A power transformer is used to set up an alternating e.m.f of `220V` to `4.4 kV` to transmits `6.6 kW` of power. The primary coil has `1000` turns. What is current rating of secondary? (Assume efficiency is `100%)`

Options

A. `2.5 A`
B. `1.5 A`
C. `1.0 A`
D. `0.75 A`

Answer

B

Explanation

`N_(s)=((E_(s))/(E_(p))). N_(p)=(4.4xx1000xx1000)/220`
`=20,000`
Current in primary is
`I_(p)E_(p)=6.6xx10^(3)`
`I_(p)=(6.6xx10^(3))/220=30A`
But `(I_(s))/(I_(p))=(N_(p))/(N_(s)) =1000/20,000=1/20`
`:. I_(s)=1/2xxI_(p)=30/20=1.5 A`