Using the equation of power for an ideal transformer, prove `(I_(p))/( I_(s)) = ( V_(s))/( V_(p)) = ( N_(s))/( N_(p))`

Explanation

If the transformer is an ideal one, its efficiency is 100% so no energy losses. The power input is equal to the power output and since.
p = IV
`:. `In put power = Output power
`I_(p) V_(p) = I_(s) V_(s) ` …(1)
Some energy is always lose, since a well designed transformer may have an efficiency of more than 95%.
For an ideal transformer,
`(V_(s))/(V_(p)) = ( N_(s))/( N_(p)) ` ...(2)
Where `V_(s)` and `V_(p)` voltage across secondary and primary coil respectively and `N_(s)` and `N_(p)` are the number of turns in secondary coil and primary coil respectively.
From equation (1) and (2),
`(I_(p))/( I_(s)) = (V_(s))/(V_(p)) = ( N_(s))/( N_(p))` ...(3)
Hence, I and V both oscillate with the same frequency as the ac source equation (3) also gives the ratio of the amplitude or rms value of corresponding quantities.