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It is desired to make a `20.0Omega` coil of wire whose temperature coefficient of resistance is zero. To do this, a carbon resistor of resistance `R_(1)` is placed in series with an iron resistor of resistance `R_(2)`. The proportion of iron and carbon are so chosen that `R_(1) +R_(2)= 20 Omega` for all temperatures near `20^@C`. Find the values of `R_(1)` and `R_(2)`. Temperature coefficient of resistance for carbon, `alpha_(C)= - 0.5 xx 10^(-3)//^(@)C` and that of iron is `alpha_(Fe) =5 xx 10^(-3)//^(@)C`. |
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Answer» Let `Delta t^@C` be the rise in temperature. As per question we need, `R_(1) ( 1 +alpha_(C) Delta t) +R_(2) (1 + alpha_(Fe) Delta t)=20` Since, `R_(1) +R_(2) = 20`, therefore, `R_(1) alpha_(C) Delta t +R_(2) alpha_(Fe) Delta t=0` or `R_(1) +alpha_(C) = - R_(2) 1 + alpha_(Fe)` or `R_(1) xx (+0.5 xx10^(-3)) = - R_(2) xx 5xx10^(-3)` or `R_(1) = 10 R_(2)` So, `10 R_(2) + R_(2) = 20 or 11 R_(2) = 20` or `R_(2) = 20//11=1.82 Omega` and `R_(1)=10 xx 1.82 = 18.2 Omega` |
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