 Hi, I'm Zora. Welcome to a new Zora education. I would like to solve a few problems related to Ohm's law and the law of Joe's lands. Very, very simple problems, but I think they illustrate basically both laws. And I would say it's probably the most important part which you have to take from this electricity-related things for direct current. Okay, now this lecture is part of the course, Physics for Teens, presented on Unisor.com. It's part of the course which means if you will go to the website, you will have the complete set of lectures starting from mechanics and going on in physics. Also, the same site has a prerequisite course called Math for Teens. And I do consider it's very important for you to be comfortable in mathematics. So that course or any other equivalent course is a necessary component before you start doing physics. Okay, so let's go to the problems. The first one is very easy. You have, by the way, all these problems are related to some simple device which everybody knows about. It's incandescent lamps. So if I'm talking about, okay, here is the lamp. It means a regular old-fashioned incandescent lamp, and I will talk about its characteristics. So let's consider you have a lamp which has a resistance of 500 ohm, and it's connected to a source of electricity with voltage 100 volts. Now, my first question is what is the power consumption of this lamp? Well, in other words, you can say how much energy a source of electricity is sponging to basically keep this lamp in a circuit. Or in another way, how much heat is generated by this lamp? Well, when I'm saying heat in relationship to lamp, it's usually both. The heat, as we know it, basically the temperature, warm temperature, and also the radiation of light, because if it's very, very hot, obviously we know that the lamp actually is lit and it lights up the room, etc. So it's a combination. All of these is basically heat. Now, speaking about power, so what do we know about power? In the previous lecture, I spent some time basically explaining that the power is basically, the amount of power consumed is derived from the concept of energy, and energy is related to the voltage, because the voltage, as you remember, is basically the amount of energy to bring one cologne of electricity from one place to another. So in this particular case, we have derived a very simple formula for power consuming, it's voltage times current. Current gives you how much electricity is going in a circuit, coulons per second, and this is basically the amount of energy necessary to bring from one terminal to another one cologne. So their product is basically the amount of energy per second needed. So we explained this many times before. Now, using the law which was formulated by Ohm's law, and you remember that Ohm's law is this, that the current is equal to voltage divided by resistance. Actually, that's how we defined resistance some time ago. From this you can see that this can be expressed as u squared divided by r, or if you write it in this way, then it would be i squared times r. Okay, so this is a very fundamental thing which I will use in all the problems today. Now, in this particular case, we have u and we have r, we have voltage and we have resistance. So we can obviously use this formula and have this p is equal to u squared, which is 100 squared divided by 500, which is what, 20. So this is the power which is consumed by this particular lamp in this particular case. Now, let me remind you that the characteristic which we talk about this, the resistance, it's an internal characteristic of a lamp. It doesn't really depend on anything, it doesn't depend on what kind of voltage you put it, you connect this lamp, or what kind of a current going is going through this lamp. The resistance is characteristic of the lamp. Okay, next problem. You know what, let me write this. Power is equal to u times i equals to u squared times r times i squared times r. That's the formula which we will be using in all the problems, now the same. Okay, now, my next problem is more of a practical kind of thing. If you know in many, many cases, at least in old places where I know, the power, electric power which you are consuming is supposed to be paid for. And usually the bill which comes from the electric company expresses the energy which you have consumed, not in joules as we used to in the sea system in physics, but in kilowatt hours. So what I'm asking right now is to express the power which is consumed by this particular lamp, not in joules, not in watts actually, which is joule per second, but in kilowatt hours. So the energy which is consumed by this particular lamp, and now we need the time. So we know the power, the power is, it was what, 50, 500, the power was 20 watts. That's what we have calculated in the previous problem. So if this is the power, now my question is how much energy it will be produced, it will consume in how much, in eight hours. So in regular sea system of units where the joule actually is the measure of the power, joule is watt times time in second, right, because the watt is joule per second. So in joules it will be 20 watt times eight hours times 3,600 seconds per hour, and that would be 576,000 joules. So that's 576,000 joules will be consumed in eight hours. But that's kind of a very inconvenient, too big a number, joule is a very small amount of energy. So electric companies used to measure the electricity consumed by whatever the consumers in kilowatt hours. So in kilowatt hours it will be what? 20 watts is 0.02 kilowatt and times eight hours, which is 0.16 kilowatt hours. So if you have a unit kilowatt hours, that's the measurement how electric companies are usually measuring how much electricity you have consumed. And they're saying, okay, it's 20 cents, for instance, or something like this in the United States per kilowatt. Well, you multiply your consumption and you will see basically how much you're paying. So that's my second problem. That's basically for one only reason, to introduce kilowatt hours and very simple conversion from joules to kilowatt hours. My third problem is, so you have a lamp, which, yes, I have to again make some practical introduction. If you take any lamp, incandescent lamp, usually it's written on the lamp. Two things, the normal voltage it's supposed to be connected to and power it consumes in watts. For instance, you have lamp which says 220 volts, 55 watts. So you know that this lamp is supposed to be connected to the source of electricity which has this voltage and this will be the power this particular lamp will be consuming. And by the way, you know that if you will buy the lamp with a higher wattage, it will basically light the room brighter, right? So because it consumes more electricity and it will produce more energy, obviously. It will radiate more light. So anyway, so here is the lamp which I have bought in the store. It's supposed to be connected to this voltage and it's supposed to consume so much power. So what's my question? My question is, what is the current which goes through this lamp and what is its resistance? Well, this is the problem which is exactly coming from this equation which I have written before. So from the current standpoint, current is equal to power divided by voltage. Power is 55 divided by 220 which is what? 0.25. Amperes. Okay, that's my current. What's my resistance? Well, if I know the current and I know the voltage and we know the Ohm's law, right? The I is equal to U divided by R or R is equal to U divided by I. So we can define R as U which is 220 divided by I which is 0.25 which is 880 Ohm. Resistance is measured in Ohms, right? Yes, that's right. Okay, so that's my third problem. You see the problems are very easy and they are all basically related to this formula and the Ohm's law. Okay, what else? We have a lamp again. What lamp in the store which has the characteristics of voltage? It's supposed to be connected to 110 volts and it has power consumption of 55 watts. So my first question is what kind of a current will go through this? Well, again the same thing. Current I is equal to power divided by voltage. In this case it's 55 divided by 110 which is 0.5 ampere. But now by mistake I have connected this lamp into 220 voltage. What happens? Well, before doing that let's define the real characteristic of this lamp which is its resistance. And resistance we can find from this. We know the voltage, normal voltage and normal amperage. So resistance is equal to U divided by I which is 110 divided by 0.5 which is 220 Ohm. Okay, so I know that we have resistance. Now what happens if my U is equal to 220 volts? If I know the voltage and I know the resistance I can use the Ohm's law again to find the current. So what would be the current? The current let's call it high because we have connected to a high voltage. Now I high would be U divided by R which is 220 divided by 220 which is 1 ampere. Twice as big. You see the current which is going through this lamp. If we have increased the voltage by the factor of 2 the current will also be increased by the power of 2. Now the question is what will be the power consumption in this particular case? Well, the power consumption can be done from here because we know the current and we know the resistance. So it would be power high. Now normal power is 55 watts, right? But the high power would be I square times R which is 1 square times 220. You see we have increased voltage by the factor of 2 from 110 to 220. But power has increased by the power of 4 because this is a square. So voltage is twice as big as before. Current therefore is also twice as big as before. But the power where the current is in square would be increased by the power of 4. Now with this amount of heat basically which is this particular lamp is supposed to emit it will probably burn very very soon because it's not really designed for this amount of heat going through it. Okay, next. Next we have two lamps. Lamp 1 is 220 volts, 55 watts. Lamp 2 is 110 volts, 55 watts. So two different lamps. Now we have two problems actually. Problem number 5 and problem number 6. They are the same problem except in problem number 5 we have connected these two lamps in a series. L1, L2. Now my question is what would be the current and what would be the powers each lamp would be consuming. Now in theory if the lamps are connected according to their specification both will be 55 watts producing or consuming and producing 55 watts. But in this particular case we didn't really connect them each to corresponding source of electricity. We have connected them both in a series to 220 volts. So what happens? First of all we have to define the current. Now the current is obviously if you have a voltage and you have two lamps each resistor connected in a series the total resistance is the sum of these resistances. Now what is resistance of each one? First of all let's just take from here R. If you have P and U then R is equal to U squared divided by P. So in this case R1 that's the nominal resistance which is a characteristic of a lamp. Not a connection. So in this case it's U squared which is 220 squared. 220 times 220 divided by 55 so it's 800 it's 880 ohm. And this lamp has 110 times 110 divided by 55 so it's 220. So we have resistances so in this case we have 880 plus 220 is 1100 ohm. From this knowing the total resistance we can find the I. I is equal to 220 divided by total resistance which is the ohms law. So it's 220 divided by 1100 which is what 0.2 ampere. So we have amperage. Now since we have an amperage we have the same current going through both of them. So I is the same here and here and here everywhere. So using the I and the resistance of each we can find the power using this particular formula I squared. So P1 is 0.2 squared times 880 which is 35.2 watt. Less than nominal you see. Why? For very simple reason. It's designed to be connected to 220 by itself. But now we put another load onto the same circuit which reduce the amount of current going through and therefore it's reduced amount of power which this lamp consumes. Now in case of the second lamp P2 is equal to 0.2 squared times its resistance which is 220. Which is equal to 8.8. Now this is significantly less than the power which is supposed to go through. Because again we have reduced the, even we have increased the voltage we have reduced the current proportionally. But the amount of energy is square of the power of the current. So if we have reduced the current by the factor of 2 or whatever then the power is reduced by the factor of 4. So that's why we have such a low level of energy which is emitted by this particular lamp. Well the moral of this story is lamps are not supposed to be connected in series. Because it's completely defeating the whole purpose. One is actually putting additional load onto another and both of them are suffering of lack of power basically. So the power which is supposed to go to one of them spreads among two of them and they don't really light up the way how it's supposed to be. So that's why all electric appliances, lamps, refrigerators, whatever they are all connected parallel to the source of electricity. So let's go to the second problem and the second problem is a parallel connection. So this is the connection which is used everywhere in households. Parallel connection of all appliances. So we have a couple of wires from the source of electricity and they are going to, they are branching or whatever, they're going to all the house and they're branching in such a way that every appliance has two independent wires. This and this which go straight into the source of electricity without anything else. But my problems are exactly the same. So I have to define what's my current here and maybe in each of those guys and I also have to find the power consumed. So this is not correct now in this case. So what is correct? Well, first of all, let's define this current I. Well, it depends on the resistance of this parallel connection, right? You remember the formula for resistance in the case of parallel, right? So what is this? This is 1 over 880 plus 1 over 220, which is 4 times 880, which is 5 times 880. So my R is equal to 880 divided by 5, which is, what, 176 or something like this? 166, something like this, or 176. 88 times 2, 176, yeah, 176. So this is my resistance ohm, 176 ohm. You see how interesting it is. This lamp has 880. This lamp has 220. But together, if they're connected in parallel, they have a less resistance. Why is that? Did I calculate correctly? Yeah, it looks like it. Okay. Well, it's because we are branching. So this one has the same resistance basically as it's supposed to be, but this one has a lower resistance. And that's why the current actually will go more into the lower resistance than into the higher resistance. And together they're actually allowing more current than as if each separator was connected. Okay. They're consuming more current and more power together, obviously. Okay. So we have this total resistance. Next. If we have total resistance now, how can we calculate the power consumed and electric current in each one of them? I1 and I2. Well, first of all, the current can be very simply defined using the ohm's law because the difference in electric potential between these two points is exactly with these two points. So each one is basically connected to 220 volts. And knowing resistance, we can have that I1 is equal to U divided by R1, which is 220 divided by 880, which is 0.25 ampere. And that's how it's supposed to be because it's designed for 220, right? And its power, P1, would be again U times I1, which is 220 times 0.25, which is 55 volt. And that's how it's supposed to be because it's connected correctly to 220. Now, the second one, that's unfortunately would be different because it's the same U, the same 220, divided by its resistance, which is 220. And I will have 1 ampere. So the power would be equal to U times I2, which is 220 times 1, so it's 220 watt. 4 times greater than the one which is supposed to be, which it's manufactured for. So this one will really burn really quickly because it's not designed for this. And that's basically the end of this set of problems. Again, the problems are very easy and there are basically two laws, the Ohm's law and Joel Lentz law, which is related to the heat. And everything else is just general common sense, I would say. Oh yes, we have to really know how the resistance is calculated in case of a series connection when the resistances are adding together and in case of parallel connection when inverse of resistances is added together. Now, why is that? We did talk about this before. Now, what I suggest you to do is to solve all these problems yourself. Go to Unisor.com, open these problems, solve them yourself and check against the answers provided. Well, that's it for today. Thank you very much and good luck.