Question
Explain the working of a transformer with a neat, labelled diagram.
✓
Answer
Working of Transformer:
$(1)$ The transformer works on the principle of Faraday's law of electromagnetic induction and mutual induction.
$(2)$ There are usually two coils primary coil and secondary coil on the transformer core.
$(3)$ The core laminations are joined in the form of strips.
$(4)$ The two coils have high mutual inductance.
$(5)$ When an alternating current pass through the primary coil it creates a varying magnetic flux.
$(6)$ As per faraday's law of electromagnetic induction, this change in magnetic flux induces an emf $($electromotive force$)$ in the secondary coil which is linked to the core having a primary coil. This is mutual induction.
$(7)$ Let $\phi$ be the magnetic flux linked per turn with both the coils at an instant $t . N_p$ and $N_s$ be the number of turns
in the primary and secondary coil respectively.
Then induced emf in primary and secondary coil will be,
$e_p=-\frac{d \phi_p}{d t}=-N_p \frac{d \phi}{d t}$ and $e_s=-\frac{d \phi_s}{d t}=-N_s \frac{d \phi}{d t}$
$\therefore \frac{e_s}{e_p}=\frac{N_s}{N_p} \ldots \ldots . .(1)$
$(8)$ The ratio $\frac{N_s}{N_p}$ is called turn ratio $($transformer ratio$)$ of the transformer.
$(9)$ For ideal transformer,
input power $=$ output power
$\therefore \quad e_p i_p=e_s i_s$
$\therefore \frac{e_s}{e_p}=\frac{i_p}{i_s}...(2)$
$(10)$ From $(1)$ and $(2),$
we get, $\frac{e_s}{e_p}=\frac{i_p}{i_s}=\frac{N_s}{N_p}$
When $N_s > N_p$ then $e_s > e_p ($step up transformer$)$ and $i_p > i_s$.
When $N_s < N_p$ then $e_s < e_p ($step down transformer$)$ and $i_p < i_s$.
$(1)$ The transformer works on the principle of Faraday's law of electromagnetic induction and mutual induction.
$(2)$ There are usually two coils primary coil and secondary coil on the transformer core.
$(3)$ The core laminations are joined in the form of strips.
$(4)$ The two coils have high mutual inductance.
$(5)$ When an alternating current pass through the primary coil it creates a varying magnetic flux.
$(6)$ As per faraday's law of electromagnetic induction, this change in magnetic flux induces an emf $($electromotive force$)$ in the secondary coil which is linked to the core having a primary coil. This is mutual induction.
$(7)$ Let $\phi$ be the magnetic flux linked per turn with both the coils at an instant $t . N_p$ and $N_s$ be the number of turns
in the primary and secondary coil respectively.
Then induced emf in primary and secondary coil will be,
$e_p=-\frac{d \phi_p}{d t}=-N_p \frac{d \phi}{d t}$ and $e_s=-\frac{d \phi_s}{d t}=-N_s \frac{d \phi}{d t}$
$\therefore \frac{e_s}{e_p}=\frac{N_s}{N_p} \ldots \ldots . .(1)$
$(8)$ The ratio $\frac{N_s}{N_p}$ is called turn ratio $($transformer ratio$)$ of the transformer.
$(9)$ For ideal transformer,
input power $=$ output power
$\therefore \quad e_p i_p=e_s i_s$
$\therefore \frac{e_s}{e_p}=\frac{i_p}{i_s}...(2)$
$(10)$ From $(1)$ and $(2),$
we get, $\frac{e_s}{e_p}=\frac{i_p}{i_s}=\frac{N_s}{N_p}$
When $N_s > N_p$ then $e_s > e_p ($step up transformer$)$ and $i_p > i_s$.
When $N_s < N_p$ then $e_s < e_p ($step down transformer$)$ and $i_p < i_s$.
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