(a) With the aid of a simple diagram, explain how a step down transformer works. (b)(i) State three ways by which energy is lost in a transformer (ii) Menti...

Answer Details

(a) How a step-down transformer works

A transformer consists of two coils, a primary (input) coil and a secondary (output) coil, wound on a common laminated soft-iron core. In a step-down transformer the primary coil has more turns than the secondary coil (\(N_p > N_s\)).

When an alternating voltage is applied to the primary coil, the alternating current flowing in it sets up a continuously changing magnetic flux in the soft-iron core. The core carries this changing flux round to the secondary coil, so the secondary is linked by the same changing flux. By electromagnetic induction, the changing flux induces an alternating e.m.f. in the secondary coil. Because the secondary has fewer turns than the primary, the e.m.f. induced in it is smaller than the applied primary voltage, i.e. the voltage is stepped down (while the output current is correspondingly stepped up).

The simple diagram below shows the arrangement:

(b)(i) Three ways energy is lost in a transformer

1. Eddy-current loss in the iron core.
2. Hysteresis loss in the iron core.
3. Copper (resistance/heating) loss in the windings.

(b)(ii) How each loss is minimized

Energy loss	How it is minimized
Eddy-current loss in the core	Use a laminated core (thin iron sheets insulated from one another) to break up the induced eddy-current paths.
Hysteresis loss in the core	Make the core of soft iron, which is easily magnetized and demagnetized, so little energy is used up in each magnetization cycle.
Copper (heating) loss in the windings	Use thick, low-resistance copper wire for the coils so that \(I^2R\) heating is reduced.

(c) Calculations

Efficiency \(=95\%=0.95\); lamp (secondary side) rated \(P_{out}=60\,\text{W},\; V_s=220\,\text{V}\); supply \(V_p=4400\,\text{V}\).

(i) Ratio of primary turns to secondary turns.

For a transformer the turns ratio equals the voltage ratio:

\[ \frac{N_p}{N_s}=\frac{V_p}{V_s}=\frac{4400}{220}=20 \]

Therefore \(\;N_p:N_s = \mathbf{20:1}.\)

(ii) Current taken from the main circuit.

Using efficiency \(=\dfrac{\text{output power}}{\text{input power}}=\dfrac{V_s I_s}{V_p I_p}\):

\[ 0.95=\frac{P_{out}}{V_p\,I_p}=\frac{60}{4400\times I_p} \]\[ I_p=\frac{60}{0.95\times 4400}=\frac{60}{4180}=0.0144\,\text{A} \]

The current taken from the mains is \(I_p \approx \mathbf{0.014\,A}\;(1.44\times10^{-2}\,\text{A},\ \text{about }14.4\,\text{mA}).\)