Question

What is a transformer? Explain the principle, construction, working and theory of a transformer.

Answer 1

Hint – You can start by defining what a transformer is. Then move on to describe the principle behind the transformer. Then describe the basic setup of a transformer. Then finally write how a transformer works.

An electrical device that can change the A.C. current is known as a transformer.
Principle – A transformer works on the principle of mutual induction. Mutual induction is the phenomenon by which when the amount of magnetic flux linked with a coil changes, an E.M.F. is induced in the neighboring coil.

Construction –
A transformer is made up of a rectangular iron core. Two coils, a primary $(P_{})$ coil with two sides $P_1$ and $P_2$, and a secondary $(S)$ coil with two sides $S_1$ and $S_2$. Both these coils are insulated from the Ferro-magnetic iron core. The source of the alternate current is connected to the primary winding and the output is obtained through the secondary winding which is connected in parallel to a resistance $R$.

Working -
For an ideal transformer, we consider that resistances of the primary and secondary coils are negligible.
Let the $E.M.F.$ of the alternate current supplied by the A.C source be
$E_P=E_0\sin \omega t$
Let’s assume that the primary winding to be a pure inductance, so here $I_p$will lag behind the voltage $E_P$ by ${90}^{\circ }$. Thus the power factor for primary coil becomes,$\cos \varphi =\cos {90}^{\circ }=0$.
Let that the number of turns in primary wire be $N_P$ and secondary wire be $N_S$
According to faraday law, the induced $E.M.F.$ through one turn of both the coils will be the same.
Let the flux through one turn be $\varphi$, the flux through the primary coil be ${\varphi }_p$ and the flux through the secondary coil be ${\varphi }_S$.
So ${\varphi }_p=N_P\varphi$
${\varphi }_S=N_S\varphi$

We also know by Faraday’s law

$E={\displaystyle \frac{d\varphi }{dt}}$
So for the primary coil this equation becomes
$E_S={\displaystyle \frac{d{\varphi }_S}{dt}}$(Equation 1)
And for the secondary coil this equation becomes
$E_P={\displaystyle \frac{d{\varphi }_P}{dt}}$(Equation 2)

Dividing equation 1 by equation 2
${\displaystyle \frac{E_S}{E_P}}={\displaystyle \frac{d{\varphi }_S}{d{\varphi }_P}}={\displaystyle \frac{N_S\varphi }{N_P\varphi }}$
$E_S=E_p{\displaystyle \frac{N_S}{N_p}}$(Equation 3)
We know that
$P=VI$
Here
$P=$Power
$V=$Voltage
$I=$Current
For primary coil this equation becomes
$P_P=E_P{I}_P$(Equation 4)
For secondary coil this equation becomes
$P_S=E_S{I}_S$(Equation 5)
For an ideal transformer no energy is lost, so
$P_p=P_S$
$E_P{I}_P=E_S{I}_S$
$I_S=I_p{\displaystyle \frac{E_p}{E_s}}$$(\because {\displaystyle \frac{E_p}{E_S}}={\displaystyle \frac{N_p}{N_S}})$
$I_S=I_p{\displaystyle \frac{N_p}{N_S}}$

Note – A transformer that increases the A.C. voltage is known as a step up transformer ($N_S>N_p$) and the transformer that decreases the A.C. voltage is known as a step down transformer ($N_S<N_p$). Additionally an iron core is used because it is a ferromagnetic material which helps in increasing the strength of the magnetic field.