## 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 V_{1} 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. E_{1} in primary and E_{2} in secondary side. According to **Lenz’s law** V_{1} and E_{1} are equal at every instant and in opposition. Magnitude of E_{1} and E_{2} depend upon the number of turns in primary and secondary. E_{1} and E_{2} 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; V_{1} 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

E_{1} = – N_{1} dΦ / dt = – N_{1} d (Φ_{m} sin ωt) / dt

= – ω N_{1 }Φ_{m} cos ωt

= – 2π *f * N_{1 }Φ_{m} cos ωt

= 2π *f* N_{1 }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 * N_{1} Φ_{m}

r.m.s value of primary emf is E_{1} = E_{m1} / √2 = 2π*f* N_{1 }Φ_{m} / √2 = 4.44 *f* N_{1 }Φ_{m}

Similarly for E_{2} = 4.44 *f* N_{2 }Φ_{m}

For an ideal transformer E_{1} = V_{1} and E_{2} = V_{2}

Equation (i) shows that induced primary emf E_{1 }lags behind the flux Φ by 90˚. Since the same flux Φ induces secondary emf E_{2} so E_{2} 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 E_{1} and secondary e.m.f E_{2} lag behind the flux Φ by 90˚. E_{1} and V_{1} are equal and 180˚ out of phase with it. E_{1} and E_{2} 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 E_{2} will cause a current I_{2} through the load Z_{L}. There is no voltage drop in ideal transformer so E_{2} = V_{2}

E_{2} = I_{2} Z_{L} and V_{2} = I_{2} Z_{L}

I_{2} = E_{2 }/ Z_{L} = V_{2} / Z_{L}

When the secondary is loaded the magnitude and phase of I_{2} with respect to V_{2} is determined by the characteristics of the load. If the load is non inductive current I_{2} is in phase with V_{2}, if the load is inductive it lags, if the load is capacitive it leads.

The secondary current I_{2} set up its own m.m.f N_{2}I_{2} and it has own flux Φ_{2} which is opposition to the main primary flux Φ which is due to primary current I_{1}. The secondary ampere turns N_{2}I_{2} are known as demagnetizing amp-turns. The secondary flux weakens the primary flux. In this way primary back emf E_{1} 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 N_{2}I_{2}.

Primary current I_{1} must follow that

N_{1}I_{1} = N_{2}I_{2}

I_{1 }= N_{2} I_{2} / N_{1} = KI_{2}

When a transformer has load in secondary side and I_{2} current flows then a primary current I_{1} 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 I_{2} lags behind V_{2} or E_{2} by Φ_{2}. It is the cause of primary current I_{1} = K I_{2} = 1.I_{2} which is antiphase with it.

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

Φ_{1} = Φ_{2}

or cos Φ_{1} = cos Φ_{2}

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

V_{1} I_{1} cos Φ_{1} = V_{2} I_{2} cos Φ_{2}