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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

`40A,16A`

B

`16A,40A`

C

`20A,40A`

D

`40A,20A`

Text Solution

Verified by Experts

The correct Answer is:
A
`eta=("outepur power")/("input power")=(E_(s)I_(s))/(E_(p)I_(p))implies(80)/(100)=(200xxI_(s))/(4xx10^(3))`
`impliesI_(s)=(80)/(100)xx(4xx1000)/(200)=16A`
Also,`E_(p)I_(p)=4KWimpliesI_(p)=(4xx10^(3))/(100)=40A`
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