Electricity. So what I thought welook at today, and this can finish loading in the background. The two things which I think probably are the most challenging for electricity are internal resistance and potential dividers. So that's what I wanted to look at today, those two things. Then we look at how they can ask things about that on questions. I wonder if I reduce this, will it be okay? So internal resistance. So you know. It's still going. It's a big I copied lots of questions. So internal resistance is the resistance of the power supply. So not all of the energy is available to do work in your circuit. So capital e is usually the symbol we give for internal resistance. Little R is the resistance of the power supply. Just reduce all them. We can get rid of this now. So basically, I'm not sure you're given this equation in your exams, but it's good to be familiar with it. So conservation of energy, the emf, which is given a symbol capital e emf. Measured in volts is equal to the voltage drop across this internal resistance vr plus the voltage drop across the external resistor. So we have I'm going to just make this a bit now. Yeah. So. The voltage drop across our in external resistor. So this obeys the law of conservation of energy. So the v times pi is equal to the voltage. The emf from the power supply times the current is equal to the voltage drop across the external resistor plus the voltage drop across the internal resistance. So divide by I, we get this. So e is equal to ir plus I little R. So the resistance using this equation, we could find the value of the external resistor if we know everything else, or the internal resistance. For example, a current of 0.2 amps flows when a cell is connected across a five ohm resistor. Calculate the emf of the cell, even its internal resistance is one ohm. So. So we're asked to find capital e, the emf of the power supply. So that will be the current times, this resistance plus this resistance. So we get an emf of 1.2 volts. So if we have an open circuit voltage that will give us the emf of the power supply. So it can also be called the open circuit voltage. When we close the switch, this will probably drop this reading. So then we have kurchoslaws. Kurruthov's, two laws allow us to work out the current and potential difference across any bit of a circuit. So the first law is that the sum of the currents entering a junction must equal to the sum of the currents leaving a junction. So basically, if we know five amps is coming in and four amps is coming in and that remember that is current is rate of flow of charge. So the charge ges coming into a junction must equal to the charges leaving the junction. So five plus four is nine minus two us one, so nine minus three is six, so x must equal to six amps. And then kurchev 's second law allows us to use loops in a circuit to work out a particular current in any branch of our circuit. The sum of the emf's is equal to the sum of the I times rthe voltage drops across the external parts of the circuit. So the total emf is equal to the sum of the I times R plus I times R plus I times R. So if we know the emfs and if we know the value of the resistors, we can use kurchoff's first and second law to work out the currents in the separate branches. Okay? And we in any closed loop, the sum of the emf's is equal to the sum of the irso. A closed loop would be going. That would be one loop. And this could be a second loop. As long as we go in the same direction. So clockwise is positive and anti clockwise would be pushing the electrons in the opposite direction. So. Emf E1 plus e two is e to ir one plus I one R2. That's one equation. And then a second equation for the second loop, loop b would be the total emf's is equal to I two times R three. And also we could say. T is equal to I one plus I two. So we could have three different equations. And if we know the values, we can work out the occurrence flowing through the different branches in this parallel circus. So I wplus side two. That's using kurcho's first law, the sum of the currents flowing into a circuit junction must equal to the sum of the currents flowing out of the junction. So if we have this example. If we know this battery has a value of four volts emf, this has a value of six fose emf, we're told that the voltage drop across this resistor is two. This is three. So we what is the voltage drop across this resistor? So four plus six is ten volts is equal to two plus three plus vc. So two plus three plus five volts will give us ten volts. So that's the law of conservation of energy. So very often we have resistors arranged like this, which can provide a potential divider circuit. We can divide the voltage output by changing the value of these resistors. So we could divide the output voltage of its twelve volts depending on the value of these resistors. So. If we have. A twelve volt power supply and we have two identical resistors. So say they're both ten ohms. Channel resistors. So the output voltage. Across each one, you would expect to be six volts V1, be two. Like that. So a potential divider circuit. That's twelve votes. So we have an equation. R equals b. Over、I. So the same current is going through both. So. The total resistance is 20 ohms. The current we can work out as. So I equals pht. Over R so twelve divided by 20. Is 0.6 amps. But if we were to make this 20o's and this ten oes. We could use this equation if this became 20 oms. So this would take a voltage of a ratio, so V2 over V1. Is equal to R2 over R one. 嗯。Or we could say feed two. Yes I think this question Yeah we can if that is 20 the total resistance is 30 ohms and twelve divided by 30 so that is zero 0.25 yes so that that is because they are in serious so they have the same current yes they have the same current so 0.25 with my. Oh. Give me a second, I find the calcuate. Okay. 1212 divided by 30 and that is, sorry, not 24. Yeah that is 0.4. So 0.4 aps, that is the total current and there is the current of each resistance. So the first one, ten ohms. So the voltage is ten times 0.4 equal to four. So the 20 ohms that is twelve divided by, sorry, twelve minus four equal to eight. It is eight. And this will be four phones. Good. So so essentially we have this relationship. If we want to find V2, we put the V2 compared to the total voltage is equal to the resistance, compared to the total resistance rt. So vt is the emf, whatever is here. So the ratio of the resistor to the total resistance is equal to the ratio of the voltage across it compared to the total voltage. So we can write this equation slightly differently. V out. The output voltage across this bit of the circuit is equal to R2. Over. So this would be R2. This would be R one. R2 over R one plus R2. Times v in. So V V in is our total voltage available to do work in our circuit. So that's v in. And this is v out often. So using this equation. We can. Play around with the output voltage. And very often, potential divider questions don't show the battery. They just put it like that. So this is ten volts here and this rail is not volts here. And this time we have a fixed resistor and a variable resistor, a thermistor. So. If you have a thermister like this in your circuit, you can change. You can if if it gets cold, you can get the circus to switch on a heater here. If the voltage drop across it. So. The resistance increases in a thermistor as temperature decreases. So the colder it gets, the higher the resistance. And the higher the resistance, the higher the voltage drop across. Across this thermistso at a particular voltage, if the temperature drops, we can get the heater to be switched on because the voltage drop across here, this will take a greater ratio of this ten volts. So the resistance of a thermistor decreases as its temperature increases. As the resistance decreases, the potential difference across t falls and the pd across b increases, so that the potential difference will go up. Calculate the potential difference across R one and the resistance of tea falls to 120 ohms. So the. Out is equal to R one over R one plus R2 times V N. So if we know R one is 100 ohms, 100, and the total resistance, if we know R2 is 120 plus R one, so, and the input voltage is ten volts, the out is that over that times that gives us 4.5 volts. So if if you wanted to use this circuit to. Switch on a fan. If it got too hot, you could do that. So you could. So we know that. Resistance increases as temperature decreases, but. Resistance across this bit will decrease. If it gets harsher. So this bit will take a smaller share of the voltage. So as R decreases, v across this bit will decrease, which means that the v across this bit will go up. So what you can do is put a fan. In this part of the circuit to cool the room. Similarly, a heaisher could go across this bit of the circus. So if the voltage output across this decreases, it means the ratio of the voltage across this bit will go up. So the potential difference across here will cause the fan to be switched on. And sometimes instead of fixed resistors, we have our sliding contact. We have just a continuous resistance wire. And we can have different amounts of that wire. So the current will go through this bit of the circuit. So a slide, a uniform resistance with a sliding contact as the contact is moved along the wire. The potential difference between a and c and that between b and C, A and c and b and c vary, but they always add up to what the total emf is. So. Okay. So let's have a look at this question. Here is a potential divider circuit. What is the value of V1 and V2? V1V2 the value of V1 and V2 so six to. Wait six times. Six over 24. 1.5, that is, the voltage of V1 and six divided, and six -1.5 is equal to 4.5. So v ve one is 1.5 and V2 is 4.5. Good. Okay. How would we calculate the current going through the battery in this circuit, Jackson? Battery, okay, circuit a and b. So where's a and b? Featin, circuit a and b. This is circuit a Jackson, and this is a whole separate circuit b. Okay. So we'll just do this one first. The current, what is the current flowing through? The main part of the circuit, okay. So the resistance in service. Well, the resistance in the service is one over four plus one over four. And so that the total resistance is six good, and twelve divided by six is two. So the current is six. Twelve over six is two. So you said it. So it's two amps. Yeah, two amps. Yeah. So twelve divided by six. Yeah, two as. Very good. Well, what about this circuit? One over four plus one over. One over six plus one over eight. And so the total. One of and so the total resistance is 1.85. Good plus the four oso, the branching 1.85 divided by this number 1.03. Very good. Yeah. So 1.03. Can you find the currents going through these separate resistors? Current or separate resistors current. Okay. 1.03 emperres is this is a total, the current of the separated resistance. 44 ohms, eight ohms and six ohms. This three. Twelve plus 6:18, so 1.03 divided by 18 equals to 0.57, so 0.57 times four equal to 2.28. 0.456. How are you doing this, Jackson? Did you work out the voltage across this bit and the voltage across this bit? No. I just separate it. Oh, you're doing ratios of currents. Yeah the current that is final current. Are six bofour ohms and that 1.03 so six divided by one point. No 1.03 what you can do times 1.03 equal to 4.12. Apairs sorry votes. Yes. 1.85. Yes, so one point. So in parallel, the voltage is the same for each of them, you know the resistance, then you can find the current ge. The the out so the voltage drop I'm working out the voltage drop across this branch if this is the total resistance. So and I get 1.87, 89, I mean, 1.89 volts. So if I know the voltage drop, if I know the resistance, I can find the current I equals v over R. For each branch. So for the 1.89. Divided by four for the first one. 2.37 so that gives me. Two points. Now that can't be right. 1.89 divided by four is. 0.47. Ampso that's pi one. And then you do the same for the other ones. So 1.89 divided by eight, 1.89 divided by six. So you get the different currents in the different branches. Okay. So again, you just do by four why it's divided by four? So we work out we know the total resistance. We know we want to work out the voltage drop across the branches. Because when they're in parallel, they have the same potential difference across them, don't they? So v out the v, the voltage drop across this branch is what I'm trying to work out. So the voltage drop fee out. Is R2s. We'll make all this R2, but we know that 1.85 over the total resistance, 5.854 plus 1.85, 5.85 times the vn times six. So 1.89. If we know. The 1.9 volts will make it so. If I want to find R2, then I know R, I know v, so I too will be equal to 1.9 boats divided by eight. So are I two. The current going through the second resistor here will be v over R, so 1.9 derided by eight, which will be 0.24, no point 24. Amps and the same to get I three itbe 1.9 divided by six. Which is 3.20 point 32. And we should find that 0.47 plus 0.24 plus 0.32 should add up to our total. 0.47 plus 0.42 plus 0.32 should give me my total current, because my current comes in and divides into three branches and then joins back up and goes back in. Okay. Okay. Okay. Let's try question 16. Jackson, what is the value of the output voltage in each of the potential dividers below this one? And this one, remember, we can use this equation. V out is R2 over R one plus R2 times v in. I always show this R one and this R2. So. Would you like to try this question? Yeah. Be out. So that's R2. That's our one over R one plus R2 total resistance. The n is our emf across our power supply. So R2, so if that is sixteenth of R2, R one, okay, without equal to R2, so that is equal to ten. 40 then divided by 40. And then times 82 be up equal to two. And what about this one? So five kiloohms. Over 15. Times 300. Okay. R2 over R one plus R2 times V N, Yeah. So you get 100 volts, 100 volts. So essentially, the ratio of this voltage compared to the total voltage is the same as the ratio of this resistor to the total resistance. Let's have a look at question 17. So we have a variable resistor so you can vary the output voltage. It's a bit like changing the setting on your heating supply. You want your heater to come on if it reaches a certain low temperature. So you can change the settings in the potential divider shown. What is the range of values of v out as the variable resistor v is adjusted over its full range from north to 50 oes? How do you think youwork that out? Zero, two, 50 oms, that is a variable resistance. So the way out will have change. If that is zero ohms, that means there is no vote. There is no way out because zero divided by ten ohms, that is zero. So times v in that is still zero, zero. So. But but if if that is 50 ohms, that is five over six times twelve equal to ten votes without equal to ten be over 60 times twelve. Yeah, so noto ten folds good. So this one. This diagram goes with this question in the potential divider shown here, the thermister has a resistance which varies from 200 ohms at 100 degrees centigrade. To five kiloohms at zero degrees centered so to 5000 ohms. Going from 100, so warm is low voltage to cold is high resistance. Calculate the value of v hours at 100 degrees and at zero degrees. Yeah the way out. So when 100 degrees win 100, let me see, that's v in, v out. So 100 degrees with 20 ohms, but that is 1000, 1000, 1020Oh, 200. And times. Times v in. Yeah v in v in in that is 665. And that is five. So the first one is five volts. And the zero degrees, that is one, that is 5000 ohms. So that is 1000 divided by 6000 times six. 11 vote tes Yeah one vote five votes, 21 vote good. So and. Have you looked at physics and maths? Because it has past papers. I'm. Trying to find the questions that we were going to look at. That's internal resistance. I think tomorrow we'll look at some past paper questions, okay, on these two topics, potential dividers and internal resistance. And then if you can do those, I think we're good on electricity. Okay, Jackson, how does that sound? Okay. Okay. So I'll say good morning and hope you had to have the rest of the day. Good. Bye bye bye miss.