Expression of voltage $V_S$ induced in secondary coil of a transformer: Let $V_P$ be the applied alternating emf in the primary coil of a transformer. Alternating current produced in the primary $=I_1$. Induced emf in the primary will be equal to the applied emf for ideal transformer. So, $V_p = N_P \; \frac{N \phi}{dt}$ (where $N_P =$ number of turns in primary, $\; \frac{d\phi}{dt} =$ rate of change of flux) Since there is no loss of flux in an ideal transformer, Induced emf in secondary $= E_S = N_S \; \frac{d\phi}{dt}$ $\therefore$ Taking the ratio, $\; \frac{E_S}{E_P} = \; \frac{N_S}{N_P} = n$ (where $n =$ turns ratio or transformer ratio) Sources of power loss: (i) Hysteresis loss: Hysteresis loss is due to reversal of magnetization in the transformer core. This loss depends upon the volume and grade of the iron, frequency of magnetic reversals and value of flux density (ii) Copper loss is due to ohmic resistance of the transformer windings. Copper loss for the primary winding is $I_1^2 R_1$ and for secondary winding is $I_2^2 R_2$. Where, $I_1$ and $I_2$ are current in primary and secondary winding respectively, $R_1$ and $R_2$ are the resistances of primary and secondary winding respectively.