EMF Equation of a transformer

The emf equation of a transformer is derived by considering a single-phase transformer. When a single-phase AC supply is given to a transformer, the emf gets induced in the secondary winding by means of electromagnetic induction.

Let,

N1 be the number of turns in the primary winding.

N2 be the number of turns in the secondary winding.

ϕ_m be the maximum flux in core in webers.

f be the frequency of AC input voltage in Hertz.

Emf induced - emf equation of a transformer — *EMF induced in the transformer winding*

As shown in fig, flux ϕ_mincreases from zero value to maximum value in one-quarter of the cycle, i.e., in 1/4f second.

$\text{Average Rate of change of flux} = \frac{\phi_m}{1/4f} = 4f\phi_m$

By faraday’s law of electromagnetic induction, rate of change of flux is nothing but induced emf,

$\text{Average emf per turn} = 4f\phi_m \, \, \text{volt}$

If the flux varies sinusoidally, then the rms value of induced emf is obtained by multiplying the average value with form factor.

$\text{Form Factor} = \frac{\text{RMS Value}}{\text{Average Value}} = 1.11$

$\text{RMS Value of emf per turn} = 1.11 \times 4f\phi_m = 4.44f\phi_m \, \, \text{volt}$

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

$= \text{Induced emf per turn} \times \text{No. of Primary Winding}$

(1) $\begin{equation*} E_1 = 4.44fN_1 \phi_m = 4.44fN_1 B_m A \,\, \text{volt} \end{equation*}$

Similarly, rms value of induced emf in the secondary winding is given by,

(2) $\begin{equation*} E_2 = 4.44fN_2 \phi_m = 4.44fN_2 B_m A \,\, \text{volt} \end{equation*}$

The above two equations are the emf equation of a transformer in the primary and secondary winding respectively.

From equations (1) and (2), we get the emf induced per turn in the primary and secondary winding of a transformer.

$\frac{E_1}{N_1} = \frac{E_2}{N_2} = 4.44f\phi_m$

Dividing the equations (1) by (2), we get,

$\frac{E_1}{E_2} = \frac{N_1}{N_2} = K$

where K is a constant, called transformation ratio.

For an ideal transformer, E₁ = V₁ & E₂ = V₂

where

V₁ – AC Supply voltage of transformer in the primary side.

V₂ – Terminal voltage of transformer in the secondary side.

Thus, the transformation ratio K of a transformer is

$\frac{E_1}{E_2} = \frac{N_1}{N_2} = \frac{V_1}{V_2}= K$

In the above equation,

If N₂ > N₁ & E₂ > E₁ then the transformer is a step up transformer.

If N₁ > N₂ & E₁ > E₂ then the transformer is a step down transformer.

Solved Problems

A 25 kVA transformer has 250 turns on the primary and 40 turns on the secondary winding. The primary is connected to 1500 V, 50 Hz supply. Find the secondary e.m.f., the full-load primary and secondary currents, and the maximum flux in the core. Neglect leakage drops and no-load primary current.

Given: Rated Capacity = 25 kVA, N₁ = 250 turns, N₂ = 40 turns, V₁ = 1500 V, f = 50 Hz.

To find: V₂, I₁, I₂, ϕ_m

Solution: Using the transformation ratio,

$\frac{V_2}{V_1} = \frac{N_2}{N_1}$

$\frac{V_2}{1500} = \frac{40}{250}$

$\boxed{\text{Secondary emf, }V_2 = 240 \, V }$

Primary and secondary currents are calculated using rated capacity (kVA), as shown below.

$I_1 = \frac{kVA}{V_1} = \frac{25000}{1500}$

$\boxed{I_1 = 16.67 \, A}$

$I_2 = \frac{kVA}{V_2} = \frac{25000}{240}$

$\boxed{I_2 = 104.2 \, A}$

We know, the emf equation of a transformer is given by,

$E_1 = 4.44f\phi_m N_1$

$1500 = 4.44 \times 50 \times \phi_m \times 250$

$\phi_m = \frac{1500}{4.44 \times 50 \times 250}$

$\boxed{\phi_m = 0.027 \, Wb}$

A single phase transformer has 400 primary and 1000 secondary turns. The net cross-sectional area of the core is 60 cm². If the primary winding be connected to a 50 Hz supply at 520 V, calculate (i) the peak value of flux density in the core (ii) the voltage induced in the secondary winding.

Given: N₁ = 400 turns, N₂ = 1000 turns, A= 60 cm², V₁ = 520 V, f = 50 Hz.

To find: B_m, E₂

Solution: We know, the emf equation of a transformer is given by,

$E_1 = 4.44f\phi_m N_1$

$E_1 = 4.44fB_m AN_1$