# EMF Equation of Transformer – Turns & Transformation Ratio of Transformer

Digital ElectronicsElectronElectronics & Electrical

## EMF Equation of Transformer

Consider an alternating voltage is applied across the primary winding of the transformer and the frequency of the supply voltage is f. This applied voltage produces a sinusoidal flux φ in the core of the transformer, which is given by,

$$\mathrm{\varphi = \varphi_{𝑚} sin \omega t}$$

Due to this sinusoidal flux, an EMF is induced in the primary winding, whose instantaneous value is given by,

$$\mathrm{𝑒_{1} = −𝑁_{1}\frac{𝑑\varphi}{𝑑𝑡} = −𝑁_{1}\frac{𝑑}{𝑑𝑡}(\varphi_{𝑚} sin \omega t)}$$

$$\mathrm{⇒ 𝑒_{1} = −𝑁_{1}\omega\varphi_{𝑚} cos \omega t = 𝑁_{1}\omega\varphi_{𝑚} sin (\omega t −\frac{\pi}{2})}$$

$$\mathrm{\because \: \omega = 2\pi𝑓}$$

$$\mathrm{\therefore 𝑒_{1} = 2𝜋𝑓𝑁_{1}\varphi_{𝑚} sin (\omega t −\frac{\pi}{2})}$$

$$\mathrm{⇒ 𝑒_{1} = 𝐸_{𝑚1} sin (\omega t −\frac{\pi}{2})}$$

Where, Em1 is the maximum value of induced EMF in the primary winding.

From this expression it is clear that the induced EMF in the primary winding (E1) lags behind the flux by 90°.

$$\mathrm{𝐸_{𝑚1} = 𝑁_{1}𝜔\varphi_{𝑚} = 2𝜋𝑓𝑁_{1}\varphi_{𝑚}}$$

Now, for the sinusoidal wave, the RMS value of the primary EMF (E1) is given by,

$$\mathrm{𝐸_{1} =\frac{𝐸_{𝑚1}}{\sqrt{2}}=\frac{2𝜋𝑓𝑁_{1}\varphi_{𝑚}}{\sqrt{2}}}$$

$$\mathrm{⇒ 𝐸_{1} = 4.44 𝑓\varphi_{𝑚}𝑁_{1} … (1)}$$

similarly, the RMS value of the secondary EMF (E2) is given by,

$$\mathrm{𝐸_{2} = 4.44 𝑓\varphi_{𝑚}𝑁_{2} … (2)}$$

The expression in the eqns. (1) & (2) is known as the EMF equation of the transformer.

Therefore, in general, the EMF equation of the transformer is represented as,

$$\mathrm{𝐸 = 4.44 𝑓\varphi_{𝑚}𝑁 … (3)}$$

Now, taking the ratio of eqn. (1) and (2), we get,

$$\mathrm{\frac{𝐸_{1}}{𝐸_{2}}=\frac{𝑁_{1}}{𝑁_{2}}}$$

$$\mathrm{⇒ \frac{𝐸_{1}}{𝑁_{1}}=\frac{E_{2}}{𝑁_{2}}… (4)}$$

It is clear from the eqn. (4), that the induced EMF per turn in primary winding as well as in the secondary winding is the same.

## Turns Ratio of Transformer

The turns ratio of a transformer is defined as the ratio of the number of turns in the primary winding to the number of turns in the secondary winding i.e.

$$\mathrm{Turns\:Ratio =\frac{Number \:of \:turns \:in\: primary\: winding (𝑁_{1})}{Number\: of\: turns \:in\: secondary \:winding (𝑁_{2})}}$$

Hence, from the equations (1) and (2), we have,

$$\mathrm{⇒ Turns \:Ratio =\frac{𝑁_{1}}{𝑁_{2}}=\frac{𝐸_{1}}{𝐸_{2}}}$$

For an ideal transformer, E1 = V1 and E2 = V2, thus,

$$\mathrm{Turns\:Ratio =\frac{𝑁_{1}}{𝑁_{2}}=\frac{𝐸_{1}}{𝐸_{2}}=\frac{𝑉_{1}}{V_{2}}}$$

Also, for an ideal transformer, the input volt-ampere is equal to the output volt-ampere i.e.

$$\mathrm{𝑉_{1}𝐼_{1} = V_{2}I_{2}}$$

$$\mathrm{⇒\frac{𝑉_{1}}{V_{2}}=\frac{I_{2}}{𝐼_{1}}}$$

Therefore,

$$\mathrm{Turns\:Ratio =\frac{𝑁_{1}}{𝑁_{2}}=\frac{𝐸_{1}}{𝐸_{2}}=\frac{𝑉_{1}}{V_{2}}=\frac{I_{2}}{𝐼_{1}}\:… (5)}$$

## Transformation Ratio of Transformer

The transformation ratio of a transformer is defined as the ratio of the output voltage to the input voltage of the transformer, i.e.,

$$\mathrm{Transformation Ratio,\: 𝐾 =\frac{Output\: Voltage (V_{2})}{Input \:Voltage (𝑉_{1})}}$$

Also, from the eqns. (1) and (2), we get

$$\mathrm{\frac{𝐸_{2}}{𝐸_{1}}=\frac{𝑁_{2}}{𝑁_{1}}}$$

For an ideal transformer,

$$\mathrm{\frac{V_{2}}{𝑉_{1}}=\frac{𝐸_{2}}{𝐸_{1}}}$$

And

$$\mathrm{𝑉_{1}𝐼_{1} = V_{2}I_{2}}$$

$$\mathrm{⇒\frac{V_{2}}{𝑉_{1}}=\frac{𝐼_{1}}{I_{2}}}$$

Hence, the transformation ratio of the transformer is given by,

$$\mathrm{𝐾 =\frac{V_{2}}{𝑉_{1}}=\frac{𝐸_{2}}{𝐸_{1}}=\frac{𝑁_{2}}{𝑁_{1}}=\frac{𝐼_{1}}{I_{2}}\:… (6)}$$

From the eqns. (5) & (6), it can be seen that,

$$\mathrm{Turns\:Ratio =\frac{1}{Transformation\: Ratio (𝐾)}}$$

Updated on 20-Aug-2021 07:45:41