Write the working procedure of a transformer.

(N/A) When an alternating voltage is applied to the primary coil,the resulting current produces an alternating magnetic flux which links the secondary coil and induces an electromotive force (emf) in it.
The value of the induced emf depends on the number of turns in the secondary coil.
Let us consider an ideal transformer in which the primary coil has negligible resistance and all the magnetic flux in the core links both the primary and secondary windings.
If no load is connected to the secondary coil and $N_{p}$ and $N_{s}$ are the number of turns in the primary and secondary coils respectively,then the induced emf in the primary coil is:
$\varepsilon_{p} = -N_{p} \frac{d \phi}{d t} \quad \dots (1)$
And the induced emf in the secondary coil is:
$\varepsilon_{s} = -N_{s} \frac{d \phi}{d t} \quad \dots (2)$
Since $\varepsilon_{p} = V_{p}$ and $\varepsilon_{s} = V_{s}$,where $V_{p}$ and $V_{s}$ are the voltages across the primary and secondary coils respectively,equations $(1)$ and $(2)$ can be written as:
$V_{p} = -N_{p} \frac{d \phi}{d t}$ and $V_{s} = -N_{s} \frac{d \phi}{d t}$
Therefore,$\frac{V_{s}}{V_{p}} = \frac{N_{s}}{N_{p}}$.
The above relation is based on three assumptions:
$(1)$ The primary resistance and current are small.
$(2)$ The same magnetic flux links both the primary and secondary coils as very little flux escapes from the core.
$(3)$ The secondary current is small.
The ratio of the number of turns in the secondary coil $N_{s}$ to the number of turns in the primary coil $N_{p}$ is called the transformation ratio $\lambda$.
If the secondary coil has a greater number of turns than the primary $(N_{s} > N_{p})$,the voltage is stepped up. This is called a step-up transformer.
If the secondary coil has fewer turns than the primary $(N_{s} < N_{p})$,then $V_{s} < V_{p}$. This is called a step-down transformer,where the voltage is reduced.