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A transformer with efficiency 80% works ...

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

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`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))`
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