When a straight wire is heated, its resistance changes...

When a straight wire is heated, its resistance changes according to the equation $$ R=R_{0}\left[1+\alpha\left(T-T_{0}\right)\right] $$ (Eq. 17.7), where $\alpha$ is the temperature coefficient of resistivity. (a) Show that a more precise result, which includes the length and area of a wire change when it is heated, is $$ R=\frac{R_{0}\left[1+\alpha\left(T-T_{0}\right)\right]\left[1+\alpha^{\prime}\left(T-T_{0}\right)\right]}{\left[1+2 \alpha^{\prime}\left(T-T_{0}\right)\right]} $$ where $\alpha^{\prime}$ is the coefficient of linear expansion. (See Chap$\begin{array}{ll}\text { Ler } 10 .) & \text { (b) Compare the two results for a } 2.00-\mathrm{m} \text { -long }\end{array}$ copper wire of radius $0.100 \mathrm{~mm}$, starting at $20.0^{\circ} \mathrm{C}$ and heated to $100.0^{\circ} \mathrm{C}$.

When a straight wire is heated, its resistance changes according to the equation
$$
R=R_{0}\left[1+\alpha\left(T-T_{0}\right)\right]
$$
(Eq. 17.7), where $\alpha$ is the temperature coefficient of resistivity. (a) Show that a more precise result, which includes the length and area of a wire change when it is heated, is
$$
R=\frac{R_{0}\left[1+\alpha\left(T-T_{0}\right)\right]\left[1+\alpha^{\prime}\left(T-T_{0}\right)\right]}{\left[1+2 \alpha^{\prime}\left(T-T_{0}\right)\right]}
$$
where $\alpha^{\prime}$ is the coefficient of linear expansion. (See Chap$\begin{array}{ll}\text { Ler } 10 .) & \text { (b) Compare the two results for a } 2.00-\mathrm{m} \text { -long }\end{array}$ copper wire of radius $0.100 \mathrm{~mm}$, starting at $20.0^{\circ} \mathrm{C}$ and heated to $100.0^{\circ} \mathrm{C}$.

Key Concepts

Electrical Resistance and Conductor Geometry

The electrical resistance of a conductor depends not only on the intrinsic resistivity of the material but also on its geometry, specifically its length and cross-sectional area. The standard relation R = ?L/A shows how resistance increases with length and decreases with an increase in area. When temperature changes occur, both the resistivity (through the temperature coefficient) and the geometry (through thermal expansion) are affected, necessitating a more precise formulation that combines these effects for accurate results.

Temperature Coefficient of Resistivity

This coefficient quantifies how the intrinsic resistivity of a material changes with temperature. In most conductors, resistivity increases with temperature due to increased lattice vibrations that scatter charge carriers. The linear approximation is often expressed with a coefficient ?, such that the resistivity at a given temperature is proportional to the resistivity at a reference temperature plus a term that linearly increases with temperature difference.

Coefficient of Linear Expansion

This coefficient describes how the dimensions (length, width, and area) of a material change as its temperature changes. Materials typically expand upon heating. The coefficient of linear expansion (?') quantifies the fractional change in length per unit change in temperature. This is crucial when dealing with physical dimensions that directly affect properties like resistance, since expansion can alter both the length and the cross-sectional area of a conductor.

Transcript

0:00 Hi there.

00:01 So for this problem, we need to demonstrate the expression that is given in dirt that relates the resistance.

00:10 So we're going to start that at a temperature t, the resistance is given by the product between the resistivity, the length over the area.

00:26 Where we know that the resistivity depends on the temperature by the following equation, that is, the initial resistivity times one plus the constant alpha times the temperature minus the initial temperature, t -subs -0.

00:46 So we also note that the length depends on the temperature, that is going to be the initial length l -sub -0, times one, plus alpha prime times the temperature minus the initial temperature t 0.

01:08 And finally, we also know that the area depends on the temperature, and that is the initial area is times 1 plus 2 times alpha and times the temperature minus the temperature t sub 0.

01:27 So we compute all of this into this expression, right here.

01:32 So we're going to have the following.

01:33 We're going to have that the resistance is equal to the resistance r sub zero times one plus alpha times t minus t sub zero.

01:50 And this times one plus alpha times the temperature minus the temperature t sub zero.

02:00 And this divided by one.

02:07 Plus two times alpha prime times the temperature minus the temperature t sub zero.

02:14 So that's the expression that we wanted to obtain.

02:20 So for part b of this problem, we are told that we need to compare the two results for a two meters long cooper, two meters long cooper, with a radius of 0 .1 .1.

02:41 100 millimeters and starting at a temperature, initial temperature, i'm going to call this initial temperature of 20 celsius degrees to a temperature of 100 celsius degrees.

03:04 So for this we use the equation from before, so we will find that the initial resistance is, equal to the resistivity that in this case is going to be 1 .7 times 10 to the minus 8 alms times meters this times the length the initial length which is 2 meters and this divided by the cross -sectional area that in this case we know that that is pi times the radius and the radius is 0 .100 times to the minus 3 meters and that elevated to this 2...

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