Question

The primary and secondary coils of a transformer have $50$ and \[1500\] turns respectively. If the magnetic flux linked with the primary coil is given by \[\phi = {\phi _0} + 4t,\;\] where $\phi $ is in webers, t is time in seconds and \[{\phi _0}\] ​ is a constant, the output voltage across the secondary coil is:
A. 30 volts
B. 90 volts
C. 120 volts
D. 220 volts

Answer 1

Hint: To solve the above given problem we should know how we can calculate the induced emf in any coil, if the flux is given. Principle of transformer is based on the principle of mutual induction and as we all know that the Faraday's law states that the Induced EMF is equal to the rate of change of magnetic flux.

Complete answer:
So, from the faraday’s law, we get
Since the rate of change of magnetic flux is the induced emf,
So, Induced emf= \[e = \dfrac{{d\phi }}{{dt}}\], where $e = $induced emf, $\phi = $flux associated with the coil.
So, for the transformer we can get the induced emf in the primary coil from the given magnetic flux linked with the primary coil, that is \[\phi = {\phi _0} + 4t,\;\]
So, the induced emf in the primary coil= ${e_p} = \dfrac{{d\phi }}{{dt}} = \dfrac{d}{{dt}}({\phi _0} + 4t)$
$ \Rightarrow {e_p} = (0 + 4)$
$ \Rightarrow {e_p} = 4V$--------equation (1)
Hence the induced voltage in the primary coil is ${e_p} = 4V$.
In the question number of the turns in the primary coil and the secondary coil is given, So from this we can get the transformation ratio, $k$
So, transformation ratio, $k = \dfrac{{{N_s}}}{{{N_p}}}$
$ \Rightarrow k = \dfrac{{{N_s}}}{{{N_p}}} = \dfrac{{1500}}{{50}}$ (putting the values ${N_s} = 1500$ and ${N_p} = 50$)
$ \Rightarrow k = 30$-----equation (2)
Also, the relation between the transformation ratio and the induced emfs in the primary and secondary is as follows,
$ \Rightarrow k = \dfrac{{{e_s}}}{{{e_p}}}$
Putting the value of the ${e_s}$ from equation (1), we get
$ \Rightarrow k = \dfrac{{{e_s}}}{4}$-----equation (3)
Now from equation (2) and equation (3), we get
$ \Rightarrow 30 = \dfrac{{{e_s}}}{4}$
$ \Rightarrow {e_p} = 120V$

So, the correct answer is “Option C”.

Note:
Since Induced EMF is equal to the rate of change of magnetic flux, and the magnetic flux is equal to Magnetic field strength multiplied by the Area =$BA$.
Therefore,\[Induced{\text{ }}EMF\; = \dfrac{{\left( {change{\text{ }}in{\text{ }}Magnetic{\text{ }}Flux{\text{ }}Density{\text{ }}x{\text{ }}Area} \right)}}{{change{\text{ }}in{\text{ }}Time}}\]
Therefore, \[Induced{\text{ }}EMF\; = \dfrac{{(B\pi {r^2}n)}}{t}\]