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A transformer having efficiency 90% is working on `100 V` and at `2.0 kW` power. If the current in the seconary coil is `5 A`, calculate (i) the current in the primary coil and (ii) voltage across the secondary coil. |
Answer» Here, `eta = 90% = (9)/(10), I_(s) = 5A, E_(p) = 100 V` (i) `E_(p) I_(p) = 2 kW = 2000 W` `I_(p) = (2000)/(E_(p))` or `I_(P) = (2000)/(100) = 20 A` (ii) `eta = ("Output power")/("Input power") = (E_(s) I_(s))/(E_(P) I_(P))` or `E_(s) I_(s) = eta xx E_(p) I_(P)` `= (9)/(10) xx 2000 = 1800 W` `:. E_(s) = (1800)/(I_(s)) = (1800)/(5) = 360` volt |
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