Let's go over the whole of Alevel physics electricity. This video is applicable to all exam boards. We're going to start off with electric current. This is defined as the rate of flow of charge. The equation for it is that the electric current I is equal to delta Q over delta T. It is also worth noting for your A levels that anytime you have the rate of change of something, then you have the change in that quantity divided by the change in time. It's also worth noting down the units. On the left here, we have current, which is measured in just amps, which is a standard SI base unit. Q stands for charge, which is measured in kum C. And delta T is just good old time measured in seconds. And here is an example problem. If 5 millum of charge flow in 3 minutes at a constant rate for a conductor, we need to figure out the electrical current. Okay, so we can just use I is equal to delta Q / delta T. We'll have 5 * 10 ^ minus 3. Remember millie stands for 10 -3 / 3 minutes. So that' be 3 * 60 seconds. We're going to get around 2.8 * 10 ^ of - 5 amp. One more example. What is the SI base unit of charge? Well, because I is equal to delta Q over delta T, we can just rearrange this for the charge, giving us that it's going to be the current multiplied by the time. So the current is measured in amps and the time is measured in seconds. So that means that the SI base unit of charge is simply amps seconds. Moving on to potential difference. This is the work done per unit charge. And there is a incredibly useful equation for it. Let me write this down over here in big letters because V is equal to W over Q. V is the potential difference measured in volts. W is just energy transferred or work done measured simply in jewels and Q stands for charge measured simply in kum. A nice example problem to get us used to this equation. 5 jewels of energy trans for every two kum of charge. We need to find the potential difference. Well, V is equal to W over Q. So, we're going to have 5 JW divided by 2, which is just going to give me 2.5 vol or 3 vol up to one significant figure. Another example problem. Show that the SI base unit for the volt is kg m^ 2 s ^ of - 3 a^ of -1. Well, remember voltage is given by W over Q. W is just measured in jewels and for the base unit for the draw, it's worth picking any energy formula. So, for instance, on the side, I can just write that kinetic energy for instance, which is also measured in jewels, is a half m. So therefore, one jewel will be the half has no units. So that'll be kilogram. And then anytime you have a square with units, put some brackets in will be m/s squared. So the jewel will be kilogram m squar s^ of minus2. W is just measured in jewels. Q is measured in kum. Okay. So we have the top which is going to be kilog 2 s^ of minus2. Remember though that charge is simply equal to I t. So that would be amps seconds i.e. we're going to divide by a s and let's tidy this up. So that's going to give me kilog m squared s ^ -2 / s yet again will give me s ^ of minus 3 and then divid amp^ minus1 which is exactly what we wanted. On to resistance. It is defined as the potential difference across the conductor divided by the current going through it. And we have the famous equation that R is equal to V / I. R is of course the resistance measured in ohms. The SI base unit of resistance can be found out from the base unit of potential difference because R is simply V / I. And just a minute ago we showed that the SI base unit for the volt is given by this expression. So all we need to do is actually divide this by the base unit for the current which is a base unit in itself which is just amps. So that means that the base unit for the ohm will be kilogram m^ 2 s ^ minus 3 a -1 / a will give us a to the power of -2. We measure the current in a circuit with an ammeter connected in series and an ideal ammeter has zero electrical resistance. On the other hand we measure the potential difference across two points with a voltmeter. This voltmeter must be connected in parallel and an ideal voltmeter has infinite resistance. Let's do an example problem. We have a resistor of resistance 100 ohms is connected to a 50W power supply. We need to show that the charge flowing through the resistor in 20 minutes is 850 kms. We can use some of the equations that you probably remember from your GCC's, but the amount of power is given by I^ 2 multiply by R. This would mean that we could figure out what the current is. So that's going to be the square root of P / R. So P is just 50, R is 100. And then we could just figure out the charge using the fact that Q is equal to IT. So this here will be 50 over 100 square rooted. This here will just our current. Multiply this by the time. So we just need to be careful to convert the 20 minutes to 60 seconds. Plugging this into a calculator, we're going to get around 848 kum, which is right about 850. Onto Ohm's law. For an omic conductor, the current is directly proportional to the potential difference across it at a constant temperature. This is often given as a two marker with the temperature part being the second mark. Mathematically, we can express this as the famous law that V is equal to I * R. Very often though, we're given IV graphs for the IV characteristics. For an omic conductor, the graph will be a straight line through the origin showcasing that the current is directly proportional to the potential difference. In this type of graph, the gradient, let's call it M, is going to be equal to 1 over the resistance. Well, if we have a graph of I against V, then I is simply equal to V / R, which we can also write as I is equal to 1 / R multiply by V. Well, if you write underneath the equation of a straight line that Y is equal to MX + C, we can write down that I is on the Y ais, V is on the X ais, and what's left for our gradient is 1 / R. And the graph should be a straight line through the origin. Example problem. The figure shows the voltage current characteristics of two resistors A and B. Which one will have the higher resistance? Well, in a V against I graph because V is simply equal to I * R. Another way to write this is R * I. And if V is on the Y ais that means that our gradient will simply be the resistance. So the one with the steeper gradient will have the higher resistance. In other words, resistor A will have the higher resistance. But what if we have an I against Vgraph? Now my gradient M is the inverse of the resistance. So that means that resistor B has a lower gradient. meaning that it will have the higher resistance. Sometimes we may have described and explained questions with regards to IV characteristics. We need to know them for these components. Number one is the standard resistor. Notice how this graph is a straight line through the origin. This shows that the current will be directly proportional to the potential difference. This means that this here is an omic conductor and the resistor obeys Ohm's law on the constant temperature. Because the gradient is constant, this would mean that the resistance is actually constant. The resistance is also the same regardless of the polarity. We can still see that proportional relationship even for negative PDs. Next up is the filament lamp. The current is not proportional to the potential difference and the lamp does not obey Ohm's law. You can see that the PD is increasing at a faster rate compared to the current. We can see that because the graph is curving away towards the potential difference axis which is the x axis. This means that the resistance will be increasing. You can confirm this in questions by applying r is equal to v / i at individual points. If you had to explain why the resistance is increasing because it's current increases, the temperature increases, that means that it's going to be more collisions between the conducting electrons and the positive ions. In practice, this means that the metal ions will have more kinetic energies in their vibrations. It will be harder for the electrons to pass through. The next one is the negative temperature coefficient thermister. As you can see, the graph is also not a straight line for the origin, which tells us that the current is not proportional to the potential difference. So, the thermister does not obey Ohm's law. In this case, the graph is curving towards the y-axis, meaning the current is increasing at a faster rate compared to the potential difference. Because of that, the resistance will decrease as the potential difference increases and as the current increases. Once again, you can confirm this by calculating R is equal to V / I at different points on the actual graph. Now, why does it happen? The resistance decreases because as the current temperature increase, the number density increases as more conducting electrons are actually freed from the latice and they're joining the current increasing the current thus decreasing the resistance. And now let's have a look at the semiconductor diode. Before we even look at the IV characteristics though, we need to mention that the only way that the diodes conduct is if they're connected from the positive end to the negative end. Remember, if we had a cell or a power supply, the bigger side is the positive one. This one here is the negative one. So current goes through here from positive to negative. And that's the correct direction for a diode or a light emmitting diode such as an LED. In terms of IV characteristics, you can clearly see that the PD is not proportional to the current. The diode does not obey Ohm's law. The diode behavior depends on the polarity. So it will not conduct. If you reverse the PDI connected the other way, point one right over here. The resistance is very very high. M scheme sometimes would even accept infinite and there is no current. At point two, well this is known as the threshold PD. It will start conducting. then the resistance decreases because the current is going to rise at a faster rate compared to the potential difference. And now let's consider resistivity. So the resistance of a material depends on the shape and the exact properties of it. For instance, if I had a wire which was a lot longer, the electrons will have to pass through through a lot more vibrating ions. So the longer the wire, the harder it will be to conduct electrons from point A to point B. So resistance is proportional to the length. If the cross-sectional area is larger, then I'll be able to pass more charge carriers through. So the larger the area, the lower the resistance. There's also going to be a constant of proportionality between the resistance and those variables that depends solely on the material. And this constant of proportionality is known as resistivity. It's given the Greek letter row. Let's rearrange this equation for it. We can see that the units of resistivity, well, they're just going to be ohms. And then we have area is measured in me squar / length, which is just in meters. So the units of resistivity will simply be ohm m. Let's do an example question. Two cylindrical resistors of uniform resistance are made from the same material and have the same length. Here's a little clue in analyzing electricity questions. So, anytime you have resistance and length or area mentioned, that means to use essentially the resistivity equation. Another clue, the same material is a clue that's telling us that the resistivity is equal for both of them. So, resistor A has resistance R and diameter D. Resistor B has a diameter 3D. We want to figure out the ratio of resistance A to resistance B. Another exam technique is that when whenever we're dealing with ratios, things are going to cancel. So, essentially what we want to do is divide R A by RB. Well, R A, well, that's just given to be R. This might end up being useless, but that's just equal to row L over A. And then that's going to be divided by RB, which is going to be row L / the area of B. Let's just call it A B. So this here will be row L over the area, which is going to be pi D over 2 squared. And then B is going to have an area which is going to be 3D over 2 squared. Notice how I've included the brackets here because it's so easy to lose some factors here. Now we're ready to do some cancellations. So all of this is going to cancel. The pi is going to cancel. The divided by two is going to cancel. So in a way what we're left with is 1 / d^ 2 divide that by 1 / 3d squared. Then I'm going to say that dividing by a fraction is the same as multiplying by the inverse. So this here will be multiply by 3d and then all of it squared. Notice how I'm keeping the brackets right until the very end. So this here will be 9 d ^2 / d^2. We'll just cancel these out and the correct answer is nine. Classic resistivity example question. We're given the length of a constant wire which is 1.6 m and it has a diameter of 0.15 mm. We're given the resistivity and we need to show that the resistance of this wire is 44 ohms. So resistance is equal to row L / A. So this here will then be given by row L over the cross-sectional area which is p<unk> D / 2 squared. And now we're ready to plug in some numbers. The resistivity is just 4.9 ultip 10 -7. Our length is 1.6. And then we have Be careful with the diameter. It's 0.15 * by 10 -3 and then I'm going to square this. I'm going to be dividing it by 2 because this is the diameter. Typical mistakes would include forgetting the two and also forgetting to square everything. So 10 the -3 the millimeters need to be squared as well. Putting this into a calculator we are going to get 44.4 4 ohms which is around 44. And now let's talk about superc conductivity. So this is a topic that is in the AQA specification. So if you're not doing AQA, I recommend looking over your specification. And if superc conductivity is not on there, feel free to move on to the next chapter. Unless of course you're interested about superconductivity. It sounds kind of cool. So some materials have zero resistivity at and below a critical temperature. So the word critical here is pretty important. So I'm just going to underline that that critical temperature is actually dependent on the materials. There's certain applications of superconductors. They produce strong magnetic fields. Uh they're widely used at the large hydron collider in Switzerland for instance. And also the reduction of energy loss in the transmission of electrical power. difficulty in using superconductors in the transmission of electrical power over long distances is that it's just simply difficult to maintain low temperature over a longer distance. What is in all exam boards is the experiment to determine resistivity. So if we want to determine the resistivity, we're going to need the resistance of a wire. So remember that resistance is simply voltage divided by current. So, we're going to need to be measuring the PD with a voltmeter in parallel and then the current with an ampmeter. We will also need to measure the length of the wire. And this typically done with a ruler with a wire next to it. So, right over here, you would have a ruler with some graduations like this. You'll be able to determine what length it is. You can vary the length of the ruler with some crocodile clips. to determine the area. Well, the area is just simply given by pi d / 2 squared. The diameter is too small to measure with a ruler. So, we're going to use a micrometer. And the answer to this, if this was to appear, how we would measure this um accurately would be that we'll take use a micrometer and then we'll take multiple readings along the length. The word along is typically underlined in the mark scheme. and then we would find the mean. We would then calculate the area using pi d /2^ squ. In this particular version of the experiment, we are varying the length of the wire by changing the position of a crocodile clip. Then we're measuring the potential difference with this voltmeter across the wire with an ammeter which is in series. For each point, we're going to calculate the resistance using R is equal to V / I. Then we're going to plot a graph of R against L. Remember R in this column we've also calculated that separately is just V / I. And because R is equal to resistivity L / A. If we compare this to linear analysis, Y is equal to MX plus C. If R is on the Y- axis, if L is on the X axis, that means that our gradient M is going to be the resistivity over the area. Meaning that row will be our gradient multiplied by the area. Something that also often appears is a systematic error. And this might be given by this line not actually passing through the origin but being shifted upwards. If that's the case, do not worry. We've only used the gradient and a systematic error does not affect the gradient, just the y intercept. And now let's look at the rules for electrical circuits. We're going to start off with resistors in series and parallel. If we were to combine resistances in series, well then the total resistance R total is simply R1 + R2. If we combine them in parallel then 1 / r total will be equal to 1 / r1 + 1 / r2. The way I personally prefer to write this is as r total will be 1 / r1 + 1 / r2 all of this raised to the power of minus1. But why are those rules the way they are? Let's derive them using a couple of simple circuits. In the serious case, let's say that we have some total voltage V_T, V_sub_1 and V2 across resistors R1 and R2. And let's say that the total resistance is also R total. Well, the current is simply I. The trick with those derivations is to always think what is being added in the circuit. So in this case, it is the voltage, the potential difference that is added by Kirkov's second law, which we'll cover properly later. But the total voltage is equal to V_sub_1 + V2. I.e. the sum of the emfs is equal to the sum of the PDS. Now we can just apply Ohm's law. Total voltage is simply equal to I total which is equal to I R1 + I * R2. But the current in a series circuit is constant and it can be cancelled leaving us with R total is equal to R1 + R2. In the parallel case, let's draw a circuit. And for simplicity, let's assume that there's no internal resistance. So if the emf here is V, we're going to have V across here and V across here in terms of the voltages. The current though will split. So this here is I total. Let's say that we have I1 through the first resistor and I2 through the second resistor. So this here is R1. This here is R2. We always start off with what's being added. So in this case the current is being added in a parallel circuit. So I total will be equal to I1 + I2. Since I is equal to V / R, we can say that V / R total is equal to V / R1 + V / R2. The potential difference across all of the resistors and the EMF will be exactly the same. So we get the good old result that 1 / r total is equal to 1 / r1 + 1 / r2. Let's also have a look at the energy and power equations. So remember that power is in general just energy over time. We also know that power is simply equal to the potential difference multiplied by the current. Now since v is equal to iir we can plug this into here and we are going to get that the power will be equal to i 2 * r or we can just simply rearrange this equation right over here for i giving us i is equal to v / r and then we can plug this into here and this here will just give us the other equation equation for power that power is just equal to v ^2 / r. Actually, there's one more that we can do. There's lots of different variations, but if we take this and if we plug that into here, let's maybe write these all the way across here. And that's the equations for electrical energy. So, electrical energy is just equal to power multiplied by time. And power is equal to vi. So, we can get that electrical energy is equal to vit. If you're doing OCR, I'm pretty sure that it's written in the formula booklet as W for work done for electrical power uh which is equal to VIT, but it's exactly the same equation. And let's talk about Kirkov's laws. Starting off with the first one, the sum of the currents entering a junction is equal to the sum of the currents exiting a junction. And this here is a statement of conservation of charge. In practice, the junction is typically a little circle in the diagram. And for instance, let's say that we had 5 amps going into a junction. And let's say that we also had 7 amps going into a junction. If we had one more pathway for the current to go, and let's say that this one here was leaving the junction, we would know that 12 amps would be leaving the junction. In practice, Kirkov's first law also explains how the current is constant in a series circuit. If I had 5 amps going through this circuit, we know that the current through here and at any point throughout the circuit will be equal to exactly 5 amps because no charge is going to be lost or gained. Hugov's second law says that the sum of the emfs in a closed loop is equal to the sum of the PDS. And this here is a statement of conservation of energy. In practice, that means that if I had a series circuit, then that the PDS are actually shared. So, let's say that the emf is 5 volts and let's say I had 3 volts across here. This one here would have to be 2 volts. In other words, 5 will just be equal to 3 + 2 in terms of voltage. If we had a parallel circuit though, we get something quite interesting. So, let's say that we had one resistor across here. Let's say we had 5 volts. And let's say this one here, we also had a couple of them. One closed loop is going to be this one across here. Across this blue closed loop, the sum of the PDS will equal the sum of the EMFs. There's only one source of emf. So that means that this one here must be 5 volts. But this here is not the only loop that the electricity can take. They can also take this let's say this red loop across here across it and let's say that if one of them here was let's say 3 volts that means that they also have to add up to 5 volts because across this individual loop as well the sum of the PDS will be equal to the sum of the emf so this one here will have to be 2 volts let's do an example we have a 9V cell with negligible internal resistances shown and it's connected in series with those two resistors we We need to show that the current in the circuit is about 0.02 amps. The total current I will be equal to our emf divided by the total resistance. So the total resistance will just be 300 + 100. So our emf or total voltage across the circuit will be 9.0 0. Divide that by 300 + 100, which is going to be 9 over 400. And this here gives me 0.0 225 amps, which is pretty close to 0.02 amps. We also want to calculate the power dissipated in the 300 ohm resistor. There are multiple different ways of doing that. For instance, we could use the power is equal to I^2 * R. So this here will be 0.0225. I'm just going to use the more the version of my answer with more significant figures just to avoid rounding error. Although in the exam you typically get full marks even if you don't do that. Then we're going to square that and then multiply by the resistance which is just 300. And this here is going to give me approximately 0.15 watts. And let's do an example on current. All we need to do is use the fact that the sum of the currents entering a junction is equal to sum of the currents exiting a junction. We're going to start off over here on the right because current is going the other way along the horizontal line. And then we'll check each individual junction. For instance, in the first junction, we have 5 + 2 going in and then seven coming out. So that means that there won't be any current in this part of the circuit. Okay. Over on the next part, we have 3 amps going in on this side, but then also 5 amps also coming in. So that means that over here, we must have 8 amps coming out of this junction. And if we have eight amps going into this junction, we also have one amp going in and then one amp going in. Well, this means that we are going to have 10 amps at point A. Hopefully those make sense. Let's do a multiple choice question. The current in a circuit with a cell of emf is I when a resistor of resistance R is connected. Internal resistance is negligible. Find the current I2 in terms of I if there are three resistors of the same resistance connected in parallel and we have some options over here. Okay. Well, we know that V is equal to IIR meaning that the current I is just equal to the emf divided by the total resistance. In the first case that was just R. In our second case, we know that I2 will be equal to V because all of these resistors will have the same potential difference V by Kirkov's second law divided by the total resistance. Okay. Now, R total is just going to be 1 / R + 1 / R + 1 / RA to -1. R total will then just be 3 / RA to the -1. R total will be R over 3. I've shown where this comes from. However, in an exam situation, if the resistors are connected in parallel and if there's three of them, the total resistance will decrease by a factor of three. There was two of them, it will decrease by a factor of two. So on and so forth. So this here means that I2 will be V over R total, which is R over 3. So this here will be equal to V / R. And then we can bring the three up here. 3 * V / R which is just going to be 3 * I. Correct answer is B. Please note that in an exam situation it is totally fine just to use proportionality. What do I mean by this? In general in this type of question V is equal to IIR. Meaning that I is V / R. So if the resistance decreases by a factor of three that means that the current will increase by a factor of three as well. But I just wanted to show you where these rules actually come from. And something which is really important state and explain what happens to the current drawn from the cell as more resistors are added in parallel. As we can see the current increases because the total resistance in the circuit decreases and this type of situation is often tested in Alevel physics exams. Whenever we have to find the voltage between parallel branches. This is a very common mistake in Alevel physics. So let's go over it. We have a 12vt battery. Let's say that we are ignoring internal resistance. Then we have 20 and a 10 in terms of resistors and then a 17 a 50. The way we deal with those problems is in terms of steps. Our first step will be simply to determine all the potential differences across all of the resistors. In this case, this is relatively straightforward because they need to add up to 12 and they will be in a ratio of 10 to 20 which is the same as 1:2. So, which two numbers are in a ratio of 1:2 and add up to 12? Well, that's simply going to be 8 and four. So, this one here will be 8 volts and this one here will be 4 volts. Down here, we have a 17 to 50. Once again, pretty easy. There will be in a ratio of 7 to 5, meaning that this one here will be 7 vol and this one here will be 5 volt. If the ratio is not nice, what we can do is figure out the total resistance and then the current. Once we have this, our final step is nice and straightforward. All we need to do is go in a complete loop around the voltmeter and find the potential difference. Let's say that I started at this 0.1 and then I'm going to go all the way across here to 2. And we can see that we've gone from 4 vol to 5 volt i.e. the potential difference between those two points is exactly 1 volt. We're looking for the magnitude in most cases in Alevel physics. I have an example from a Cambridge admission questions where polarity is also considered which I will link this into the description of this video should you like to have a look at a more in-depth example of this. You could have also gone the other way. So we could have gone this way as well. And we can see that in the other loop across here from 8 volts the potential difference between them is also 1 volt in magnitude. Let's cover the condition for no potential difference between parallel branches. I've only seen this as a prerequisite in AQA and for this exam board is going to make some questions infinitely easier. I've not really seen this for other exam boards. So you're welcome to skip to the next chapter if you're doing a different exam board. But for AQA, let's draw a little circuit to illustrate this. So we have a power supply. And then let's connect a resistor which I'm going to call R1. Then let's have another one R2. Apologies for my handwriting and I'm not drawing a very beautiful circuit, but it will illustrate the point. Let's have R3 here. And then we're also going to have R4. The condition for the potential difference between them to be zero is the following. R1 / R3 has to be equal to R2 / R4. You can derive this using Kirkov's laws. I think this is actually technically called a bridge circuit and it's something that you might end up doing at university deriving this relationship for zero potential difference across this bridge between the circuits. But if you just remember that before your paper one in AQA, this might make some problems much easier. For instance, if I was to replace R3 with an LDR and then the voltmeter reading was initially equal to zero, the light intensity is then increased. We need to state and explain a change in the circuit that will restore the voltmeter reading to zero. Well, exactly as we have discussed above, the voltmeter reading is zero when R1 over R3, which is the resistance of the LDR, is equal to R2 over R4. So, line intensity is increased and the graph of resistance against light intensity for an LDR was something like this. So that means that the resistance of the LDR has actually decreased. And now it's just a question of fractions. If we decrease the denominator by some amount, then a way of restoring the potential difference back to zero would be to also decrease R4. There are multiple other things that we could have said. We could have also said if we haven't decreased R4 then we could have also well decreased the amount of resistance in R1 by exactly the same amount and that would have also produced the same effect. If this ever appears in a question or a multiple choice question all we need to do is consider the change that will keep the fraction of R1 / R3 to be equal to R2 / R4. This rule can turn intimidating looking problems to quite simple ones. For instance, in this complicated looking circuit, we're being asked to determine what is the potential difference between the points one and two. Well, the first thing to check is R1 over R2. Well, this circuit here will basically be equivalent to a circuit which kind of looks like this. The 100 and 100 are going to add up to 200. Okay. Then we have our 0.1. Then we have our 10 ohm resistor. And then under here we also have 200 ohms. Then it's our 2. And then the 30 the 30 and the 30 when they are added up together they're just going to decrease by a factor of three. So they're their equivalent resistance is also going to be just 10 ohms. So that means that the potential difference between 1 and 2 will actually be just equal to zero because remember it's zero whenever R1 / R3 is equal to R2 over R4 where this one here is our R1. This one here is our R3. This one here is our R2. This one here is our R4. Because 200 over 200 is the same as 10 / 10. So the voltmeter reading if there was a voltmeter there would be zero. We're going to talk about this more once we cover internal resistance, but we can have cells in series and also in parallel. Let's imagine that I have a couple of cells like this. And let's say that each one of them was five volts. Well, if they're connected positive to negative, positive to negative, this combination will have an emf of 10 volt. If they were connected the other way around though, so let's say positive to negative and then if I had a negative here and a positive here, then my equivalent emf will just be 0 volts. If one of them had a larger emf, well then we're going to take the one which is connected negative to positive as negative. So the total emf will be 5 volts take away 3 volts which will be just 2 volts. If I have identical cells in parallel and we're only going to get cases with uh identical cells in parallel then what we're going to get is essentially the same PD. So let's say I had two identical cells with so 5 volts and 5 volts. then my PD across this resistor will also be 5 volts. And now let's talk about potential dividers. What is a potential divider? Well, if we have some input voltage, maybe a cell, so for instance, something like 6 volts, maybe we want the part of the circuit in which we have a different voltage. A way of controlling and reducing the voltage, a way of dividing the voltage is with a combination of resistors. So let's say I have an R1 and then an R2. One of the most useful formulas that we have is that the output voltage i.e. the potential difference that a voltmeter across here essentially where I've put V out would measure is given by the following formula that V out is R2 over R1 + R2 multiply by the input emf assuming no internal resistance V in. So V out is R2 / R1 + R2 * V in. Where does this equation actually come from? Let's show that this equation is true. Well, if we have some current I, V out will be given by the current multiplied by R2. But on the other hand, the current will be equal to the total PD in the circuit which is VN divided by the total resistance which is R1 + R2. So if we plug this into here, we get that V out is equal to V in R2 over R1 + R2, which is our original expression. Here's a little example to get us used to this equation. We want to calculate the reading on the voltmeter. So we know that V out will be equal to R2 / R1 + R2. multiply this by VN and our R2 in this case is 55 ohms / 25 + 55 and our input voltage is simply 12 and this here will simply give us 8.25 volt. Another incredibly useful equation in potential dividers is that if we have a typical potential divider let's say that the potential difference across this one is V_sub_1 and across this one here is V2. The equation is that the potential difference across the first resistor divided by the potential difference across the second resistor V1 over V2 will be equal to R1 over R2. And of course we can turn this into a really nice show problem. In order to derive this expression we must consider the current. Because this is a series circuit we can say that the current through R1 is the same as the current through R2. We can even call these I1 has to be equal to I2. But in general current is equal to potential difference divided by the resistance. This is because V is equal to IIR and I is simply V / R. So V_sub_1 / R1 will be equal to V2 over R2. And just simply rearranging this we get that v_sub_1 over v2 will be equal to r1 / r2. Potential dividers often involve the use of light dependent resistors or LDRs or thermisters. We need to remember the general shape of the graphs for resistance against light intensity and resistance against temperature. Luckily for us they're very very similar. We can see that as light intensity or temperature increase, resistance tends to decrease. A good rule of thumb is if the environment is increasing, the resistance is decreasing. Or if the environment, for instance, the temperature is decreasing, the resistance is increasing. Or if the line intensity is decreasing, the resistance is also increasing. Let's do a potential divider example. We have a resistor which obeys Ohm's law and that is connected in series with an LDR and this cell. We want to state and explain what will happen to the reading of the voltmeter if the light intensity on the circuit increases. Something which really helps me in those problems is to simply write down riff. And what does that mean? R stands for just resistance. What happens to the total resistance in the circuit? Well, light intensity increases and the graph for resistance against light intensity was something like this. Therefore, the resistance has decreased as light intensity has increased. We can also say that the total resistance in the circuit has decreased. Our next bullet point will relate to I which is just the current. The current will always do the opposite to what the resistance does. So if the resistance has decreased, this means that the current in the circuit has increased. F stands for the fixed resistor. Across this fixed resistor, V is equal to I * R. Now notice how R is actually fixed meaning that it doesn't change. And if the circuit is increased, that means that the potential difference across the fixed resistor has increased and hence the reading on the voltmeter will increase. This problem can then be really easily extended to asked what has happened to the PD across the LDR. Well, if the PD across the fixed resistor is increased, that means that the PD across the LDR must decrease as they must add up to the fixed emf. A variation of this problem, a variation of this problem appears quite often. And following this structure of explanations will generally give us full marks. What we don't want to do is try and apply Ohm's law. So I'm going to write this in red. This is what not to do. We cannot really apply V is equal to IR. Number one, the LDR is not an omic component. But also if we really really think about it, while the current in the circuit has increased, the resistance here has decreased. So we can't make any judgments on what happens to the PD across the LDR using Ohm's law. It does not apply across it. So we shouldn't use it. It does apply to the fixed resistor though. That's why we use it there. And now let's talk about the difference between EMF and potential difference. We're going to try and illustrate this with a simple circuit. Let's draw the world's simplest circuit. Even something as simple as a cell. And let's say that we have connected this to a filament lamp. EMF or electromotive force is the energy transferred per unit charge from other forms to electrical. For instance, in the cell, we're talking about chemical energy being transferred to electrical energy. If you're doing AQA, this is often defined as the chemical energy transferred to electrical when unit charge passes through the cell. Potential difference on the other hand occurs when electrical is transferred to other forms and is defined as the energy transferred per unit charge from electrical to other forms. So for instance here we've got a transfer of electrical to light and heat energy. Both emf and potential difference can really be expressed in terms of voltage being equal to energy per unit charge or work done per unit charge. You could also write the W as E. Very often though, the actual emf is written with this symbol over here just to differentiate them. A classic written problem is to state what is meant by emf. For instance, if we have a cell of 5 volts, state what is meant by the cell having an emf of 5 volts. The way we would answer this is that five jewels of energy converted from other forms to electrical per unit charge are actually required. And this here is the work done in moving one kum or a unit charge. Just write one column with charge through the electrical cell. This is typically a two marker and those should give us the marks. And now let's discuss a crucial concept and that is internal resistance. So some of the emf will be lost by charge carriers as they're moving through the cell. This energy will typically be dissipated as heat. It is often really convenient to think of internal resistance as a little resistor which is in series with the original cell. It is often really convenient to think of internal resistance as this little resistor of resistance R which is in series with the cell. But this entire region here is actually the cell if that makes sense. In practice, if the cell has an emf of let's say 3.0 0 volts. But if we were to connect something to a cell, something like a filament lamp for instance, we might not measure 3.0 exactly, but we might measure something, let's say 2.9 volt. This would mean that 0.1 volts will be lost and they're lost due to internal resistance. And hence the definition of internal resistance, it is the lost volts per unit of current. There are several different formula which are very very useful. I tend to quite like to start using that emf is just equal to the terminal PD plus V lost. This here is just the emf. This here is the terminal PD. This here is obviously the lost volts. We can then apply Ohm's law and we can say that E is equal to V + IIR like so where R is the internal resistance. We can also apply Ohm's law to V and we can get yet another formula. E will be I R + I little R meaning that E will be I R + R. Sometimes this formula is written rearranged for V as well which is simply the emf take away the lost volts and this is where it comes from. By the way, this is something that occasionally comes up. How do we measure the emf of a cell? We need to connect a voltmeter across the cell with no external resistance. So in practice, if this here is our cell, then we just simply connect a voltmeter like so. And this shouldn't be connected with anything. Sometimes there might be a switch over here for instance. And let's say if we had a filament lamp or something like that, if the switch is open and the voltmeter is there, we will be measuring the EMF. And here's a very typical experiment to determine the internal resistance of the cell. We're going to use a variable resistor, an ammeter connected in series, and then a voltmeter connected in parallel. The voltmeter can also be positioned across the variable resistor. We're going to vary the resistance of the variable resistor to change the resistance in the circuit. Then we're going to measure the current with the ammeter in series. Then we're going to measure the terminal PD using the voltmeter. We're going to take multiple readings. an average to improve the accuracy of this experiment. We're also going to plot a graph of V against I. The graph might look something like this. It would typically have a negative gradient. But how can we determine the internal resistance? Well, let's do some linear analysis. We know that E the emf is equal to the terminal PD plus the lost volt iir. Then I'm going to rearrange for whatever is on the y ais which is just v which is going to be equal to this minus iir. And then let's do just a little bit of rearranging. So I'm going to say that v will be equal to minus r ult* i + d emf. Then directly underneath I'm going to write y is equal to mx + c. Look at this. If V is on the Y ais, if I is on the X axis, this means that my gradient M is equal to the negative of the internal resistance. So the magnitude of the gradient will just give us the internal resistance. Our intercept C will be our emf. Different variations of this experiment can appear in exam situations, but this is the general method. Let's do an example problem. We have a practical cell of EMF 12 volts and internal resistance of 10 ohms. If you're doing AQA, by the way, very often if the question says a practical cell, that means that internal resistance cannot be ignored. The cell has some internal resistance. And then we've connected it to two resistors R1 and R2. The PD across R1 is 5 volt. Okay, let's put that onto the diagram. We also have the emf of 12. And then the current in the circuit is 75 milliamps. What we need to figure out is the value of R2. So we can probably use our good old equation that our emf is equal to I multiply by the external resistance plus the internal resistance. In this case, our external resistance is actually the sum of those two resistors because they're just connected in series. Let's do a little bit of rearranging. So we can go that the emf divided by the current is equal to R + R. But R is just equal to R1 + R2. Okay. Well, we need to figure out what R1 is. Well, luckily for us, we can just use Ohm's law. And we can say that V is equal to I * R1 across it. Therefore, R1 will just be equal to V / I. So this here will just be 5. Divide that by 75 * 10 -3 giving us around 66.67 ohms. Oops. Let's not forget the internal resistance R here. I was just checking if you were paying attention. Well done. Okay, we have pretty much everything that we need in this equation. We have the emf in the question. That's just 12. We have the current. We've now calculated R1. We have the internal resistance as well. We're ready to calculate our R2 which is just going to be equal to this take away R1 take away little R. So this here just going to be 12 over 75 over 75 * 10 -3 takeway 66.67 take away 10 giving us an answer of around 83 ohms. Here's a very counterintuitive question on internal resistance. We need to make sure that we get this one right. We have a practical cell with internal resistance connected as shown over here. Then the resistance of the variable resistor is increased. We want to state and explain what happens to the readings of the ampmeter and the readings of the voltmeter. Okay. So let's take it by steps. The first thing that we need to mention is that the resistance in the circuit the total resistance will be increasing. Because of that the current in the circuit will need to decrease. Hang on a minute though. If there is current in the circuit and if there is internal resistance, let's draw a little resistor across here. The amount of lost volts is simply equal to I * the internal resistance. R is fixed the internal resistance. And if the current goes down, then the amount of lost volts must come down as well. If the amount of lost volts has decreased, that means that the if the amount of lost volts has decreased, that means that the terminal PD must increase because they need to add up to the fixed emf. In other words, our emf is just simply equal to the terminal PD plus the amount of lost volts. And if the lost volts have decreased, then V must increase to compensate for that. This will not be true if there was no internal resistance. Another example, we have two identical cells of emf6 volts and internal resistance 20 ohms each are connected to a resistor of resistance 250 ohms. What we want to calculate is the power dissipated through the resistor. Anytime we have an internal resistance problem, I tend to just draw these internal resistors. Each of them is 20 and they are connected in parallel. Remember if we connect cells in parallel the emf is unchanged. So this circuit here it will be equivalent to a circuit of a single cell of emf 6 vol and an internal resistance which is going to be 10 ohms. Why has it halfed? Well, remember our total will simply be equal to 1 / 20 + 1 / 20 raised to the -1, which is 2 over 20. And if we flip that, we get 20 / 2, which is equal to 10 ohms. And it's this circuit here that is connected to this 250 ohm resistor. Well, what will the current for the circuit be? We could probably use our good old trusty equation that the emf is equal to I external plus internal resistance. Current will be E over R + R. Our emf will just be six. Our external resistance is 250 plus the internal resistance which is just 10. And we're going to get around 0.0.23 amps. To figure out the power dissipated, we could use for instance P is equal to I^2 * R. Or we could find the voltage across this and then multiply the two. Essentially, it's the same thing giving us 0.0.23. Now this is squared. Multiply this by 250 around [Music] 0.13 watt. This next topic at the time of this recording is not on the AQA specification. However, it is on the OCR specification and it is on ed Excel, WJC, Educas and that is number density. What is number density? Well, it is simply the amount of charge carriers per unit volume. It is typically given the letter lowerase N and that's defined as capital N which is the amount of charge carriers per unit volume. We're going to be dividing that by the volume. The number of charge carriers, this is unitless. However, the unit of volume is meters cubed. So therefore, the units of number density are simply m to a power of minus3. These brackets are sometimes used in physics to signify the units. In practice, I mean, if we think about something like a unit volume, let's say that we had a region which is a meter meter by meter. And let's say that we had, I don't know, let's say we had a few electrons. Let's say we had three or four electrons across here. If there's four of them per meter cubed, that would mean that our number density will just be 4 m to the power of minus 3. Obviously though n number density is normally a large number for a conductor. Numa density can often be used in questions to explain the different resistivities of conductors, semiconductors and insulators. So conductors have a very high number density and insulators will have a relative very low number density. Semiconductors well they'll be somewhere in between. onto drift velocity. It is defined as the average displacement per unit time of charge carriers along the conductor. So let's say that this here is our conductor. Then our drift velocity will let's say we had some electrons which are moving along here. The drift velocity will simply be their average velocity along the length. And that word here is typically underlined in mark schemes. This is because the charge carriers will be vibrating and oscillating. that will have an RMS speed in a way in all different directions but will be slowly drifting towards this direction the direction of the current. The equation for the drift velocity is given by the famous nave equation sometimes called that the current is equal to the number density multiply by the cross-sectional area multiply by the charge sometimes it's written as Q and then multiply by V which is the drift velocity. Rearrange for V. We get that V is equal to I over N A E for electrons in a wire. This is typically a very small number. And let's derive the nave equation for the current and the drift velocity. So imagine that I have some sort of a conductor with some electrons which are going across like this at a drift velocity V. Our first job would be to see if we can figure out the number of electrons that pass in terms of the number density and other quantities. Remember number density n is equal to the number of charge carriers per unit volume. We're going to need an expression for the volume in a time t. The distance covered will just be given by v * t. And the total volume, let's say that this here was a cylindrical conductor will be given by the cross-sectional area a multiplied by x. The number density will just be given by N over a X where a is once again the cross-sectional area. X is the distance traveled along the conductor and X was simply given by VT. So we have our number of electrons which is simply going to be given by the number density multiplied by AVT. If we have our number of electrons, then we can find our total charge. Assuming that there are all electrons, the total charge will simply be the number of electrons multiply by the elementary charge i.e. N A Vt multiplied by E. Okay. And remember current is given by delta Q over delta T. So that will just be N A V T E / T. Well, look at this. The time is just going to cancel out and what we're left with is that I is just equal to N A V E. Okay folks, I hope this video has been useful. Well done. Your next step is to start practicing questions based on electricity and have a look at this question right over here and see if you can solve