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A room has AC run for 5 hours a day at a voltage of 220V. The wiring of the room consists of Cu of 1mm radius and a length of 10m. Power consumption per day is 10 commerical units. What fraction of its goes in the joule heating in wires? What would happen if the wiring is made of aluminium of the same dimensions. |
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Answer» SOLUTION :Power consumption in 1 day (5 hours) P. = 10 unit `therefore `Power consumed in 1 hour, P ` = (P.)/(5) = (10)/(5) = 2 `unit. `therefore P = 2 ` kW `"" [ because 1 " unit = 1000 Watt" ] ` `therefore P = 2000 (J)/(s)` Power dissipated in resistor P = VI `therefore 2000 (J)/(s) = 220V x I` `therefore J = (2000)/(220) = 9` A Resistance of wire R = `(rho l)/(A)` A = AREA of cross section of wire power consumed in CURRENT carrying conductor, P = `I^(2) R = I^(2) (rho l)/(pi r^(2))` `therefore P = (9)^(2) xx 1.7 xx 10^(-8) xx (10)/((3.14 xx 10^(-3))^(2))` `therefore P = 81 xx (17)/(3.14 ) xx 10^(-1)` `therefore P= 4.38` W `therefore P approx 4 ` W ` therefore `Power used in conductor, `= ("power used")/("total power") xx 100 % ` = `(4)/(2000) xx 100 %` 0.2% Power consumed in aluminium wire = 4 `xx (rho_(Al))/(rho_(Cu))` P. = `4 xx (2.7 xx 10^(-8))/(1.7 xx 10^(-8))` `= 4 xx 1.588` `approx 4 xx 1.6` `= 6.4 (j)/(s)` Power consumed in aluminium in percentage, `Delta P = (P.)/(P ) xx 100 `% = `(6.4)/(2000) xx 100 %` = 0.32 % |
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