Question

In the transformer, the power of the secondary coil is:
A) Less than the primary coil.
B) More than the primary coil.
C) More in step up and less in step down than the primary coil.
D) More in step down and less in step-up than the primary coil.

Answer 1

We know that whenever there is a change in magnetic flux linked then there is induced emf produced. And also depending upon the number of turns in the primary and secondary coil, an emf is induced.

Complete step by step answer:
Alternating voltages can be changed into the desired value with a help of a device called a Transformer. It works on the principle of mutual induction. In construction, it consists of a closed laminated iron core of low hysteresis loss. The laminations are insulated from one another to minimize eddy current losses. Two coils are wound over the limbs of the core. One of the coils is called the primary and the other secondary. The primary coil connected to the source of alternating voltage and an output voltage appears across the terminals of the secondary coil due to induction.
When an alternating voltage is applied to the primary, current flows through it and the core is magnetized. The alternating magnetic flux produced by this current links the secondary coil and induces an emf in it. As a result, an alternating voltage appears across the secondary coil as output voltage. The output voltage across the secondary coil depends on the input voltage across the primary coil and the ratio of the number of turns in the secondary coil to that in the primary coil (also known as turns ratio).
The induced emf in the secondary with ${{\text{N}}_{\text{s}}}$ turns is ${{{\varepsilon }}_s} = - {N_s}\dfrac{{d\phi }}{{dt}}$
The alternating magnetic flux $\phi$ also includes a back emf in the primary and it is, ${\varepsilon }_p = - {N_p}\dfrac{{d\phi }}{{dt}}$
Where, ${N_p}$ is the number of turns in the primary coil. If ${{\text{V}}_{\text{p}}}$ is the applied voltage across primary and ${{\text{V}}_s}$ is the output voltage across secondary, we have ${\varepsilon _p}{\text{ = }}{{\text{V}}_{\text{p}}}{\text{ and }}{\varepsilon _s}{\text{ = }}{{\text{V}}_s}{\text{ }}$
That is, ${V_s} = - {N_s}\dfrac{{d\phi }}{{dt}}$ and ${V_p} = - {N_p}\dfrac{{d\phi }}{{dt}}$
$\Rightarrow \dfrac{{{V_s}}}{{{V_p}}} = \dfrac{{{N_s}}}{{{N_p}}}$
Thus,
$\dfrac{{{I_p}}}{{{I_s}}} = \dfrac{{{V_s}}}{{{V_p}}} = \dfrac{{{N_s}}}{{{N_p}}}$
For practical transformers, there is always some energy loss and the output power is smaller than the input power. Means, power in secondary coil is more compared to primary coil.
There is no $100\%$ efficiency in the transformer.
$\Rightarrow Efficiency = \dfrac{{{\text{output power}}}}{{{\text{input power}}}}$
In general, power is a product of current and voltage that is, $P = VI$
$\Rightarrow \eta = \dfrac{{{V_S}{I_S}}}{{{V_P}{I_P}}}$

$\therefore$ The correct option is (A).

Note:
Alternating voltages can be changed into the desired value with a help of a device called a Transformer.
Mutual induction is the working principle of the Transformer.
If the secondary coil has a greater number of the turns than primary coil, $\left( {{N_s} > {N_p}} \right)$, the output voltage across secondary is more than the input voltage across primary $\left( {{{\text{V}}_{\text{s}}}{\text{ > }}{{\text{V}}_{\text{p}}}} \right)$. This type of transformer is called a step-up transformer.
If the secondary coil has less turns than the primary coil, $\left( {{N_s} < {N_p}} \right)$, then the output voltage across the secondary is less than input voltage across the primary, $\left( {{{\text{V}}_{\text{s}}}{\text{ < }}{{\text{V}}_{\text{p}}}} \right)$. This type of transformer is called a step-down transformer.