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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.

ideal transformer no load

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 Im. As we know in pure inductive circuit the current lags behind the voltage by 90˚. Hence magnetizing current Im 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 = Φm sin ωt

Instantaneous value of emf in primary side is

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

                                                                                           = – ω N1 Φm cos ωt

                                                                                           = – 2π f  N1 Φm cos ωt

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

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

                                                             Em1 = 2π f  N1 Φm

r.m.s value of primary emf is E1 = Em1 / √2 = 2πf N1 Φm / √2 = 4.44 f N1 Φm

Similarly for E2 = 4.44 f N2 Φm

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

ideal transformer phasor on no loadAbove 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

ideal transformer with load

Upper figure shows an ideal transformer. It has a load ZL 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 ZL. There is no voltage drop in ideal transformer so E2 = V2

          E2 = I2 ZL and V2 = I2 ZL

          I2 = E2 / ZL = V2 / ZL

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

ideal transformer on load phasor

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

Md Ebrahim Shah

Electrical Engineer, teacher, professional article writer and founder of His passion is to describe electrical technology at easy way as people can get clear conception and upgrade the technology for the better human life.

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simon - 01/12/2016

very nice explanation


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