*Ideal Transformer-Properties,Phasor Diagram,Working*

*Ideal Transformer-Properties,Phasor Diagram,Working*

**What is an Ideal Transformer?**

A transformer which is free from all the losses is called an ideal transformer. It has zero copper loss, zero iron loss and therefore 100% efficient. In an ideal transformer, the output power is equal to the input power in an ideal transformer. An ideal transformer can not exist in real condition.

The resistance and reactance of the winding must be as minimum as possible to have low power losses. However, it is only imaginary that winding has zero resistance and reactance.

## Properties of an Ideal Transformer

The proprieties of an ideal transformer are follows.

**Zero Copper loss**– Copper loss of an ideal transformer is zero because the winding resistance of the primary and the secondary is zero. This means that both the winding have zero resistance and thus, copper loss I^{2}R is zero**Zero leakage flux**– The leakage flux is part of the main flux which does not link to secondary winding of the transformer. In an ideal transformer, we assume that the entire flux produced in the primary links to the secondary, and thus the entire flux produced in the primary links to the primary and secondary winding. In an ideal transformer main flux is restricted in the core only and there is no flux leakage.**Zero Iron loss**– It is assumed that an ideal transformer has a core of infinite permeability and have infinite resistance. Therefore, no load loss is zero.**Low Magnetizing Current-**An ideal transformer draws very low current to set up magnetic flux in the core. An ideal transformer has infinite permeability core.**Zero % Voltage Regulation-**There is no variation in the secondary output voltage with current from zero to full load secondary current. Therefore, an ideal transformer has zero % voltage regulation.**100% Efficiency**– If a transformer fulfills above three criteria the losses- eddy current, hysteresis and copper loss is zero and the transformer delivers output power equal to the input power, and efficiency of the transformer is 100 %. However, this is hypothetical condition and transformer or any equipment can not be 100 % efficient.

Therefore, we can define an ideal transformer as a transformer which has no copper loss, no iron loss and therefore, 100 % efficient.

**Equivalent Circuit of Ideal Transformer**

The equivalent circuit of an ideal transformer is as given below.

**An ideal transformer has following value of quantities.**

- Primary resistance – R1=0
- Primary reactance – X1 =0
- Magnetizing reactance – X
_{0}Pure inductive - Core loss component -R
_{0}=∞ [ Core loss- I_{w}^{2}R_{0 }= 0 , because I_{w}=0 ] - Secondary resistance – R2=0
- Secondary reactance – X2 =0

## Working principle of Ideal Transformer

We can redraw the equivalent circuit diagram of an ideal transformer taking all the transformer quantities as mentioned above for an ideal transformer.

_{1}to primary of a transformer, the primary winding induces back EMF E

_{1}. In an ideal transformer the magnitude of E

_{1}is equal to the applied voltage V

_{1}because transformer has zero winding resistance and leakage reactance. The alternating voltage cause alternating current to flow in the primary. The rate of change of current cause magnetic flux production. The magnetic flux flow in the core and it links to the secondary. The voltage induced in the primary and secondary winding depends on;

- Number of turns in the primary and secondary
- Frequency of the supply voltage
- Rate of change of flux

**E**

_{1}=4.44 N_{1}f Ф_{m}————–(1)**Voltage induced in the primary winding is;**

**E**

_{2}=4.44 N_{2}f Ф_{m}————-(2)**E**

_{2}/E_{1}= N_{2}_{ }/N_{1}—————–(3)**Where,**

E

_{1}= Induced EMF in primary winding

E

_{2}= Induced EMF in secondary winding

N

_{1}= Number of turns in primary winding

N

_{2}= Number of turns in the secondary winding

f = Frequency

Ф

_{m}= Flux in the core

**Ideal Transformer Equations**

The losses in an ideal transformer is zero. Therefore input power is equal to the output power.

Input Power = Output Power

P_{1} = P_{2 }E_{1} I_{1} Cos Ф = E_{2} I_{2} Cos Ф

E_{1} I_{1} = E_{2} I_{2 }**E _{2} /E_{1} = I_{1} / I_{2 }———–(4)**

From equation(3) & (4)

**E _{2} /E_{1} = I_{1} / I_{2 } = N_{2}_{ }/N_{1} ——-(5)**

## Phasor Diagram of Ideal Transformer

When we apply voltage V_{1 }to primary, the back EMF (E_{1}) induced across the primary which opposes the primary voltage. The magnitude of E_{1} is equal to V_{1} and polarity of E_{1} is just 180 degree opposite to V_{1}. The back EMF E_{1 }oppose the applied voltage. The phasor diagram of V_{1 }and E_{1} is given below.

The transformer draws more current at the instant of switching on of the transformer because the back EMF induced in the primary is zero.

The transformer draw magnetizing current to set up magnetic field in the transformer core. The magnetizing current lags the applied voltage by 90 degree. This current is called the magnetizing current I_{μ }. The phasor diagram of V_{1 }and I_{μ} is given below.

The alternating magnetizing current set up the alternating flux in the core, and the flux produced is proportional to the magnitude of the magnetizing current and it is in the phase with the current. The flux produced in the primary gets linked to the secondary and induce E_{2 }voltage. The induced voltage E_{2} in the secondary is equal to the output voltage V_{2} of the secondary. The phasor diagram of V_{1}, E_{1}, E_{2} and I_{μ} is given below.