Transformer On No Load
For an ideal transformer, we have assumed that there are no core losses and copper losses. For practical transformers, these two losses cannot be neglected. At no-load condition, the primary current is not fully reactive and it supplies (i) iron loss in the core, that is, hysteresis loss and eddy current loss and (ii) very small amount of copper loss in the primary. There is copper loss in the secondary because it is an open circuit. The no-load current lags behind V1 by an angle θ0, which is less than 90° (around 80°–85°). The no-load input power is given by
where cosθ0 is the no-load power factor. Figure 1.23 shows the no-load phasor diagram of a practical transformer. From Figure 1.23, the no-load primary current (I0) has the following two components:
- One component of I0, that is Iw = I0 cosθ0 is in phase with V1. Since Iw supplies the iron loss and primary copper loss at no load, it is known as active or working or iron loss component.
- The other component of I0 that is, = I0 sinθ0 is in quadrature with V1. It is known as magnetizing component. Its function is to sustain the alternating flux in the core and it is wattless.
From Figure 1.23, we have
The following points are most important:
- The no-load primary current is 1–5 per cent of full-load current.
- Since the permeability of the core varies with the instantaneous value of exciting or magnetizing current, the waveform of exciting or magnetizing current is not truly sinusoidal.
- Since I0 is very small, the no-load copper loss is negligible. Hence, no-load input is practically equal to the iron loss in the transformer.
- Since core loss is solely responsible for shifting the current vector I0, the angle θ0 is known as hysteresis angle of advance.
Figure 1.23 Phasor Diagram at No-Load
Figure 1.24 shows the equivalent circuit of the transformer at no load.
Figure 1.24 Equivalent Circuit of Transformer at No-load
Example 1.5 The no-load current of a 4,400/440 V, sinlge-phase, 50 Hz transformer is0.04. It consumes power 80 W at no load when supply is given to LV side and HV side is kept open. Calculate the following:
(i) Power factor of no-load current.
(ii) Iron loss component of current.
(iii) Magnetizing component of current.
W0 W, I0 =0.04 A, V1 = 4,400 V
- Since W0 = V1I0cosθ0
The no-load power factor is 0.454 (lagging).
- IW = I0 cosθ0 =0.04 × 0.454 = 0.0187 A
- ∴ Iμ = I0 sinθ0 = 0.04 × 0.891 = 0.0356