An ideal transformer is an imaginary transformer. The proprieties of the practical transformers are different than the ideal transformers. The operation of the practical transformer is close to the ideal transformer.

Table of Contents

**Properties of an ideal transformers**

- Primary and secondary winding resistance are negligible.
- The core has infinite permeability (μ). Hence, very small amount of MMF required to establish the flux in the core.
- Leakage flux and leakage inductance are zero. The entire flux linked with the core and windings.
- The efficiency of the ideal transformer is 100%. Because we considered that there are no losses produced due to resistance, hysteresis and eddy current.

**Make note that above properties assumed for the ideal transformers only. It is not possible to achieve these properties in the practical transformer.**

An ideal iron core transformer consists of two coils wound on the common magnetic core. These coils wound in the same direction. The winding which connected with the supply voltage V_{1}, known as the primary winding. The winding which connected with the load, known as the secondary winding.

**Phasor diagram **

As we have assumed in the properties, the primary and secondary winding has zero impedance. Therefore, the applied voltage V_{1} is same as the voltage induced in the primary winding E_{1}. Similarly, the secondary voltage V_{2} is same as the voltage induced in the secondary winding E_{2}. The primary current I_{1} is sufficient to produce mutual flux ф_{1}. This flux is enough to produce MMF I_{1}T_{1} to overcome the demagnetizing effect of the secondary MMF I_{2}T_{2} as a result of load.

According to the Lenz’s law, E_{1} is equal and opposite to V_{1}. E_{1} and E_{2} induced by the same mutual flux. Therefore, E_{2} is in the same direction of E_{1} and opposite to the V_{1}.

The magnetizing current I_{μ} lags V_{1} by 90 degree. It produces the magnetizing flux ф_{m} in phase with I_{μ}. E_{1} and E_{2} produced by the ф_{m} and lag ф_{m} by 90 degree. V_{2} is equal in magnitude to E_{2} and is opposite to V_{1}.

Below figure shows the phasor diagram of ideal transformers in no load condition.

For an ideal transformer,

a = transformation ratio (turns ratio)

According to the equation (2), the demagnetizing MMF of secondary are equal and opposite to the magnetizing MMF of primary for an ideal transformers.

According to the equation (3), the apparent power drawn from the primary supply is equal to the apparent power transferred to the secondary without any loss in the ideal transformers.

**Input Power = Output Power**

Therefore, the input kVA is equal to the output kVA for the ideal transformer. Thus, the efficiency of is 100%.

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