Operation of Ideal transformer

CASE NO:1 IDEAL TRANSFORMER AT NO-LOAD

 

In figure;

                 V1= Infinite voltage source 

                 Io = No load current

                 Φ = Fulx due to no load current 

                 I2 = Load current

                 S = Switch

• "Infinite source means it is capable to deliver any value of load power"

✴️ OPERATION:

⇾Initially Switch "s" is open and transformer is energing through Infinite voltage source and Setup the magnetic flux Φ in the core (Φ∝N1*Io).

⇾The value of Io is 2 to 5% of full load current. But for ideal transformer its value is tends to zero because in ideal transformer the permeability of core is very high (tends to Infinite) and due to high permeability the Reluctance will also tends to Zero ( Reluctance= L/μA).

CASE NO:-2 IDEAL TRANSFORMER AT ON-LOAD

Fig. Vector Diagram

Fig(b) Transformer at ON-LOAD

✴️ OPERATION:

⇾Now switch "S" is closed and load is connected to the transformer secondary side at that time load draws current I2.

⇾Due to I2 current,  Flux Φ2 are produced in the opposite direction to the main flux Φ, and Φ2 weaken the main flux Φ. Due to this main flux will try to maintain its value constant so it will draws extra current I1 from the source.

⇾Due to I1 current another flux Φ1∝ N1*I1 is produced in the direction of main flux Φ. So due to flux Φ1 effect of flux Φ2 is compensated to each other and main flux Φ is maintained in the Core of transformer.

• From m.m.f balanced equation;

                 N1*I1 = N2*I2

                  

                 N1/N2 = I2/I1

So for ideal transformer,

          V1/V2 = N1/N2 = I2/I1 = a

Where ; a= Turns ratio.

               ................Thank you 🙂.................