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Question

You are working in a laboratory that studies the effects of currents in various crystals. One of the experiments involves a requirement for a steady current of I = 0.500 A in a wire that delivers the current to the crystal. Both the wire and the crystal are in a chamber whose interior temperature T will vary from 40.0C-40.0^{\circ} \mathrm{C} to 150C150^{\circ} \mathrm{C}. The wire is made of tungsten and is of length L = 25.0 cm and radius r = 1.00 mm A test run is being made before the crystal is added to the circuit. Your supervisor asks you to determine the range of voltages that must be supplied to the wire in the test run to maintain its current at 0.500 A.

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Answered 1 year ago
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The voltage at the lowest temperature is determined from the given current and the resistance expressed through the temperature difference:

Vmin=RI=IR0(1+α(TT0))=Iρ0lr2π(1+α(TT0))=0.55.61080.25(103)2π(1+4.5103(4020))103mV=1.63mV\begin{align*} V_{\text{min}}&=RI\\ &=IR_{0}(1+\alpha(T-T_{0}))\\ &=\dfrac{I\rho_{0}l}{r^{2}\pi}(1+\alpha(T-T_{0}))\\ &=\dfrac{0.5\cdot5.6\cdot10^{-8}\cdot0.25}{(10^{-3})^{2}\cdot\pi}(1+4.5\cdot10^{-3}(-40-20))\cdot10^{3}\:\text{mV}\\ &=\boxed{1.63\:\text{mV}} \end{align*}

The voltage at the highest temperature then is:

Vmax=RI=IR0(1+α(TT0))=Iρ0lr2π(1+α(TT0))=0.55.61080.25(103)2π(1+4.5103(15020))103mV=3.53mV\begin{align*} V_{\text{max}}&=RI\\ &=IR_{0}(1+\alpha(T-T_{0}))\\ &=\dfrac{I\rho_{0}l}{r^{2}\pi}(1+\alpha(T-T_{0}))\\ &=\dfrac{0.5\cdot5.6\cdot10^{-8}\cdot0.25}{(10^{-3})^{2}\cdot\pi}(1+4.5\cdot10^{-3}(150-20))\cdot10^{3}\:\text{mV}\\ &=\boxed{3.53\:\text{mV}} \end{align*}

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