Ideal Transformer on No Load

       Consider an ideal transformer on no load as shown in the Fig. 3. The supply voltage is and as it is V₁ an no load the secondary current I₂ = 0.

       The primary draws a current I₁ which is just necessary to produce flux in the core. As it magnetising the core, it is called magnetising current denoted as I_m. As the transformer is ideal, the winding resistance is zero and it is purely inductive in nature. The magnetising current is I_m is very small and lags V₁ by 30^o as the winding is purely inductive. This I_m produces an alternating flux Φ which is in phase with I_m.

Fig.  1  Ideal transformer on no load

      The flux links with both the winding producing the induced e.m.f.s E₁ and E₂ , in the primary and secondary windings respectively. According to Lenz's law, the induced e.m.f. opposes the cause producing it which is supply voltage V₁. Hence E₁ is in antiphase with V₁ but equal in magnitude. The induced E₂ also opposes V₁ hence in antiphase with V₁ but its magnitude depends on N₂. Thus E₁ and E₂ are in phase.

       The phasor diagram for the ideal transformer on no load is shown in the Fig. .2.

Fig.  2  Phasor diagram for ideal transformer on no load

       It can be seen that flux Φ is reference. I_m produces Φ hence in phase with Φ. V₁ leads I_m by 90^o as winding is purely inductive so current has to lag voltage by 90^o.

       E₁ and E₂ are in phase and both opposing supply voltage .

       The power input to the transformer is V₁ I₁ cos (V₁ ^ I₁ ) i.e. V₁ I_m cos(90^o) i.e. zero. This is because on no load output power is zero and for ideal transformer there are no losses hence input power is also zero. Ideal no load p.f. of transformer is zero lagging.

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