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The following observations were recorded on a platinum resistance thermometer: Resistance at melting point of ice = 3.70 ohm, resistance of boiling point of water at normal pressure Omega and resistance at t”^(@)C = 5.29 ohm. Calculate (i) Temperature coefficient of resistance of platinum, (ii) Value of temperature t |
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Answer» SOLUTION :(i) TEMPERATURE coefficient of resistance is givenby `aplha=(R_100-R_0)/(R_0xx100)=(4.71-3.70)/(3.70xx100)` `(1.01)/(370)=2.73xx10^(-3)PER"^(C )` For temperature t we have `t=100"^(@)Cxx(R_1-R_0)/(R_100-R_0)` `100^(@)Cxx(5.29-3.70)/(4.71-3.70)=100^(@)Cxx(1.59)/(1.01)=157.4"^(@)C` |
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