This video is about internal resistance. Internal resistance refers to the resistance inside of a
battery. No battery is ideal, and all contain at least a little resistance inside of them.
In IB, we use this symbol to show internal resistance. So we've got this battery. This
dotted line represents the kind of barrier of the battery. Anything inside of that dotted
line is inside of the battery, which we symbolize with those cell symbols, and
it also contains an internal resistance which we symbolize just like any other resistor with that
resistor symbol with the lower case r. We call the voltage that the battery delivers to the rest of
the circuit the "terminal voltage" so we'll talk about that more in just a moment, and the variable
for internal resistance is lowercase r. So before this we were using capital R for resistances. For
internal resistance specifically we use lowercase r. In all previous problems the voltage we've been
talking about has really been the terminal voltage of the battery, just the voltage that the battery
gives to the rest of the circuit excluding the voltage drop that occurs within the battery
itself. Here's an example problem with internal resistance: what is the terminal voltage of this
battery? Wo we can see that this battery contains this voltage of 10 and the internal resistance is
2 ohms and it also has a 5 and 3 ohm resistor outside of the battery, so I can see that the
equivalent resistance in the circuit is going to be equal to 10 ohms because the internal
resistance behaves like any other resistor in the circuit. It just happens to be inside the battery,
so it's also in series with those two other resistors. So those add together to make 10 ohms
using v = ir. I can see that the current in the circuit is 1 amp, so if this current is 1 amp
each of these resistors is experiencing one amp of current, and I can use that to calculate the
voltage drop in each resistor. So in the 3 ohm resistor there's a voltage drop of 3, and in the
5 ohm resistor there's a voltage drop of 5 using v = ir, and inside of the battery there's a voltage
drop of 2, so what that means is that if we just pay attention to what's happening outside of the
battery, the battery is delivering eight volts to the outside of the circuit. So we would say that
the terminal voltage of this battery is 8 volts. From one terminal of the battery to the other
it delivers 8 volts. There's an eight volt drop in electrical potential. In internal
resistance problems we refer to the voltage source inside the battery as the electromotive force,
or "emf." This can be confusing because emf is actually not a force, it's a voltage causing
electrons to move. It's measured in volts and can be used as the voltage in the circuit when you are
considering the internal resistance. So when we say emf we mean that full total voltage inside of
the battery, whereas when we just use the word voltage we're usually really referring to the
terminal voltage. So there's a difference here: emf refers to the total voltage experienced by
everything in the circuit including the internal resistance, whereas voltage is referring to the
terminal voltage just the voltage drop that's delivered outside of the battery.
The IB symbol for emf is epsilon. That's a lower case epsilon specifically. It's a Greek letter
so we would say in this problem here that the emf epsilon is equal to 10 volts,
but the voltage or the terminal voltage is equal to 8 volts, so that's the difference between these
two things. So going back to this problem here I know that the current flowing through this is
1 amp, so that means that the voltage drop inside of this battery is 2 volts, so I can see here that
the emf is going to be equal to that voltage drop inside of the battery plus the terminal voltage of
the battery, so basically whatever voltage drop happens outside of the battery plus whatever
voltage drop happens inside the battery, those two things added together are equal to the emf
of the battery. So in this case the outer voltage drop is 8 and the inner voltage drop is 2,
so altogether the emf is clearly 10 volts. If I rearrange this equation and call v the
current times the equivalent resistance outside of the battery I end up with this equation. The
emf is equal to the current multiplied by the sum of the resistance in the circuit plus the
internal resistance of the battery. This equation appears in your IB data booklet. So this is the
information that you're given in your data booklet about emf and internal resistance. In general
going back to this first equation, if I rearrange this equation to look like this and then create a
graph of the voltage in the circuit on the y-axis and the current in the circuit on the x-axis,
I can see that this equation fits y = mx + b where m is the slope and b is the y-intercept.
So this is going to yield an important pattern that's going to show up on a few different IB
problems, where if you make a graph of the voltage on the y-axis and the current on the x-axis of a
circuit with internal resistance, the y-intercept of the graph is going to be the emf of the battery
and the slope of the graph is going to be the negative internal resistance of the battery.
So if you're ever asked a question about which graph shows a greater internal resistance,
it's going to be the one with a steeper slope and the one with a greater emf is going to be
the one with the greater y-intercept. So basically using a voltage current graph
you can find the emf using the y-intercept and the internal resistance using the slope.