# Emf Equation Of A Transformer

Figure 1.22 shows the representation of alternating flux, varying sinusoidally, which increases from its zero value to maximum value (*Φ _{m}*) in one-quarter of the cycle, that is in one-fourth of a second where

*f*is the frequency of AC input in hertz.

The average rate of change of flux is given by , that is 4*fΦ _{m}* Wb/s or V.

**Figure 1.22** Representation of Alternating Flux

This rate of change of flux per turn is the induced emf in V.

Therefore, average emf/turn = 4*fΦ _{m}*m V.

Let *N*_{1} and *N*_{2} be the number of turns in primary and secondary.

The rms value of induced emf in primary winding is given by

*E*_{1} = (4.44*fΦ _{m}* m) ×

*N*

_{1}= 4.44

*fΦ*mN = 4.44

_{m}*f B*

_{m}A_{r}N_{1}(1.1)

where is the maximum value of flux density having unit Tesla (T) and *A _{r}* is the area of cross-section.

Similarly, RMS value of induced emf in secondary winding is

*E* _{2} = (4.44*f**Φ _{m}* )x

*N*

_{2}= 4.44

*fΦ*

_{m}*N*

_{2}= 4.44

*f*

*B*

_{m}A_{r}N_{2}(1.2)

From Equations (1.1) and (1.2), we have

i.e.,

where ‘*a*’ is the turns ratio of the transformer,

i.e.,

Equation (1.3) shows that emf induced per turn in primary and secondary windings are equal.

In an ideal transformer at no load, *V*_{1} = *E*_{1} and *V*_{2} = *E*_{2}, where *V*_{2} is the terminal voltage of the transformer. Equation (1.3) becomes

**Example 1.1** The voltage ratio of a single-phase, 50 Hz transformer is 5,000/500 V at no load. Calculate the number of turns in each winding if the maximum value of the flux in the core is 7.82 mWb.

**Solution**

Here

*E*_{1} = *V*_{1} = 5,000 V

*E*_{2} = *V*_{2} = 500 V

φ_{max} = 7.82 m Wb = 7.82 × 10^{−3} Wb, *f* = 50Hz

Let *N*_{1} and *N*_{2} be the number of turns of the primary and secondary windings, respectively.

Since

*E*_{1} = 4.44 *f* φ_{m}*N*_{1}

i.e.,

Again,

∴