# 20.3 Resistance and resistivity  (Page 3/6)

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The resistance of an object also depends on temperature, since ${R}_{0}$ is directly proportional to $\rho$ . For a cylinder we know $R=\mathrm{\rho L}/A$ , and so, if $L$ and $A$ do not change greatly with temperature, $R$ will have the same temperature dependence as $\rho$ . (Examination of the coefficients of linear expansion shows them to be about two orders of magnitude less than typical temperature coefficients of resistivity, and so the effect of temperature on $L$ and $A$ is about two orders of magnitude less than on $\rho$ .) Thus,

$R={R}_{0}\left(\text{1}+\alpha \Delta T\right)$

is the temperature dependence of the resistance of an object, where ${R}_{0}$ is the original resistance and $R$ is the resistance after a temperature change $\Delta T$ . Numerous thermometers are based on the effect of temperature on resistance. (See [link] .) One of the most common is the thermistor, a semiconductor crystal with a strong temperature dependence, the resistance of which is measured to obtain its temperature. The device is small, so that it quickly comes into thermal equilibrium with the part of a person it touches.

## Calculating resistance: hot-filament resistance

Although caution must be used in applying $\rho ={\rho }_{0}\left(\text{1}+\alpha \Delta T\right)$ and $R={R}_{0}\left(\text{1}+\alpha \Delta T\right)$ for temperature changes greater than $\text{100º}\text{C}$ , for tungsten the equations work reasonably well for very large temperature changes. What, then, is the resistance of the tungsten filament in the previous example if its temperature is increased from room temperature ( $\text{20ºC}$ ) to a typical operating temperature of $\text{2850º}\text{C}$ ?

Strategy

This is a straightforward application of $R={R}_{0}\left(\text{1}+\alpha \Delta T\right)$ , since the original resistance of the filament was given to be ${R}_{0}=0\text{.}\text{350 Ω}$ , and the temperature change is $\Delta T=\text{2830º}\text{C}$ .

Solution

The hot resistance $R$ is obtained by entering known values into the above equation:

$\begin{array}{lll}R& =& {R}_{0}\left(1+\alpha \Delta T\right)\\ & =& \left(0\text{.}\text{350 Ω}\right)\left[\text{1}+\left(4.5×{\text{10}}^{–3}/\text{ºC}\right)\left(\text{2830º}\text{C}\right)\right]\\ & =& \text{4.8 Ω.}\end{array}$

Discussion

This value is consistent with the headlight resistance example in Ohm’s Law: Resistance and Simple Circuits .

## Phet explorations: resistance in a wire

Learn about the physics of resistance in a wire. Change its resistivity, length, and area to see how they affect the wire's resistance. The sizes of the symbols in the equation change along with the diagram of a wire.

## Section summary

• The resistance $R$ of a cylinder of length $L$ and cross-sectional area $A$ is $R=\frac{\mathrm{\rho L}}{A}$ , where $\rho$ is the resistivity of the material.
• Values of $\rho$ in [link] show that materials fall into three groups— conductors, semiconductors, and insulators .
• Temperature affects resistivity; for relatively small temperature changes $\Delta T$ , resistivity is $\rho ={\rho }_{0}\left(\text{1}+\alpha \Delta T\right)$ , where ${\rho }_{0}$ is the original resistivity and $\text{α}$ is the temperature coefficient of resistivity.
• [link] gives values for $\alpha$ , the temperature coefficient of resistivity.
• The resistance $R$ of an object also varies with temperature: $R={R}_{0}\left(\text{1}+\alpha \Delta T\right)$ , where ${R}_{0}$ is the original resistance, and $R$ is the resistance after the temperature change.

## Conceptual questions

In which of the three semiconducting materials listed in [link] do impurities supply free charges? (Hint: Examine the range of resistivity for each and determine whether the pure semiconductor has the higher or lower conductivity.)

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The reason why a photon can go at this speed is BECAUSE it had no mass. nothing can go this speed or faster because it needs to have no mass or negative mass. that's why it's called the constant.
when a photon hits something that is opaque, this is the only way to "stop"it. it isn't merely stopped but absorbed and turned into heat energy, then the remaining energy is reflected in different wavelengths. that reflection is what we call color. the darker something is, the less photons are ther
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