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The resistance of wire in a heater at roo temeperature is `65Oemga`. When the heater is connected to a `220V` supply the current settles after a few seconds to 2.8A. What is the steady temperature of the wire. (Temperature coefficient of resistance `alpha=1.70xx10^(-4) ^(@).C^(-)`)A. `955^(@)C`B. `1055^(@)C`C. `1155^(@)C`D. `1258^(@)C` |
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Answer» Correct Answer - D `"Here", T_(1)=27^(@)C, R_(1)=65Omega` `R_(2)=("Supply voltage")/("Steady current")=(220)/(2.8)=78.6Omega` Now using the relation `R_(2)=R_(1)[1+alpha(T_(2)-T_(1)) ` `therefore T_(2)-T_(1)=(R_(2)-R_(1))/(R_(1))xx(1)/(alpha)=(78.6-65)/(65)xx(1)/(1.7xx10^(-4))` `T_(2)-T_(1)=1231` `T_(2)=1231+T_(1)=1231+27=1258^(@)C` |
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