Electrical Engineering ⇒ Topic : Ideal Transformer

When the primary and secondary winding resistance drops are neglected, the secondary induced voltage E₂ will be equal to its terminal voltage V₂.Likewise, on the primary side, the induced emf E₁ will be equal to the applied voltage V₁. This is an ideal transformer. However, V₁ and E₁ will oppose each other. The phasor diagram shown in Figure (a)  shows these details along with the load currents

Consider a load on the transformer secondary such that it gives a lagging power factor Φ₂. Then the load current in the secondary /₂ will lag the secondary voltage V₂ by an angle Φ₂. /₂ in the secondary causes a current I to flow in the primary such that  neutralises the demagnetising effect of /₂. In addition to /₂', the no-load current /₀ (discussed in the previous section) also flows in the primary (see Fig (A)). The primary current I₁ will 

               FIGURE  (A) Phasor diagram of an ideal transformer on load

now be the phasor sum of I₀ and I₂^'. The power factor of the primary side will be cos Φ₁ where  Φ₁ is the phase angle difference between V₁ and I₁.However, as far as the magnetisation of the core is considered,the magnetising current(I₀) is same on load as on noload.

IDEAL TRANSFORMER

It is an imaginary transformer having the following properties:

The resistance of primary and secondary windings is negligible.

There is no leakage flux and no leakage inductance.

The core has infinite permeability so that the m.m.f. required to establish the flux in -is negligible.

The entire flux is confined to the core.

There are no losses, i.e., no ohmic loss, hysteresis, and eddy current losses. Therefore,the efficiency in this case is 100 percent.

The practical transformers do not satisfy the above conditions.

An ideal transformer is shown in Figure (a). The voltage applied to coil 1 is V₁ and the secondary is connected to a load 4. The voltage across the load is V₂. Since the ideal transformer has zero primary and zero secondary impedance, the voltage induced in the primary, E₁ is equal to the applied voltage V₁ and the output voltage V₂ is equal to the secondary induced voltage E₂. Here E₁ is called counter e.m.f. or back e.m.f. of the primary and E₂ is called the mutually induced e.m.f. in the secondary. Both are in anti-phase with V₁. Figure (b) shows the phasor representation of an ideal transformer

                                      .............. (1)

The current in the primary is just sufficient to provide the required m.m.f. I₁N₁ to overcome the demagnetizing effect of secondary m.m.f. I₂N₂ which is a result of connected load

                  ................ (2)

From Eqs. (1) and (2), it can be written as:

                              ............... (3)

Thus, in an ideal transformer input kVA and output kVA are identical.

For ideal transformer, there is no copper loss and core loss. But for practical transformer when secondary is put on load, there is iron loss in core and copper loss in winding and these losses are not negligible.