Actually in practical transformer iron core causes hysteresis and eddy current losses as it is subjected to alternating flux. While designing the transformer the efforts are made to keep these losses minimum by,

- Using high grade material as silicon steel to reduce hysteresis loss.
- Manufacturing core in the form of laminations or stacks of thin lamination to reduce eddy current loss.

Apart from this there are iron losses in the practical transformer. Practically primary winding has certain resistance hence there are small primary copper loss present.

Thus the primary current under no load condition has to supply the iron losses i.e. hysteresis loss and eddy current loss and a small amount of primary copper loss. This current is denoted as I

_{o}.Now the no load input current I

_{o }has two components :- A purely reactive component I
_{m }called magnetising component of no load current required to produce the flux. This is also called wattless component. - An active component I
_{c }which supplies total losses under no load condition called power component of no load current. This also called wattful component or core loss component of I_{o}.

Th total no load current I

The phasor diagram is shown in the Fig. 1. It can be seen that the two components I

This is magnetising component lagging V

This is core loss component which is in phase withV

The magnitude of the no load current is given by,

While Φ

The total power input on no load is denoted as W

a) Magnetising component of the no load current

b) Iron loss and c) Maximum value of flux in the core.

Assume primary winding turns as 500.

a) I

Φ

b) P

= W

= 1000 W

c) On no load, E

Now E

_{o }is the vector addition of I_{m }and I_{c}. In practical transformer, due to winding resistance, no load current I

_{o }is no longer at 90^{o }with respect to V_{1}. But it lags V_{1 }by angle Φ_{o }which is less than 90^{o }. Thus cos Φ_{o }is called no load power factor of practical transformer.Fig 1. Practical transformer on no load |

_{o }are,This is magnetising component lagging V

_{1 }exactly by 90^{o }.This is core loss component which is in phase withV

_{1}.The magnitude of the no load current is given by,

While Φ

_{o }= no load primary power factor angleThe total power input on no load is denoted as W

_{o }and is given by, It may be denoted that the current is very small, about 3 to 5% of the full load rated current. Hence the primary copper loss is negligibly small hence I

_{c}is called core loss or iron loss component. Hence power input W_{o }on no load always represent the iron losses, as copper loss is negligibly small. The iron losses are denoted as P_{i }and are constant for all load conditions.**Example 1**: The no load current of a transformer is 10 A at a power factor 0f 0.25 lagging, when connected to 400 V, 50 Hz supply. Calculate,

b) Iron loss and c) Maximum value of flux in the core.

Assume primary winding turns as 500.

**Solution**: The given value are, = 10 A, cos = 0.25, = 400 V and f = 50 Hza) I

_{m }= I_{o }sin Φ_{o }= magnetising componentΦ

_{o }= cos^{-1}(0.25) = 75.522^{o }**.**I^{.}._{m }= 10 x sin (75.522^{o }) = 9.6824 Ab) P

_{i }= iron loss = power input on no load= W

_{o }= V_{1 }I_{o }cos Φ_{o }= 400 x 10 x 0.25= 1000 W

c) On no load, E

_{1 }= V_{1 }= 400 V and N_{1 }= 500Now E

_{1 }= 4.44 f Φ_{m }N_{1 }_{ }**.**400 = 4.44 x 50 x Φ^{.}._{m }x 500**.**Φ^{.}._{m }= 3.6036 mWb

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