Operation of Transformer on No load condition

A transformer is said to be at no load when the primary of a transformer is connected to the source of ac supply and the secondary is kept open. Similarly, a transformer at load condition indicates that the secondary is connected to the load.

In the previous article, we discussed the ideal transformer, which doesn’t have a core loss or copper loss. But in actual transformers, there will be a considerable amount of losses associated with the windings and core when it is operating at load conditions.

Before analyzing the transformer at load condition, it is important to understand the operation of the transformer on no load condition.

Operation and Phasor diagram

Consider the following practical transformer, where the primary winding is connected to the alternating voltage source and the secondary winding is kept open. It means the transformer is operated on no load condition.

When the transformer is connected on no-load, a small amount of current flows through the primary winding which is not purely reactive. This primary current under no-load conditions has to supply two losses.

iron loss in the core, which includes both hysteresis loss and eddy current loss
a very small amount of copper loss in primary winding (no copper loss in secondary as it is open).

Hence, the no-load primary input current I₀ does not lag behind the applied voltage V₁ by 90⁰ but lags behind V₁ by an angle φ₀ < 90⁰.

No-load input power is given by

$W_0 = V_1 I_0 cos \phi _0$

where cos φ₀ is the power factor under no-load conditions. The phasor diagram of the transformer under no-load condition is shown in the following Figure.

phasor diagram of Transformer at No load condition

As you can see in the above phasor diagram, the no-load current or the exciting current I₀ has the following two components.

(i) First is the active or working or wattful or iron loss component. It supplies the iron loss plus a small quantity of copper loss in the primary winding.

$I_w = I_0 cos \phi _0$

(ii) The other component is the magnetizing or wattless component. Its function is to sustain the magnetic flux in the transformer core.

$I_m = I_0 sin \phi _0$

Thus the no-load primary current is obtained from the vector sum of the active component and magnetizing component.

$I_0 = \sqrt{I_w^2 + I_m^2}$

Power factor lagging angle,

$\phi _0 = tan^{-1} \frac{I_m}{I_w}$

Points to Remember

The following points should be kept in mind while analyzing the transformer on no load.

The no-load primary current I_o is very small compared to the full-load primary current. Its value ranges from 2 to 5 percent of the full-load primary current.
The no-load current I₀ is made up of a relatively large magnetizing component I_m, and a comparatively small iron loss component I_w, so the power factor of a transformer on no load is very small.
Since the no-load primary copper loss is very small, it may be neglected. Thus the no-load primary input power is practically equal to the iron loss in the transformer (I₀ = I_w).
As core loss is the cause for the shift in the current phasor, the phase angle φ₀ is also known as the hysteresis angle of advance.
Since the permeability of the core material varies with the instantaneous value of the no-load primary current, the waveform of the magnetizing current is not truly sinusoidal. As such, it should not be represented by a phasor because only sinusoidally varying quantities are represented by rotating phasors. But, in practice, it makes no appreciable difference.

Solved Problem #1

A single-phase transformer of 3,300/220 V, 50 Hz takes a no-load current of 0.8 A and 500 W. Calculate (i) the active and magnetizing current (ii) its power factor.

Given: I₀ = 0.8 A, V₁ = 3300 V, W₀ = 500 W

Solution:

(i) The active component of no-load current is obtained from the no-load input power.

$W_0 = V_1 I_0 cos \phi _0$

$500 = 3300* I_0 cos \phi _0$