Ideal Transformer

 Definition : 

An ideal transformer is a theoretical transformer or an imaginary transformer that has no losses in it, meaning no copper losses, no core losses, and any other transformer losses. In case of Ideal transformer the efficiency of the transformer is considered as 100%. Let's see what is the expectation from an ideal transformer. 

Properties of Ideal Transformer :

1. The core of an ideal transformer must have high permeability (μ→∞) and low reluctance (R→0).  

As R = L/μA where R = Reluctance, L=Length of the winding, A= Cross sectional area of the winding. 

In ideal transformer the permeability is high and the reluctance is low because we want the entire flux should linked to the core. In case of electrical circuit we have seen that when we do short circuit then entire current flows through the short circuit, similarly here if we make the reluctance as zero then entire flux will flow through the core. 

The other reason to keep reluctance as zero because the magnetising(Iμ) current should be zero, so the power factor will be improved which is always desirable. 

We know NIμ = ΦR. If Φ is constant and R = 0 then Iμ = 0. 

This implies that without application of any current, flux is generated, that's why it is ideal transformer. 

2. There will be no losses in the transformer.

   ∴ Hysteresis Loss = 0, Eddy current Loss = 0, So the Core losses = 0. 

  Resistance of winding (Rwdg) must be zero, so that the copper losses will also be zero. That implies both coils are purely inductive in nature.

3. There should be no leakage of the flux from core to the air.

                            Φleakage = 0

4. Efficiency of the transformer should be 100% as there is no losses. That means the ideal transformer gives the power as output exactly equal to input power. But in practical it is not possible.

4. Magnetization curve must be linear, that B is directly proportional to H (B ∝ H) and there must be no saturation. 

Ideal Transformer Working :

Case 1 : Under No load Condition -

 

 Here no load condition means the switch is opened or secondary side is open circuited.  As it is ideal transformer, so the Reluctance (R) = 0 and Permeability (μ) = ∞. 

Due to i(t), the flux Φ(t) will be set up in primary winding and this entire flux is linked with secondary winding through the core. and there will be an induced emf in the primary side, E1(t).

             If i(t) = Imsin(ωt)  then Φ(t) = Φmsin(ωt)

According to lenz's law :

 As the transformer is in no load condition, that's why here E1(t) is known as Self induced emf and E2(t) is known as Mutual induced emf.

 

Phasor Diagram - 

The Value of E1(t) and E2(t) that means which will be more, is depend upon N1 and N2.

If  N1 > N2 ⇒ Step Down Transformer,

    N1 < N2 ⇒ Step Up Transformer,

     N1 = N2 ⇒ Isolation Transformer

Case 2 : Under load Condition -

Here load condition means the switch is closed or secondary side is connected with load. In the primary winding there will be an induced emf E1 and the entire flux will be linked with secondary winding and an emf will also be induced in the secondary winding E2(t). As switch is closed, so there will be a current i2(t) in the secondary side. 

So, i2(t) flows due to E2(t) and E2(t) is produced due to Φ(t). As i2(t) flows in the secondary winding then one another flux Φ2(t) will also be produced due to  i2(t) as well as E2(t). The direction of flux Φ2(t) will determine by the right hand thumb rule which will try to oppose the main flux Φ(t). But we know that transformer is a constant flux device, so to maintain the constant flux, Φ(t) will increase that means we require more current in primary side.

 

The value of Φ2(t) is - 

In the primary side the value of MMF will be 

As i0(t) = 0 in under no load condition so

Phasor Diagram -

The Value of E1(t) and E2(t) that means which will be more, is depend upon N1 and N2.