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Chapter 18: Problem 28

The resistance of a conductor is \(19.8 \Omega\) at \(15.0^{\circ} \mathrm{C}\) and \(25.0 \Omega\) at \(85.0^{\circ} \mathrm{C} .\) What is the temperature coefficient of resistance of the material?

Short Answer

Expert verified
Answer: The temperature coefficient of resistance of the material is approximately \(3.72 \times 10^{-3} \ \mathrm{K}^{-1}\).

Step by step solution

01

Identify the given values

We are given the resistances of the conductor at two temperatures: - \(R_1 = 19.8 \Omega\) at \(T_1 = 15.0^{\circ} \mathrm{C}\) - \(R_2 = 25.0 \Omega\) at \(T_2 = 85.0^{\circ} \mathrm{C}\)
02

Convert the temperatures to Kelvin scale

To calculate the temperature coefficient of resistance, it is essential to use the Kelvin scale for temperature. The conversion from Celsius to Kelvin is as follows: \(K = ^{\circ}\mathrm{C} + 273.15\). So, the temperatures in Kelvin are: - \(T_1 = 15.0^{\circ} \mathrm{C} + 273.15 = 288.15 \mathrm{K}\) - \(T_2 = 85.0^{\circ} \mathrm{C} + 273.15 = 358.15 \mathrm{K}\)
03

Substitute the given values into the temperature coefficient of resistance formula

Now that we have the resistances and corresponding temperatures in Kelvin, we can calculate the temperature coefficient of resistance (\(\alpha\)) using the following formula: \(\alpha = \dfrac{(R_2 - R_1)}{R_1(T_2 - T_1)}\) Substitute the given values into the formula as follows: \(\alpha = \dfrac{(25.0 \Omega - 19.8 \Omega)}{19.8 \Omega(358.15 \mathrm{K} - 288.15 \mathrm{K})}\)
04

Simplify and solve for the temperature coefficient of resistance

First, simplify the equation: \(\alpha = \dfrac{5.2 \Omega}{19.8 \Omega(70 \mathrm{K})}\) Next, cancel out the units and solve for \(\alpha\): \(\alpha = \dfrac{5.2}{19.8 \times 70}\) \(\alpha \approx 3.72 \times 10^{-3} \ \mathrm{K}^{-1}\)
05

Write down the final answer

The temperature coefficient of resistance of the material is approximately \(3.72 \times 10^{-3} \ \mathrm{K}^{-1}\).

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Most popular questions from this chapter

A source of emf \(\&\) has internal resistance \(r\), (a) What is the terminal voltage when the source supplies a current \(I ?\) (b) The net power supplied is the terminal voltage times the current. Starting with \(P=I \Delta V,\) derive Eq. \((18-22)\) for the net power supplied by the source. Interpret each of the two terms. (c) Suppose that a battery of emf 8 and internal resistance \(r\) is being recharged: another emf sends a current \(I\) through the battery in the reverse direction (from positive terminal to negative). At what rate is electric energy converted to chemical energy in the recharging battery? (d) What is the power supplied by the recharging circuit to the battery?
A 1.5 -horsepower motor operates on \(120 \mathrm{V}\). Ignoring \(l^{2} R\) losses, how much current does it draw?
(a) Given two identical, ideal batteries (emf \(=\mathscr{E}\) ) and two identical light-bulbs (resistance \(=R\) assumed constant), design a circuit to make both bulbs glow as brightly as possible. (b) What is the power dissipated by each bulb? (c) Design a circuit to make both bulbs glow, but one more brightly than the other. Identify the brighter bulb.
(a) In a charging \(R C\) circuit, how many time constants have elapsed when the capacitor has \(99.0 \%\) of its final charge? (b) How many time constants have elapsed when the capacitor has \(99.90 \%\) of its final charge? (c) How many time constants have elapsed when the current has \(1.0 \%\) of its initial value?
Poiseuille's law [Eq. (9-15)] gives the volume flow rate of a viscous fluid through a pipe. (a) Show that Poiseuille's law can be written in the form \(\Delta P=I R\) where \(I=\Delta V / \Delta t\) represents the volume flow rate and \(R\) is a constant of proportionality called the fluid flow resistance. (b) Find \(R\) in terms of the viscosity of the fluid and the length and radius of the pipe. (c) If two or more pipes are connected in series so that the volume flow rate through them is the same, do the resistances of the pipes add as for electrical resistors \(\left(R_{e q}=R_{1}+R_{2}+\cdots\right) ?\) Explain. (d) If two or more pipes are connected in parallel, so the pressure drop across them is the same, do the reciprocals of the resistances add as for electrical resistors \(\left(1 / R_{\mathrm{eq}}=\right.\) $\left.1 / R_{1}+1 / R_{2}+\cdots\right) ?$ Explain.
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