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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

A

40 A, 16 A

B

16 A, 40 A

C

20 A, 40 A

D

40 A, 20 A

Text Solution

Verified by Experts

The correct Answer is:
A

`P_(P)=I_(P)*e_(P)`
`I_(P)=(P_(P))/(e_(P))=(4xx10^(3))/(100)=40A`
`eta=(P_(S))/(P_(P))=(I_(S)e_(S))/(P_(P))`
`I_(S)=(eta*P_(P))/(P_(P))=(0.8xx4xx10^(3))/(200)=16A`
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