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

W₀=V₁I₀cosθ₀ (1.8)

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 I_w = 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
  
  and

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.

Solution

W₀ W, I₀ =0.04 A, V₁ = 4,400 V

1. Since W₀ = V₁I₀cosθ₀
   
   The no-load power factor is 0.454 (lagging).
2. I_W = I₀ cosθ₀ =0.04 × 0.454 = 0.0187 A
3. 
   ∴ I_μ = I₀ sinθ₀ = 0.04 × 0.891 = 0.0356