1.

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`

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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