Question

A transformer with efficiency `80%` works at `4 kW` and `100 V`. If the secondary voltage is `200 V`, then the primary and secondary currents are respectively

Answer 1

`eta = 80%, E_(P) = 100 V, E_(P) I_(P) = 4 kW = 4000 W`
`E_(s) = 240 V, I_(P) = ?, I_(s) = ?`
`I_(P) = (E_(P) I_(P))/(E_(P)) = (4 xx 1000)/(100) = 40 A`
To calculate `I_(s)` use `eta = (E_(s) I_(s))/(E_(P) I_(P))`