Using the equation of power for an ideal transformer,prove $\frac{I_p}{I_s} = \frac{V_s}{V_p} = \frac{N_s}{N_p}$.

(N/A) For an ideal transformer,the efficiency is $100 \%$,meaning there are no energy losses. Therefore,the input power is equal to the output power.
Since power $P = VI$,we have:
$\text{Input power} = \text{Output power}$
$I_p V_p = I_s V_s$
$\frac{I_p}{I_s} = \frac{V_s}{V_p} \quad \dots (1)$
For an ideal transformer,the ratio of voltages is equal to the ratio of the number of turns in the coils:
$\frac{V_s}{V_p} = \frac{N_s}{N_p} \quad \dots (2)$
Combining equations $(1)$ and $(2)$,we get:
$\frac{I_p}{I_s} = \frac{V_s}{V_p} = \frac{N_s}{N_p}$
This equation shows that the ratio of currents is inversely proportional to the ratio of voltages and turns.