**Construction of single-phase transformers-** A single-phase transformer consists of two winding, called primary and secondary windings mounted on a magnetic core. To confine flux to a definite path magnetic flux is used. Transformer cores are made from thin sheets (called lamination) of high-grade silicon steel. The laminations reduce eddy-current loss and the silicon steel reduces hysteresis loss. The laminations are insulated from each other by heat resistant enamel insulation coating. L-type and e-type laminations are used. The laminations are built up into

These are two basic types of transformer constructions, the core type and the shell type.

**IDEAL TRANSFORMER:**

An ideal transformer has the following properties:

- Its primary and secondary winding resistances are negligible.
- The core has infinite permeability(µ) so that negligible mmf is required to establish the flux in the core.
- Its leakage flux and leakage inductances are zero. The entire flux is constricted to the core and links both windings.
- There are no losses due to resistance, hysteresis and eddy currents. Thus, the efficiency is 100 percent.

It is to be noted that practical (commercial) transformer has none of these properties in spite of the fact that its operation is close to ideal.

An ideal iron-core transformer is shown in figure below. It consists of two coils wound in the same direction on a common magnetic core. The winding connected to the supply V_{1}, is called the primary. The winding connected to the load, Z_{1}, is called the secondary.

Since the ideal transformer has
zero primary and zero secondary impedance, the voltage induced in the primary E_{1}
is equal to the applied voltage V_{1}. Similarly, the secondary voltage
V_{2}, is equal to the secondary induced voltage E_{2}. The
current I_{1} drawn from the supply is just sufficient to produce mutual
flux ɸ_{M} and the required magneto-motive force (mmf) I_{1},T_{1},
to overcome the demagnetizing effect of the secondary mmf I_{2},T_{2},
as a result of connected load.

By Lenz’s law E_{1}, is equal and opposite to V_{1}, Since E_{2}, and E_{1}, are both induced by the same mutual flux, E_{2}, is in the same direction’s E_{1}, but opposite to V_{1}, The magnetizing I_{µ} lags V_{1 }by 90˚ and produces ɸ_{M} in phase with I_{µ}_{1}_{2} lag ɸ_{M} by 90° and are produced by ɸ_{M} .V_{2}, is equal in magnitude to E_{2}, and is opposite to V_{1}.

For an ideal transformer, if

a = transformation ratio = turn ratio

then, a = T_{1}/T_{2 }= E_{1}/E_{2}
V_{1}/V_{2} = I_{2}/I_{1} ……(1)

I_{1} T_{1} = I_{2} T_{2} …….(2)

E_{1}
I_{1} = E_{2} I_{2} = S_{2} = S_{1 } ……..(3)

V_{1}
I_{1} = V_{2} I_{2} = S_{2} = S_{1 }………(4)

Equation (2) states that the demagnetizing ampere turns of the secondary are equal and also opposite to the magnetizing mmf of the primary of an ideal transformer.

Equation (3) shows that the volt-amperes (apparent power) drawn from the primary supply is equal to the volt-amperes (apparent power) transferred to the secondary without any loss in and ideal transformer. In other words,

input volt-amperes = output volt-amperes

Also, V_{1}I_{1}/1000
= V_{2}I_{2}/1000

(kVA)_{1}
= (kVA)_{2 ………….}(5)

Or,

input kilovolt amperes output kilovolt amperes

Thus, the kVA input of an ideal transformer is equal to the kVA output. That is, kVA is the same on both the sides of the transformer.

Thanks for sharing your thoughts on rabatine. Regards

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