The resistance of a conductor at 20,^{circ}C | vedclass

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(D) The resistance of a conductor at temperature $t$ is given by $R_t = R_0(1 + \alpha t)$.Given $R_1 = 20\,\Omega$ at $t_1 = 20\,^{\circ}C$ and $R_2

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$80$

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(D) The resistance of a conductor at temperature $t$ is given by $R_t = R_0(1 + \alpha t)$.Given $R_1 = 20\,\Omega$ at $t_1 = 20\,^{\circ}C$ and $R_2 = 60\,\Omega$ at $t_2 = 500\,^{\circ}C$.Using the ratio: $\frac{R_1}{R_2} = \frac{1 + \alpha t_1}{1 + \alpha t_2} \Rightarrow \frac{20}{60} = \frac{1 + 20\alpha}{1 + 500\alpha}$.Solving for $\alpha$: $\frac{1}{3} = \frac{1 + 20\alpha}{1 + 500\alpha} \Rightarrow 1 + 500\alpha = 3 + 60\alpha \Rightarrow 440\alpha = 2 \Rightarrow \alpha = \frac{1}{220}\,^{\circ}C^{-1}$.Now,to find the temperature $t$ where $R = 25\,\Omega$,we use: $\frac{R_1}{R} = \frac{1 + \alpha t_1}{1 + \alpha t}$.$\frac{20}{25} = \frac{1 + 20(1/220)}{1 + t(1/220)} \Rightarrow 0.8 = \frac{1 + 1/11}{1 + t/220} \Rightarrow 0.8 = \frac{12/11}{1 + t/220}$.$1 + \frac{t}{220} = \frac{12}{11 	imes 0.8} = \frac{12}{8.8} = \frac{120}{88} = \frac{15}{11}$.$\frac{t}{220} = \frac{15}{11} - 1 = \frac{4}{11} \Rightarrow t = \frac{4}{11} 	imes 220 = 80\,^{\circ}C$.

The resistance of a conductor at $20\,^{\circ}C$ and $500\,^{\circ}C$ is $20\,\Omega$ and $60\,\Omega$ respectively. At what temperature will the resistance be $25\,\Omega$?

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