Alright, on this slide here, I just want to kind of go over something I've kind of mentioned it, but I want, I want to make sure it's clear. All right. So, here we get we have our capacitor. So the circuit, okay. And we stated before that Q is in cool ohms. All right.

And I also stated that one column of charge is defined as having an imbalance of electrons on a capacitor of 6.25 times 10 to the 11 18th electrons. Okay, that's, that's wonderful. Okay, so what really does that mean? Well, It means this and I think I've stated before and I just want to review it. Since this is my negative terminal of the battery, we know that electrons will flow this way. Alright, so what happens is, there's this many electrons additionally on this side of the capacitor, so on that side, there's a surplus.

Alright, right there, there's a surplus. And on this side, there's a shortage. Okay, there's a shortage of electrons. All right on this side of the plate. Where do we get those excess electrons we get those excess electrons from this side of the plate of the capacitor. All right.

So now what happens if I disconnect the capacitor from the circuit? It's still gonna hold that charge. Okay, there's still gonna be a surplus of electrons by that number 6.25 times 10 to the 18th electrons on the negative plate if that electron is charged up, or has a charge of one colome and a shortage of electrons on the other side of the plate. All right, so now I've disconnected that capacitor. All right. Now, before we go on, I'm going to clear the slide as I talk here, before we go on, what does anybody remember what that number represents that 6.25 Five times 10 to the a D. Well, if you've taken some of my earlier circuit theory, courses 6.25 times 10 to the 18th electrons is one amp.

And if you remember, what was stated previously in that course, that if I have a conductor and 6.25 times 10 to the 18th electrons flow by a point and that conductor that is one ampere or one amp, all right. So, now, think about this, I charge you up this capacitor I have a koulos of charge on that which is 6.25 times 10 to the 18th electrons I put a shot or a conduction path or between the plates of that conductor. If I have one cool arm of charge, I can supply one amp of current for one second. And again one amp represents that number there 6.25 times 10 to the 18th electrons. And one cool home also represents that same number of 6.25 times 10 to the 18th electrons. So a cool arm.

If you look at it, one colome is capable of supplying one amp of current for one second. Now let's take a let's go one step further and look at this. One. What happens if I have two cools and then I put a conduction path between the plates of the capacitor that I charged up. Well, if I have two cool ohms, I can supply one amp of current for two seconds, or I can supply two amps of current for one second. All right, so the point I'm trying to make here is one amp is the amount of electronic flow through one point in a conductor and an AMP is made is made up of or an ampere, I should say, is made up of 6.25 times 10 to the 18th electrons.

And a colome is a charge on a capacitor in this example, where I have 6.25 times 10 to the 18th electrons, where one plate of the capacitor as a surplus by that number, and the other plate has a deficiency or a shortage of electrons by that i. So, I'm trying what I'm trying to do here is I'm trying to give you the relationship of ampere and a cool and that you can think of a cool arm as a potential current path. And obviously, an ampere is what is what's happening in Real Time, the amount of current that's happening in real time. All right, so let's let's look at these problems here. All right, let's look at those problems. And we know that we've got this little formula here.

Q, the charge on a capacitor equals capacitance times voltage. All right, so now I charge it charge stored in a two micro farad capacitor at 50 volts DC across it, basically, you know that micro is 10 to the minus six, so q would be two times 10 to the minus six times 50 volts. Okay. So 100 times that is 100. Cool. ohms are 100 micro cool ohms because 10 to the minus six is micro.

So we have 100 Micro cool ohms All right. And then so, if I increase the capacitance, so the capacitance is going in this example here is going from two micro ferrets to 40 micro ferrets because again 10 times 10 to the minus six is micro. So we just we just do the math, and we get 2000. Micro cool ohms All right, because that's micro, and my charge is labeled in cool ohms. So what I do the math 40 times 10 to the minus six times 50 volts. I do the math and I get two times micro 2000 Thousand micro cools.

Alright, so obviously, this tells us that we kept the voltage the same, but we increase my capacity on my capacitor and obviously, it's a direct relationship. So Q is gonna go walk by the same factor. All right, so let's clear the slide and go on. Alright, on this slide here, we're going to introduce a new formula. This one here. I equals q over t, where is current?

T is time and q is charge in cool ohms. All right. So let's let's just do the this this problem here okay. Two micro amperes of current charges a capacitor for for 20 seconds, find Q. Well, q will equal what I times t, because if i transpose this where i equals q over T, if I divide out a T, T i equals q right there, all right? So, we know that this is right q equals Ikki.

We just do a multiplication, two times 10 to the minus six which is current to micro hampers All right, time times time, which is 20 seconds. And and basically I just do the math and I get 40 micro cool ohms. So, if I have this if I have two micro amps of current and I tragic capacitor for 20 seconds. All right. That's what I will get. All right.

40 Micro cool. ohms. All right. So let me Let's clear off the squat slide. And I want to just again, I want to harp on something. All right, I just wanted to kind of mention this again, we know that i equals q over T. We looked at that in the previous slide, I've widened out this slide.

So one app, remember one ampere equals one cool overcharge divided by one second. All right, so I couldn't believe However, what that saying is I can deliver one amp of current. If I have a charge of one cool arm, and I can do that for one second. That's what that means here. Looking at it another way. If I have one amp of current right there, and I supply a capacitor, all right, supply one amp of current for one second to a capacitor, I have one cool long of charge.

So the relationship is very similar. So if I can charge a capacitor up to one cool ohm, I can supply one amp of current to my load for one second. That's the relationship I want. I kind of want to bring home here and then obviously, like I Said I can supply if I've got two cool ohms of charge. I can supply one in one amp of current for two seconds. If I have two cool amps of charge, I can supply a half an AMP for four seconds.

So I'm trying to give you that relationship. All right. That's the point I'm trying to make here. All right, we're just going to finish up here. What we can do is we can use this formula and this formula to find current and voltage. All right, because this this will give us q right there.

So let's just look at this here. We did this, this one here in the previous slide. So what we're asking is find the value of capacitance from the above problem if voltage equals 20 volts DC, alright, so if we know that we know that C equals Q divided by V i, so we know what Q is from the previous problem up there. Okay, we defined voltage here. So now we just do our math. And we we go through it, we come up with two micro ferrets.

And again at this point, you should know the powers of 10 how they add how they change when I divide. You should know that by now, so if you're a little bit laxed on that you need a refresher. I put a course up math electronics or basic math skills. You can get again you can get a barque. You can go on the internet and find a website. But at this point you should, you should be able to manipulate numbers, especially powers of 10.

All right, with that said, so now let's see here, count of five milliamp hours charges a 10 micro farad capacitor for one second find v across the capacitor. Alright, so now I'm going to find q right up here using this formula. Okay, q equals i times t again, I just do my math. Okay? Five times 10 to the minus three is five milli amperes times let's see, times one equals five times 10. to the to the minus three, cool ohms are five micro cool. ohms.

So let me just put that in here. I didn't, so that would be five micro cool homes. All right. And then What are we looking for? We're looking for V. So we use this formula here. Q divided by C equals v. Well, we know what Q is.

All right? Q is five micro cool. ohms right there. We solve that over here. And let's see is, let's see, did we get did he give us see over here. And we were given see right there.

10 Micro ferrets. And now again, we just do the math. And when all is said and done, we got to get 500 volts across that capacitor. That's it. That's it. So these are two good formulas.

I can I can find everything with those two formulas. What are the things that I need to do is bounce back and forth between one on the other. Now I'm going to give you some problems. Okay. And it'll probably it'll be In attachment at the end of this section with the solutions, download them, do them, you should be okay. And again, you can always get ahold of me.

And we if you have some problems, I can clarify them for you. All right, so I'm going to clear the slide here and we're going to go to the last slide in this section. Okay, and this slide, all I want to do here is explain what happens if I have capacitors in series. Or if I have capacitors connected in parallel, so this one here, we've got three capacitors in series. Okay, I want to find CT. I use this formula here, one over, in this case, one over c one, plus one over one over c two, plus one over one over c three.

All right. And I just do the math like I show you here and I come out with 2.5 microphones alright. So, what I can say is when I have capacitors in series CT will always be less than the smallest capacitor. In this case the smallest capacitor is five micro ferrets and when I find my capacitance total within this case with three series capacitors, the total is 2.5 micro ferrets. So, I guess we can say capacitors in series of very add like resistances in parallel Hmm, same deal. And when I have capacitors in parallel, they add directly So, is my circuit or his How am I capacitors are connected, what would CTB?

Well, they just add, so I just add 1010 and five, I got 25 micro farads. So those three capacitors that are that are connected in parallel, they add directly and they act like 125 micro farad capacitor. Alright, that pretty much ends this section. We're going to move on and we're going to look at what I call capacitive reactance or with the books called capacitive reactance. And we're going to take a look at that and then we're going to put a resistor in there in the next section. So with all that said, let's keep on trucking.

Ladies and gentlemen, we'll get there