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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 (Φm) 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 images, that is 4fΦm Wb/s or V.
images
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 = 4fΦmm V.
Let N1 and N2 be the number of turns in primary and secondary.
images
The rms value of induced emf in primary winding is given by
E1 = (4.44fΦm m) × N1 = 4.44fΦm mN = 4.44f BmArN1 (1.1)
where images is the maximum value of flux density having unit Tesla (T) and Ar is the area of cross-section.
Similarly, RMS value of induced emf in secondary winding is
E 2 = (4.44fΦm )x N2 = 4.44fΦmN2 = 4.44f BmArN2 (1.2)
From Equations (1.1) and (1.2), we have
images
i.e., images
where ‘a’ is the turns ratio of the transformer,
i.e., images
Equation (1.3) shows that emf induced per turn in primary and secondary windings are equal.
In an ideal transformer at no load, V1 = E1 and V2 = E2, where V2 is the terminal voltage of the transformer. Equation (1.3) becomes
images
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
E1 = V1 = 5,000 V
E2 = V2 = 500 V
φmax = 7.82 m Wb = 7.82 × 10−3 Wb, f = 50Hz
Let N1 and N2 be the number of turns of the primary and secondary windings, respectively.
Since
E1 = 4.44 f φmN1
images
i.e., images
Again, images
∴ images
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