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A wire has a resistance of 2.5 Omegaat 100^@ C. Temperature coefficient of resistance of the material alpha = 3.6 xx 10^(-3) K^(-1)at 25^@ C. Find its resistance at 25^@ C. |
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Answer» Solution :`R = 2.5 Omega , t_2 = 100^@ C` `ALPHA = 3.6 xx 10^(-3) K^(-1) , R_1 = ?` `t_2 = 100+ 273 = 373 K . T_1 = 25^@ C = 273 + 25 = 298 K` `alpha =(R_2 -t_1)/(R_1 (t_2 -t_1))` `3.6xx10^(-3) = (2.5 - R_1)/(R_1 (373-298)) = (2.5 - R_1)/(R_1 xx 75)` `3.6xx10^(-3) xx 75 xx R_1 = 2.5 - R_1` `0.270 xx R_1 = 2.5 - R_1` `0.270 xx R_1 + R_1 = 2.5` `1.27 R_1 = 2.5 impliesR_1 = 1.96 Omega` Resistance at `25^@ C , R_1 = 1.97 Omega` |
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