# EMF Equation of a Transformer | Voltage and Current Ratio of Transformer

In today’s lesson, we will learn about the EMF equation of a transformer. We will understand how electromotive force (EMF) is related to different factors in a transformer. We will also discuss the Transformation ratio of the Transformer. Join us as we explore this important equation that helps us understand how transformers work.

## EMF Equation of a Transformer

For deriving the EMF equation of a transformer refer to the figure that we have mentioned below. The primary winding is connected across the AC supply. This forces an alternating current through the primary winding to produce an alternating flux (Φ) in the core.

This varying flux gets linked with the secondary and primary windings to induce the mutually induced and self-induced EMFs in the secondary and primary windings respectively.

### Derive the EMF Equation of a Transformer

Let us now obtain the expression for the induced voltages in the primary and secondary windings.

Step 1: Expression for the instantaneous flux Φ

As shown in the figure, the instantaneous flux changes in a sinusoidal manner with respect to time. Its frequency ‘f’ is the same as that of the AC voltage applied to the primary winding.

\Phi = \Phi_m \sin \omega t

Where Φm= Maximum value of the instantaneous flux, ω = 2πf where f is the frequency of the flux waveform

Step 2: Obtain the expression for induced voltage

According to Faraday’s lows of electromagnetic induction, the induced emf due to varying flux is given by,

\mathbf{e = - N \frac{d \Phi}{dt} \;volts}

where e = emf induced due to varying flux

Step 3: Obtain the maximum value of ‘e’ per turn

The value of induced emf per turn can be obtained by substituting N =1 in the above equation then,

\therefore \mathbf{e = - \frac{d \Phi}{dt} }

Substitute \Phi = \Phi_m \sin \omega t,

\mathbf{e = - \frac{d }{dt}[\Phi_m \sin \omega t] }

\therefore \mathbf{e = - \Phi_m \omega . \cos \omega t}

The maximum value of induced voltage per turn is given by,

\therefore \mathbf{e_{max}= \omega \Phi_m = 2\pi f \Phi_m \, volts}

Step 4: Obtain the RMS value of ‘e’ per turn

RMS value of ‘e’ per turn = \mathbf{\frac{e_{max}}{\sqrt{2}} = \frac{2\pi f \Phi_m}{\sqrt{2}} = \sqrt{2}\pi f \Phi_m = 4.44\,f\, \Phi_m}

Step 5: Obtain the expression for induced voltage E1 and E2

Let E1 be the RMS-induced voltage in the primary winding with N1 turns and E2 be the RMS-induced voltage in the secondary winding having N2 turns.

The RMS value of induced voltage in the primary winding is,

E1 = RMS value of ‘e’ turn X Number of primary turns

\therefore \mathbf{E_1 = 4.44\,f\, \Phi_m \times N_1}

\therefore \boxed{\mathbf{E_1 = 4.44\,f \cdot N_1\cdot \Phi_m\, volts}}

Similarly, the RMS value of induced voltage in the secondary winding is,

E2 = RMS value of ‘e’ turn X Number of secondary turns

\therefore \boxed{\mathbf{E_2 = 4.44\,f \cdot N_2\cdot \Phi_m\, volts}}

## Voltage and Current Ratio of Transformer

For obtaining the voltage and current ratio of the transformer, consider the elementary transformer shown in the figure below.

### Voltage Ratio of Transformer

The ratio of the primary and secondary terminal voltages (i.e. V1 and V2) is called the voltage ratio.

\mathrm{Voltage\;Ratio = \frac{Primary\;terminal\;volatage (V_1)}{Secondary\;terminal\;volatage (V_2)}}

#### Voltage Ratio of Transformer with Load

Let N1 be the number of turns in the primary winding and let N2 be the number of turns in the secondary winding. Also, let the RMS-induced voltage in the primary winding be E1 volts and the RMS secondary-induced voltage be E2 volts.

\mathbf{E_1 = 4.44\,f \cdot N_1\cdot \Phi_m\, volts}

and \mathbf{E_2 = 4.44\,f \cdot N_2\cdot \Phi_m\, volts}

Taking the ratio of these expressions we get,

\boxed{\mathbf{\frac{E_1}{E_2} = \frac{N_1}{N_2} \;OR\; \frac{E_2}{E_1} = \frac{N_2}{N_1}}}

These are the ratios of EMFs induced in the two windings of the transformer.

#### Voltage Ratio of Transformer without Load

Assume the load on the secondary winding is now disconnected. Hence the secondary current I2 = 0. Hence the load terminal voltage V2 is equal to the secondary induced voltage E2.

∴ V2 = E2

The primary current I1 on no load is very small. Hence the AC assupply voltage V1 and primary induced voltage E1 can be equated.

∴ V1 ≈ E1

Hence,

\boxed{\mathbf{\frac{V_1}{V_2} = \frac{N_1}{N_2} \;OR\; \frac{V_2}{V_1} = \frac{N_2}{N_1}}}

Thus the primary and secondary terminal voltages are proportional to the number of turns of the respective windings.

### Transformation Ratio (K)

The transformation ratio for voltage is defined as the ratio of the secondary voltage to the primary voltage of a transformer. It is defined by K.

\boxed{\mathbf{Transformation\; ratio\;K = \frac{V_2}{V_1} = \frac{E_2}{E_1} = \frac{N_2}{N_1}}}

### Turns Ratio of the Transformer

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

\therefore \boxed{\mathbf{Turns\;Ratio = \frac{N_1}{N_2}}}

### Current Ratio of Transformer

The transformer transfers electrical power from one side to the other (primary to secondary) with a very high efficiency (η). If we assume that the power loss taking place in the transformer is very low (η≈100%) then, we can write that

Power input = Power output

\therefore\;\mathbf{V_1 I_1 \cos \Phi_1 = V_2 I_2 \cos \Phi_2}

Where I1 and I2 are the RMS values of the primary and secondary currents of the transformer.

cos Φ1 and cos Φ2 are the power factors of the primary and secondary sides of the transformer. Practically they are of the same values.

\therefore\;\mathbf{ \cos \Phi_1 = \cos \Phi_2}

\therefore\;\mathbf{V_1 I_1 = V_2 I_2}

\therefore\;\mathbf{\frac{I_1}{I_2} = \frac{V_2}{V_1}}

But , \therefore\;\mathbf{\frac{V_2}{V_1} = \frac{N_2}{N_1} = K}

\therefore\; \boxed{\mathbf{\frac{I_1}{I_2} = \frac{N_2}{N_1} = K}}

This expression shows that the primary and secondary currents are inversely proportional to the number of turns of the corresponding windings. Hello friends, my name is Trupal Bhavsar, I am the Writer and Founder of this blog. I am Electronics Engineer(2014 pass out), Currently working as Junior Telecom Officer(B.S.N.L.) also I do Project Development, PCB designing and Teaching of Electronics Subjects.

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