Two heating elements which is 500 ohms and 250 ohms are connected in series with temperature coefficients of 0.001 and 0.003 ohms per °C, respectively at 20°C. Calculate the effective temperature coefficient of the combination.
Resistance at t=0 °C
R0=RT(1−αTT)R0(500 Ω)=500(1−0.001×20)=490 ΩR0(250 Ω)=250(1−0.003×20)=235 Ω(490+235)=(500+250)(1−αT×20)725=750(1−20αT)1−20αT=0.9667αT=0.033320=0.00167R_0=R_T (1- α_TT) \\ R_0(500 \; \Omega) = 500(1-0.001 \times 20) = 490 \; \Omega \\ R_0(250 \; \Omega) = 250(1 -0.003 \times 20) = 235 \; \Omega \\ (490 + 235) = (500+250)(1-α_T \times 20) \\ 725 = 750(1 -20α_T) \\ 1-20α_T = 0.9667 \\ α_T = \frac{0.0333}{20} = 0.00167R0=RT(1−αTT)R0(500Ω)=500(1−0.001×20)=490ΩR0(250Ω)=250(1−0.003×20)=235Ω(490+235)=(500+250)(1−αT×20)725=750(1−20αT)1−20αT=0.9667αT=200.0333=0.00167
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