Are you ready to learn about the ideal transformer? Join us in this lecture as we break down the fundamentals in a simple and straightforward way. Discover how transformers work, their importance in electrical systems, and how they help us daily. Perfect for students wanting to grasp the basics of ideal transformers.

## What is an Ideal Transformer

An ideal transformer is a theoretical concept used in electrical engineering to explain the behavior of transformers in a simplified manner. It is a theoretical model that assumes the transformer has no losses and operates at 100% efficiency.

## Properties of Ideal Transformer

An Ideal transformer is a transformer having the following properties:

- The losses are Zero ( No iron loss, No copper loss)
- The primary and secondary winding resistances are zero.
- The leakage flux is zero. Therefore all the flux produced by the primary winding is coupled with the secondary winding.
- A small current is required to develop flux inside the core. This happens because the permeability of the core is very large.
- The external voltage applied to the primary, V
_{1}is the same as the primary induced voltage E_{1}. This is because the primary winding resistance is zero and so there is no voltage drop across it.

**∴ E _{1} = V_{1}**

- Similarly, the voltage induced in the secondary winding E
_{2}will be equal to the load voltage V_{2}because the secondary winding resistance is zero.

**∴ E _{2} = V_{2}**

- The transformation ratio for an ideal transformer is given by,

\mathbf{K = \frac{E_2}{E_1} = \frac{V_2}{V_1}}

- The efficiency of an ideal transformer is 100%. This is because there are no losses taking place.
- The voltage regulation is 0%. That means the secondary voltage will remain constant irrespective of the load current.

Now we will learn the Ideal transformed on No Load and Ideal transformer on Load with their circuit and phasor diagram in detail.

Also Read:Losses in Transformer

## Ideal Transformer on No Load

An Ideal Transformer with No load condition is shown in the figure below. An ideal transformer is one that has no power losses. In order to have a transformer with zero loss, the following conditions are to be satisfied.

**Conditions for an Ideal Transformer**:

- The primary and secondary windings do not have any resistance. i.e. winding resistance of primary and secondary should be zero.
- The losses taking place in the core i.e. hysteresis loss and eddy current loss should be zero.
- There should not be any leakage flux.

**Phasor Diagram of Ideal Transformer**

**Operation of ideal Transformer on No Load**

An AC voltage V_{1} is applied across the primary winding of the transformer. As the load on the transformer is zero i.e. R_{L} = ∞, ideally the primary current I_{1} = 0. But practically a small current called the magnetic current I_{mag} flows through the primary winding.

The magnetic current is used for magnetizing the transformer core. As the primary winding is assumed to be purely reactive (R = 0) the magnetizing current lags behind the primary induced voltage by 90^{o}, as shown in the phasor diagram of Ideal transformer.

Due to the sinusoidal magnetizing current, a sinusoidally varying magnetic flux is produced in the iron core. The flux is in the time phase with I_{mag}, as shown in the above figure.

Due to this varying flux, EMFs are induced in the primary (self-induced voltage) E_{1} and secondary (mutually induced voltage) E_{2} respectively.

**E _{1} = – V_{1}** and

**E**

_{2}= V_{2}The magnitudes of E_{1} and E_{2} are proportional to the number of turns N_{1} and N_{2} respectively.

The secondary induced voltage E_{2} will also oppose V_{1}. So E_{2} also appears in phase opposition with V_{1}. The magnitude of E_{2} however is dependent on the turns ratio N_{2}/N_{1}.

E_{1} and E_{2} appear in phase with each other and in phase opposition with V_{1}.

**Power Input on No Load**

The input power to the ideal transformer on no load is given by,

\therefore \mathbf{P_o = V_1 I_m \cos \Phi_0}

Where Φ_{0}= Angle between V_{1} and I_{m} which is 90^{o}.

\therefore \mathbf{P_o = V_1 I_m \cos 90^{\circ} = 0\;W}

Thus the input power to an ideal transformer on no load is zero. The output power is zero and there are no losses taking place in the ideal transformer.

Also Read:EMF Equation of a Transformer

## Ideal Transformer on Load

When some load is connected between the secondary terminals of the transformer, the transformer is said to be loaded or on load.

Due to the load on the secondary, a finite secondary current starts flowing. If the load is (R + L) type then I_{2} will lag behind V_{2} by an angel Φ_{2} as shown in the phasor diagram shown in the below figure.

As per the Len’z law, the secondary current I_{2} will oppose the cause producing it. Hence it opposes the magnetic flux. This is called the demagnetizing effect of I_{2}.

Due to demagnetizing, the flux is weakened and it reduces the amount of seld-induced voltage E_{1}. Due to the reduction in E_{1}, the difference between V_{1} and E_{1} will increase and the additional primary current I’_{2} called as load component starts flowing as shown in the below figure.

\mathbf{I'_2 = I_2 \times \frac{N_2}{N_1} = K I_2}

The current I’_{2} = K I_{2} and it is 180^{0} out of phase with the current I_{2}. The net primary current I_{1} is the phasor sum of I’_{2} and Imag as shown in the below phasor diagram.

\therefore\; \mathbf{\overline{I_1} = \overline{I_2'} + \overline{I_{mag}}}

Thus due to the load on the secondary side, the primary current of the transformer increases to supply the additional power to the load.

The angle between V1 and I1 is Φ_{1} as shown in the figure. hence the primary power factor is cosΦ_{1}.

## Why does the primary current increase when the load current is increased?

When the transformer is loaded, the load current I_{2} will start flowing. Due to the increase in load current I_{2} the secondary ampere turns N_{2}I_{2} will also increase.

This increased secondary mmf (N_{2}I_{2}) will increase the flux Φ_{2} set up by the secondary current.

This flux opposes the main flux Φ_{1} set up in the core by the current flowing through the primary winding. Hence the secondary mmf N_{2}I_{2} is called the demagnetizing ampere turns.

Due to the reduction in the main flux Φ_{1}, the induced emf in the primary winding E_{1} will also reduce. Hence the difference between V_{1} and E_{1} will increase and the primary current will increase.

Also Read:Voltage and Current Ratio of Transformer

## FAQs on Ideal Transformer

**What is an ideal transformer?**

A transformer that doesn’t have any losses like copper and core is known as an ideal transformer.

**An ideal transformer is one which has**

An ideal transformer is one which has no losses and magnetic leakage. A transformer that is free from all types of losses can be called an ideal transformer.

**What is an ideal transformer and its properties?**

An ideal transformer is a transformer that has no copper losses, no iron loss in the core, and no leakage flux. In other words, an ideal transformer gives output power exactly equal to the input power.

**Where is the ideal transformer used?**

An ideal transformer is used on a 220 V line to deliver 2 A at 110 V.

**What is the EMF of an ideal transformer?**

E_{1} = V_{1} and E_{2} = V_{2}

**What is the ideal efficiency of a transformer?**

100% efficiency