In any circuit there are components that put energy in to the circuit and components that take energy out. From now on, we will say that any device putting energy into a circuit is providing an electo-motive force (emf) and any device taking it out has a potential difference (pd) across it.
Both emf and pd are measured in volts, V, as they describe how much energy is put in or taken out per coulombof charge passing through that section of the circuit.
The best way to think of them is:
Emf - is the amount of energy of any form that is changed into electrical energy per coulomb of charge.
pd - is the amount of electrical energy that is changed into other forms of energy per coulomb of charge.
Sources of emf:
Cell, battery (a combination of cells), solar cell, generator, dynamo, thermocouple.
Cells and batteries are not perfect (what is - apart from the moment your last exam finishes, of course?). Use them for a while and you will notice they get hot.
Where is the heat energy coming from?
It's from the current moving through the inside of the cell. The resistance inside the cell turns some of the electrical energy it produced to heat energy as the electrons move through it.
It is easy to explain if you imagine that each cell is perfect except that for some bizarre reason (probably part of a plot to take over the world, masterminded by Dr Evil) the manufacturers put a resistor in series with the cell inside the casing.
Therefore, inside the cell, energy is put into the circuit by the cell (the emf) but some of this energy is taken outof the circuit by the internal resistor (a pd).
So the pd available to the rest of the circuit (the external circuit, as some questions may refer to it) is the emf minus the pd lost inside the cell:
V = E - Ir
Where:
V = pd across the external circuit (V)
E = emf of the cell (V)
I = current through the cell (A)
r = value of the internal resistance (Ω)
(Ir = the p.d. across the internal resistor)
Note: V is sometimes called the terminal pd as it is the pd across the terminals of the cell
Example 1:
What is the terminal p.d. for a cell of emf 2V and internal resistance 1 ohm when it is connected to a 9 ohm resistor?
Answer:
Just pretend the internal resistance is one of the normal resistors in the circuit. Draw it in the circuit diagram next to the cell so that all the current that goes through the cell also goes through the resistor.
To find V, the terminal pd (or the voltage available to the external circuit), calculate the current, I, for the whole circuit:
Note: VT and RT are the voltage and resistance for the whole circuit, including external and internal resistance.
Therefore, the 9Ω resistor gets V = IR = 0.2 x 9 = 1.8V
So this 2V emf cell actually supplies 1.8V to the external circuit.
Example 2:
Now, swap the 9Ω resistor in the last example for a 1Ω resistor.
Answer:
Find V, the terminal pd, using the same method again:
Now the 2V emf cell only supplies 1V to the external circuit!!! The other 1 V is lost making the cell hot. Not very efficient.
Note: You need to consider the internal resistance when deciding if a cell is appropriate to use in a particular circuit. For the greatest efficiency the external resistance must be much greater than the internal resistance of the cell. However, for the maximum power to be delivered to the external circuit the internal resistance must be equal to the resistance of the external circuit, although the cell will only be 50% efficient.
Power supplies which deliver low voltages and higher currents, like a car battery, need to have a low internal resistance, as shown above. High-voltage power supplies that produce thousands of volts must have an extremely high internal resistance to limit the current that would flow if there was an accidental short-circuit.
As V = E - Ir, if you plot a graph of terminal pd, V, against current, I, the gradient of the graph will be equal to the internal resistance of the cell. (negative because the graph slopes down)
By recording values of current and terminal pd as the external resistance changes you can plot the graph and find the internal resistance and the emf of the cell.
If there is more than one cell in series the internal resistances of the cells must be added.