How are the e.m.f. in the primary and secondary coils of a transformer related with the number of turns in these coils?

E_{s} / E_{p} = (N_{s} / N_{p})^{2}

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E_{s} E_{p} = N_{s} N_{p}

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E_{p} / E_{s} = N_{s} / N_{p}

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E_{s} / E_{p} = N_{s} / N_{p}

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Solution

The correct option is D $E_{s} / E_{p} = N_{s} / N_{p}$
In a transformer, source of alternating current is applied to the primary coil. Due to this, the current in the primary coil (called as magnetizing current) produces alternating flux in the core of transformer. This alternating flux gets linked with the secondary coil, and because of the phenomenon of mutual induction an emf gets induced in the secondary coil.
Mutual induction:
An alternating current flowing in a coil produces alternating magnetic field around it. When two or more coils are magnetically linked to each other, then an alternating current flowing through one coil causes an induced emf across the other linked coils. This phenomenon is called as mutual induction.
From these principles, it can be derived that emf divided by number of turns is same for both primary and secondary winding.
That is, $\frac{E 1}{N 1} = \frac{E 2}{N 2}$
where,
E1 is the induced EMF in the primary coil
E2 is the induced EMF in the secondary coil
N1 is the number of turns in the primary coil
N2 is the number of turns in the secondary coil

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E_{p}

E_{s}

n_{p}

n_{s}

\frac{E_{p}}{E_{s}} = \frac{n_{s}}{n_{p}}

\frac{N_{s}}{N_{p}} = 4

200 V A C

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\frac{N_{S}}{N_{P}} = 4

200 V

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