00:01
So in this question this is a circuit that we're given.
00:04
Right, in position, we have two possible positions for the switch over here, switch as, can be position one, where it'll be connected to just the resistor.
00:17
So this is kind of like the discharging position.
00:21
And the other position would be number two, which is, across the emf battery.
00:29
So across the battery, we would be, charging our capacitor.
00:34
So these are the two different modes.
00:36
So by given that at first it starts off at 1, where the capacitor is completely uncharged, and then we move on to 2.
00:44
And we allow it to stay for very, very long time on 2.
00:49
We want to find what is the total charge that has accumulated on the capacitor for a very long time.
00:56
So basically, we can use the capacitor equation q is equals to cv where v would be the emf of the battery because after a very very long time the emf of the battery will be just the potential drop across the capacitor because the capacitor will be acting like an open circuit after a very long time so the voltage across it just the emf this case the capacitance given to be 5 .9 micro ferrets and the voltage emf is 28 volts.
01:34
Therefore, the charge is given as such.
01:41
Now of course, what we want to find next is actually the charge, but only after 3 milliseconds.
01:51
It is considered to be 110 times 10 power minus 6 columns.
01:57
This is for t equals to 3 milliseconds.
02:03
You want to find what is the resistance r based on this.
02:09
So what we actually need is the equation that governs the increase in charge, where q is equals to q0, multiplied by 1 minus exponential negative t over rc.
02:28
So this tells us what is the amount of charge at time t for here...