(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 lo...

(a) Step-down transformer

A step-down transformer has more turns in the primary coil than in the secondary coil, that is, \(N_p > N_s\). When an alternating e.m.f. is applied to the primary coil, an alternating current flows in it and produces a changing magnetic flux in the laminated soft-iron core. This changing flux links the secondary coil and induces an alternating e.m.f. in it by electromagnetic induction.

Since the induced e.m.f. is proportional to the number of turns,

\[\frac{E_s}{E_p}=\frac{N_s}{N_p}\]

As \(N_s < N_p\), \(E_s < E_p\). Hence, the output voltage is lower than the input voltage.

(b) Energy losses and their reduction

1. Copper loss: Heat is produced in the windings because of their resistance. This is minimized by using thick, low-resistance copper wire.
2. Eddy-current loss: Eddy currents are induced in the iron core and produce heating. This is minimized by laminating the core into thin insulated sheets.
3. Hysteresis loss: Energy is dissipated during repeated magnetisation and demagnetisation of the core. This is minimized by using soft iron or a soft magnetic material with low hysteresis loss.

(c)

Given: \(V_p=4400\text{ V}\), \(V_s=220\text{ V}\), lamp power \(P_{out}=60\text{ W}\), and efficiency \(\eta=95\%=0.95\).

(i) Turns ratio

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

Therefore, the ratio of primary turns to secondary turns is:

\[\boxed{N_p:N_s=20:1}\]

(ii) Current taken from the mains

\[\eta=\frac{P_{out}}{P_{in}}\]

\[0.95=\frac{60}{P_{in}}\]

\[P_{in}=\frac{60}{0.95}=63.16\text{ W}\]

\[I_p=\frac{P_{in}}{V_p}=\frac{63.16}{4400}=0.0144\text{ A}\]

\[\boxed{I_p=0.0144\text{ A}=14.4\text{ mA}}\]