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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, 40A

D

40A, 20A

Text Solution

Verified by Experts

The correct Answer is:
A
`n%=(E_(s)I_(s))/(E_(p)I_(p))xx100 " Now" E_(p)I_(p)=4xx10^(3)`
`80=(200xxI_(s))/(4xx10^(3)) " " I_(p)=(4xx10^(3))/(E_(p))=40A`
`:.I_(s)=16A`
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