Ideal transformer

Ideal transformer on no load

 A transformer is ideal if it has

• No leakage flux
• No windings resistance
• No iron loss in core

Ideal transformer can not possible physically. But it gives powerful tool in the analysis of a practical transformer. Properties of ideal and practical transformer is very close each other.

Above figure shows an ideal transformer on no load. Secondary side is open circuited. According to ideal transformer definition primary coil is simply a coil of pure inductance. When we apply V₁ alternating voltage in primary side it creates a small magnetizing current I_m. As we know in pure inductive circuit the current lags behind the voltage by 90˚. Hence magnetizing current I_m lags behind the applied voltage by 90˚. This current creates an alternating flux Φ which is proportional to and in phase with it. The alternating flux links both winding so it is common for two sides and induces e.m.f. E₁ in primary and E₂ in secondary side. According to Lenz’s law V₁ and E₁ are equal at every instant and in opposition. Magnitude of E₁ and E₂ depend upon the number of turns in primary and secondary. E₁ and E₂ both e.m.f.s lag behind flux Φ by 90˚. E.m.f. equation of transformer proves that primary and secondary e.m.f.s lag behind flux Φ by 90˚.

emf equation of transformer

 Consider the above figure; V₁ is alternating voltage applied to primary side with frequency f. The sinusoidal flux Φ is produced by primary side

                                                                         Φ = Φ_m sin ωt

Instantaneous value of emf in primary side is

                                                                        E₁ = – N₁ dΦ / dt = – N₁ d (Φ_m sin ωt) / dt

                                                                                           = – ω N₁ Φ_m cos ωt

                                                                                           = – 2π f  N₁ Φ_m cos ωt

                                                                                           = 2π f  N₁ Φ_m sin (ωt – 90˚) ———— (i)

This equation clears that maximum induced e.m.f value in the primary side is

                                                             E_m1 = 2π f  N₁ Φ_m

r.m.s value of primary emf is E₁ = E_m1 / √2 = 2πf N₁ Φ_m / √2 = 4.44 f N₁ Φ_m

Similarly for E₂ = 4.44 f N₂ Φ_m

For an ideal transformer E₁ = V₁ and E₂ = V₂

Equation (i) shows that induced primary emf E₁ lags behind the flux Φ by 90˚. Since the same flux Φ induces secondary emf E₂ so E₂ also lags behind the flux Φ by 90˚.

Ideal transformer on no load phasor diagram

Above figure shows the phasor diagram of ideal transformer on no load. Since the flux Φ is same for both windings so it is taken as reference phasor. We know from equation (i) primary e.m.f E₁ and secondary e.m.f E₂ lag behind the flux Φ by 90˚. E₁ and V₁ are equal and 180˚ out of phase with it. E₁ and E₂ are in phase.

Ideal transformer on load

Upper figure shows an ideal transformer. It has a load Z_L in secondary side. As it has a load in secondary side the secondary coil e.m.f E₂ will cause a current I₂ through the load Z_L. There is no voltage drop in ideal transformer so E₂ = V₂

          E₂ = I₂ Z_L and V₂ = I₂ Z_L

          I₂ = E₂ / Z_L = V₂ / Z_L

When the secondary is loaded the magnitude and phase of I₂ with respect to V₂ is determined by the characteristics of the load. If the load is non inductive current I₂ is in phase with V₂, if the load is inductive it lags, if the load is capacitive it leads.

The secondary current I₂ set up its own m.m.f N₂I₂ and it has own flux Φ₂ which is opposition to the main primary flux Φ which is due to primary current I₁. The secondary ampere turns N₂I₂ are known as demagnetizing amp-turns. The secondary flux weakens the primary flux. In this way primary back emf E₁ tends to be reduced. The secondary flux will change the main flux value. However, the original value of the main flux in the core should not change. To fulfill this condition primary side should have an m.m.f which exactly counterbalances the secondary m.m.f N₂I₂.

Primary current I₁ must follow that

                                                        N₁I₁ = N₂I₂

                                                             I₁ = N₂ I₂ / N₁ = KI₂

When a transformer has load in secondary side and I₂ current flows then a primary current I₁ must flow to maintain the m.m.f balance. It can be said in another word to neutralize the demagnetizing effect of secondary current a primary current must draw so that the common flux Φ remain constant it is not changed.

Ideal transformer on load phasor diagram

Figure shows the phasor diagram of ideal transformer on load. Here the value of K is assumed unity so primary phasors are equal to secondary. Secondary current I₂ lags behind V₂ or E₂ by Φ₂. It is the cause of primary current I₁ = K I₂ = 1.I₂ which is antiphase with it.

Thus the power factor of primary side is equal to secondary side.

                                                    Φ₁ = Φ₂

                                              or cos Φ₁ = cos Φ₂

Since ideal transformer has no losses, input primary power is equal to output secondary power.

                                            V₁ I₁ cos Φ₁ = V₂ I₂ cos Φ₂