# Drive the EMF equation of a transformer

Transformer. Hmm

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## 1 Answer

### Answer :

EMF equation of the transformer

• The EMF induced in the primary winding of

the transformer is given as:

Ep= -Np(dф/dt)..........(1)

• The flux can be given as:

Ф= фmax cosωt

Put value of ф in equation (1), we will get:

Ep= -Np. d/dt(фmax cosωt)

= Np.ωфmax sinωt

• The induced EMF will be maximum, if sin ωt=1

So, Epmax.= Np .ωфmax

• The RMS value of the induced EMF in primary is

Eprms= Epmax./√2

= Np .ωфmax./√2

= Np.(2πf)фmax./√2

Ep= 4.44 f Np фmax Volts

Ep= 4.44 f Np BmaxA Volts ....(2)

• Similarly, for the secondary windings

Es= 4.44 f Ns фmax Volts

Es= 4.44 f Ns BmaxA Volts ......(3)

• The induced EMF in the primary and

secondary windings is in the phase. So divide

eq. (2) by eq. (3):

Ep/Es = Np/Ns = n =Turn ratio or transformer ratio

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