A transformer having efficiency of `90%` is working on 200 V and 3kW power supply. If the current in the secondary coild is 6, then the voltage across the secondary coil and the current in the primary coil respectively are

Explanation

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