# Is the supply voltage and the induced voltage same in a loaded (practical) transformer?

For a loaded transformer not considering the winding resistances and inductance, induced voltage = Applied/terminal voltage in primary and secondary sides. But given a practical transformer (with winding resistance and inductance) this will result in a drop in the voltage and applied voltage > induced voltage (for primary side).

We know, $$\frac{N_{primary}}{N_{Secondary}} = \frac{E_1}{E_2}$$ where

$$E_1$$ = primary induced voltage

$$E_2$$ = secondary induced voltage

In case of Ideal transformers, $$\frac{N_{primary}}{N_{Secondary}} = \frac{E_1}{E_2} = \frac{V_1}{V_2}$$ where

$$V_1$$ = Applied Voltage (Primary)

$$V_2$$ = terminal voltage (Secondary)

But when considering a practical transformer, clearly the above relation should not hold. My book says that $$V_1, E_1$$ and $$V_2, E_2$$ can be used interchangeably everywhere. But in a practical transformer, those quantities will not be equal. $$V_1 = E_1 + I(R_1 + jX_1)$$ $$V_2 = E_2 - I(R_2 + jX_2)$$ So only the first relation should hold. Is my book wrong, or am I missing something here?

Any help would be appreciated.