Magnetic Fields Tutorial 7 - Transformers

We can use a magnetic field to induce a voltage in two ways:


1. Relative movement.  The size of the voltage depends on:

2.  Changing a magnetic field.  We donít have to make the magnetic field move.  If we turn the current on or off, there is a change in the magnetic field, and that induces a voltage in a second unconnected coil.  This is called the transformer effect or mutual induction.


Electric locomotives take their power from a variety of systems.  In most modern electrified railways the power is supplied to the locomotive by a 25 000 V overhead line at 50 Hz ac.  However the motors work at about 1500 V, so there needs to be a way of reducing the voltage.  This could be done using a resistor, but it would be extremely wasteful.  So it's done with a transformer.



The transformer is a machine that is simplicity itself.  It consists of:


The two coils are electrically completely different circuits.  Either of the coils can act as a primary. 


The laminated core is made up of layers of soft iron separated by an insulating layer of varnish or glue.  This reduces losses from eddy currents.  Soft iron is NOT soft like putty; it is heavy and hard.  However "soft" means that it loses its magnetism immediately the current is turned off.  Therefore the magnetic field can change forwards to backwards as the current changes.


The ratio of the input voltage to the output voltage is the same as the ratio of the number of turns on the primary to the number of turns on the secondary.  We can write this as:


Number of turns on the primary    = Primary Voltage

Number of turns on the secondary    Secondary voltage


In Physics code:


N1 = V1

N2    V2


If N1 is greater than N2, we have a step-down transformer, because the voltage is reduced.  A step-up transformer increases the voltage.


If a transformer is 100 % efficient (and it nearly is) we can say that:


 power in = power out


V1I1 = V2I2


Therefore we can say that when the voltage is lower, the current is bigger.  We can rewrite the transformer equation in terms of current to give us:


N1 = I2

N2    I1


In practice, the transformer is about 97 % efficient.  When a large transformer is transferring a lot of energy, even 3 % losses produce a fair amount of heat.  Therefore the transformer is cooled with oil which is pumped to heat exchangers.



The picture above shows the huge transformers used at a power station.  You can see the massive cables coming out from the generator (they are like pipes).  In this power station, the generator voltage is 15 000 V, and this transformer steps the voltage up to 132 000 V.


Question 1

A power station generator generates 500 MW at a voltage of 15 000 V.

(a) What is the current?

(b) The voltage is stepped up to 275 000 volts.  Assuming that the transformer is 100 % efficient, what is the current in the secondary?

(c) What is the turns ratio in the transformer?




What are the sources of inefficiency in a transformer?

A few points to note:


Hysteresis, please, NOT hysteria!


Transformers can only work with alternating current; they cannot work with direct current. 


If an electric locomotive passes from 25 000 V ac to 3000 V dc overhead power line (as they do when passing from France to Belgium), a different solution is needed to step down the voltage.  The 1500 V motors are put in series.


Question 2

A transformer has a primary of 3600 turns and a secondary of 150 turns.  It takes 1.5 amps from the 240 V mains. 

(a) What is the turns ratio?

(b) What is the output voltage and current?



In the exam you may well be asked to discuss the causes of inefficiency in a transformer.  Question 3 asks you to do that.


Question 3

Transformers are very efficient machines, but are not 100 % efficient.  The best efficiency for a large power station transformer is about 97 %.  Outline the reasons for this and discuss whether there is a limit to the power input and output of such a transformer.  

Question 4

A large industrial transformer steps an input voltage of 132 kV to an output voltage of 1000 V to provide power to an electric arc furnace.  The furnace takes a current of 40 000 amps.  The transformer is 95 % efficient.

(a) Calculate the power taken by the furnace.

(b) Calculate the input current.

(c) Calculate the power lost when the furnace is running.

(d) Suggest how the lost power is removed from the transformer to avoid damage due to overheating.