What is an Ideal Transformer?

Digital ElectronicsElectronElectronics & Electrical

An ideal transformer is an imaginary transformer which has the following characteristics −

  • The primary and secondary windings have negligible (or zero) resistance.
  • No leakage flux, i.e., whole of the flux is confined to the magnetic circuit.
  • The magnetic core has infinite permeability, thus negligible mmf is require to establish flux in the core.
  • There are no losses due winding resistances, hysteresis and eddy currents. Hence, the efficiency is 100 %.

Woking of Ideal Transformer

Ideal Transformer on No-Load

Consider an ideal transformer on no-load, i.e., its secondary winding is open circuited (see the figure). Thus, the primary winding is a coil of pure inductance.

When an alternating voltage V1 is applied to the primary winding, it draws a very small magnetising current Im to establish the flux in the core, which lags behind the applied voltage by 90°. The magnetising current Im produces an alternating flux ϕm which is proportional to and in phase with it. This alternating flux (ϕm) links the primary and secondary windings magnetically and induces EMF E1 in the primary winding and EMF E2 in the secondary winding.

The EMF induced in the primary winding E1 is equal to and in opposition to the applied voltage V1 (according to Lenz’s law). The EMFs E1 and E2 lag behind the flux (ϕm) by 90°, although their magnitudes depend upon the number of turns in the primary and the secondary windings. From the phasor diagram of the ideal transformer on no-load, it is clear that the flux is common to both the windings, hence it can be taken as the reference phasor. Also, the EMFs E1 and E2 are in phase with each other, but E1 is equal to V1 and 180° out of phase with it.

Ideal Transformer On-Load

When load is connected across the terminals of secondary winding of the ideal transformer, the transformer is said to be loaded and a load current flows through the secondary winding and the load.

Consider an inductive load of impedance ZL is connected across the secondary winding of the ideal transformer (see the figure). Then, the secondary EMF E2 will cause a current I2 to flow through the secondary winding and the load, which is given by,

$$\mathrm{I_{2}\:=\:\frac{E_{2}}{Z_{L}}\:=\:\frac{V_{2}}{Z_{L}}}$$

Since, for an ideal transformer, the EMF E2 is equal to secondary terminal voltage V2.

Here, the load is inductive, therefore, the current I2 will lag behind the E2 or V2 by an angle ϕ2. Also, the no-load current I0 being neglected because the transformer is ideal one.

The current flowing in the secondary winding (I2) sets up an mmf (N2I2) which produces a flux ϕ2 in opposite direction to the main flux (ϕm). As a result, the total flux in the core changes from its original value, however, the flux in the core should not changes from its original value. Therefore, to maintain the flux in the core at its original value, the primary current must develop an MMF which can counter-balance the demagnetising effect of the secondary mmf N2I2. Hence, the primary current I1 must flow such that,

$$\mathrm{N_{1}I_{1}\:=\:N_{2}I_{2}}$$

$$\mathrm{⇒\:I_{1} =\:\frac{N_{2}}{N_{1}}\times I_{2}\:=\:KI_{2}}$$

Therefore, the primary winding must draw enough current to neutralise the demagnetising effect of the secondary current so that the main flux in the core remains constant. Hence, when the secondary current (I2) increases, the primary current (I1) also increases in the same manner and keeps the mutual flux (ϕm) constant.

It is clear from the phasor diagram of the ideal transformer on-load that the secondary current I2 lags behind the secondary terminal voltage V2 by an angle of ϕ2.

raja
Published on 18-Aug-2021 07:15:36
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