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A loss free transformer has 500 turns on...

A loss free transformer has `500` turns on its primary winding and `2500` in secondary. The meters of the secondary indicate `200` volts at `8` amperes under these condition. The voltage and current in the primary is

A

100 V, 16 SA

B

40 V, 40 A

C

160 V, 10 A

D

80 V, 20 A

Text Solution

Verified by Experts

The correct Answer is:
B
`(e_(P))/(e_(S))=(n_(P))/(n_(S))`
`therefore e_(P)=(n_(P)*e_(S))/(n_(S))=(500xx200)/(2500)=40V`
`I_(P)=(n_(S))/(n_(P))xxI_(S)=(2500)/(500)xx8=40A`
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