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Transformer with Losses but no Magnetic Leakage

We shall consider the follow two cases:

1.When transformer is on no-load.

2.When transformer is on load.

Transformer on no-load. A transformer is said to be on no-load if its secondary side is open and primary is connected to a sinusoidal alternating voltage V1 The alternating applied voltage will cause flow of alternating current in the primary winding which will create alternating flux. This primary input current (I0) under no-load conditions supply:

  1.         i.            Iron losses in the core (i.e. hysteresis loss and eddy current loss).
  2.      ii.            A very small amount of copper losses in primary (there being no copper loss in the secondary as it is open).
  3.    iii.            Thus I0 is not at 900 behind V1, but lags it by an angle ɸ0 < 900

No-load power input, P0 = V1I0 cos ɸ0

where cos ɸ0 = primary power factor under no-load conditions.

As is evident from Fig. 19, primary current I0 has the following two components:

i.                    Active or working or iron loss component Iw, This component is in phase with V1 and mainly supplies the iron loss plus small quantity of primary copper loss.

Iw = I0 cos ɸ0                                                      … (i)

ii.                 Magnetising component Im. This component is in quadrature with V1 and its function is to sustain the alternating flux in the core. It is wattless. 

The following points are worthnoting :

  • The no-load primary current Iw is very small as compared to the full-load primary current.
  • As 10 is very small, the no-load primary copper loss is negligibly small which means that no-load primary input is practically equal to the iron loss in the transformer.
  • Since, it is primarily the core loss which is responsible for shift in the current vector, angle ɸ0 is known as hysteresis angle of advance.

Example 7. A 3300/300 V single phase transformer gives 0.6 A and 60 W as ammeter wattmeter readings when supply is given to the low voltage winding and high voltage winding is kept open, find:

(i)                           Power factor for no-load current

(ii)                        Magnetising component.

(iii)                      Iron loss component.

Example 8. Find (i) active and reactive components of no-load current; and (ii) no-load current of a 440/220V single-phase transformer if the power input on no- load to the high voltage winding is 80 W and power factor of no- load current is 0.3 lagging.

Solution. Primary,             . E1 = 440 V

Secondary, E2 = 220 V

Power factor, cos ɸ0 = 0.3 (lagging)

No-load losses, P0 = 80 W

Example 9. A 3300/220 V, 30 kVA, single-phase transformer takes a no-load current of 1.5 A when the low voltage winding is kept open. The iron loss component is equal to 0.4 A find:

(i)                           No-load input power.

(ii)                        Magnetising component and power factor of no-load current:

Solution. Rating of transformer = 30 kVA

Primary, E1 = 3300 V

Secondary, E2 = 220 V

No-load current, I0 = 1.5 A

Iron loss component, Im = 0.4 A

(i) No-load input power,

P0 = V1I0 cos ɸ0 = V1Iw

= 3300 × 0.4 = 1320 W. (Ans.) 



Transformer on load. The transformer is said to be loaded when the secondary circuit of a transformer is completed through an impedance or load. The magnitude and phase of secondary current I2 with respect to secondary terminal voltage will depend upon the characteristic of load, i.e. current I2 will be in phase, lag behind and lead the terminal voltage V2 respectively when the load is purely resistive, inductive and capacitive.

The secondary current I2 sets up its own ampere-turns (= N2I2) and creates its own flux ɸ2 opposing the main flux ɸ0 created by no-load current I0, The opposing secondary flux ɸ2 weakens the primary flux ɸ0 momentarily hence primary counter or back e.m.f. E1 tends to be reduced. V1 gains the upper hand over E1 momentarily and hence causes more current to flow in primary. Let this additional primary current be I2‘. It is known as load component of primary current. The additional primary m.m.f. N1I2‘ sets up its own flux ɸ2‘ which is in opposition to ɸ2 (but is in the same direction as ɸ0) and is equal to it in magnitude. Hence they cancel each other. Thus we find that the magnetic effects of secondary current I2 are immediately neutralised by the additional primary current I2’ which is brought into existence exactly at the same instant as I2,

Fig. 20. An ideal transformer on load.

From above discussion it can be concluded that:

  • Whatever be the load conditions, the net flux passing through the core is approximately the same as at no-load. .
  • Since the core flux remains constant at all loads, the core loss almost remains constant under different loading conditions. 

The total primary current is the vector sum of I0 and I2‘ ; the current I2‘ is in antiphase with I2 and k times in magnitude.

The total primary current can be shown as below:

The vector diagrams for transformer on non-inductive, inductive and capacitive loads are