## What is Ideal Transformer?

An ideal transformer is one that has no losses i.e., its windings have no resistance, there is no magnetic leakage, and hence it has no iron and copper losses. It may, however, be noted that it is impossible to imagine such a transformer in practice, yet for convenience, we have to assume such a transformer.

## Ideal Transformer Model :

Let us consider an ideal transformer as shown in the figure. The primary of the transformer is connected to sinusoidal alternating voltage V_{1} and its secondary is open. This V_{2} causes an alternating current to flow in the primary.

Since the primary coil is purely inductive and the secondary is open (no output) the primary draws the magnetizing current I_{µ} only. The function of this current is merely to magnetize the core, it is small in magnitude and lags V_{1} by 90°.

This current I_{µ} produces an alternating flux Φ which is, at all times proportional to the current and hence, is in phase with it as shown in the phasor diagram. This flux Φ links with both primary and secondary windings.

Therefore, it produces self-induced emf E_{1} in the primary. This E_{1} is at every instant equal and opposite to V_{1}. It is also known as counter emf or back emf of the primary.

Similarly, an emf E_{2} is induced in the secondary winding and is known as mutually induced emf due to E_{1}. This E_{2} is in anti-phase with V_{1} (and is proportional to the number of turns in the secondary). Check out the below video for more detailed explanation,

## Difference Between Ideal and Practical Transformer (non-ideal transformer) :

Ideal Transformer | Practical Transformer |
---|---|

The permeability of an ideal transformer is always infinity, i.e., µ = ∞. | There exists a finite permeability in practical transformer, i.e., µ ≠ ∞. |

In an ideal transformer the core losses i.e., iron losses are equal to zero. | The iron losses of a practical transformer are given I_{w}. Where no-load component I_{o} = √I_{w}^{2} + I_{µ}^{2} |

The winding resistance of an ideal transformer, i.e., R = 0. Hence, copper losses in the windings of the transformer are absent. | In practice, the winding resistance cannot be zero in a transformer, i.e., R ≠ 0. The total resistance when referred to the primary side of a transformer is given by R_{01} = R_{1} + R_{2} / K^{2}. |

In an ideal transformer flux produced by the primary winding will link with the secondary winding without any leakage of flux. | In the case of a practical transformer, there will be a wastage of flux due to leakage (not linking total flux produced by primary with secondary). |

It is impossible to make an ideal transformer. All the transformer made will consist of losses in its. | The transformers which we see in our daily life are practical transformers. |