Step Up Your Knowledge: Induction Principles in Transformers

Introduction

Transformers are at the heart of modern electrical systems, silently ensuring that the right voltage reaches your home, your phone charger, and even massive industrial machines. But how exactly do these devices step voltage up or down with such efficiency and precision? The answer lies in the fascinating phenomenon of electromagnetic induction, a principle discovered by Michael Faraday that revolutionized electrical engineering.

In this blog, we’ll break down the core principles of electromagnetic induction and show you how they are applied in the design and operation of transformers. Whether you’re a student, an aspiring engineer, or simply curious about how the world stays powered, this is your step-by-step guide to understanding one of the most essential components of electrical technology.

Transformer

A transformer is a device which uses the phenomenon of mutual induction to change the values of alternating voltages and currents. In fact, one of the main advantages of a.c. transmission and distribution is the ease with which an alternating voltage can be increased or decreased by transformers. Losses in transformers are generally low and thus efficiency is high. Being static they have a long life and are very stable. Transformers range in size from the miniature units used in electronic applications to the large power transformers used in power stations; the principle of operation is the same for each. A transformer is represented below as consisting of two electrical circuits linked by a common ferromagnetic core. One coil is termed the primary winding which is connected to the supply of electricity, and the other the secondary winding, which may be connected to a load. A circuit diagram symbol for a transformer is also shown.

Transformer Principle of operation

When the secondary is an open-circuit and an alternating voltage V₁  is applied to the primary winding, a small current, called the no-load current I₀  flows, which sets up a magnetic flux in the core. This alternating flux links with both primary and secondary coils and induces in them e.m.f.’s of E₁ and E₂  respectively by mutual induction. The induced e.m.f. E in a coil of N turns is given by: 

E = ^{-N dφ} / _dt

where ^dφ / _dt is the rate of change of flux.

In an ideal transformer, the rate of change of flux is the same for both primary and secondary and thus E₁ / N₁ = E₂ / N₂  i.e. the induced e.m.f. per turn is constant. Assuming no losses, 

E₁ = V₁ and E₂ = V₂

^V₁/_N2 = ^V₂ / _N2 or ^V₁/_V2 = ^N₁/_N2

^V₁/_V2 is called voltage ratio and ^N₁ / _N2 is the turns ratio, or the ‘transformation ratio’ of the transformer. If N₂  is less than N₁ then V₂  is less than V₁. Hence, the device is termed a step-down transformer. If N₂ is greater than N₁  then V₂  is greater than V₁  Hence, the device is termed a step-up transformer. When a load is connected across the secondary winding, current I₂  flows. In an ideal transformer losses are neglected and a transformer is considered to be 100 percent efficient.

Input power = output power or V₁I₁ = V₂I₂  in an ideal transformer, the primary and secondary ampere-turns are equal.

^V₁/_V2 = ^I₂ /_I1

Combining equations gives:

^V₁/_V2 = ^N₁/_N2 = ^I₂/_I1

The rating of a transformer is stated in terms of the volt-amperes that it can transform without overheating. The transformer rating is either V₁I₁ or V₂I₂  where I₂  is the full-load secondary current. 

Example 1: A transformer has 500 primary turns and 3000 secondary turns. If the primary voltage is 240V, determine the secondary voltage, assuming an ideal transformer.  

Solution

For an ideal transformer, voltage ratio = turns ratio

^V₁/_V2 = ^N₁/_N2

Therefore,

²⁴⁰/_V2 = ⁵⁰⁰ / ₃₀₀₀

V₂ = ^{(240 x 3000)} / ₅₀₀ = 1440V or 1.44kV

Example 2:An ideal transformer has a turns ratio of 8:1 and the primary current is 3A when it is supplied at 240V. Calculate the secondary voltage (V₂)  and the secondary current (I₂)

Solution

A turns ratio of 8:1 means ^N₁ / _N2 = ⁸/₁; so  a step down transformer.

^V₁/_V2 = ^N₁/_N2

or secondary voltage

V₂ = ^N₂V₂ / _N1 = ^{(1 x 240 )}/₈ = 30V  

 Also ^N₁ / _N2 = ^I₂/_{I1 I2} = ^{I₁ N₁} / _N2 = ^(3×8) / ₁ = 24V

Transformer no-load phasor diagram

The core flux is common to both primary and secondary windings in a transformer and is thus taken as the reference phasor in a phasor diagram. On no-load the primary winding takes a small no-load current I₀ and since, with losses neglected, the primary winding is a pure inductor, this current lags the applied voltage V₁  by 90º. In the phasor diagram shown below assuming no losses, current I₀ produces the flux and is drawn in phase with the flux. The primary induced e.m.f. E₁ is in phase opposition to V₁ (by Lenz’s law) and is shown 180º out of phase with V₁ and equal in magnitude.

The secondary induced e.m.f. is shown for a 2:1 turn ratio transformer. A no-load phasor diagram for a practical transformer is shown below. If current flows then losses will occur. When losses are considered then the no-load current I₀  is the phasor sum of two components.

(i) I_m  is the magnetising component in phase with the flux, and

(ii) I_c is the core loss component (supplying the hysteresis and eddy current losses).

 

No load current I₀ = √I_m² + I_C² 

Where I_m = I₀ sin φ₀ and I_c = I₀cos φ₀

Power factor on no-load = cos φ₀  = ^I_c/_I0  

The total core losses (i.e iron losses) = V₁I₀ cos φ₀

Example: A 2400V/400V single-phase transformer takes a no-load current of 0.5A and the core loss is 400W. Determine the values of the magnetising and core loss components of the no-load current. Draw to scale the no-load phasor diagram for the transformer.

Solution

V₁ = 2400V, V₂ = 400V, I₀ = 0.5A

Core loss (i.e.iron loss) = 400W = V₁I₀ cos φ₀ 

400 = 2400 x 0.5 x cos φ₀  

Hence,

cos φ₀ = ⁴⁰⁰ / _{( 2400 x 0,5)} = 0.33

Φ₀ = cos^-1(0.33) = 70.53º

The no-load phasor diagram is then shown below:

Magnetising component,

I_m = I₀sin φ₀  = 0.5 sin 70.53 = 0.471A

Core loss component,

I_c = I₀ cos φ₀ = 0.5 cos 70.53 = 0.167A 

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