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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 secondary coil is 5A, calculate (i) the current in the primary coil and (ii) voltage across the secondary coil. |
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Answer» Solution :Here `eta= 90% = (9)/(10), I_(s) = 5A` `E_(p) = 100 V`, `E_(p)I_(p) = 2kW = 2000 W` (i) `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` `THEREFORE E_(s) = (1800)/(I_(s)) = (1800)/(5) = 360` volt |
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