Consider an ideal transformer on no load as shown in the Fig. 3. The supply voltage is and as it is V

_{1 }an no load the secondary current I_{2 }= 0. The primary draws a current I

_{1 }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 V_{1 }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 E

The phasor diagram for the ideal transformer on no load is shown in the Fig. .2._{1 }and E_{2 }, 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 V_{1}. Hence E_{1 }is in antiphase with V_{1}but equal in magnitude. The induced E_{2 }also opposes V_{1 }hence in antiphase with V_{1 }but its magnitude depends on N_{2}. Thus E_{1 }and E_{2 }are in phase.Fig. 2 Phasor diagram for ideal transformer on no load |

It can be seen that flux Φ is reference. I

E_{m }produces Φ hence in phase with Φ. V_{1 }leads I_{m }by 90^{o}as winding is purely inductive so current has to lag voltage by 90^{o}._{1 }and E

_{2 }are in phase and both opposing supply voltage .

The power input to the transformer is V

_{1 }I_{1 }cos (V_{1 }^ I_{1 }) i.e. V_{1 }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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