Cells, EMF, terminal voltage & internal resistance

Let's explore the meaning of EMF, terminal voltage, and internal resistance of a cell. And see the difference between these two, which is always confusing for me. And finally, we'll be able to write an expression connecting all of them. So, let's begin. Let's start by looking at a situation. Imagine you have a battery connected to a bulb. What happens? It glows, no surprise. But what will happen if I were to connect another identical bulb, in parallel to this circuit. What will happen? Well, if you do that, you'll find that the bulbs will become dimmer. If you attach one more bulb in parallel, it will become even dimmer. This is something you may have experienced if you have inverters at home. When the power goes off and your house is running on inverter, you may have seen that as you switch on more and more tube lights, the brightness decreases. But why is this happening? Or what does it mean? Well, let's think about this. As I start attaching more bulbs, I hope you agree that more current starts getting drawn from the battery. Does that make sense? Because every bulb will start drawing current. So, the thing that is happening as I attach more bulbs in parallel, is that more current... Let me write that down over here. More current gets drawn from the battery. Another way to see as to why more current gets drawn from the battery is... Think about this. When you attach more bulbs in, more resistors in parallel, that effectual resistance decreases Right? And so, as the effectual resistance decreases, more current gets drawn. And as you see, more circuits, more current gets drawn from the battery. Alright, what happens because of that? It turns out, because of that, the voltage provided by the battery starts reducing. Again, let me clarify what I mean. When I attach one bulb, I am drawing little current - high voltage. As I attach another bulb, more current starts drawing from the battery - the voltage reduces. Put another bulb, even more current - voltage reduces even further. So what's happening over here, for some mysterious reason, which we have to figure out now is, as you draw more current, the voltage across the battery drops. And this is really confusing. Why does that happen? To answer this question we need to understand exactly what EMF, terminal voltage, and internal resistance are. So let's look at just one battery and a bulb. Let me get rid of this. And let's bring in just one battery and a bulb. And start with a question of... Let me just get rid of the bulb for now. Let's start with the question of what is the meaning of EMF. EMF is a number that is written on the battery, like 1.5V or 9V. Let me just write that down. So that is the EMF. For example this could be a 9 volt battery. So that 9 volt is the EMF. And what does it mean? EMF stands for Electromotive force. And you can immediately see there is a problem with the wordings over here. It's a force, it's named as a force, but it's not really a measure of force. It's a measure of energy. The name is stuck, it's a misnomer, but it's okay. So let's think about what does this 9 volt mean? What exactly is this? For that we need to know what a battery does. What does the battery do? If I bring back that bulb... What the battery really does is that it pushes charges. Let me get an example of what I mean. You might know that there are electrons all around. But electrons are negative. I don't like negative. So let me imagine there are positive charges. Let's pick one positive charge. With that positive charge over here, it gets repelled by this positive or attracted by this negative, and as a result it falls like this. Now, once inside the battery, notice the charge will not automatically go from here to here. No, no, there is a repulsion. Now, what the battery does is the battery pushes this charge against this electric repulsion. And it pushes it, pushes it, pushes it and brings it over here. Again, the whole thing falls. Again, the battery pushes it, pushes it, and so on, and so forth. In doing so... Why are the batteries pushing this charge? I hope you agree, it's transferring energy into that charge. This is very, very similar to how when parents are pushing the child up onto a slide, it transfers energy into that child and then the child comes all the way down. And again, the parental push... The parent is like the battery Pushing the child up transfers energy into that child. Similarly over here, the battery transfers energy into this child. Not child, charge. And the EMF is a measure of that. So what does it mean to say that the EMF is 9 volt? It just means that the battery transfers 9 joules of energy per coulomb. So what that means is: if this was a one coulomb charge, when it goes from here to here the battery transfers 9 joules of energy to it. So, are we clear with the word EMF? Alright. Now, let me ask you this question. If this was one coulomb, then when it goes from here to here how much energy do you think the coulomb has gained? It might be reasonable to think that because the battery is transferring 9 joules of energy the charge must have gained 9 joules of energy. Just like over here. When the parent does work, whatever energy it transfers, gets transferred to the child as potential energy. Maybe same thing is happening. Right? This is where things get interesting. Although... and this is where we need to be very careful. Although the battery is transferring 9 joules of energy into the charge, one thing to remember is that inside the battery, there is a lot of chemical, a lot of material. There is a lot of medium that can be wet, it can be dry, but there is a lot of chemicals. And as a result, as the charge moves through that chemicals, the chemicals offer some resistance to it. Just like when you move your hand through water, it causes resistance. Just like when you move through air, there is some resistance. Similarly the chemicals inside the battery themselves resist the flow of charges. And by the way, this resistance is called internal resistance because it's a resistance inside the battery. Internal to the battery. Anyways, coming back to our story, as the charge moves through this there is heat generated. Think about it. Whenever you move through any medium because of the resistance there will be heat generated. And as a result, some of the energy, some energy is dissipated as heat. The battery becomes hot. For the sake of example, in this case let's say out of that 9 joules transferred to the charge, 2 joules per coulomb, goes out as heat. My question is: by the time the charge comes over here how much energy does it have? Well, it got 9 from the battery, but 2 got wasted as heat. So what's remaining now is only 7 joules. Only 7 joules per coulomb gets transferred eventually. Therefore, every coulomb, when it comes from here to here, only gains 7 joules. And therefore, we now say that the voltage across the battery is 7 volt. And this is what we call the terminal voltage. This is the terminal voltage. It's how much energy the coulomb finally gains. You know? When it goes from here to here. And hopefully now you see why the terminal voltage need not be equal to EMF, or won't be equal to EMF - because of the internal resistance there will be some heat loss. Long story short, think of terminal voltage, or the voltage of the battery, as the energy gained by 1 coulomb as it goes from here to here, that will be equal to the EMF, that is the energy transferred by the battery per coulomb, minus the energy that is lost due to heat due to the internal resistance. If you understood this, I have a question for you. My question now is: what if we had the same battery, with the same EMF, with the same chemistry inside, meaning the same internal resistance, but this time, what if we made the charge go faster through the battery. What will then be the terminal voltage? Would it be the same, 7 volt? Would it be more than 7 or less than 7? Can you pause the video and think about it? Remember, same EMF, same internal resistance, same chemistry, the charge is moving faster. What happens with terminal voltage? Pause and think about this. Alright, if you've tried - because the EMF is the same, the energy transferred when the coulomb goes from here to here is the same. It doesn't matter whether you move it slowly or you move it fast. The energy transferred by the battery is the EMF. That remains the same. That's 9 joules per coulomb. Alright. The internal resistance is the same but the charge is moving faster through the medium. Because of that, what happens to the heat energy formed? When you move faster through any medium, I hope you agree that the heat generated now is more. And to give you an example, think about space shuttles that are entering into our atmosphere. You may have seen, maybe in movies, that when they're going in an incredibly high speed, they start heating up and burning. Because of the high speed, the heat generated is so high, they almost start burning. But that doesn't happen for an airplane. Why? Because they are going very slowly. So when you go faster through any medium, and the medium is the same air, but when you go faster through it, the heat generated is more. So the same idea is applicable here as well. As the charge moves faster through this battery, more heat gets generated. Maybe this time not 2, maybe, I don't know, 5 joules per coulomb is dissipated as heat. More heat. So now, what would be the new terminal voltage? Out of 9, 5 got lost so the new terminal voltage would only be 4. So let me just write that over here. 4. There are a lot of numbers but I hope you see what is going on over here. What did we see? As the charge goes faster, the terminal voltage drops. Now can you answer our original question? When you are drawing more current from the battery, you're forcing the charges to go faster. The energy supplied to the charge stays the same, but as you go faster, the heat loss is more, and therefore the terminal voltage drops. Does that make sense now? This is the reason why, as you attach more and more bulbs in parallel, you are drawing more current, more heat loss, and as a result, the voltage across the battery was dropping, and therefore the bulb was going dimmer. Similarly, if the charges move slower and slower, then, I hope you agree, less heat will be lost and more of that EMF will be over there, but as terminal voltage. So the terminal voltage will increase. What if the charges go very, very slow, incredibly slow. At a snail's pace. Almost not moving at all. Then, almost no heat energy would be lost. Then the terminal voltage would be almost equal to EMF. Ah! So, do you now agree, do you now see why if the current is zero, there will be no heat loss at all, then all of the EMF would be available as terminal voltage. So with current at zero, terminal voltage equals the EMF. Does that make sense now? Hopefully, we now have a clearer understanding of the difference between the EMF and the terminal voltage, and why they differ, and how internal resistance comes into the picture. The last thing that I want to do is write this entire story into an equation. Okay? For that, let's assume that the battery contains a tiny resistor inside of it which represents its internal resistance. There isn't any resistor, that's a model we like to work with. So let's assume that there is a tiny resistor which represents the resistance, our internal resistance of the battery. The question now is: if the current through this battery... Let's give it some name. If the current is I what is the equation that connects all these variables? The internal resistance, the current, the EMF, and the terminal voltage. Use the same logic that we have used so far. Alright, let me write it over here. We don't need this anymore. The way I think about this is: I know that the terminal voltage is basically EMF minus the heat loss. Let me write that over here. The terminal voltage equals the EMF minus the heat loss. And when I am saying heat loss, I am talking about heat lost per coulomb. Right? Heat lost per coulomb. So what I need to figure out is how do I calculate this heat loss per coulomb. If you think carefully, this heat loss per coulomb is the voltage across the resistor. Think about it. Whenever a current flows through the resistor, we say there is a voltage drop. That voltage drop represents the energy dropped per coulomb. And that energy dropped per coulomb is the heat energy per coulomb. So this is basically the voltage across this resistor. And from Ohm's law we know voltage across any resistor would be just I times R. And so we can now write our equation as VT, the terminal voltage, equals E, the EMF, minus I times R. And hopefully, this equation now makes a lot of sense to you. This is the energy transferred by the battery per coulomb, this is the heat energy lost per coulomb, and the terminal voltage is the net energy gained per coulomb as it goes from here to here.