Transformer at no0load and with load

Transformer on No-load

In the above discussion, we assumed an ideal transformer i.e. one in which there were no core losses
and copper losses. But practical conditions require that certain modifications be made in the foregoing
1126 Electrical Technology
theory. When an actual transformer is put on load, there is iron loss in the core and copper loss in the
windings (both primary and secondary) and these losses are not entirely negligible.
Even when the transformer is on no-load, the primary input current is not wholly reactive. The primary
input current under no-load conditions has to supply (i) iron losses in the core i.e. hysteresis loss and eddy
current loss and (ii) a very small amount of copper loss in primary (there being no Cu loss in secondary as
it is open). Hence, the no-load primary input current I0 is not at 90° behind V1 but lags it by an angle ?0 <
90°. No-load input power
W0 = V1I0 cos ?0
where cos ?0 is primary power factor under no-load conditions. No-load
condition of an actual transformer is shown vectorially in Fig. 32.16.
As seen from Fig. 32.16, primary current I0 has two components :
(i) One in phase with V1. This is known as active or working or
iron loss component Iw because it mainly supplies the iron loss plus
small quantity of primary Cu loss.
Iw = I0 cos ?0
(ii) The other component is in quadrature with V1 and is known as
magnetising component I? because its function is to sustain the
alternating flux in the core. It is wattless.
I? = I0 sin ?0
Obviously, I0 is the vector sum of Iw and I?, hence I0 = (I? + I?).

The following points should be noted carefully :
1. The no-load primary current I0 is very small as compared to the full-load primary current. It is
about 1 per cent of the full-load current.
2. Owing to the fact that the permeability of the core varies with the instantaneous value of the
exciting current, the wave of the exciting or magnetising current is not truly sinusoidal. As such it
should not be represented by a vector because only sinusoidally varying quantities are represented
by rotating vectors. But, in practice, it makes no appreciable difference.
3. As I0 is very small, the no-load primary Cu loss is negligibly small which means that no-load
primary input is practically equal to the iron loss in the transformer.
4. As it is principally the core-loss which is responsible for shift in the current vector, angle ?0 is
known as hysteresis angle of advance