Answer:

Ideal Transformer on No Load

• Let us consider an ideal transformer on no load as shown in the Fig. The supply voltage is V, and as it is on no load the secondary current$\mathrm{I_2 = 0}$
• The primary draws a current$\mathrm{I_1}$, which is just necessary to produce flux in the core. As it is magnetising the core, it is called magnetising current denoted as$\mathrm{I_m }$ As the transformer is ideal, the winding resistance is zero and it is purely inductive in nature. The magnetising current In, is very small and lags$\mathrm{V_1 = 0}$ by 90° as the winding is purely inductive. This$\mathrm{I_m }$produces an alternating flux so$\mathrm{\phi}$, and$\mathrm{I_m }$ are in phase.
• The flux links with both the windings producing the induced e.m.f.s$\mathrm{E_1 \space and \space E_2}$, 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$\mathrm{V_1}$ Hence$\mathrm{E_1}$is in antiphase with$\mathrm{V_1}$ but its magnitude depends on$\mathrm{N_2}$. Thus$\mathrm{E_1 \space and \space E_2}$ are in phase.
• The phasor diagram for the ideal transformer on no load is shown in the Fig

• It can be seen that flux$\mathrm{\phi}$is reference.$\mathrm{I_m}$ produces$\mathrm{\phi}$ hence in phase with$\mathrm{\phi}$.$\mathrm{V_1}$leads$\mathrm{I_m}$by 90° as winding is purely inductive so current has to lag voltage by 90°.
• $\mathrm{E_1 \space and \space E_2}$are in phase and both opposing supply voltage$\mathrm{V_1 \space I_1}$The power input to the transformer is$\mathrm{V_1 \space I_m \space cos(90^0)}$ 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.