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- Construction of Transformer
- EMF Equation of Transformer
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- Introduction to 3-Phase Synchronous Machines
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- Losses and Efficiency of 3-Phase Alternator
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EMF Equation of Transformer
For electrical transformer, the EMF equation is a mathematical expression used to find the magnitude of induced EMF in the windings of the transformer.
Consider a transformer as shown in the figure. If N1 and N2 are the number of turns in primary and secondary windings. When we apply an alternating voltage V1 of frequency f to the primary winding, an alternating magnetic flux $\phi$ is produced by the primary winding in the core.
If we assume sinusoidal AC voltage, then the magnetic flux can be given by,
$$\mathrm{\mathit{\phi }\:=\:\phi _{m}\:\mathrm{sin}\:\mathit{\omega t}\:\cdot \cdot \cdot (1)}$$
Now, according to principle of electromagnetic induction, the instantaneous value of EMF e1 induced in the primary winding is given by,
$$\mathrm{\mathit{e_{\mathrm{1}}}\:=\:\mathit{-N_{\mathrm{1}}}\frac{\mathit{d\phi }}{\mathit{dt}}}$$
$$\mathrm{\Rightarrow \mathit{e_{\mathrm{1}}}\:=\:\mathit{-N_{\mathrm{1}}}\frac{\mathit{d}}{\mathit{dt}}\left ( \phi _{m}\: \mathrm{sin}\:\mathit{\omega t}\right )}$$
$$\mathrm{\Rightarrow \mathit{e_{\mathrm{1}}}\:=\:\mathit{-N_{\mathrm{1}}}\:\mathit{\omega \phi \:cos\:\omega t}}$$
$$\mathrm{\Rightarrow \mathit{e_{\mathrm{1}}}\:=\:-\mathrm{2}\mathit{\pi fN_{\mathrm{1}}}\:\mathit{\phi_{m} \:cos\:\omega t}}$$
Where,
$$\mathrm{\mathit{\omega \:=\:\mathrm{2}\pi f}}$$
$$\mathrm{\because -\mathit{cos\:\omega t}\:=\:\mathrm{sin}\left ( \mathit{\omega t-\mathrm{90^{\circ}}} \right )}$$
Therefore,
$$\mathrm{\mathit{e_{\mathrm{1}}}\:=\:\mathrm{2}\mathit{\phi fN_{\mathrm{1}}}\:\mathit{\phi_{m}\:\mathrm{sin}\left ( \mathit{\omega t-\mathrm{90^{\circ}}} \right )}}\:\cdot \cdot \cdot (2)$$
Equation (2) may be written as,
$$\mathrm{\mathit{e_{\mathrm{1}}}\:=\:\mathit{E_{m_{\mathrm{1}}}}\mathrm{sin}\left ( \mathit{\omega t-\mathrm{90^{\circ}}} \right )\:\cdot \cdot \cdot (3)}$$
Where,$\mathit{E_{m_{\mathrm{1}}}}$ is the maximum value of induced EMF $\mathit{e_{\mathrm{1}}}$.
$$\mathrm{\mathit{E_{\mathrm{m1}}}\:=\:\mathrm{2}\mathit{\pi fN_{\mathrm{1}}}\:\mathit{\phi_{m}}}$$
Now, for sinusoidal supply, the RMS value $\mathit{E_{\mathrm{1}}}$ of the primary winding EMF is given by,
$$\mathrm{\mathit{E_{\mathrm{1}}}\:=\:\frac{\mathit{E_{m\mathrm{1}}}}{\sqrt{2}}\:=\:\frac{2\mathit{\pi fN_{\mathrm{1}}}\phi_{m}}{\sqrt{2}}}$$
$$\mathrm{\therefore\mathit{E_{\mathrm{1}}}\:=\:4.44\:\mathit{f\phi _{m}N_{\mathrm{1}}}\:\cdot \cdot \cdot (4)}$$
Similarly, the RMS value E2 of the secondary winding EMF is,
$$\mathrm{\mathit{E_{\mathrm{2}}}\:=\:4.44\:\mathit{f\phi _{m}N_{\mathrm{2}}}\:\cdot \cdot \cdot (5)}$$
In general,
$$\mathrm{\mathit{E}\:=\:4.44\:\mathit{f\phi _{m}N}\:\cdot \cdot \cdot (6)}$$
Equation (6) is known as EMF equation of a transformer.
For a given transformer, if we divide the EMF equation by the supply frequency, we get,
$$\mathrm{\frac{\mathit{E}}{\mathit{f}}\:=\:4.44\:\phi _{m}\mathit{N}\:=\:\mathrm{Constant}}$$
Which means the induced EMF per unit frequency is constant but it is not same on both primary and secondary side of the given transformer.
Also, from equations (4) and (5), we have,
$$\mathrm{\frac{\mathit{E_{\mathrm{1}}}}{\mathit{E_{\mathrm{2}}}}\:=\:\frac{\mathit{N_{\mathrm{1}}}}{\mathit{N_{\mathrm{2}}}}\:or\:\frac{\mathit{E_{\mathrm{1}}}}{\mathit{N_{\mathrm{1}}}}\:=\:\frac{\mathit{E_{\mathrm{2}}}}{\mathit{N_{\mathrm{2}}}}}$$
Hence, in a transformer, the induced EMF per turn in the primary winding is equal to the induced EMF per turn in the secondary winding.
Numerical Example
A single phase 3300/240 V, 50 Hz transformer has a maximum magnetic flux of 0.0315 Wb in the core. Calculate the number of turns in primary and secondary windings.
Solution
Given data,
$$\mathrm{\mathit{E_{\mathrm{1}}\:=\:\mathrm{3300}\:\mathrm{V}\:\mathrm{and}\:\mathit{E_{\mathrm{2}}\:=\:\mathrm{240}\:V}}}$$
$$\mathrm{\mathit{f}\:=\:50\:Hz;\:\phi _{m}\:=\:0.0315\:Wb}$$
The EMF equation of the transformer is,
$$\mathrm{\mathit{E}\:=\:4.44\:\mathit{f\phi _{m}N}}$$
Therefore, for primary winding,
$$\mathrm{\mathit{N_{\mathrm{1}}}\:=\:\frac{\mathit{E_{\mathrm{1}}}}{4.44\:\mathit{f\phi _{m}}}\:=\:\frac{3300}{4.44\times 50\times 0.0315}}$$
$$\mathrm{\mathit{N_{\mathrm{1}}}\:=\:471.9\:=\:472}$$
Also, for secondary winding,
$$\mathrm{\mathit{N_{\mathrm{2}}}\:=\:\frac{\mathit{E_{\mathrm{2}}}}{4.44\:\mathit{f\phi _{m}}}\:=\:\frac{240}{4.44\times 50\times 0.0315}}$$
$$\mathrm{\mathit{N_{\mathrm{2}}}\:=\:34.32\:=\:35}$$
It is not possible for a winding to have part of a turn. Thus, the number of turns should be a whole number.