Basic Electricity transcript

Let's you and I talk about the electrifying subject of electricity. And by the way, they ask some pretty tough questions these days on the&mp general exam about electricity. But never fear, because by the time you and I get done, we're going to be able to answer every question [music] with ease. We're going to take care of it in a flash here. All right. First of all, let's you and I get some definitions out of the way.

What is electric current? Well, electric current is electrons in motion. Now, when the current flows in one direction only, it's called direct current. But current that reverses itself periodically is called alternating current. All right. Now, let's take a look at how you measure current flow. And it's measured in ampers. That's a unit of measurement of electrical current flow. How about, you might ask, little teeny tiny current flows?

Well, that would be a milliamper, 1,000th of an amper. That's written 0.001. Now if you have a wire, a wire is typically called a conductor. Something that limits the flow of electricity is called resistance. So anything that limits the flow of electricity is called resistance. A wire is a conductor. Sometimes a wire has a little bit of resistance in it. However, and so what determines the amount of resistance is the length and the cross-sectional area of the wire.

Now if you decrease the length of the wire, you decrease the resistance. But if you make that wire bigger, give it a bigger cross-sectional area, you also decrease the resistance. [snorts] So what makes low resistance in a wire is using a big wire that makes sense? Or a and a short wire. And those things, two things decrease the resistance in any kind of conductor, particularly a wire. All right. A resistor then is something that limits the flow of electricity.

You can have various kinds of resistors and sometimes you actually build a resistor into an electrical circuit. Now, here's what a resistor looks like that could be built into an electrical circuit. And what you're trying to do again is limit the flow of current with this resistor. Now, resistors can be of all different values, low value or high value. And that value is measured in ohms. We'll talk more about ohms later on.

And resistors are generally classed as fixed or adjustable or variable depending on their construction and use. And a typical fixed resistor like this one consists of a small rod of carbon compound. So here's a typical fixed resistor. Now an adjustable resistor is usually a small wire wound type with a metal collar that can be moved along the resistance wire and it'll let you vary the value of resistance placed in the circuit.

And a variable resistor is arranged so that it can be changed in value any time by the operator generally a a knob. And variable resistors are commonly known as riostats or potentiometers. Now let's take a look here on figure 17 and take a look at how a potentiometer is shown. Now a potentiometer is a specific type of variable resistor. It's one in which the contact can be moved along to a different location and you can pull off the current at that particular point thereby changing the amount of resistance.

And that's a potentiometer. That's how it's shown. It's actually shown with a variable contact like so. Now a potentiometer is a type of variable resistor. If you did not want to specify a potentiometer and just wanted to say variable resistor, you might take a look at figure 21. And here is a schematic indication of any kind of variable resistor, not specifically a potentiometer. And this is a variable resistor. So anytime you see an arrow going through an an electrical component like you have here, that means it's a variable component.

And in this particular case, it's a variable resistor. Now, resistors have to absorb a lot of electrical power. That's what they're doing is absorbing and dissipating electrical power. And electrical power is measured in watts. We'll talk more about that later on. So, what determines the wattage rating of a resistor? Well, the wattage rating of electrical resistor is determined by its size. The bigger the resistor physically, the more power it can absorb.

Now, what causes electrons to flow from one place to the other. Well, what it is is called electromotive force. And this potential difference between two conductors is measured in volts. So, how do you measure electromotive force? You measure it in volts. So, that's the potential difference we're going to be talking about. It's measured in volts. That's what causes electrons to flow. And since emf is measured in volts, you might guess that the FA might want you to know exactly what a volt is.

And they do. A volt is the amount of emf that causes a current rate of 1 ampere through a resistance of one ohm. You've got that nothing to it. It's one volt. Causes a current rate of 1 ampere through a resistance of 1 ohm. Everything is one. Now, let's assume that you wanted to say, "I've got a,000 volts." And you want to be scientific about it. You wouldn't say a,000 volts. You'd say, "I have a 1 kilovolt." And that's a,000 volts.

And that prompts the FA to ask you a question. Let's assume the FA says you have 0.002 kilovolts. How many volts is that? All you have to do is multiply that. 0.2* a,000. Use your electronic calculator and you find out that comes out to two volts. Now the basic mathematical formula that describes how electricity works is referred to as Ohm's law. Let's you and I take a look at Ohm's law. First of all, I in this equation stands for the current the flow of electrons in ampers.

E stands for potential difference in volts and R stands for resistance in ohms. And by the way, the electrical unit of resistance is an ohm or ohms. And I remember one time an instructor said to me, "John, what's the electrical unit of resistance?" And I said, he says, "You're right." Okay, let's take a look at it. First of all, the relationship goes like this. The current flow is dependent on the voltage. The more volts, and E stands for electromotive force, and that's E is for volts.

In this particular case, it's measured in volts. And the current flow, which is I, depends on the volts. If you have more electromotive force or the volts are higher, there will be more current flow. What limits current flow is resistance. And the more resistance you have in ohms, the less current flow. So you can see the greater the ratio of volts to ohms, the more the current will be in amps. Now, you're going to need to know this electrical formula right here, this formula.

And when you walk in to the FA written exam to take the general test for the A&P, you should have this formula memorized because you're going to need to use that. What I recommend you do is sit down and write it down on a piece of paper when you walk in there right off. They will not let you carry this formula in on a piece of paper, but they will give you some blank sheets of paper, and they'll let you write it down.

So, I recommend you write it down on a sheet of paper as soon as you get in there. All right. First of all, we've talked about it. The current flow is determined by the ratio of volts to ohms right there or electromotive force if you want to get fancy about it to resistance. Now you can work around with this formula mathematically. Multiply both sides by R and you get E= I * R. Now what does that tell us? Well, what that tells us if you're crossing a resistance, the voltage drop across that resistance is equal to the current times the resistance.

We'll talk more about that later on. But the voltage drop across the resistance is equal to the current. There's the voltage drop. That's electromotive force E is equal to the current times the resistance. And then also you can play around with this some more. And you can divide each side for instance by I. And you get R equals E / I. Now what does that tell us? Well, that tells us that the resistance is equal to the volts divided by or the electromotive force divided by the current in amps.

So, you'll need to know these three formulas if you're good at algebra. Knowing the first one will automatically give you the other two, but you need to be able to use those three formulas. Now, as you might guess, the FA is going to ask you some questions about this. And once again, when you walk into that test, make sure you have that written down. Now we said earlier the voltage drop across the resistance is dependent on what?

Well, it's dependent on the amperage of the circuit. Why? Once again, it's the same formula. The voltage drop, which is E, or the voltage difference or potential across the resistance is equal to the amperage times resistance. So, it's dependent on amperage. It's also dependent on resistance, but they ask you this on the test. So, the correct answer on the test is it's dependent on uh amperage. All right. Now, you can measure resistance with a device called an ohm meter.

Now, here is an example of an ohm meter. And the way you use the ohm meter is you set the scale to the right level you want. Let's assume we have it set on R* 10. Well, what you do is read the resistance off the meter and then multiply it times 10. So, let's assume we've done that. We're measuring a resistance like the resistance we showed you earlier. And let's assume the needle reads 50 and we've set it to times 10. and what is the resistance?

And the answer is it'll be 500 ohms. Now, let's take a look at figure seven here and we'll talk about how to use an ohm meter in a circuit to measure resistance. Well, we have a circuit here. There's a battery in it, but we have an open switch right here. So, the this is not a continuous circuit through the battery because that switch is open. And we have a few resistances. We've got a resistance over here and another resistance here and another one here.

Now, you could put an ohm meter right here. And that's what this is, an meter. And if you connect it like this, you'd be measuring all the way around these circuits like so. And you'd be measuring the total resistance of these three resistors right here. If you want to measure only the resistance of this resistor right here, what you could do is disconnect the terminal at D. If you had the terminal disconnected at D, then this would be the path or the circuit that the ohm meter would be measuring.

Now, in this particular case, they're going to ask you a question on a test. They're going to say, take a look at figure seven here. You have disconnected the ohm meter here at D. And what you have done are disconnected that terminal at D. And now you're measuring this particular resistor, but this resistor has a break in it. What would it read? Well, if it has a break in it, it would have no current going through at all because it's not possible.

It would read infinite resistance. Now, there are two basic kinds of electrical circuits. One is a parallel circuit, which gives several paths for electricity to follow. And one is a series circuit. And here's a good example of a series circuit. A series circuit gives only one path for the electricity to follow. And you have resistances in that circuit and the current must go through one then another and another in series.

That's called a series circuit. It only gives one path for the electricity to follow. Now no matter how many components are included in a series circuit, the current is the same intensity all throughout that circuit. So no matter how many components are included in that series circuit, the current is the same intensity all throughout the entire circuit. That's one of the things you need to know. The same current goes all the way throughout the entire circuit.

Now let's take a look at how you determine the total resistance in a series circuit. It's really pretty easy. All you do is add up all of the resistances individually. In a series circuit, to find the total resistance, we're going to call that R subt. All you do is add up the resistances one at a time. There's resistance one, resistance two, resistance 3, four, five, however many they are. Add them up and that gives you the total resistance.

Let's take a look at a way they're going to ask us a question about this on the FA written exam. They're going to say a circuit has an applied voltage of we'll say 30 volts in a load consisting of 10 ohms 10 ohms in series with a 20 ohm resistor. Now the question is what is the voltage drop across the 10 ohm resistor? Now how can we find that out? Well the first thing we need to know to find that out is to find out the current going across that 10 ohm resistor.

Now the important thing to remember here is no matter how many components there are in this series circuit, the current going through each component is the same all the way through. So we can find the current for the total circuit real easy by using Ohm's law. Once we find the current then we know the current for that particular 10 ohm resistor. Then we found the current for that 10 ohm resistor by using Ohm's law again.

We can find the voltage drop across that resistor. Let's take a look at it. First of all let's find the current for the entire circuit. We'll use Ohm's law. And the current is equal to the electromotive force in volts divided by the resistance in ohms. The volts they told us in this circuit was 30 volts. There were two resistors. All we have to do is add them together with a 10 ohm resistor and a 20 ohm resistor. And the total resistance was 30 ohms.

Now the volts is 30 volts. The resistance is 30 ohms. You divide and the current is 1 amp or amper depending on how you want to pronounce it. It's optional by the way. So we have one amp going through that circuit. Now that one resistor also has one amp going through it. The 10 ohm resistor also has one amp going through it because all the components have the same amperage going through them, the same current. All right, so let's find out for that one resistor what the voltage drop is.

Now remember we said earlier the voltage drop for a resistor is equal to the current times the resistance. And the current is 1 amp. The resistance is 10 * 10 is 10. So the voltage drop across that resistor is 10 volts. So you see once you have Ohm's law nailed you can solve these kinds of questions they're going to ask you on the test. Let's do another one. This is obviously too easy for everybody. And so let's assume you have three resistors of 3 ohms, 5 ohms, and 22 ohms.

They're all in series. And let's assume you have a 28V circuit. Now what is the current that will flow through the 3 ohm resistor? What's the current that will flow through that 3 ohm resistor? Well, once again, we need to figure out the total resistance and then we're going to find the current flying through the whole circuit. Once we know the current through the whole circuit, it's the same all the way through each of those components in a series circuit.

So, let's find out the total resistance. All we have to do is add the resistances because remember that in a series circuit, the total resistance is the sum of the individual resistances. And so we do this and we get a total of we've got see it's 3 + 5 + 22 and I get a total resistance of 30 ohms in that circuit. Now once we know the total resistance in that circuit now we can use Ohm's law and find the current in that circuit and that's all we need to know because the current for each resistor is exactly the same.

That's one of the rules. Okay, here we are. Ohm's law. The current is equal to the electromotive force in volts divided by the resistance in ohms. The electrootive force in volts was 28. They told us that. We just figured out the total resistance for the circuit is 30 ohms. So the current for the entire circuit is 93 amps. And that's the answer to the question. Let's do another one. This time they're getting a little bit tricky.

They say you have a lead acid battery with 12 cells and each cell is 2.1 volts per cell and they're connected in series. And then they're going to say that you have a 10 amps furnished by that battery to a 2 ohm resistance, a circuit with a 2 ohm resistance. Now, if that's the setup, the question they're asking you is, what is the internal resistance of the battery? Oh my god, never fear. It's a lot easier than that.

Let's take a look at it. First of all, to find the total voltage of the battery when the cells are connected in series, you take the volts for each cell, which is 2.1, and multiply times the number of cells. We'll talk more about that later on, but if the battery is connected in series that cells are, that's the formula for finding the voltage of the battery, you use your electronic calculator and take the 2.1 times the 12 cells, and you're going to get 25.2 volts.

So the entire circuit uh the electromotive potential of that battery is 25.2 volts. Now let's figure out the resistance for the entire circuit. Formula for resistance is resistance is equal to volts electromotive force expressed in volts divided by the current in amps. And that gives us the resistance. Now if you remember the voltage we just figured out is 25.2 and they told us that circuit had 10 amps in it. They told us that in the question.

So what is the resistance? The total resistance of the circuit was 2.52 ohms. Now wait a minute. They told us the resistance in the circuit was 2 ohms. Now we figured out mathematically the total resistance of the circuit is 2.52 ohms. What's the difference? And the difference is the internal resistance of the battery. And that's the question they're asking in this case. So if the total circuit works out to 2.52 ohms and the resistor in this circuit is 2 ohms then what's left must be the total resistance of the battery the internal resistance of that battery and it's 0.52 ohms.

All we have to do is subtract and we've got it figured out. All right let's do another one. This time however let's talk about parallel circuits. Now parallel circuits are different in that a parallel circuit provides several paths for the current to go through. So here we have a case which the current can go this path or this path or this path and that is a parallel circuit. Things are just sometimes a little bit more complicated in a parallel circuit.

But the thing you need to know is that the voltage drop across any resistance in a parallel circuit in in that group is equal to the voltage drop across any other resistance. In other words, in a parallel circuit, the voltage across any resistance is equal to the voltage across any other resistance. All of the voltage drops across resistors in a parallel circuit are equal to each other. Now, let's take a look at another question.

And to do that, let's take a look at figure 11 again. Now, let's take a good look at this before we get started. First of all, notice you have a 24volt battery here. And you have three resistors, and they are in parallel. In other words, there are several paths for the current to take. Three different paths in this particular case for the current to take. Now, the thing you need to know is the voltage drop across the entire circuit is the same throughout for the total circuit and the voltage drop for each resistor.

They're all equal. And so, what is the voltage drop across the entire circuit? 24 volts. That means there's also the same voltage drop across resistor one, resistor 2, and resistor 3. Now, let's take a look at the question they're going to ask you. It's going to be very simple. The question is, what is the voltage AC drop across the 8 ohm resistor? Well, let's take a look at it one more time. It's 24 volts across each resistor.

So, for the 8 ohm resistor, it's got to be 24 volts. That's unbelievably easy. Now, here's another basic rule you need to know about parallel circuits. And that is the total current in that circuit is equal to the sum of the currents for the individual branches. So the total current is equal to the sum of these currents. So let's take a look at this drawing in more detail again. First of all, you take the current through this branch, the current through this branch, the current through this branch.

Total them together and that is the total current. The current is a sum of the currents. Now let's you and I take a look at figure 13. And they're going to ask you what is the total current flow in this circuit in figure 13 on page six. Well, what we have to do is figure out the current on each of these different paths. There are three paths to go. You figure out the current for each one of those paths and then add them up for the total current because there are three paths for the current to go.

Now, let's take a look at it. How many volts is the circuit? The circuit is 12 volts. So, there is a 12vt drop across each one of these resistances. We know that to begin with. We also know how many ohms each one of these resistances is because it tells us here 30, 60, and 15. So to figure out the current for each path, all we have to do is use Ohm's law. So let you and I take a look at that. Remember there were three resistances.

They were 30 ohms, 60 ohms, and 15 ohms. And to find the current you take the volts and the volts for all of them were 12 volts because that was the voltage drop across each resistor in that particular case. Take the volts the or electromotive force the volts and divide it by the resistance and that gives you the current. That's the basic ohms law that we said you need to know. All right. So all you have to do is do that for each one of these. 12 divided by 30, 12 divided by 60 and 12 divided by 15.

And those are the three currents, the three different paths provided in that parallel circuit. Well, with your electronic calculator, 12 divided by 30 comes out to 4 amps. 12 divided by 60 comes out to 2 amps. 12 / 15 comes out to8 amps. So to get the total current through the entire thing, you add them up. 8 and 2 is 10. And four makes 1.4 amps. So the total current through that circuit was 1.4 amps. Now, that's the easy part.

Now, let's figure out the formula for the total resistance in a parallel circuit. And this is another formula, by the way, that you need to write down. So, let's take a look at it. This is a little bit complicated, but never fear. By the time we get done, you're going to find it easy. Okay. The total resistance in the circuit is equal to a fraction that looks like this. It's one over these three individual fractions.

And these three individual fractions consist of a numerator of one in each case and a denominator of the resistance for that particular resistor we're talking about. So the total resistance for the circuit is equal to 1 divided by this these uh three fractions added together. And it's resistance one, one over resistance one, one over resistance two, and one over resistance three, etc. Depending on how many resistances you have.

Now, you need to remember that the total resistance is equal to one over these three fractions. It's one over resistance one, one over resistance two, and one over resistance three. And by the way, that's one of those things that you should write down when you take the written exam. And as soon as you walk down into there, write this formula down. We'll talk about this later on, which ones you should write down. And once again, they won't let you sneak them in.

You can sneak them in in your brain and as soon as you get there, write them down. Now, there are some things that the mathematics of that formula will tell you. And one of them is this. If you have a parallel circuit, the total resistance of that circuit will be smaller than the smallest resistance in that circuit. So, if you have three resistances in the circuit, the total resistance is going to be smaller than any one of those three.

And there's another thing that I'll tell you, and let's use light bulbs as an example. Let's assume you have a parallel circuit and there's three light bulbs in it and one of the three bulbs is removed. What's going to happen to the total resistance? And the answer is the total resistance will become greater because as you add more resistance, the total resistance becomes smaller. And that's mathematical formula tells you that.

Now, let's you and I take a look at figure 8, appendix two, the back of the book. And let's take a look at the question we're going to ask you here. First of all, you have this circuit with a battery in it, we'll say. Okay. And by the way, this is a fuse right here, and it has no resistance to it. And that's a switch up here. We'll talk more about these later on. All right. Now, what they're going to do is they're going to put an ohm meter in this circuit.

And notice that there is a break right here. So, if you put an ohm meter in this circuit at this particular point, you're basically going to be reading the total resistance of the parallel circuits consisting of resistance one and resistance two. Resistance three has a break in it, so it's not involved in that. And so basically the question is what is the total resistance of these two uh with this circuit consisting of these two parallel resistances.

All right let's take a look at it. The formula goes like this. The total resistance is equal to one over the these two fractions and it's 1 over resistance 1 divided by resistance 1 plus 1 divid resistance 2. What are the numbers we're going to put in here? Well, each of these resistances on figure 8 is 20 ohms. So now let's go back and take a look at the formulas and it becomes the total resistance is 1 divided by these two fractions. 1 over 20 + 1 over 20.

Well, 120 plus another 120th comes out to what? Well, let's take a look at it. Comes out to 220ths. So now the total resistance is equal to 1 over 220ths. Wait a minute. What is 220th equal to? You got it. it's equal to one/10th. So now its resistance total is equal to 1 / 110th. There is another way we could express that and that is 1 / with a division sign 110th. Now if you want to get rid of the division sign, all you have to do is flip-flop the one10th.

And so the total resistance is 1 is equal to 1 * 10. And what's the total resistance of the circuit? The answer is the total resistance is 10 ohms. You can see it right there. 10 * 1 is 10 divided by one is still 10 and the answer is 10 ohms. All right, let's take a look at another one. They're going to say, "All right, let's assume you have a circuit that contains five lamps in parallel and three of those lamps have 6 ohms each.

Two lamps have 5 ohms each and it's a 28vt circuit. How many amps are in the circuit?" Well, first of all, we need to find the total resistance in the circuit. Then we'll use Ohm's law and find out how many amps are in the circuit. You need to be able to figure out what you need to know. Let's figure out the total resistance of the circuit. First and here you are. The total resistance is equal to one over these fractions.

And the fractions are three resistances of 6 ohms. So it's 1 over 6 + 1 over 6 plus 1 over six and two resistances of five ohms. So you got two more. the one over uh plus one over five plus one over five. That is your formula. Boy, if you can remember that, this will be just duck soup for you. All right. Now, we have three of these sixes. So, it's three six. So, 1 16 plus 1 six plus 1 six comes out to 36. And 1/5 plus 1/5 comes out to two fifths.

So, now the resistance total is equal to 1 divided by these two fractions. 36 plus two fths. Well, 36 is a half and two fifths still comes out to two fifths. So, let's find out where we're going from there. Who knows where we're going from there? Well, where we're going is we're going to use our electronic calculator. And 1/2 is 0.5. Two- fths is 4. So now the total resistance is 1 /.5 +4. And the total resistance is now is 1 / 0.9.

You can see where we're going. Use your electronic calculator. And the total resistance is 1.11 ohms when you use your electronic calculator. Now we've got the total resistance. Let's use Ohm's law and figure out what the current is because that's what the question is. What's the current? All right. The Ohm's law for current says you take the volts or electromotive force divided by the resistance and that will give you the amps or the current.

Now, they told us in the problem that it was a 28 volt circuit, and we just figured out the ohms was 1.11 in resistance. And so, you take the 28 / 1.11 and you get 25.23 amps. All right, let's you and I do another one. And to do that, let's take a look at figure six in appendix 2 in the back of the book. And they're going to say, "All right, let's assume you put the ometer in the circuit right here, like so." And what the ohm meter will be reading then is these parallel circuits.

But they say, "Wait, hold the phone." Because if you look at R5, you'll notice it's disconnected. This resistance is disconnected at the junction with R3 and R4. So R5 is out of the picture. So the question is, what will this ohm meter read? Well, if you take a look at this, here's what the meter will read. It will read the parallel circuits of R1, R2 and an R3 and 4 are in series. They will be treated as one resistance.

R5 is out of the picture. So the ohm meter will read these parallel circuits with R5 out of the picture and R3 and four actually being in series with each other. We'll treat them as one resistance. All right. Now, let's take a look at the mathematics of what we're going to have to do to do this. And here's the way it's going to work. The total resistance, and that's what we're asking here is what will the meter read?

It'll read total resistance. The total resistance is equal to 1 divided by and it's R1 over R1 because R1 was one resistance path. One over R2 because R2 was another resistance path. And remember, R3 and four are in series. We'll treat them as one resistor. So all we have to do is to add up the value of those two together. So it's 1 over R3 plus R4. Now let's go back to that drawing one more time on figure six and get the numbers to put in there.

And we have to have the numbers to do that. So let's you and I go back to figure 6, appendix 2, and get those numbers. And we do that. We can look in here real close. And I think R1 comes out to let's take a little tighter look at that. R1 comes out to 12 ohms. R2 is 6 ohms. R3 is 6 ohms. R4 is 6 ohms. And we can ignore R5 because it's out of the picture. Now, let's go back to the math and fill in those things. R1 is 12 ohms.

R2 is 6 ohms and R3 and four each 6 ohms. And so now you have the total resistance is 1 over the fraction 112th + 16 plus the fraction 1 over 6 + 6. You can see what's going to happen next. We'll add that 6 plus 6 together and everything's the same. Now it becomes 1 over2. Let's take a look at the next one down here that we're going to. And we have now RT is equal to 1 over the fraction 112th + 16. But that's 21 12th + 11 12th because that became 112th.

And so 1 + 2 + 1 comes out to 4 12th. So now the total resistance is equal to 1 over the fraction 4 / 12. Let's take a look at the next item down there again. And 1 over 4 / 12. Well, 4 / 12 is 1/3. So now it becomes 1 over 1/3. Now another way to express 1 over 1/3 would be the total resistance equal to 1 / 1/3. And if you want to get rid of that division sign, all you have to do is flip-flop these two and the total resistance becomes 1 * 3 over 1 and the total resistance is 3 ohms.

Now, the thing you need to remember about that question is that you use the formula for parallel uh resistors, but if you have resistors in series, you don't have to use that formula. You just add them together as far as those resistors that are in series are concerned. Now, let's make life a little more complicated because after all, that's what the FA is going to do to you on the test. So, we might as well do it now.

So, let's take a look at figure 14, appendix 2 in the back of the book. And you see a circuit that looks like this. Oh my god, this is complicated. This is a combination of series and parallel circuits. First of all, you start off with this circuit right here, this resistance in series. Then you have the path split out and you have three paths that the electricity can go with a resistance in each one of them. then it goes back to one path and another series resistance there.

So what we're going to have to do is convert this parallel circuit into one total resistance and then all we have is three series resistances right there. So let's convert this one parallel resistance first of all this parallel circuit into one single total resistance. And to do that let's get the numbers. Resistance two, if you look at the ohms there, I believe comes out to 4 ohms. And resistance three comes out to 6 ohms.

And let's take a look over here and see what resistance 4 comes out to 12 ohms. So what we want to do is take that path right now and convert that into a total resistance. And then when we get done with that, we're going to add it to these series resistances over here. All right, let's convert that path we just talked about into, if you don't mind, a total resistance. And total resistance for a parallel circuit looks like this.

It's one over these fractions and the resistances were 4 ohms, 6 ohms and 12 ohms. So the fractions become 1/4 + 16 + 112th. All right. So now the total resistance for those three uh resistors in parallel become 1 over 1/4 can be converted to 31 12th. We're converting these all to 12th so we can add them together. And 1 16 becomes 2 12ths and 11 12th still stays 112th. Now let's add them together. 3 + 2 + 1 comes out to 61 12ths.

So now the total resistance is 1 over 61 12ths. Back in the old days 62ths used to be 1/2. I think it still is. So now we'll call that 1 over 1/2. But let's express that a different way. We'll call it 1 / 1/2. It's the same thing. Now if you want to get rid of the division sign, all you do is flip-flop this fraction. It's 1 * 2 over 1. And the resistance for those three resistors in parallel is 2 ohms. Now let's go back and look at the drawing again and see where we go from here.

What we did is we figured out the total resistance for these parallel resistors was 2 ohms. Now we take that 2 ohms and the and the the path of the current goes like this. It goes through this resistance up here and then through the 2 ohms and this resistance down here and then back. So let's figure out what the resistance is total. And all we have to do is add these up. So, let's take a look at that resistance right there and see if we can read the number off that chart.

And if you look at resistance one, we get in there real tight and take a look at it and see what it says. And I think it says that's 5 ohms. And now let's take a look at the bottom resistance down there and see how many ohms that is. And that is 10 ohms. So now we have 2 ohms for the parallel circuit. And and the series resistance of 2 ohms and 5 ohms. Let's take a look at the math. It's 2 ohms plus 5 ohms plus 10 ohms.

And so the total resistance of the entire circuit is 17 ohms. All right, that was complicated, wasn't it? Let's do it more complicated. And to make it more complicated, let's take a look at figure 12 in appendix 2 in the back of the book. And here is another even more complicated circuit. And the circuit looks a little bit like this. First of all, you start off here and you have a resistance in series. And then you have a parallel path.

You can go two different directions. And then you have another two paths you could take here. And then you come back here, go through a switch, go through a fuse, and then back. And the fuse, of course, has no resistance at all as far as we're concerned. All right. Now, the way we're going to do this is we're going to have to try and resolve these parallel paths into a total resistance. So the first one is this one on the right hand side over here.

So let's take these two uh resistances as a parallel path and figure them out as a total resistance. Let's take a look here and see if we can read these numbers off these resistances. And I believe the uh resistance number four is 12 ohms. Resistance number five is 6 ohms. So let's you and I first of all mathematically convert those into a total resistance. And so the total resistance is one over and it was 12 ohms and 6 ohms.

So the fractions 112th + 16 and of course we can convert that 16 to 22. So we can add these together. So now the total resistance is 1 over 112 + 2 12th and I believe if you look down there a little further that's going to come out to 31 12th. Let's see if it does. And it does. The total resistance is 1 over 31 12th. And 31 12ths is 1/4. So the total resistance is 1 over 1/4. Do you think that there's going to be another way to express that?

There sure is. And it's 1 / 1/4. If you want to get rid of the division sign, you just simply flip-flop the fraction and it's 1 * 4 over one. So that path, the total resistance is 4 ohms. Let's you and I go back and look at figure 12 one more time. And what we have just done, we have just made this path right here. We've made a total resistance out of that of 4 ohms. So now if you look at where this current can go, current starts out here.

We have a resistor here. And now we have a resistor here and a resistor here. We have parallel paths or more than one path. But we've reduced this to one path. So what we have to do now is add this resistance to this resistance. And that gives us a path here. And then we'll have to figure out these two paths right here. This resistance R2 is 12 ohms. and we just figured out that the resistance we were working on just now was 4 ohms.

So let's take a look at the math of adding those two together. And all we have to do is add those two in series together. It's 4 ohms plus 12 ohms. And then it comes out to 16 ohms. So now let's go back and look at the drawing and see where we're going from there. All right. First of all, we have said now that we have this first path. We have a resistor here and we have a path here. The total path is 16 ohms. And we have another path here that it could go which in this case if you look at that carefully is 4 ohms.

So now we need to convert these two paths these two paths here and this path into one total resistance. And so what we have is if you remember 16 ohms and 4 ohms. So let's take a look at the math for converting those paths into one total resistance. And we have one over and it was 16 ohms and 4 ohms. So the fractions are 1 over 1 / 16 plus 1 / 4. And we can convert that 1/4 into 16. That's 416. Now we can add them together.

And so it becomes 1 over 516. Do you think there's going to be another way of expressing that? Well, I bet if you look down there, you'll find that works out to 1 / 516. If you want to get rid of the fraction, you flip-flop it. It's 1 * 16 fths or 3.2 ohms. Now, let's take a look at where we're going to go from there. And to do that, let's you and I take a look at the drawing one more time. So, here we are looking at that drawing one more time.

And the paths we have just figured out is we have 3.2 ohms for this entire pathway from here to here all the way. So now all we have is this resistance here plus this entire mess now has been resolved to 3.2 ohms. So all we have to do is add them together because they're series circuits. So let's take a look at the mathematics for adding those two together. And when you do that, you get 3.2 + 18 and you get 21.2 ohms.

Let's one more time look at that drawing to see what we did on figure 12. And what we did is we tried to resolve these parallel paths into a total resistance. We said this is a series resistance. There's only one path here. But now we have two paths. And if we go over here, we have a path with these two together. So we resolve these two right here into a total resistance. Then when we got done with that, we said, "All right, now this is a series." So we added the total resistance here to this resistance here which is in series added them together and we made them series.

But now you have to take this whole meth mess and and take these two paths and make a total resistance out of them. And we did that and now all we had then was two series resistances to add together. See there's nothing to that stuff. We've been doing pretty tough questions. It's time for you and I to do an easy one. So let's take a look at figure 11. And the question is going to be, what is the total current flowing in the wire between points C and D?

Well, let's take a look at figure 11. First of all, we have a 24volt battery. And then we have a fuse right here, which has no resistance as far as we're concerned. And then C and D is what we're talking about. So we'd have the current going. It could go this way or the current goes this way. Of course, it goes both ways. And then we have these two parallel resistances and then back. So all we have to do to find out the total current, we have to figure out the resistance of this circuit that involves a wire between C and D.

Now we know the volts and if we know the resistance for that path, then we'd know the current because we'd use Ohm's law to figure that out. So all we have to do is resolve this parallel circuit into a total resistance. And we have 10 ohms and 40 ohms. Let's figure out the resistance. Then we'll use Ohm's law to figure out the current. See, I told you this was an easy one. So, let's take a look at the math on it. Resistance two and resistance three.

So, the total resistance for resistance two and three goes like this. One over the fractions 1 / resistance 2 plus 1 / resist resistance 3. And the resistances were 10 ohms and 40 ohms. So, it becomes 1 over 110th + 140th. Now, let's take a look at how we're going to solve the rest of this down here. To add these two together, we'll convert that one/10 into 440th. So we can add them together. So the total resistance is 1 over 440th plus 140th.

And we add those two together and the total resistance is 1 over 540th. Now let's take a look at what 540th works out to. Well, first of all, instead of dividing it 1 over 540ths, let's express it another way. 1 / 540ths. And to get rid of that division sign, all we have to do is flip-flop the fraction. And it's 1 * 40 / 5. And we can use a calculator for that. And 40 / 5 comes out to 8. And the resistance is 8 ohms.

Now, Ohm's law says if we know the volts, which was 24, we know the resistance, which is 8. To get the current, all you have to do is divide the volts by the resistance. So 24 / 8. and we'll take a look and see what the current is. And the current on the wire between C and D was three amps. And that is back to an easy one again. Now, this one drawing is going to cover three concepts the FA wants you to know about, and that is dodes, rectifiers, and transformers.

Let's talk about diodes first. A diode is basically an electrical check valve. It allows the current to flow in only one direction. So, you have the current going like so. goes past the diode but it can't go back and that is a diode and that's the purpose of it. A diode can either be a tube or a transistor and this is a symbol for a transistor diode like so. Now a diode then you can think of as a check valve and they're used in rectifiers.

Now a rectifier is a circuit that converts alternating current which reverses itself periodically to direct current. You can see how it would work. If you have alternating current that's reversing itself. This diode lets the current go this direction but not that direction. So what you have now is intermittent direct current. Now if you had opposite phase alternating current down here reversing itself. This diode lets it only go in one direction.

And if it's the opposite phase, it's going at the different time than this is. So that these two join to be direct current going out this direction. And that's a rectifier. And that's the whole purpose of it. And diodes are used in rectifiers. Now here is a special kind of diode known as a xener diode. Now a xener diode is one that will conduct electricity only under certain voltage conditions. And a typical application for a xener diode would be as a voltage regulator.

So that is a special kind of diode known as a xener diode. Now here is a transformer. Now a transformer actually is two coils of wire normally wrapped around the same bar or wrapped over the top of each other. And if you have a small coil in the primary circuit and more wraps or more turns in the secondary circuit, you step up the voltage. And by the way, if you step up the voltage, we'll say from one to four here, you would step down the current by the same ratio.

That's one of the things you need to know about a transformer. When it steps up the voltage by a ratio, it steps the current down by that same ratio. Now, here's a question they're going to ask you about this on a test. They're going to say, "All right, let's assume you have a transformer with a step up ratio of 5:1 and the primary voltage is 24 volts. Well, the secondary voltage would be five times that. But the primary amperage is what they're going to ask you what it was because they're telling you the secondary amperage is 0.2 amps." Well, if the secondary amperage is 02 amps, it gets stepped down by the same ratio that the voltage got stepped up.

So, the primary amperage had to be five times. 2 amps or the answer is 1 amp. And that's a typical way they're going to ask you questions about this on a test. All right. Now, let's talk about my favorite subject, power. And the question is, what is power? Well, power is the rate of doing work. And there is a formula for electrical power. And here it is. Electrical power. And by the way, it's measured in watts. We'll talk about that in just a few seconds.

Electrical power is the current flow in amps times the electrootive force in volts. It's amps times volts and that gives you power. Now mathematically you can play around with this a little bit and you can divide each side by e and you find that the current is equal to the power divided by the electromotive force in volts. You could also play around with this and divide each side by I and you'd find out that the volts or electromotive force is equal to the power divided by the current.

Now the two that you need to know for the FA exam are these top two and we'll work with those for quite a bit. Now once again, what is the unit used to express electrical power? And the answer is the unit of electrical power is a watt. Now I had an instructor ask me that question. Johnny said, "What is a unit of electrical power?" I said, "What?" He says, "You're right." All right. Now, let's take a look at a question the FA is going to ask you about this on the test.

They're going to say, "Let's assume you have a 24volt source and it furnishes 48 watts to a parallel circuit and there are four resistors of equal value. What is the voltage drop across each resistor?" Now I am sure that you have heard of the concept of blowing smoke. This is a wonderful example of blowing smoke because the 48 watts in this question has nothing whatsoever to do with it. If you have a parallel circuit, the voltage drop across each resistor is identical to the total voltage drop.

What is the total total voltage drop? 24 volts. What's the voltage drop across each resistor? 24 volts. What did the 48 watts have to do with it? Absolutely nothing. And here is yet another case of blowing smoke on the same subject. The FA just loves to blow smoke. Here you have a parallel circuit. You have a cabin entry light that is 10 watts and a dome light that is 20 watts and they are wired in parallel with each other across a 30volt source.

Now the question is if you measure the voltage drop across the 10 watt bulb what will that tell you about the voltage drop across the 20 watt bulb and the answer is it will be equal because these are just resistances in the circuit and the voltage drop across all resistances that are parallel to each other in a circuit is the same. And what did the wattage have to do with it? Absolutely nothing. Now let's take a look at another question that the watts does have something to do with it.

You have a 48vt source, 192 watts in a parallel circuit, and you have three resistors of equal value. What's the value of each resistor? Now, let's stop right here and take a look at this and see how we're going to figure this out. First of all, we need to know the total current through the circuit. Then after we figured out the total current through the circuit, then we're going to figure out what the resistance is for the entire group.

So let's figure out first of all the total current through the circuit. To do that, we'll use this law right here. You've seen that before. It's the power law. Okay, first of all, the current is equal to the power in watts divided by the electromotive force in volts. The power in watts was 192 and the volts was 48. And so the current in ampers was 4 ampers. So now we know the current in ampers. All right. Now let's take a look at what the total resistance is.

The total resistance for the entire circuit is equal to the electromotive force which is E which is volts divided by the current in amps. The volts was 48. The current in amps was four. The total resistance was 12 ohms. Now that total resistance is made up of three identical resistors. So if we want to know the formula for three identical resistors providing a total resistance, we know that this total resistance is 12 ohms.

Here's how it got there. It's one over the the three fractions of resistance one plus resistance two plus resistance three. One over those as fractions. Okay. Now if the denominators are all the same, you can add fractions together. So that 12 ohms as a total resistance can also be expressed as 1 over the fraction 3 divided by the resistance value for each one of those resistances. Now we can simplify that expression just a little bit.

That 12 ohms total resistance is equal to 1 divided by three time three over the individual resistances. Now we want to get rid of the division sign. All we have to do is flip-flop that fraction as you know. So that total resistance is equal to 1 times the individual resistances divided by three. And now we can begin to see that we can figure out the value of each individual resistance. And so that total res resistance of 12 ohms is equal to R.

That's the individual resistance of each resistor divided by three. Multiply each side by three and you find out the value of R is 36 ohms. The value of each one of those individual resistors is 36 ohms. All right. So what we did first of all is figured out the current for the entire circuit. Then we figured out the resistance here current for the entire circuit then the resistance for the entire circuit. Then we used the formula for parallel resistance parallel circuits.

And then we found out the value of R, which is each one of those resistances. You guys, you guys have got it made. You're doing fantastic. All right. What's the operating resistance? The next question says, of a 30 W bulb in a 28 volt system. And the question is, what is the resistance? Well, first of all, let's figure out the current and then we'll figure out the resistance. That makes it pretty easy. The current works out at a formula like this.

Here is the current is equal to the power in watts divided by the volts. Power in watts is 30. Volts is 28. The current is 1.07 amps. Once we know the current, then we can figure out the resistance with Ohm's law. And resistance equals the volts divided by the amps. The volts was uh uh 28. The amps is 1.07. And the resistance is 26.17. And that is the answer to that question. Let's take a look at another question involving parallel circuits.

The FA is trying to make your life complicated with these parallel circuits and they're doing it. Let's take a look at the question they're going to ask. First of all, they say that you have a 24volt source and that's the voltage that's required to furnish 48 watts to a parallel circuit. And this circuit contains two resistors of equal value. Now the question is what is the value of each resistor? Oh my god, we can figure it out and never fear.

Let's take a look at another bit of information they give you that's extremely valuable. They tell you that the total resistance in this circuit, they give you this formula. The total resistance is equal to the voltage squared divided by the power. So they give you that formula. That's a tremendous help. So now all we have to do is plug in some numbers. We know that the total resistance is equal to voltage squar divided by power.

And the voltage we know was 24 and the power was 48 watts. So you take the 24 * 24 / 48. You do that with your electronic calculator and you're going to get 12 ohms. And so right now we know the total resistance in that circuit is 12 ohms. Now we talked about earlier the formula for the resistance when they're in parallel in a circuit. And it's 1 / 1 / R + 1 / R. And in this case, these resistors are equal to each other.

So this value of R is the same in each case. So that 12 ohms resistance, that's that total resistance is equal to 1 divided by and you can factor out 1 / R here and it's 1 + 1 times the quantity 1 / R and it's the same as 1 + 1 that quantity divided R. So we factored out that 1 / R and it's 1 + 1 that quantity times 1 divided R which is the same as anything divided by R over R. And now we have the total resistance that was 12 ohms is equal to 1 over and this is 1 plus one that comes out to two.

Two divided R. So that's all there is to it. And so the total resistance is 1 / 2 / R. Well to multiply all we have to do is flip this. So it's total resistance at 12 ohms is equal to 1 * R / 2. Okay. one times both of these means the total resistance is equal to R / two. That's all there is to it. And the total resistance was 12 ohms. And so 12= R / 2. Multiply both sides by two and R comes out to 24 ohms. The value of each resistor comes out to exactly 24 ohms.

Well, let's do it a little bit more complicated. question is how much more how much power must a 24volt generator furnish to a system with the following loads and this is the following loads right here we have a 24volt generator you have one motor that's 75% efficient and it's 1/5 horsepower you have three position lights in the airplane that are 20 watts each you have one heating element that is 5 amps and you have one anti-colision light that is 3 amps How much power is required?

Well, we have watts here. We have horsepower here. We have amps here. Let's convert this all into watts. First of all, let's figure out the motor. Now, there is something they give you that you absolutely have to have. And don't panic because you don't know this because they tell you this. What they tell you is that one horsepower is 746 watts. And you do not need to memorize that. They tell you that when they give you the question.

Now, this motor was only 75% efficient. So, in order to get one horsepower, you actually have to have 746 watts divided by 75 or 75%. That would come out actually closer to about a,000 watts. And we weren't actually looking for one horsepower in this case. We were looking for 15 horsepower. So, what we'll do is we'll take 1/5* 746ide by 75. and you find out this motor requires 199 watts. Now, they told us also we they had three position lights and each one of those was 20 watts.

So, three times 20 is 60 watts. And they also told us we had a heating element, but they said the heating element was 5 amps, not watts. And so, we need to convert that into watts, which is power. And so what you do is use this formula for power and you take the current times the voltage or the electromotive force and that gives you the power in watts. So was 5 amps times the 24 volts gives us 120 watts. All right. Now let's take a look at the rest down there.

The anti-colision light was 3 amps and it's 24volt system. So we can do the same thing. Uh 3 * 24 or 72 watts. And all we have to do is total those up. We had 199 for the motor. The position lights was 60 watts. The heating element was 120 watts. The anti-colision light was 72 watts. And the total of that is 451 watts. So, if you know these formulas, you got it nailed. They're just asking you to be able to multiply volts times amps to get watts.

And that's all we had to do in that particular case. Now, here's a question the FA is going to ask you that I think is actually kind of fun. It involves three different types of electrical systems. And it says, "Which one of these do you think will require the most electrical power?" Well, let's take a look at each of these three situations one at a time. In the first one here, you have four 30 watt lamps and they're arranged in a 12vt parallel circuit.

Now, they ask you about power. And what's the term you use to describe power? It's watts. So, they've already told you you have four 30 watt lamps. So the total power is going to be 4 * 30 or 120 watts right there. And that is deceptively simple. So the first circuit is 120 watts. Well, that's pretty easy. Let's do another one. Now you have a 24volt anti-colision light circuit. It consists of two light assemblies which require three amps each during operation.

Well, let's take a look at this. First of all, power as you remember is amps times volts. Well, we have 3 amps and 24 volts. So, uh, this is 3 * 24 in each circuit. And that's the power required each circuit. It's the amps times the volts. And we have two of those circuits. So, to figure out the power here, 2 * 3 * 24, that's the power in each circuit equals 144 watts. So, that's a little bit more. It's 144 watts. So, let's take a look at the third situation they're going to ask you about.

Here you have a motor that produces 1/5 horsepower. It's a 24volt motor that's only 75% efficient. And they tell you that one horsepower is equal to 746 watts. That's a key bit of information you have to know, but they already told you that. So, let's take a look at it. First of all, we have one horsepower is 746 watts, but this motor is only 75% efficient. So to produce a horsepower, it's 746 divided by 75. So that would give you one horsepower, but we're only talking about 1/5 horsepower.

So we'll multiply that times 1/5. So the formula will be 1/5 times the the power required for one horsepower. 746 watts for for one horsepower, but it's only 75% efficient. And then you multiply that out and you get 199 watts. That's the one that requires the most power. And I think that's a fun question. Now, here's another question involving power they're going to ask you about on the test. They're going to say, "Let's assume that you have a 12volt electric motor and that it has a 1,000 watt input and a 1 horsepower output." So, you can see it's pretty efficient.

Now, let's assume that you had a 24volt motor with one horsepower output that had the same efficiency ratio. The question is how many watts would be required? And the answer is the same 1,00 watts. If it has the same efficiency ratio, watts to horsepower, it's going to be the same down here. St,00 watts. And whether it's 12vt or 24volt doesn't make any difference. Now, let me ask you a question. Do you think that even in the most remote stretches of your imagination that the FA could ever be deceptive?

You bet they can. And here is a case in point. They have a question like this. They give you this information. They tell you that you have a one horsepower 24volt DC electric motor and that this electric motor is 80% efficient and it requires 932 1/2 watts to produce that one horsepower. They give you that information. Then they ask you this question. They're going to ask you, "How much power will a 1 horsepower 12volt DC electric motor require if it's 75% efficient?" How much horsepower or watts will be required to produce that one horsepower?

And they also tell you that one horsepower equals 746 watts. Now, you might look at this and say, "Hold it, wait a minute. What does this top part have to do with this question? I've got the information I need right here, don't I?" And the answer is this top part had absolutely nothing to do with the question. They just threw that in there to confuse you. It works, doesn't it? Well, now if this engine, this motor were 100% efficient, it would take 746 watts to produce this one horsepower.

But in this particular case, we're told it's 75% efficient. And so all you have to do is take that 746 watts divided by 75 and you're going to find out that it's going to take about 1,000 watts to produce one horsepower. Actually 994.7 watts and you have produced one horsepower. And that is a case of being very tricky. Let's do another one. This time they're going to say you have a 14 ohm resistor to be installed in a series circuit carrying. 05 amps.

Now the question is what is the power required to be dissipated by that resistor. All right let's take a look at it. First of all we want to know how many volts are in the circuit before we can multiply volts times amps to get watts. All right to find out how many volts are in the circuit you use this formula. Volts is electromotive force equals the current which is I times the resistance. The current was 0.05 >> [snorts] >> 05 amps and the resistance was 14 ohms and you multiply that out and we have.7 volts.

Now we have the information to find out power and so you take the volts which was 7 and multiply it times the current which is 05 and 05 * 7 equals watts and that's 35,000th of a watt. Another way to express that is 35 mwatts because 1 millatt is a thousandth of a watt. All right, let's do another one. This time, let's take a look at figure 8, appendix 2, the FAA's book, and they give you a circuit like this. And they say, how much power is being furnished to this circuit?

Well, once again, first of all, you have to find the voltage before you can multiply the volts times the amps. Well, let's call this a generator. Let's call this a 23 amp meter here. It's an ampmeter and it's measuring 23 amps and a resistance of 5 ohms. And so we know the amps and we know the resistance. So now let's take a look at what we need to do to find volts. Well, if you remember the formula, it's very simple.

Volts equals the amps times the resistance. And we have 23 amps and resistance of 5 ohms. It's 115 volts. When you multiply that out to find the power, you multiply the current times the voltage. And the current was 23 amps. The voltage was 115. Multiply that out and it's 2,645 watts. All right, it's getting simpler and simpler and simpler, isn't it? Let's do another one. Let's assume that you have a 30volt 1/2 horsepower motor that is 85% efficiency.

The question is how many amps will it draw? Well, first of all, let's find out how many watts this motor is and then we can figure out how many amps it is. All right, to find out how many watts, we have a half horsepower. They will tell you once again that one horsepower is 746 watts. You do not need to memorize that. Don't worry about that. Saying it's 85% efficient. So to produce 1 horsepower at 85% efficiency, it would be 746 /.85, we only want to produce 1/2 horsepower.

So we multiply that times 1/2. So it's 1/2 * 746 /85. And it's 438.8 watts for this motor. But hold the phone. They ask us for amps. And the formula for amps is uh the current equals the power in watts divided by the electromotive force in volts. And the power is 438.8 divided by the 30 volts. And so the current is 14.6 amps. And that's the answer to that question. Now before we wrap up DC current, I want to cover those formulas that you need to have in your mind when you walk into that test room.

Let's take a look at them. First of all, if you remember, you need to know Ohm's law. And Ohm's law is current in amps is equal to electromotive force in volts divided by resistance in ohms. Now, if you know that, you can get the other two in Ohm's law if you're good with algebra. But the other two are that the electromotive force in volts is equal to the current in amps times the resistance in ohms. and the resistance in ohms is equal to the electromotive force in volts divided by the current in amps.

You need to know those three. If you got algebra nailed, then all you have to do is know the first one and you can figure out the other two. Now, let's take a look at the formulas for resistance. First of all, resistance in series. To find the total resistance in a circuit, just add up the individual resistances, however many there are. Resistances in parallel. To find the total resistance, you use this formula. It's 1 divided by these fractions. 1 over resistance 1 plus 1 over resistance 2 plus 1 over resistance 3.

If you know those two resistance formulas, you've got that taken care of. And finally, let's take a look at the formula for power. Power is equal to, and it's in watts, by the way, equal to the current in amps times the electromotive force in volts. And the other formula for power, you can figure it out algebraically. If you know this one, you can figure this one out. But the other one is is that the current in amps is equal to the power in watts divided by the electromotive force in volts.

Now, if you've got everything that we've talked about, you have everything you need to know about DC current on the general exam. Now, let's see if you and I can figure out AC current or alternating current. Now, alternating current is different than direct current. Direct current goes only in one direction. Well, alternating current goes in one direction, then it reverses itself, goes the other direction, then it reverses itself, goes the other direction, then it reverses itself and goes the other direction.

I think you get the idea. Alternating current periodically changes in direction and in continuously changes in its magnitude. Now, if you look at it in a sine wave, you'd see what I'm talking about. Here's what it would look like in an oscilloscope. And this is a sine wave on an oscilloscope. And it describes or shows you what alternating current looks like. Now, here's the midpoint and you have the alternating current first going one direction and very low intensity and then it increases its intensity and then it begins to level off, gets weaker, then changes direction, goes the other direction, gets stronger in intensity the other direction, weakens off and so on, back and forth and back and forth.

Now, they're going to ask you on the test, how would you describe alternating current? Here's how you define alternating current. It's current that periodically changes direction and continuously changes in magnitude. And that's a picture of what alternating current looks like right there. Now, there are three values regarding alternating current that should be considered. They're instantaneous, maximum, and effective.

Let's take a look at them. First of all, the instantaneous value of alternating current is the value of the induced current or voltage flowing at any instant. And let's take a look at that in writing here. The instantaneous value is the value of voltage or current flowing at any instant. That's the instantaneous value. In other words, at any particular time, you can figure out what the value of the current is. That's the instantaneous value.

The next thing we want to see in writing is the maximum value. And the maximum value is the largest instantaneous value. It's the peaks here. That's the maximum value, the largest instantaneous value. Finally, what's really important is the effective value. And the effective value is the direct current to produce an equal heating effect. In other words, if you're trying to heat up the circuit, how much current would you have to do in direct current to produce the same values as this particular alternating current?

And that's the effective value. And by the way, unless somebody says otherwise, when they talk about the value, they're talking about an effective value. And as a rule, by the way, the effective value, would you think it would be more or less than the instantaneous value? And the answer is the effective value is less than the maximum instantaneous value because the effective value is kind of an average. It's called a root mean squared but it's really an average and it's less than the peaks or the maximum instantaneous voltage.

So the effective voltage would be less. Now once again unless otherwise specified any values given will be effective values unless somebody says otherwise. And that's either for current or voltage in an AC circuit. And that's what you assume is effective values. Let's talk about lines of force because that's a concept that'll be useful later on. Anytime you pass a current through a wire, you create a magnetic field around that wire and that's called lines of magnetic force around that wire.

And that's what we're talking about. And when you apply a voltage to put a current through a wire, first what happens is you create a magnetic field. And that actually delay delays the flow of current for just a little bit. And so the current doesn't flow for a while. Then when the magnetic field is created, the current flows. If you happen to apply alternating current, what happens is first you have a voltage applied in this direction.

There's a little delay while the field is being created. And then when that voltage drops off, the current still flows for a while while the field collapses. Then you apply a voltage in the other direction. There's a delay while the magnetic field gets created and so on. But you have a constantly collapsing and expanding magnetic field when you run alternating current through a wire. Now, if you want to intensify that magnetic field, you can do something very, very creative.

What you can do is take that wire and wrap it into a coil and you concentrate or intensify that magnetic field and get a much stronger magnetic field. If you want that magnetic field to be even stronger yet, what you can do is put a iron core in here and that really concentrates the magnetic field. And in fact, one of the things you want to know is magnetic lines of force pass most readily through iron when compared we'll say with copper or aluminum.

And therefore putting an iron core in here really concentrates the magnetic field. Now another bit of information you might want to know is that if you took an electromagnet, let's assume you don't have any current through this wire right now and you just take any kind of magnet for that matter, take a magnet and wave it by this wire. What you'll do is you as you wave the magnet by this coil of wire, you'll create a current in that wire.

Of course, that's the basis on which a generator works. And that's one way of creating a current is just take a magnet and pass it by a wire and you'll create a current in that wire. Now, let's get a little more complicated. Let's have two coils of wire. Here we have a coil of wire with an iron core and another coil of wire with an iron core through it. In fact, we'll have the iron cores be connected. But these two coils of wire will not have connected electrically in any way at all.

Now, let's assume that we put an alternating current in this coil of wire. What happens is we're going to create a magnetic field that keeps expanding and collapsing. Well, we said if you run a magnetic field past the coil of wire, you could create a magnetic you could create a current in that other coil of wire. And that's exactly what happens here. The expanding and collapsing field from this coil of wire creates a current in this coil of wire even though they are not connected to each other electrically in any way.

So transfer of electrical energy from one conductor to another without the aid of any kind of electrical connections connections between the two that is called inductance. And this arrangement is called a transformer where one coil creates a current in another coil over here. And the basis for transformer operation in the use of alternating current is what is known as mutual inductance. A magnetic field from one coil induces a current and the other and that's called mutual inductance.

So when you create a coil of wire in a circuit, you are creating what they call an inductor in the circuit. Now let's assume that we have an electrical circuit and we have a couple coils of wire. If you have one after the other then you have more inductance. In fact, when you put two inductors in a circuit, what happens is the total inductance is equal to the sum of the individual inductances. So now we have two coils, we have twice as much inductance.

That means twice as much magnetic field and in some conditions twice as much impedance to the flow of current while that field is being created. Okay. Now let's assume instead of having these in series as this is one after the other, you put these inductances in parallel with each other. Now you're going to find that there is less inductance totally. How much less inductance totally is there? Well, the total inductance is less than the lowest rated inductor of all of these of any one of these is less than in any of them.

This is putting inductances in parallel. Now we said earlier anytime you have inductance a coil of wire it in it tends to impede the flow of current. What's that called? Well, that is called inductive reactance. Let's take a look at the words for that. That's called inductive reactance. And it's measured in ohms. Why is it measured in ohms? Because it slows down the current. It impedes the flow of current just the way resistance does.

And so it's called inductive reactance. And it's measured in the same unit that resistance is measured in in a direct current. And it's called ohms. The unit is ohms. What is inductive reactance? Well, it's the inductive reactance is the opposition of the flow of a coil to the flow of alternating current. It tends to slow the current down. And remember, it does it uh 90 it makes the current flow 90 degrees later than the voltage.

Now, let's take a look at a circuit here and see if we can make some sense out of it. Here we have a circuit where we'll say with a generator in it and we have some resistance, some capacitance and some inductance. And this is a typical circuit. We'll talk more about this later on. Now, as you increase the frequency of the current going through here, you change the rate at which these lines of force expand and contract around this inductance.

And the higher the frequency of the current, the more inductive reactance there is. It tends to impede the flow of the current. And because it changes, expands and collapse the magnetic field much more rapidly. So an increase in in in frequency will cause an increase in inductive reactance and also the the the more the natural inductance the more the coils of wire around there will change and the inductive reactance.

So an increase in inductance itself and that is how many wires are wrapped around there how tightly they're wrapped whether or not you have a core in it that will change the inductive reactance and also the frequency of the alternating current changes the inductive reactants. So those are the two things that will change the inductive reactance to a current. Now there's another thing installed in this current we haven't talked about here.

We'll talk about that and that is capacitance. Now capacitance is basically two metal plates next to each other with a uh or an insulator or a dialectic it's called an insulator in between them. And what happens is remember this is current that tends to build up and you build up a charge and as the current flows it charges one of these plates. So, let's have the current coming through here charging this plate. Now, if this were direct current, because there's an insulator between the two plates, the current would charge the plate and quit flowing because couldn't get across this capacitor.

But what happens is with alternating current, the current tends to immediately charge the plate when it gets to a capacitor. It's a static, all you're doing is putting a static electricity charge on the plate here. So, we have a static electricity charge on this big plate of metal. And it actually could be aluminum foil. Two sheets of aluminum foil wrapped around each other with a with a dialectic or an insulating material between them.

But anyway, you're charging this plate first. And if it's direct current, the current will quit flowing immediately. But if it's alternating current, the current tends to reverse in direction. And then what happens? The capacitor gives up such its charge right away. And so then the current will flow back this way, charge the capacitor, then it gives up its charge when a current's going back. So this capacitor also tends to impede the flow of current.

You can see that when you have direct current, the impedance is absolute. It just completely stops the the current going through here. But if the alternating current has a frequency to it, as the frequency increases, this this capacitor tends to impede the current less and less because it charges the plate and it gives up its charge. Charges the plate gives up the charge. And the faster the current changes in frequency and direction.

In other words, the faster the frequency of the current, the higher the frequency of the current, the less impedance is imposed by this capacitor here. And by the way, since the capacitor is almost like a short at first, it gives a place for the current to go. What happens is is the current is about 90 degrees ahead of the voltage. Now impedance, you remember that the current's about 90 degrees behind the voltage. And this is going to become important.

So now what determines the amount that a capacitor can store? Well, let's take a look at it. It's directly proportional to the plate area and inversely proportional to the distance between the plates. If you have great big plates, it can store more electricity. And if you move those plates further apart, it can store less. So the closer the plates are together, the more it can store. And the bigger the plates are, the more it can store.

Now, there is a maximum working voltage of a capacitor because the dialectic is only so strong. And so if you had too much voltage, it would just jump right across the capacitor and of course ruin the capacitor. So, what they say is the working voltage of a capacitor should be at least 50% greater than the highest applied voltage you're ever really going to put on it. Otherwise, you could short out the capacitor and go right across it.

Now, sometimes in a circuit, including DC circuits, you'll see capacitors installed like this so they go to ground. Now, why do they do that? Well, they do that to smooth out fluctuations in current and voltage. Let's take a look at how it works. Let's assume you have a current going like this and the voltage increases that voltage will tend to charge this plate of the capacitor and then when the voltage decreases that capacitor will give up that charge and the net result is that it's smoothed out fluctuations in current and voltage.

So sometimes a capacitor in a DC circuit is installed to smooth out slight pulsations in current and voltage. Now, just as there are variable resistors, there are variable capacitors. And the FA is going to ask you on the test, what is the symbol for a variable capacitor. So, let's take a look at figure 17 in the back of the book. And up in the upper leftand corner here, you'll see a variable capacitor. It's E. Let's take a real good close look at that.

We get in there and look at that. The symbol for a variable capacitor is the capacitor symbol, the normal capacitor symbol with a little arrow through it. When you see that arrow on a variable resistor, that arrow is a variable resistor. On a variable capacitor, the arrow stands for a variable capacitor. So that's a symbol for a variable capacitor. Now, let's take a look at what happens when you have more than one capacitor in a circuit.

You can see right here that we have two capacitors in this circuit and we have them arranged in series. That's one after the other. Now, if you were talking a resistor or inductor and you had more than one in the circuit and in series, you just add up the resistances or inductances and you'd have it figured out. That's not the way it is for a capacitor. A capacitor is the other way. When you have capacitors in series, they behave like resistors and inductors do when they are in parallel.

So when different rated capacitors are connected in series in a circuit, the total capacitance is less than the capacitance of the lowest rated capacitor. Let's take a look at a question they're going to ask you about this on the test. And by the way, you won't have to memorize formulas. They're going to give you formulas, but the question goes like this. You have three capacitors in series. 02 microfarad, 005 microfarad, and 0.10 microfarad.

By the way, instead of working in farads, let's keep it in microfarads. All we have to do is keep it the same. Now, they're going to ask you, what is the total capacitance? Well, now capacitors behave when in series like resistors and inductors do when they are in parallel. So, they will give you the formula. You don't even need to memorize the formula. And here it is. What you do is you take the uh capacitors and it's one over this whole mess and it's one over this capacitor, one over that capacitor and one over that capacitor.

It's the same formula. They will give you that formula. So it's one divided by 02. You can use your calculator and that comes out to 50. One divided by 05 comes out to 20. And 1 divided by 0.1 comes out to 10. So it's 1 over 50 + 20 + 10 that comes out to 1 over 80. And so you use your calculator and figure that out and it's 0125 microfarads. So this is capacitors in series. Now what happens when you put capacitors in parallel?

Well, when you put capacitors in parallel, the total capacitance is equal to the sum of all the capacitors. Now all you have to do is add them up. They'll ask you another question about that. They're going to ask you one that goes a little bit like this. They're going to say you have three capacitors in parallel. 025 microfarad, 03 microfarad, and 0.12 microfarad. What's the total capacitance? Well, this is easy. Even I can do this.

You just simply add them up. So the total capacitance is the sum of all of them. You do that with your electronic calculator and you get 0.40 microfarad. Now once again, you don't need to memorize these formula. They will give them to you. But for capacitors in parallel, just add them up like that. All right, let's you and I go back and look at our drawing one more time. And we said just like inductance, capacitance impedes the flow of electricity.

What is that called? Well, the term for it is capacitive reactance. And what's it measured in? Just like inductance and impedance, capacitive reactance is measured in ohms. All right. Now the combination of resistance, capacitive reactants and inductive reactants all together is called impedance. And once again impedance a combination of resistance, inductive reactants and capacitive reactants and all of those things impede the flow of current or there an opposition to current in an AC circuit.

What is the unit of measure for impedance? You've got it figured out. The question the answer is ohm. That's correct. The unit for the measure of impedance is an ohm. All right. Now, what is the term that describes the combined resistive of for forces in an AC circuit? Well, that term is impedance. The combined resistive forces are described as impedance in an AC circuit. Now, on the test, they're going to ask you to figure out the impedance in an AC circuit.

And they're going to give you a problem that looks like this. They'll say the inductive reactance in this particular circuit is 10 ohms. The capacitive reactance is 4 ohms and the resistance is 8 ohms. Now the question is what is the impedance? As you know there is a relatively complicated mathematical formula for impedance. But the amazing thing about this question is they give you that formula in the question and they tell you what all of it means.

For instance, they tell you that Z equals impedance, R stands for resistance, X subL stands for inductive reactance and X subC stands for capacitive reactance. And they give you the formula. They say impedance is equal to the square root of this entire expression which is R squ that's resistance squared plus the quantity and XL's inductive reactance minus XC which is capacitive reactance that quantity squared. And that's the formula and they give it to you.

So all you and I have to do to solve this problem is to fill in the numbers. So let's fill in the numbers. And first of all the impedance or Z is equal to and it was the square root of the entire expression resistance squared. That becomes 8 squared plus and then that's the quantity and then the first thing in the quantity as you remember was inductive reactance and inductive reactance was 10 minus inside the quantity is capacitive reactance and capacitive reactance was four.

And then you close that quantity and square all of that. So now let's take a look at it. Z equals the square t of 8 * 8 is 64 plus 10 - 4 is 6 and that's going to be squared + 6 2. So continuing down with this z = the<unk> of 64 + 6^2. 6 * 6 is 36. So z = the<unk> of 64 + 36. Add those two together and z = the<unk> of 100. and you use your electronic calculator and Z equals 10 ohms. And that's the question they ask you about it.

And do not panic. You do not need to memorize that formula. They'll give you the formula. Now, here's another question they're going to ask you about a circuit. They'll say a circuit has resistance of 10 ohms, inductive reactance of 20 ohms, and capacitive reactance of 30 ohms. And then they say, how would you describe this particular circuit? Would you describe it as capacitive? And the answer is yes. This is a capacitive circuit because capacitance is predominant.

So you would call that a capacitive circuit. Let you and I take one more look at this circuit we've been looking at for quite a while now. I want to point out there's three things in that circuit. First of all, there's resistance as you can well see and also there's capacitance right here. Of course, there's inductance right there. Now let's talk a little bit more about inductive reactance. Let's review it because we talked about it earlier.

We said that the higher the frequency in this AC circuit, the more inductive reactance there is. And that's because the magnetic field expands and collapses more rapidly when the frequency of the AC current is higher. We also said that when there is inductive reactance, current lags voltage by 90°. But hold the phone. We also said in capacitive reactants, the current leads the voltage by 90°. And we said with capacitive reactants that when the frequency decreases we said that you get more capacitive reactants.

So these behave opposite of each other. And as you can see right now, there will be a frequency that the AC goes in this uh circuit in which inductive reactants and capacitive reactants cancel each other out and it will be as if they were not there. The only thing left in that circuit will be resistance. Now what do you call that uh frequency and the answer is you call the entire circuit and the frequency resonant. When capacitive reactants is equal to inductive reactants, the circuit is said to be resonant.

Now, here is the answer to the last question we need to cover about AC circuits. And the answer is true power is less than apparent power. Let's talk about what we mean when we say that. Well, if you have inductive reactants or capacitive reactants in an AC circuit, they tend to throw the current out of phase. And that's kind of like having three guys trying to get a car stuck in a mud puddle out of that mud puddle by rocking the car back and forth.

Well, if you have capacitive reactants or inductive reactants in the circuit, it's like two of those guys rocking at different times than everybody else. And that means that they won't get as much work done. And that's exactly what happens in an AC circuit when you have either reactive or inductive reactants and or capacitive or inductive reactants I should say. And the answer is true power then becomes less than apparent power because of either the capacitive or inductive reactants.

All right. Now let's talk about batteries. And on figure and let's take a look at that figure 10 in the FA book they ask you a question about batteries and they say what would be the voltage of the total production of this situation right here at A and B at that particular point. What's the voltage? Well the way you figure it out is we have 1 and a half volt batteries. The important part is are they connected in series or are they connected in parallel?

Now if you connect batteries in parallel that means they're connected plus pole to plus pole and minus pole to minus pole. And when you connect them in parallel and are one and a half volt batteries the result will be one and a half volts. But on the other hand if you connect batteries in series you cross the poles. The plus is connected to minus and the plus is connected to minus all along. Then you add up the voltage.

Now in this particular case we have here two pairs of batteries and notice in these pairs of batteries left and right pairs that the plus pole is connected to the minus pole and over here by the plus poles in the center and the minus poles around the outside of the battery. And again here we have the plus pole connected to the minus pole. So each pair of batteries gives you three volts. Now let's take a look at how the pairs are connected together.

And the way the pairs are connected together is that the plus pole is connected to the plus pole and the minus poles are connected to the minus poles as far as the pairs of batteries are concerned. So you have three volts on this side, three volts on that side. And since they're connected parallel, it's still 3 volts, just more capacity. And so now you find that if you put a voltmeter right across A to B, that you'd have three volts.

And that's a question to ask you about on the FA written exam. So in this case you would read three volts. Now let's talk about aircraft batteries called storage batteries inside an aircraft. This is an example of a lead acid storage battery inside an aircraft. And they want to ask you and they will ask you a few questions on the FA written exam about these batteries. Let's take a look at it. First of all, how are batteries rated?

Well, they're rated according to voltage and ampere hour capacity. How many amps can they put out for what period of time? And that's called ampour capacity. Now, the state of charge determines the current that will flow through a battery while it's being charged in a constant voltage source. Now, let me explain that just a little bit. And a constant voltage source is like a charger or a generator in or an amp alternator in in the aircraft and the generator or alternator puts out a constant voltage.

We'll say 14 volts. And when the battery is extremely low, a lot of current flows into the battery. But when it's charged, very little current flows into the battery. So, it's the state of charge that determines the current which will flow through a battery while being charged by a constant voltage source. By the way, the other way you could charge a battery, and you do this very often on certain batteries, is a constant current source of charging a battery.

We'll talk more about that later on. Now, let's talk about some specifics. They ask you about lead acid batteries. First of all, they want you to know that a sulfuric acid that's used as an electrolyte in there has a much lower freezing point than water. And what that tells you is a fully charged lead acid battery will not freeze and still extremely low temperatures are reached because most of the acids in solution and sulfuric acid has a much lower freezing point than water.

Another question they ask you about is a hydrometer reading. And what you do is you can measure the charge of a lead acid battery by measuring the specific gravity of the electrolyte with a hydrometer. And they want you to know that a hydrometer reading does not require a temperature correction at 80° Fahrenheit. And that's generally true because anywhere between 70 and 90 degrees Fahrenheit, you do not have to correct the hydrometer reading for temperature.

So 80 degrees Fahrenheit is a good temperature. Now, let's assume you're working with a lead acid battery and maybe you tilt it or something and you spill the electrolyte solution all around an aircraft. Well, the owner of the aircraft is going to think that is a very bad thing. They're going to ask you how you did it. But what the question is, what are you going to do to get out of this mess? And what you want to do is neutralize it with an alkali.

The electrolyte is very acid. You want to neutralize it with an alkali. So, you'd use something like sodium bicarbonate because that's an alkali. So you'd apply sodium bicarbonate and then of course you wouldn't want to leave that lying around the aircraft. So then you'd rinse it with water. The best thing is don't spill the electrolyte. The owner will not like it. Okay. Now let's take a look at another problem with lead acid batteries.

And sometimes you have a sediment buildup at the bottom of a lead acid battery. Builds up under the plates here in a cell container. And if that sediment builds up enough it can contact these plates and cause a short circuit. So to prevent sediment buildup from contacting the plates and causing a short circuit, they provide lead acid batteries with a space underneath the plates to hold that sediment buildup so it doesn't contact the plates.

Now let's take a look at nickel cadmium batteries or NIKAD batteries. As you know, NIKAD batteries generally are used in turbine aircraft because they have very little internal resistance and they can put out a tremendous amount of electricity in a short period of time. What is the electrolyte used in a nickel cadmium battery? It's potassium hydroxide. You need to remember that on the test. Now, the level of electrolyte in a NIKAD battery is low sometimes.

Let's take a look at when the level is low in a NIKAD battery. And the answer is when it's discharged. The level is lowest in a NIKAD battery when it's discharged. By the way, you might be interested in what we're looking at here behind us. This is a NIKAD cell, one cell out of a battery that is being strapped off to short it out to make sure that it's completely discharged before it's charged. Okay. Now, remember we said earlier the level in a NIKAD battery is lowest when it's discharged.

So, what would happen if you add water to a NIKAD battery when it's not fully charged? Well, the answer is you'd have spewing, excessive excessive spewing. The FA says if you add water to a NIKAD battery when it's not fully charged because the level automatically comes up when it's charged and then you could spew it out because at the end of a charging period a NIKAD battery tends to put out some gas. Nickel cadmium batteries or NIKAD batteries which are stored for a long period of time will also show a low fluid level and you need to know that.

Now here's a question I do not like on the FA written exam. They ask you, "How do you determine the state of charge of an IAD battery?" And the answer, the only one that can possibly be a correct answer is that the state of charge of an ICAD battery can be determined by a measured discharge. Now, what they're talking about is you discharge the battery at a specified rate and measure its ampere hour capacity, then recharge it.

Now, that might be a way to tell the condition of the cells of the battery, but I don't think it's a very good way to determine the state of charge. Doesn't matter what I think. When you look at the FA question and you go back and look at the FA sources of information, this is the only one that's the correct answer. So, this is the correct answer regardless of what I think of it on the FA exam. It doesn't matter what I think.

I guess it matters what the FAA thinks. So, the state of charge of an NIKAD battery according to the FAA on the test can be determined by a measured discharge. All right. Now, should you ever service NIKAD and lead acid batteries together in the same service area? And the answer is no, you should not because it can result in contamination of both types of batteries because the fumes are not compatible with each other.

Now, let's talk about charging batteries and the methods you can use to charge batteries. Charging batteries is a good idea. If you couldn't charge a battery, you could use them only once. That's not such a good idea. Let's take a look at the methods. One method you could use to charge a battery is the constant voltage method. Now the constant voltage method is what you're using in the air because a generator puts out a constant voltage and what happens is the rate of charge is very rapid when the battery is very discharged and the rate changes and gets less as the battery becomes charged.

So the constant voltage method gives you a constant voltage but the current or the rate of charging varies. Now the advantage of the constant voltage method is that you get a very rapid charging rate particularly when the battery is very discharged. However, the disadvantage is if you have a NIKAD battery, that very high charging rate could actually cause the battery to overheat. So, that's both the advantage and disadvantage of the constant voltage method because you're not really controlling the rate that you're charging the battery.

Now, let's take a look at the other method, the constant current method. And it does exactly that. It does control the current going into the to the battery. And it does that by varying the voltage. So, you have a constant current going into the battery. The advantage is you do have that controlled charging rate. The disadvantage is that it takes longer when you control the rate and generally you'll have a a lower rate of charge.

Now, if you're charging multiple batteries at the same time, you could use the constant voltage method. To use the constant voltage method, the batteries must be connected in parallel and they can even be of different capacities, but they must be of the same voltage because you're applying a certain voltage to the batteries and you therefore want the batteries to all be of the same voltage. On the other hand, if the batteries are of different voltages, you could use the constant current method.

Then the batteries are connected in series and they can be of different voltages, but they must be of the same capacity. So if you have them all of the same voltage, you'd probably want to use the constant voltage method. If they are of different voltages, uh then you could use the constant current method. But if they're of the same voltage and capacity, you can use either method of these two and it's your choice. Let's take a look at the next question.

When do NIKAD batteries tend to emit gas? Well, the cells of NIKAD batteries emit gas only towards the end of the charging cycle. We talked about that earlier. Now in the test, they're going to ask you what determines the cell voltage of a fully charged 19 cell nickel cadmium battery. And the answer is temperature and the method used for charging determine the cell voltage. So it can vary. Cell voltage normally varies from about 1.45 volts to 1.58 volts.

And that's determined by the method for charging and the temperature. Now, one thing about NIKAD batteries is they can develop a condition that's really bad called thermal runaway. And let's take a look at how it works. Here's an example, by the way, of thermal runaway. It's a cell that's had thermal runaway. This is just a very mild case of the result of the damage from thermal runaway. It can get a lot worse than this.

So, let's take a look at it. First of all, NIKAD batteries tend to heat up when current moves either into them or out of them. So, let's assume that you've really discharged a NIKAD battery. Say you had a a difficult engine start and now you're charging that battery back up. Well, it'll tend to heat up the battery when you charge it back up. Now, NIKAD batteries have low internal resistance. So, that allows it allows a very high rate of charging current and that of course heats up the battery further as you as you charge it.

So, you discharge it, it heats up the battery. You charge it, it heats up the battery. Now, high temperatures in a NIKAD battery cause the separating material inside the cells to break down and that even further reduces the internal resistance and that allows even more current flow which causes an even higher heat inside the cell and more separator material to break down. So that cycle is called thermal runaway. So what contributes to thermal runaway?

Well, the low internal resistance of the battery is one factor. And of course, the resistance gets even lower when you have high cell temperatures. And finally, the charging system in the aircraft makes a difference. First of all, it discharges very quickly and that tends to heat it up and start this cycle. But it also charges very quickly. And one of the reasons it charges very quickly is that we have generators and aircraft that put out a constant voltage.

That's a that's called a constant potential charging system. And what that means is when the battery is very discharged, it charges up very rapidly and that tends to heat the battery even further. So, as you can guess, thermal runaway is really a bad thing. It can cause the battery to sever the control cables or can even cause the battery to fall through the floor of the aircraft. So, what's the solution to all this?

Disconnect the battery so that these things are no longer in play. Let's assume that you find heat or burn marks on the hardware of a battery. What's that an indication of? Well, it's an indication of improperly torqued cell link connections and it means you have had some arcing there. What is the most probable cause of an excessive amount of potassium carbonate formation on a battery? And the answer is battery overcharging will cause an excessive amount of potassium carbonate formation.

All right. and nickel cadmium battery cases and drain surfaces which have been affected by electrolyte should be neutralized. What should you neutralize them with? And the answer is a solution of boric acid. And that is everything they ask you about AC current and batteries on the written exam. Now let's you and I do something really fun. Let's talk about troubleshooting in electrical diagrams for electrical circuits.

And here on figure eight in the FA book is a good example of a electrical circuit in the diagram. Now the question I want to talk about here is how do you use an ohm meter to correctly test a resistance? And what you do is you put the ohm meter parallel to the resistance. Now what the ohm meter does is it sends a little current through the resistance and checks the resistance by doing that. By the way, we have a break in this resistor.

So that doesn't factor in at all. And we're sending a current through these two resistors. In order to do that, you must have at least one end of the resistors disconnected. In this particular case, we have this switch open and therefore we can send a current around. Then the ohm meter measures how the current goes and measures the resistance. Now, if you did not have an open circuit, you would have a a voltage going through here and a current going through here and that of course mess up your reading.

So, the correct way to use an ohm meter is to put it parallel to the resistors you want to check and have at least one end of the circuit open. And that's what we have right there. Let's you and I do another one. And to do that, let's take a look at figure nine. And son of a gun, here is figure nine. They're going to ask you a question about this. But let's talk about the correct way to connect an ampmeter. And here we have an ammeter correctly connected in here.

The ammeter should actually be in series with the circuit at the point where you want to measure the amperage. And actually what you have is the current actually flows through the ammeter. And that's the case here. You've got the current flowing through the ammeter. Now, what you want to avoid when you're using an ampmeter to test a circuit is you want to avoid creating an unintended second path for the current to flow.

And that's what's been done here. You wouldn't be testing anything here because now you're creating a second path for the current. That's certainly not desirable. So, this ammeter is incorrectly installed because it's basically creating a second path for the current. Actually, could even be considered a short here and that's a bad deal. You want the ammeter in series with the circuit. So this ammeter is correctly installed.

This ammeter is incorrectly installed. Now let's talk about how to connect a voltmeter to test the circuit. Well, in a voltmeter, you want it parallel to the circuit that you're testing. In this particular case, we're probably testing the voltage drop off across this lamp. So we have the voltmeter parallel to the lamp and we can test the voltage drop off. Now, one of the things you want to be careful of is the polarity of the voltmeter.

Here you have the minus side of the voltmeter. Here's a plus side. You want to make sure each of those goes to the right side. So the plus should come down here and come to the plus end of the battery, which it does. And the minus should come over here and come to the minus end of the battery, which it does. Now, we have another voltmeter in this drawing here, and it's improperly installed. And the reason it's improperly installed is the plus goes to the minus side of the battery, and the minus goes to the plus side of the battery, and that's improperly installed.

So, they're going to ask you a question about figure six here in the test. And the question is, how many of these meters are correctly installed? How many are incorrectly installed? And the answer is one ampmeter is incorrectly installed, one ampmeter is correctly installed, one voltmeter is incorrectly installed, and one voltmeter is correctly installed. So, there's two of each there. So, you guys have got that figured out.

Now, let's you and I take a look at figure 20 and see if we can figure out the question they ask us about this. And they say, "All right, super mechanic, let's assume you're troubleshooting this open circuit." Now, the reason it's an open circuit is there's a break in a resistor right here. You're troubleshooting it. You've installed the voltmeter like so. Now, the question they're going to ask you is, will you get an accurate reading of the battery voltage with the voltmeter installed the way it is?

The answer is yes, you will. This is okay. But you say, wait a minute, hold the phone. Won't there be a voltage drop off across the resistance of this lamp? And the answer is yes. If there were a current flowing in here, there would be a voltage drop off across that resistance, the resistance of that lamp. But since there is no current flowing, then there is no voltage drop off and you will get a true measure of the voltage of the battery with no current flowing.

Had there been a current flowing, then you would have had a voltage drop off and not had a true measure of the battery. I consider that kind of a tricky question. All right, let's take a look at one ammeter question. And here it is. is an ammeter shows a full charge rate and and this is in the aircraft. We're talking about an ammeter in the aircraft and the battery remains discharged. What's the problem? Well, the most likely cause is an internally shorted battery.

The battery itself is shorted out and no matter how much you charge it, it stays discharged. There's nothing you can do about it. So, that's one of the things to check for. or if an ammeter in an aircraft shows a full charge rate, but the battery remains discharged, the likely cause is an internally shorted battery. Now, here is the fun part. Here's figure 18 out of the FA book. It is, as you figured out, an electrical diagram.

And we're going to have some fun with these. We're going to make them easy. So, don't panic. Now, whenever I look at an electrical diagram like this one, the first thing I always try and figure out is where is the electricity coming from and where is it going to? Well, where it's coming from is some kind of power source. Here's a a bus up here. It might have 24 or 28 volts. And where it's going to is ground. So, what you do is figure out where the grounds are.

Here's a ground here. Here's a ground here. Here's a ground here. And here we've figured out a relatively ciruitous path to the ground like so. So, what you're trying to do is figure out the path of the electricity from the power source to the ground. So, the first thing to do when you look at these drawings is identify the power source and identify the grounds. and that way you know where you might be going. Now, let's take a look at a question they're going to ask you about this.

And what you need to know to answer this question is this. In this case, figure 18. The control valve switch must be placed in the neutral position when the landing gears are down to prevent the warning horn from sounding when the throttles are closed. And what they're saying here is even though you have the landing gears down, the warning horn will still sound when the throttles are closed, unless unless you have the control valve in the neutral position.

Let you and I prove it. Let's start at the ground and trace the current back to the source. And here we are in the ground, the lower leftand corner, and we have wire 14. And we have the left gear switch and the and the gear is down. So, we have this switch closed because this tells you when the gear is down, that switch is closed. And so, now we have from the ground through wire 14, the the left gear switch being down.

And now we have wire three here and the right gear switch is down. We've still got a continuous uh circuit here, a continuous path. And now notice that if the control switch is in neutral, it would be over in this position. But if the control switch is not in neutral other than neutral, it would be in this position. Let's see what happens if the control switch is not in neutral. Then you would have a path for the current to wire 10.

Now let's go beyond wire 10 to wire five. And now when the throttles are closed, these switches are closed. And now you have a path through the throttle switches to wire six to the warning horn and then through the warning horn which will now sound all the way back up to the current source. So you see, let's take a look at this uh switch again. When this switch is not in neutral, the warning horn will sound even though the landing gears are down.

And you have just answered the first question about this and all we had to do was trace our way through it. By the way, it is a good idea if you've got a book that you can mark in to use the yellow pen and do exactly what we did to figure out where everything is going. Now, if the control valve switch were not in neutral position, that is the path the current would take. Let's take a look at another example and we look at figure 18.

And the question is, what you need to know is when the landing gears are up and the throttles are the warning horn will not sound if you had an open or a break in wire number four. Well, let's take a look at it. We'll start over here in the lower left hand corner with the ground. And of course, the ground is connected to wire 14, which is connected to the left gear switch. When the gear is in the up