Practical Transformer on No Load

Practical Transformer on No Load

AU: Dec.-06,14, May-08

• Actually in practical transformer iron core causes hysteresis and eddy current losses as it is subjected to alternating flux. While designing the transformer the efforts are made to keep these losses minimum by,

1. Using high grade material as silicon steel to bas reduce hysteresis loss.

2. Manufacturing core in the form of laminations or stacks of thin laminations to reduce eddy current loss.

• Apart from this there are iron losses in the practical transformer. Practically primary winding has certain resistance hence there are small primary copper loss present.

• Thus the primary current under no load condition has to supply the iron losses i.e. hysteresis loss and eddy current loss and a small amount of primary copper loss. This current is denoted as I₀.

• Now the no load input current I₀ has two components:

1. A purely reactive component I_m called magnetising component of no load current required to produce the flux. This is also called wattless component.

2. An active component I_c which supplies total losses under no load condition called power component of no load current. This is also called wattful component or core loss component of I₀.

• The total no load current I₀ is the vector addition of I_m and I_c.

• In practical transformer, due to winding resistance, no load current I₀ is no longer at 90° with respect to V₁. But it lags V₁ by angle ϕ₀ which is less than 90°. Thus cos ϕ₀ is called no load power factor of practical transformer.

• The phasor diagram is shown in the Fig. 6.7.1. It can be seen that the two components of I₀ are.

I_m = I₀ Sin ϕ₀ ............... (6.7.2)

• This is magnetising component lagging V₁ exactly by 90^o.

I_c = I₀ cos ϕ₀.................... (6.7.3)

 • This is core loss component which is in phase with V₁.

The magnitude of the no load current is given by,

I₀ = √I_m²+I_c² ......................(6.7.4)

While ϕ = no load primary power factor angle

• The total power input on no load is denoted as W₀ and is given by,

W₀ = V₁ I₀ cos ϕ₀ =V₁ I_c  ............(6.7.5)

• It may be noted that the current I₀ is very small, about 3 to 5 % of the full load rated current. Hence the primary copper loss is negligibly small hence I_c is called core loss or iron loss component. Hence power input W₀ on no load always represents the iron losses, as copper loss is negligibly small. The iron losses are denoted as P_i and are constant for all load conditions.

W₀= V₁ I₀ cos ϕ₀ = P_i = Iron loss ......(6.7.6)

Ex. 6.7.1 The no load current of a transformer is 10 A at a power factor of 0.25 lagging, when connected to 400 V, 50 Hz supply. Calculate,

a) Magnetising component of no load current

b) Iron loss and 

c) Maximum value of flux in the core. Assume primary winding turns as 500.

Sol. :

Review Question

1. Draw and explain the no load phasor diagram of a single phase transformer. AU: Dec.-06,14, May-08, Marks 8