`(i)` Draw a labelled diagram of a step-down transformer. State the principle of its working.
`(ii)` Find the ratio of primary and secondary currents in terms of turn ratio in an ideal transformer.
`(iii)` How much current is drawn by the primary of a transformer connected to `220V` supply when it delivers power to a `110V-550W` refrigerator ?

Explanation

`(i)` N/A `(ii)` If the resistance of primary coil is negligible the emf `(epsilon_(beta))` induced in the primary coil, will be equal to the applied potential difference `(V_(p))` across its ends. Similarly if the secondary circuit is open, then the potential difference `V_(s)` across its ends will be equal to the emf `(epsilon_(s))` induced in it. Therefore,
`(V_(s))/(V_(p))=(epsilon_(S))/(epsilon_(p))=(N_(S))/(N_(P))=("say")`
where, `r=(N_(S))/(N_(P))` is called the transformation ratio. If, `i_(p)` and `i_(s)` are the instantenous currents in primary and secondary coils and there is no loss of energy, then
For about `100%` efficiency, Power in primary `=` Power in secondary
`V_(p)i_(P)=V_(s)i_(S)`
`(i_(S))/(i_(P))=(V_(P))/(V_(S))=(N_(P))/(N_(S))=(1)/(r )`
`(iii) V_(P)I_(P)=V_(s)I_(s)`
`V_(p)=220V`
`V_(s)=110V`
Out put power `=550W`
Output current `I_(s)=(550)/(110)=5A`
Now put in `(i)`
`220xxI_(P)=110xx5`
`I_(P)=(5)/(2)=2.5A`