(a) Draw a simple labelled diagram illustrating the principle of a step-down transformer and explain how it works
(b) State three ways by which energy is lost in a transformer and how they can be minimized.
(c) If a transformer is used to light a lamp rated at 60W, 220V from a 4400V a.c. supply, calculate the; (i) ratio of the number of turns of the primary coil to the secondary coil in the transformer (ii) current taken from the main circuit if the efficiency of the transforme is 95%.
Explanation
(a) 
The step-down transformer has more turns in the primary turns than in the secondary turns. It works by electromagnetic induction, the magnetic flux from the primary coil in the iron core links the secondary coil. When an alternating e.m.f \(E_{p}\) is connected to the primary coil, an alternating current flows. This produces corresponding variation of magnetic flux in the iron core which links the secondary coil. An induced e.m.f \(E_{s}\) of the same frequency as \(E_{p}\) is obtained. From Faraday's law, the e.m.f induced in a coil proportional to the rate of change of the magnetic flux linking the coil. So,
\(\frac{E_{s}}{E_{p}} = \frac{N_{s}}{N_{p}}\). \(N_{s}\) - Number of turns in secondary coil; \(N_{p}\) - number of turns in primary coil.
(b) Three ways by which energy is lost and how they can be minimised are:
(1) Energy is lost through Eddy current produced in the soft — iron. This is reduced by laminating the core in thin strips. (2) Heat generation in the copper coils. This is reduced by using copper coils with low resistance. (3). Loss of energy due to reversal of iroes of force. This is reduced by using soft magnetic materials for the core.
(c)(i) Using \(\frac{E_{p}}{E_{s}} = \frac{N_{p}}{N_{s}}\)
\(\frac{4400}{220} = \frac{N_{p}}{N_{s}}\)
The ratio of the primary turns to the secondary turns is 20 : 1.
(ii) \(Efficiency = \frac{\text{power output}}{\text{power input}}\)
\(\frac{95}{100} = \frac{60}{input}\)
\(\text{Power input} = \frac{6000}{95} = 63.2W\)
\(\therefore Current I = \frac{P}{V} = \frac{63.2}{4400}\)
\(I = 0.0144A = 14.4 mA\)