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15 kW power is supplied through a line of 0.5Omega resistance under 250 V potential difference .Fimd the efficiency of the supply in percentage . |
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Answer» Solution :power , p = 15 kW = 15000 W So , current N the supply line , `I=(P)/(V)=(15000)/(250)=60A` power loss due to inactive resistance in the line `I^(2)R=(60)^(2)xx0.5=1800W`, Therefore , total power in line= 15000 + 1800 = 16800 W `therefore"Efficiency"=("effective power")/("total power")xx100%` `= (15000)/(16800)xx100%=89%` |
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