Description

If the primary coil is connected to an A.C source, the A.C. passing through the coil sets up a changing/alternating magnetic flux in the soft iron core.
The secondary coil is• linked by the changing flux which induces an emf in the secondary. Since the number of turns in the secondary is less than that in the primary, the induced emf is less than the emf in the primary. For the same reason the A.C output is less than the A.C in the primary.

(b)(i) Any three correct losses: e.g.
-copperlheatljoules losses;
-Eddy current losses;
-Flux leakages;
-hysteresis losses.
(ii) Any three methods of minimizing losses correctly paired with the type of lose

Loses	Minimizing Metods
Heat losses/lose in wire/joules losses	Low resistance Coil
Eddy current losses	Lamination of force
Flux leakages	Efficient core design
Hysteresis	Core made of soft magnetic material or soft iron

 (c) (i) Ns= Es
             Np   Ep
             =  220
               4400
        (ii)         Efficiency = Power output
                                            Power input
                            = Vs Is
                                 NpIp

           95/100   = 60/4400 x Ip
        Ip = 100/96 x 60/4400 x Ip
        Ip = 0.014A = [14mA]

This question on the transformer was attempted by many candidates. A good number of the candidates performed fairly well.-

Some of the diagrams presented in part (a) did not clearly show all the necessary components of a step down transformer. Many candidates could not explain correctly, the mode of operation of the transformer.
In part (b) the various ways by which energy is lost in a transformer were correctly listed by many candidates. However, most of the candidates could not correctly state the method by which these energy losses could be minimized.

The candidates’ reaction to the numerical problem in part (c) was quite satisfactory. Most of the candidates were able to correctly calculate the turns ratio and the efficiency of a transformer.