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## Class 12 Physics (India)

### Course: Class 12 Physics (India) > Unit 3

Lesson 5: Cell, EMF, terminal voltage, & internal resistance# Cells, EMF, terminal voltage & internal resistance

EMF is the work done by the cell in moving a coulomb of charge across its terminals. It represents the energy transferred per coulomb to the charges. Some of this energy is lost as heat due to internal energy. Hence the net energy gained by the charge = the emf - heat lost. This net energy gained per coulomb is called the terminal voltage. This is the voltage across the ends of the battery. Created by Mahesh Shenoy.

## Want to join the conversation?

- Isn't the electric field on both ends of the battery constant? W=qE, when the voltage drops, why can the battery still successfully "push" the electrons to the other end?(2 votes)
- What causes the current to run faster? The lack of resistance in the rest of the circuit? Why then does the internal resistance fail to impede the speed of the current, instead creating heat?(1 vote)
- The internal resistance does indeed impede the speed (wow, it sounds like assonance XD ) of the current carrying charge, but since the negative terminal does the work to PUSH the charge against such resistance as well as the repulsion by terminals, providing the charge enough energy, such that it restores its initial speed and no decrease in speed is visible. and hence due to all this hard work the heat is evolved. Its just like, when your car gets stuck between large rocks or in a swamp, you apply all the force you can probably by hitting that accelerator.... and as you persistently try to get out it, your engine has heated up a good deal. So its kind of analogous to our case, where we try to get over the obstacles and in the process we end up dissipating heat, which is an indication that work is being done against an external force.(3 votes)

- @3:40, i thought +ve charges can't actually move, i.e. protons can't move. isn't it actually only the electrons can move from 1 atom to the next. ?(1 vote)
- There is this IDEA of current flow that the scientists and researchers have been following since the earliest of times and hence, we too have come to accept it as the CONVENTION. This convention created, is to assume the flow of positive charges from the positive terminal to the negative terminal. But what actually happens inside a circuit is the flow of electrons in the opposite orientation as that of our convention. Now one reason for such contradiction could be that by the time the concept of electrons as current carriers was discovered, a lot of research and results had already been plotted assuming the motion of positive charges and hence in order to not disturb the then equilibrium, we simply took it into consideration as a CONVENTIONAL ASSUMPTION.(2 votes)

- The intuition that I applied to understand why the energy lost would be more if the charge moves faster was: since the charge wants to travel faster, it will have to overcome the resistance offered quicker, that is, it should have more energy than before to get past the resistance. Thus, energy lost would be more and
**terminal voltage**would be lesser. Is this a correct way to think?(1 vote) - How comes a voltmeter manages to show you the emf when you connect it to the battery in parallel,however,when you connect a resistor,the shown voltage is influenced by the internal resistance(1 vote)
- In one round around the circuit the charge looses 5v out of 9, so remaining is 4v. Then the second time around the loop does it again loose another 5 to end up in -1??(1 vote)
- So my textbook goes that VOLTAGE DROP --> is the work done per unit charge in carrying the positive test charge through the electrolyte (because it is no longer available for use to us) but according to sir here the work done (in ideal condition ) on a couloumb charge == EMF

I am confused>>>?(1 vote) - Please explain to me why electromotive force is
**independent**of the following factors?

The factors are

1. The shape of electrodes

2. The distance between the electrodes

3. The amount of electrolyte in the cell(1 vote) - Why emf is not force?From the formula of force F=m.a ,after charge move from negative to positive end does its acceleration change?(0 votes)

## Video transcript

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.