Hello, miss. Hello, Jackson. How are you? Great. Where are you? Are you in England still or have you come back to Rome? I'm still in England. Okay. Okay. So we were going to revise electricity. Yeah. Okay, so current is a flow of electrons, so. If we have a current flowing, it's a change in charge moving with change in time. So I is q over t. So the physical electrons move from the negative to the positive. And but we always show conventional flow of current out of depositive end into the negative end. So the electrons go in the opposite end, but conventional flow always shows current going from plus to minus. So if we have a current against time graph, the area under the graph will show us the total charge that has moved. Okay. So we've met this equation before. You can use that. And charges in coulooms time is in seconds. And the area under a current time graph is the total charge that has moved. And then we have kurthoff's first law. So if we have a junction in a circuit. If we have five amps coming into a circuit into a junction. And if we have two amps going that way. Then. This must be. Three amps. Because charge is conserved, so the amount of charge entering a junction must equal the amount of charge leaving a junction. So that's called kercho's first law. That's all okay with you. Yeah. Miss, can you explain it? I can explain it more because I'm not understand very well. Okay. So if we think of. George is conserved. In a circuit. Remember the charge. What is the charge flowing? Opposite chothe flow of the current. Yes. So the charches are electrons, so the electrons Carry the energy. So if you have a circuit that looks like this. So we have two bulbs. And let's say we have an amatesure here and it's reading three aps. And then. We have an ammesure here and an amatesure here. Let's use the proper. If both lamps are identical, so theyhave the same resistance in the circuit, what would I expect as the reading on this amater and this amiter? So if the current comes this way. At this junction, some of the electrons go that way, some of them go that way if both bulbs have the same resistance R. Let's say they've both got a resistance of tnomes. What fraction of the three amps will go in branch one and branch two? Yeah, there's three. So there are all three amps then also there are all three amps in a one and a no. Can't be. If you have a fixed number of electrons carrying energy. So all the electrons go through this part of the circuit, okay? Half of them, we'll go through this branch, and the other half will go through this branch. Then they meshe up again and again. If you have an ambisure here, they will have three amps. So this will be 1.5 amps. Jackson, and this will be 1.5. Why? Because they have the same amount of resistance exactly. Yeah. If this was ten amps and this was 20 sorry, ten ohms and 20 ohms, then this would change. So. What do you think the reading on your amateurs would now be? If this has lamp has double the resistance of this. Remember, electrons are like students. They take the easiest route. So which path will it be easier for electrons to go through? So which will have more. So this time this will take that is ten, that is 20, 20 ohms. There is negative proportional with current and resistance. So if that is 20 ohms, that is just the one a's, yes. Herres, Yeah. So this is all the kurcheslow is. Remember, your electrons are like porters in a station. They Carry your bags. So they Carry energy. They Carry energy. And the energy is converge in your bulb. So charge is conserved in a circuit that's kurcho's first law. So kurcho's first law so that that is that bit. So if we look at. The first equation we met, I is q over t, so q is charge, I is current time is t. So one electron has a charge of 1.6 phi, ten to the -19. Yeah and divided by time. Yeah Yeah of course the charge maybe we need to Yeah there's number of electrons we should one divided divithis number Yeah but that's not but that's not relevant with this question. Not really. Remember we're just going through the theory first. I have some questions for us to try. If we have a current time graph. Like so the area, so I times t will give me Q So the area under my line will give me the charge. That has moved in a circuit, okay because I times t. Equals q. Okay. We won't do those questions yet. So we have this equation. The next equation we've met is the drift velocity equation I equals and A, B cute. Okay. Do you remember what each of these. Bits of this equation means free electrons and conductors move around randomly when there's no potential difference, across a conductor when there is a potential difference. So when we put, say, six volts across, the electrons get pushed from the negative end to the positive end through the conductor. So they drift through the conductor with a drift velocity. So when we put a potential difference across a metal wire, the electrons drift. They move through with a certain velocity v through the conductor. They may collide and give up kinetic energy. So the I is the current n is the number of charge carriers per meter cube. So the number of free electrons per meter cubed. Flowing through your wire a is the cross sectional area. Ir squared Yeah. Ir squared v is the average drift velocity at which they move through the conductor, and e or q is to charge on an electron. So a steady current of 13 amps is maintained in a circuit for 30s. Calculate the number of electrons passing a point in the circuit in this time. So. If we know 13 amps is flowing, so we know I we know t. So I equals q over t. So the current 13 times the time 30 will give us our total charge. So have you got calculator? Work out the total charge, work out the total charge, and then divide it by the charge per electron to find the number of electrons. 30 times 13. And then is that divided by 1.6 times ten to the -19? -2.44 times ten to the 21. Good Yeah good. So we can use this equation. We know this is uncharge on an electron. Here's another question. Using this equation, I equals n number of deloccalized electrons per meter cubed, a cross sectional area drift velocity, and q. I'll stick with q for a charge on electrons. So that's q or e. Some books use e for the charge on an electron. Okay, so to solve this equation, a semiconductor has 3.8 by ten to the 15 electrons per meter. Cua current of 30 pico amps is maintained in a slice of the material with a cross sectional area of 2.5 millimesquared. Calculate the drift velocity of the electrons. So we have to do something with pico, and we have to convert millimeter squared to meters squared for a cross sectional area. Okay. Washes picko amps. That. Minus six, no, that's micro. And remember micronnano and then pickle, what is tent? What is nano? Minus twelve. So pico is ten to the minus twelve. Good. So pick mili micro nano pico and then femmpto minus twelve. So 30 pico amps is equal to n. What is the number of charge carriers. 2.5 mm square no is no, that's the cross section area. What is n. Three. Point eight times ten to the 15 electrons per second meters, 3m cube. Good, excellent. And the charge is 1.6 times ten to the -19. Yes. And this Jackson, we're going to have to convert this to millimeters from millimeters squared to mesquared. Okay, millimeter square, 2 mm is millimeter, 2m is ten to the three. So square, that is to the six, 2.5, times ten to the six, minus six, minus six. Yeah, minus six. So times 2.5 by ten to the minus six. Good. Minus six. Times v which we're trying to find the drift velocity times charge on little electron times 1.6 by ten to the -19. Okay, three, let me check three times ten to the minus eleven times 3.8 times ten to. No 3.3 point eight times ten to the 15. Times 2.5No Yeah 2.5 times ten to the minus six and then final. 呃哎呀你。So v, that is three times, three times ten to the minus eleven, divided by 1.6 times ten, two zero -19 and then times 2.5. 2.5 times ten to use a minus six final, finally, times 1.6 times ten user. 3.8. Times ten to 15. 0.020Yeah 0.02 point 197 good meters per second, meters per second. So we know that this equation and this equation are important when we're talking about what a current is. It's a flow of charges per unit time, and they move at a particular drift philosopocity. Okay. So you're happy with charge and current? Yeah. Okay. Resistance and oms law. So resistance is a reluctance, something that opposes the flow of your electrons. If you have to measure resistance in a component, you have to have your ammeter in series, your voltmeter in parallel. And o's law states that the ratio of the potential difference to the current is constant for a conductor at a constant temperature. So v over I is a constant. And that constant is the resistance of your conductor, providing that temperature stays the same. Why do you think in a wire carrying occurrent it's important for the temperature to stay the same? Jackson? The temperature yes, because once the temperature increase, the resistance will increase as well because why? Okay. Okay, this question. Because when the temperature increases, there are more coinetic energy for electrons. So electrons will viaporate more, so that will colde, that will collide the the latest irons. So the latest ions have more polation and vibration. And then there is less current will go through it because there less space. Yes, very good. So if you plot v against I, you get a straight line, and the gradient of this graph is your resistance. So your resistance is your gradient. So non, no miconductors. If you do a lamp filament, the more voted you put across your lafor lament, the more the current won't you won't get a direct proportional relationship, meaning that a lamp filament will not obey ohms law. So a lamph filament. If you think of current against voltage, it's s shaped for a lamp filament. Here is nowire at constant temperature. That erbeys ohms law. And this is a diode. What do you remember about a diode? I'll dial, just allow one direction of current. Excellent. Yes. And then if we have thermistors and light dependent resistors, a thermistor decreases resistance with temperature. And a light dependent resistor. These are not metals. These are semiconductors. So they behave slightly differently. They depend on the environmental conditions. Semiconductor. So they have a slightly different profile to these metallic components. So with a metal, free electrons collide more frequently if the temperature increases, so the resistance will increase. With semiconductors, it's the opposite. You're actually releasing more delocalised electrons to Carry a current. So the current will increase if resistance is decreasing, because resistance is inversely proportional to current. So with semiconductors, the more temperature, the more light, the more the current, the less the resistance. So there are. So calculate the resistance of a semiconductor device when the potential difference of twelve volts produces a current of 48 micamps. So there's ten to the minus six micro. So twelve divided by 48 micro gives you a resistance. So to talk about significant figures, you should always follow the pattern of the data you're using. So you have two sick Fikes in both of these. You have two significant figures for your data. So your answer should be given to two significant figures as well. Just a if finger a finger okay Yeah. And. Let's have a look at this question. Okay. So that looks okay. We'll do this question in a minute. Resistors in series and in parallel. So do you remember resistivity? Resistance is resistivity times length over ver area. Now, resistivity is the property of the material that the conductor is made of. So it doesn't matter what the length or the cross section area is, because resistance varies with cross section area we know, and resistance also varies with length. It's harder for electrons to travel a long length compared to a short length. It's harder for electrons to travel through a. Thnoir compared to a thick war. But if we know the dimensions of our wiir, we can state the resistivity as a constant for whatever length, whatever area we have. This is the way to compare materials. When calculating resistivity, we must measure the diameter of the bar to get the cross section area high R squared, so we use a micrometer screw gauge to measure this. You can measure a no meto measure resistance. L is measured in meters resistivity, so we can prove that resistivity has got units of ohms times meters. Rearranging our equation, resistivity is equal to ra over l. Jackson, what? Shes the unit for resistance? Resistance is ohms, but resistiity is ohms per meters, Oh meters ter squared and length is a niches. So if you think of it. That cancers with that. So you're left with units for resistivity as. Ohm meters, ohms times meters. So that is another equation. Talking about resistance. So resistivity is very small. Wires like copper. Okay. Are you okay so far? Do you remember using all these equations before Jackson? Because when we've gone over enough of the theory, we can start looking at the type of question that might come up, and we can do these questions together. What do you feel you need to do most to help you prepare for the exams next may? Next may. Okay. Yeah Yeah next may at a level examination. Yeah, a level examination I think. Yeah, well, I remember remember that it is different than China, so I can get the equation. Sheit is that sheet begood? Okay, Yeah. So, so so maybe I need to practice the answer. To be more logic and can get more Marks. Yes. So you have to recognize which equation to use. You're very good at the mathematical side. It's just familiarizing yourself with which equation to use when if you don't need to remember them. Okay. So series arrangement of resistors. If we have three resistors, what is the total value of the resistors in series? So we just add them together. And if we have resistors in parallel. Because the potential difference across each of these is the same, v is equal to ir. R is v over, I is over. So the equation for resistors in parallel is like this. The equation for resistors in series is like this. I know that the serious and parallel so here's a past paper question. Two resistors are connected as shown show that the resistance of the combination is about 70 oats. Oso one over R T equal to. I'm sure you guys. 72. So you're very quick, you're very good. It's just a matter of recognizing which equation to use and when the resistor combination is connected to a battery of emf and internal resistance are so we haven't revised internal resistance yet. Where does internal resistance come from in a circuit? One. Hundred 20, 180, 30, 72 ohms and are equal to 2.5 and emf equal to nine vots, but nine votes. Let me go down to the equations. So you're not given the equations of resistors in series and parallel, but you're given e equals I times v times t. You're given. You're not given the equation for internal resistance. Sorry for your eyes. Well, just go back to our question. Calculate the energy dissipated in the resistor combination, so e equals I times v. Ten two. Now internal resistance, if you remember, comes from the power supply, so little R is there. So we have 72 ohm resistance from the parallel combination, plus 2.5 ohhm resistance from the internal resistance. Calculate the energy dissipated. So that's the work done or in converting it to heat, which is current times, voltage times, time. So we need to work out the current so we know that we can work out the total resistance. Times. The voltage of the closed circuit times the times. Firstly, I need to know about the current. So the total resistance is 72 plus 2.5. So that is nine divided by seven, 4.5 and that is 0.12. Yeah, that is significant to significant significant bigger. 0.12 so 0.12 this value times nine times the time 300s so that the final answer is, sir, 326. I think there's one more thing we've forgotten. Remember, when you have an internal resistance, you can't assume the voltage of the working circuit is nine volts, so it will be less. So we might have to work out the voltage drop across the. V is ir. Yeah. Yeah. So we're fine. So what did you get for current three, 0.12? Yeah, good. And then we can use this equation to work out the energy dissipated in the circuit. So that will give you 0.12. Times nine times. Five times 60. 60s Yeah. So that will give us a measurement of the heat dissipated point one, two times, nine times, five times, 63, two, four. Jews, okay, good. Let's try this equation practicing this R is row l over a. The resistivity of nchme moire is this. So we're given the resistivity value, calculate the length of the nromoir of a diameter 3.3H millimeters. Thatwill make a resistance of 12.5 ohms. So we're practicing using this equation. So we're given the diameter. We need the cross section area in meter squared. We're trying to find the l. We know the resistivity. And we know that we need a resistance of twelve and a half ohms. So first of all, let's work out the cross section area. Multiplied by the resistance. Derided by. The resisterity. So I guess 123456. So resistance is 12.5. Resistivity 1.1I ten to the minus six. Times l. Over. North Point 19. I. Ten to the minus three. I R squared times pi. Okay, so resistance is resistivity times length over area, so this will let us work out length. Okay? And we get 1.2 measures. So then we have potential difference emf and internal resistance. So the potential difference. The electromotor force is the total energy available per unit charge, but very often you have an internal resistance, and not all the energy gets out of your power supply. The difference between electromotor force and potential difference when the circuit is open, this might read twelve volts, the voltmeter, because you are not making the electrons move yet, there is no flow of electrons. But if you close the switch, this might drop to 11.8 bots or something. So you know, there is that it because of the internal resistance, it is yes. Okay. So the internal resistance has to be considered in a working circuit. And we have this equation, which is conservation of energy. The energy in the power supply is equal to the energy loss converted across your external resistors. And the energy drop across your internal resistance. So that's the same as e. Loss volts across internal resistance plus loss volts across your external resistor. The total voltage supplied in the cell provides the voltage across the external resistor and the voltage across the internal resistance of the cell itself. So emf is energy transferred per coulum of charge. The potential difference is the energy converted from electrical energy to other forms. With resistance, the loss fads are the energy converted that is never used as useful work in your circuit. So that would be the voltage across your internal resistance. And we can measure it by rearranging this equation. Y equals mx plus c. So x is your current, y is your voltage across your. External resistor, the gradient. Is the internal resistance and this value. Is your emf. Okay. So a very high resistance fotmeter is connected across a cell and gives a reading of 1.46 volts. When a twelve volt resistor is connected in series with a cell, the vote tmeter reading falls to 1.3 nvolts calculate the emf of the cell, the internal resistance of the cell. So the emf. Emf equal to ir plus ir. Yeah, so the emf of the cell should be that. Before we close the circuit, we just take a value of a reading with the switch open, which will give you your emf, the internal resistance of the cell. So if we subtract these two, we'll get the energy drop across our internal resistance. So it was 1.42, 1.46-1.3. So that gives us. North points. One, six votes drop across the internal resistance of the cell. So 1.3 over twelve gives us the current. So that's. The voltage drop across the little resistor, the internal resistor. So if we do the working voltage divided by the. Resistance of the external resistor, we get the current. So plittle R. Is equal to v. Which is 0.16 volts v over R. V over I. The same current will be going through your internal resistance. So. Will give us the value for our internal resistance 1.48. Okay, so Jackson, shall we do more of this tomorrow or do you want to do it a different way? You tell me what's what you find most useful in our sessions? Well, I want to learn more about the circuit. Okay, so we have a bit more of this to finish and I'll find some past paper questions for us to to work through. And it's important to see how they answer some questions. How they ask some questions, okay? In past papers because they're do you want are you on holiday now or are you still at school? No, no, I'm in holiday now. Okay, do you want me to set you some questions to do before the lesson or after the lesson? After the lesson? Okay, right. Okay, so we'll finish electricity tomorrow and then I'll send you some questions. Okay, you're very good. Mathematically, it's just a matter of understanding what the questions are asking and what you need to know. Okay, Jackson will talk again tomorrow, same time. Okay, bye bye, promise.