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The primary coil of a transformer has 200 turns and the secondary has 1000 turns. If the power output from the secondary at 1000 V is 9 kW, calculate (i) the primary voltage, and(ii) the heat loss in the primary coil if the resistance of primary is 0.2 Omegaand the efficiency of the transformer is 90%. |
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Answer» SOLUTION :Here, `N_(P) = 200, N_(s) = 1000 V, V_(s) = 1000 V`, output power = 9 kW = 9000 W, and EFFICIENCY `eta = 90% = 0.9` (i) `therefore V_(s)/V_(p) = N_(s)/N_(p)`, hence, `V_(p) = (V_(s).N_(p))/(N_(s)) = (1000 XX 200)/1000 = 200 V` (ii) `therefore` Efficiency `eta = ("output power")/("input power") = ("Output")/(V_(p).I_(p))` `therefore I_(p) = (9000)/(200 xx 0.9) = 50 A` As resistance of primary coil `R_(p) = 0.2 Omega` `therefore` Power IONS in primary coil as heat `=I_(p)^(2) . R_(p) = (50)^(2) xx 0.2 = 500 W` |
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