Question

The primary and secondary coils of a transformer have 50 and 1500 number of turns respectively. If the magnetic flux $\phi $ is linked with the primary coil is given by $\phi = {\phi _0} + 4t$ where $\phi $ is in Weber’s, t is time in seconds, and ${\phi _0}$ is a constant, the output voltage across the secondary coil is;
A) 30volts
B) 90volts
C) 120volts
D) 220volts

Answer 1

Hint: 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 solution:
Induced emf, $e = - \dfrac{{d\phi }}{{dt}}$
Where $\dfrac{{d\phi }}{{dt}}$ is the rate of change of magnetic flux linked.
Given,
Number of turns in primary coil, ${N_p} = 50$
Number of turns in secondary coil, ${N_s} = 1500$
Magnetic flux linked in primary coil, $\phi = {\phi _0} + 4t$
Calculate the output voltage across the secondary coil, ${V_s} = ?$
From Faraday’s law, we know that
The induced emf produced in a coil is equal to the negative rate of change of magnetic flux linked with it.
$e = - \dfrac{{d\phi }}{{dt}}$
Emf induced(can also be represented as voltage V) in primary coil is given by,
${V_p} = - \dfrac{{d\phi }}{{dt}} = \left| {\dfrac{{d\phi }}{{dt}}} \right|$……………………….. (1)
$\left| {\dfrac{{d\phi }}{{dt}}} \right|$ represents only magnitude.
We have, $\phi = {\phi _0} + 4t$ substitute this value in equation (1), we get
${V_p} = \dfrac{{d\left( {{\phi _0} + 4t} \right)}}{{dt}}$
After differentiation we get,
${V_p} = 4volts$
Then, induced emf or voltage in a secondary coil is given by,
$\dfrac{{{N_s}}}{{{N_p}}} = \dfrac{{{V_s}}}{{{V_p}}}$
Then, ${V_s} = \dfrac{{{N_s}{V_p}}}{{{N_p}}}$
Substituting the values,
${V_s} = \dfrac{{1500 \times 4}}{{50}} = 120volts$

$\therefore$ The output voltage across the secondary coil is 120V. Hence, correct option is (C).

Additional information:
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 an output voltage. The output voltage across the secondary coil depends on the input voltage across the primary coil and the ratio of a 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
${V_p} = - {N_p}\dfrac{{d\phi }}{{dt}}$
$\therefore \dfrac{{{V_s}}}{{{V_p}}} = \dfrac{{{N_s}}}{{{N_p}}}$

Note:
A transformer is a device used to change alternating voltages to any desired value.
It works on the principle of mutual induction.
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 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 step-down transformer.