, we can just find the overall currents first.
01:35
I is equals to the total emf divided by the total resistance or effective resistance, which is 100 oms, be 1 ampios.
01:46
So the voltage across the 40 oms, just be v equals to i will i ir.
01:54
So the current times the resistance.
01:57
So 1 times 40 oms, which we get 40 volts.
02:01
And therefore, because the capacitor is in parallel to the 40 ome resistance, the charge in the final potential across the capacitance after a very long time must be 40 volts and therefore we can use this to calculate the charge or the capacitance taking the capacitance of 2 times 10 power of minus 6 ferrets this is given in the figure, multiply that by the voltage which is 40, get 8 .0, understand the power minus 5, close.
02:58
Now once the switch is open, what happens is that this now becomes an open circuit...