Electricity Tutorial 8 - EMF and Internal Resistance


Electromotive Force

Internal Resistance

Measuring Internal Resistance

Resistance of Wires

Electromotive Force

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.

I assume that an early physicist got a shock from a high voltage.  It gave him a fair belt, so he thought it was a force.


Question 1

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 zero 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.


Question 2 

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?



Internal Resistance

All 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.)


Question 3

What is meant by a perfect battery?  Why are real batteries not perfect?



In this circuit the voltmeter reads (very nearly) the emf.


Suppose we now add a load.  We will assume the wires have negligible resistance.


This time we find that the terminal voltage goes down to V.  Since V is less than E, this 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. 


Emf = Useful volts  + Lost volts


In code:

E  = V + v


Question 4 

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.


We 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


              E  =         V                        +                     v


We also know from Ohmís Law that V = IR and v = Ir, so we can write:

E = IR + Ir


E = I(R + r)


Many 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 series circuit.


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.


Worked Example

A battery of emf 12 volts and internal resistance 0.5 W is connected to a 10 W 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

Rtot = R1 + R2 = 10 W + 0.5 W = 10.5 W


Step 3: Now work out the current:

I = V/R = 12 V ų 10.5 W = 1.14 A 


Step 4: work out the voltage across the internal resistor (lost voltage):

v = Ir = 1.14 A ◊ 0.5 W = 0.57 volts 


Step 5: work out the terminal voltage:

Terminal voltage = emf - lost voltage = 12 V - 0.57 V = 11.43 volts


We can of course work out the terminal voltage by working the voltage across the 10 W resistor, assuming there are no losses.


Question 5

A battery has an internal resistance of 0.50 W.  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.  


Measuring Internal Resistance

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:


Resistance of wires (Extension for A-level Students)

We assume that wires are perfect conductors, i.e. have zero resistance.  However wires have a definite value of resistance.   Consider this extension lead commonly found at home:


It can carry a current of 13 A.  However it needs to be fully unwound to do so.  When carrying a heavy current, the cable gets warm.  If it were coiled up, the temperature rise could become excessive, which would be hazardous.  There is energy loss in the wires.


The resistances in this wire are:

These are actual figures using the extension lead in the picture above, although my ohm meter reads to 0.1 W, which represents a 25 % uncertainty.  I have done repeats as well so the data are consistent.


A tumbler drier takes 12 A from the 230 V mains.  We can represent the power supply as a perfect source in series with resistances.  It is connected to the tumbler drier like this:



Question 6

A tumbler drier of rated power 2300 W is connected to a plug using an extension lead that is 15 m long.  The resistance of each conductor is 0.7 W and there are two plugs, each of which has a fuse of resistance 0.1 W.  This is shown in the diagram above.

a.  Show that the current flowing through the tumbler drier is about 9.4 A.

b.  Calculate the voltage across the tumbler drier.

c.  Calculate the power across each of the two fuses.

d.  Calculate the power lost in the cable.

e.  Work out the actual power of the tumbler drier.



The cable itself is 15 m long, which results in a power loss of 13 W m-1 .  The fuse also gets hot.