A transformer having efficiency 90% is working on 100 V and at 2.0 kW power. If the current in the secondary coil is 5A, calculate
(i) the current in the primary coil and
(ii) voltage across the secondary coil.

Explanation

Here `eta=90%=(9)/(10),I_(s)=5A,E_(p)=100V`,
(i) `E_(p)I_(p)=2kW=2000W`
`I_(p)=(2000)/(E_(p)) or I_(p)=(2000)/(100)=20A`
(ii) `eta=("Output power")/("Input power")=(E_(s)I_(s))/(E_(p)I_(p)) or E_(s)I_(s)=eta xxE_(p)I_(p)`
`=(9)/(10)xx2000=1800W`
`therefore E_(s)=(1800)/(I_(s))=(1800)/(5)="360 volt"`