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 transformer has two winding – primary and secondary winding. The winding has resistance and the reactance. Therefore, the losses occurs in a practical transformer. However, an ideal transformer has zero losses and 100 % efficiency, which is completely hypothetical condition.

The equivalent circuit of a practical transformer is as given below.

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 I²R 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 – X₀ Pure inductive
Core loss component -R₀ =∞ [ Core loss- I_w²R₀= 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.

When we feed AC supply V₁ to primary of a transformer, the primary winding induces back EMF E₁. In an ideal transformer the magnitude of E₁ is equal to the applied voltage V₁ 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

Voltage induced in the primary winding is;

E₁ =4.44 N₁ f Ф_m ————–(1)

Voltage induced in the primary winding is;

E₂ =4.44 N₂ f Ф_m ————-(2)

E₂ /E₁ = N₂/N₁ —————–(3)

Where,
E₁ = Induced EMF in primary winding
E₂ = Induced EMF in secondary winding
N₁ = Number of turns in primary winding
N₂ = 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

P₁ = P₂E₁ I₁ Cos Ф = E₂ I₂ Cos Ф
E₁ I₁ = E₂ I₂E₂ /E₁ = I₁ / I₂———–(4)

From equation(3) & (4)

E₂ /E₁ = I₁ / I₂ = N₂/N₁ ——-(5)

Phasor Diagram of Ideal Transformer

When we apply voltage V₁to primary, the back EMF (E₁) induced across the primary which opposes the primary voltage. The magnitude of E₁ is equal to V₁ and polarity of E₁ is just 180 degree opposite to V₁. The back EMF E₁oppose the applied voltage. The phasor diagram of V₁and E₁ 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 V₁and 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 E₂voltage. The induced voltage E₂ in the secondary is equal to the output voltage V₂ of the secondary. The phasor diagram of V₁, E₁, E₂ and I_μ is given below.