Saved Bookmarks
| 1. |
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 isA. 100 V, 16 SAB. 40 V, 40 AC. 160 V, 10 AD. 80 V, 20 A |
|
Answer» Correct Answer - 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` |
|