00:01
So this question gave us this configuration, where we have two capacitors in parallel, and we need to find a lot of things.
00:09
We need to find abcde.
00:11
There are five parts to this question.
00:13
Let's just walk through them one by one.
00:16
So what we already know is that we know the potential difference between a and b is 220 volts.
00:25
And because this is a parallel circuit, that means the potential difference between each of these capacitors, between each of these capacitors will be the same.
00:37
So it will all be 220 volts.
00:40
But because you can see they have different capacitors, they will have different charges stored on them.
00:47
So the first thing we find is the total charge stored in this network.
00:52
We can find, because q equals cv, we can find cv of each of these capacitance, and add the mob or we can just find the total capacitors of this entire system in times v and i'm going to do the second thing.
01:10
If this is c1 this is c2 this equals c1 plus c2 because it's in parallel that makes the capacitans add up.
01:18
So that is 35 nanofarrant plus 75 nanofarant times 220 volts and that gives me 242002 00 nanofarrent or 24 .2 microfarrant is a better unit in this case.
01:49
Part b asks us, what is a charge on each capacitor? okay, so now we have to do this separately.
01:58
So for q1 equals c1 times v.
02:03
So this is 35 times 10 to negative 9.
02:09
Variant times 220 volts.
02:12
This is 7 .7 times 10 to the negative 6 column or 7 .7 microtolon.
02:24
And q2 equals c2 times v equals 75 times 10 to the negative 9, fahrenheit times 220 volts.
02:32
And this is 16 .5 microcolon.
02:38
That's part b.
02:40
And we can note that these two will add up to be this one...