Electricity Tutorial 8 - EMF and Internal Resistance
Batteries (or more strictly speaking cells) convert chemical energy into electrical energy. Generators turn kinetic energy into electrical energy. In doing so, they keep the negative terminal with an excess of electrons and the positive terminal with a deficiency of electrons. A battery does a job of work in pumping the electrons around the circuit. Positive charges do not move.
The early day physicists got it wrong when they said that electric current flows from positive to negative. They didn't know about electrons. When the mistake was discovered, they decided to stick to the positive to negative, so all conventional current flows from positive to negative.
A battery is said to produce Emf
(electromotive force) which is defined as
the energy converted into electrical energy when
unit charge passes through the source.
This is similar to the definition for potential difference that we saw before, except that it describes the conversion to electrical energy, rather than the conversion from electrical energy. It represents the total energy that can be supplied to a circuit. EMF is a voltage.
Note that EMF is energy per unit charge, NOT a force, which can lead to confusion. Watch out for this particular bear trap.
What is the difference between emf and potential difference?
A good working definition of emf is the open circuit terminal voltage of the battery, i.e. when there is no current flowing. Although the old text books had a complex method for measuring emf using a metre bridge, nowadays a digital multimeter will give you a good reading as it takes a very small current indeed.
The energy supplied to a circuit by a battery is given by:
No circuit at all is 100 % efficient. Some energy is dissipated in the wires, or even in the battery itself.
A battery converts 13 000 J of chemical energy into electrical energy. It does so by giving a current of 0.5 A for 2 hours. What is its emf?
batteries and generators dissipate heat internally when giving out a current,
due to internal resistance.
A perfect battery has no
internal resistance, but unfortunately there is no such thing as a perfect
battery. Nickel-Cadmium and
Lead-Acid batteries have very low internal resistance, and we can regard these
as almost perfect. These batteries
can provide very high currents.
Suppose we connect a cell to a high resistance voltmeter. (A perfect voltmeter has infinite resistance. A digital multimeter has a very high resistance, so needs a tiny current; it is almost perfect. An ordinary moving coil voltmeter has a relatively low resistance, so it takes a small but appreciable current.)
What is meant by a perfect battery? Why are real batteries not perfect?
Suppose we now add a load. We will assume the wires have negligible resistance.
time we find that the terminal voltage goes down to
Since V is less than
tells us that not all of the voltage is being transferred to the outside
circuit; some is lost due to the internal resistance which heats the battery up.
E = V + v
|How is this statement consistent with Kirchhoff II?|
So we can represent the circuit as:
So our cell is now a perfect battery in series with an internal resistor, r. You cannot open up the battery to find the internal resistor; it is part and parcel of the battery.
can now treat this as a simple series circuit and we know that the current, I,
will be the same throughout the circuit. We also know the voltages in a
series circuit add up to the battery voltage.
Emf = voltage across R + voltage across the internal resistance
We also know from Ohm’s Law that V = IR and v = Ir, so we can write:
IR + Ir
ÞE = I(R + r)
students panic at the sight of internal resistance problems.
All you have to do is turn the cell with the internal resistance into a perfect
battery in series with its
internal resistor, and treat it as a simple
look at a worked example. It will show you a problem solving strategy that
hopefully will take the anxiety that these problems cause many students.
Let's look at a worked example. It will show you a problem solving strategy that hopefully will take the anxiety that these problems cause many students.
A battery of emf 12 volts and internal resistance 0.5 ohms is connected to a 10 ohm resistor. What is the current and what is the terminal voltage of the battery under load?
Step 1: we treat the circuit as a perfect battery in series with an internal resistor. The circuit becomes:
Step 2: Work out the total resistance
R tot = R1 + R2 = 10 ohms + 0.5 ohms = 10.5 ohms
Step 3: Now work out the current:
I = V/R = 12 ÷ 10.5 = 1.14 A
4: work out the voltage across the internal resistor (lost voltage):
v = Ir = 1.14 amps × 0.5 ohms = 0.57 volts
5: work out the terminal voltage:
Terminal voltage = emf - lost voltage = 12 - 0.57 = 11.43 volts
|We can of course work out the terminal voltage by working the voltage across the 10 ohm resistor, assuming there are no losses.|
|A battery has an internal resistance of 0.5 ohms. The battery has an emf of 1.52 V. When it is connected to a resistor, the terminal voltage falls to 1.45 V. What current is flowing. What is the value of the resistor?|
Click HERE to see important definitions that you will need for the exam.
To measure the internal resistance, we set up the circuit like this:
We change the value of the variable resistance. If the resistance is zero, we get a short circuit, so the the current will be at the maximum. The voltage will be zero. When the variable resistance is at its highest, the voltage will be less than the emf. We then extrapolate the graph.
In experiments to determine the internal resistance, we get a graph like this:
The graph is a straight line, of the form y = mx + c. We can make the equation for internal resistance V = -rI + E. There are three features on the graph that are useful:
The intercept on the y-axis tells as the emf.
The intercept on the x-axis tells us the maximum current the cell can deliver when the p.d. is zero, i.e. a dead short circuit.
The negative gradient tells us the internal resistance.