 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 battery connected to a bulb. What happens? It glows No surprise, but what would happen if I were to connect another identical bulb in parallel to this circuit? What happened? Well, if you do that, you'll find that the bulbs will become dimmer if You attach one more bulb in parallel It'll become even dimmer and this is something that you may have experienced if you have inverters at home when the power goes off and your House is running on inverter. You might 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 Bulb, I hope you agree that more current starts getting drawn from the battery Does that make sense because every bubble start drawing current So the thing that's happening as I attach more bulbs in parallel is that more current? So let me write that down. Let me write that down over here more current 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 effective resistance decreases Right and so as the effective resistance decreases more current gets drawn. Anyways, you see more circuits More currents get drawn from the battery. All right. What happens because of that? Well, it turns out because of that the voltage provided by the battery that starts reducing So again, let me clarify what I mean So when I attach one bulb, I'm drawing little current high voltage as I attach another bulb More current starts drawing from the battery the voltage reduces as I put even more bulb 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 as to 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. So let me get rid of this and Let's bring in just one battery and a bulb and start with the question of let me just get rid of the Bulf for now. Let's start with the question of what is the meaning of emf? Well, emf is the number that's written on the battery like 1.5 volt or 9 volt. Let me just write that down. So that is the emf So for example, this could be a 9 volt battery. So that 9 volt is the emf. And what does it mean? Well, emf stands for Electromotive Force and you can immediately see there is a problem with the wordings over here It's a force, but it's not really 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 now let's think about it. What does this 9 volt mean? What exactly is this for that? We need to know we need to remember what a battery does. What does the battery do if I bring back that bulb? What the battery really does is that it it pushes charges. Let me get an example of what I mean So 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 and let's pick one positive charge So when that positive charge is 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, no There's a repulsion now what the battery does is the battery pushes this charge Against against this electric repulsion pushes it pushes 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 while the battery is 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 the parent will 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 the emf is nine volt? It just means that the battery Transfers nine 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 nine joules of energy to it. So are we clear with the word emf? All right Now let me ask you this question when this charge if this was one coulomb Then when it goes from here to here How much energy do you think the coulomb the yeah the coulomb has gained? Well, it might be reasonable to think that because the batteries Transferring nine joules of energy the charge must have gained nine joules of energy just like over here When the parent does work Whatever energy it transfers gets gets transferred to the child as potential energy. Maybe same thing is happening, right? Well, this is where things get interesting You see although Although and this is where you know, we need to be very careful although The batteries transferring nine joules of energy into the charge one thing to remember is that inside the battery? there is a lot of Chemical lot of material there's a lot of medium to be wet to be dry But there's a lot of chemicals and as a result as the charge moves through that chemicals It it the chemicals offer some Resistance to it just like when you how you when you 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 the resistance inside the battery Internal to the battery. Anyways, coming back to our story as the charge moves through this There's 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 Desipated as heat the battery becomes hot Let's say for the sake of example in this in this case Let's say out of that nine joules which is transferred to the charge two joules per coulomb goes out as heat Now my question is by the time the charge comes over here. How much energy does it have? Well, it got nine from the battery, but two got wasted as heat So what's remaining now is only seven joules? So only seven joules per coulomb gets transferred eventually And therefore every cool on when it comes from here to here only gains seven joules and therefore we now say that the voltage across the battery is seven 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 So long story short Think of terminal voltage or the voltage of the battery as the energy gained by one 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 So if you've 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 seven volt would it be more than seven or less than seven? Can you pause the video and think about it? Remember same emf same internal resistance same chemistry The charge is moving faster. What happens to terminal voltage pause and think about this All right, if you've tried well because the emf is the same The energy transferred from here to when the charge 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 So that remains the same. That's nine joules per coulomb All right, 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? Well, 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 it then when they're going at 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 aeroplane Why because they're going very slowly So when you go faster through any medium the the medium is the same is the same air But when you go faster through it the heat generated is more so the same idea It's applicable here as well as the charge moves faster through this battery more heat gets generated Maybe this time not not to maybe I don't know maybe say five joules per coulomb is Desperate as heat more heat. So now what would be the new terminal voltage out of nine five got lost So the new terminal voltage would only be four. So let me just write that over here Four and there are a lot of numbers, but I hope you see what's going on over here So what did we see as the charge moves 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 now the energy supplied to the charge says the same But as you go faster the heats heat lost is more and therefore the terminal voltage drops Does that make sense now? This is the reason why as we attached more and more bulbs in parallel We were drawing more current more heat loss and as a result the voltage across the battery was dropping Therefore the bulb was glowing 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 available 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 will be lost Then the terminal voltage will be almost equal to emf So do you now agree? Do you now see why if the current is zero? There'll be no heat loss at all then all of that emf will be available as terminal voltage So with current is zero terminal voltage equals the emf. Does that make sense now? Hopefully we now have a clear understanding the difference between the emf and the terminal voltage And why they differ and how internal resistance comes into the picture All right, the last thing you 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 now there isn't any resistor, but that's a model we like to work with So let's assume that there is a tiny resistor which represents the resistance or 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 All right. Let me write it over here. We don't need this anymore So the way I think about this is I know that the terminal voltage is basically emf Minus the heat loss. So let me write that over here. So the terminal voltage Equals the emf minus the heat loss and When I'm say heat loss I'm talking about heat lost per coulomb, right? So heat lost per coulomb So all I need to know is to figure out how do I calculate this heat loss per coulomb? Well, if you think carefully this heat lost 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