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Expert-verified Found in: Page 734 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # An electronic device designed to operate at any temperature in the range from $$-10.0°{\rm{C}}$$ to $$55.0°\;{\rm{C}}$$ contains pure carbon resistors. By what factor does their resistance increase over this range?

The resistance is increased by the factor $$1.03$$.

See the step by step solution

## Step 1: Concepts and Principles

The resistance of a conductor can be given in terms of the temperature as:

$$R = {R_0}\left( {1 + \alpha \left( {T - {T_0}} \right)} \right)$$

The value of $${R_0}$$ is the resistance at some reference temperature, and the value of $${T_0}$$ and the value of $$\alpha$$ is the temperature coefficient of resistivity.

## Step 2: The given data

• The temperature coefficient of resistivity for carbo is: $$\alpha = - 0.5{\rm{ }} \times {\rm{ }}{10^{ - 3}}^ \circ \;{{\rm{C}}^{ - 1}}$$
• Temperature range for which electronic device can be used from $$- {10.0^o}\;{\rm{C}}$$ to $${55.0^o}\;{\rm{C}}$$.

## Step 3: Resistance of carbon in its upper and lower limit of temperature

Resistance of the carbon resistors corresponding to the lower limit of the temperature range is found from the above equation as:

$${R_{lower}} = {R_0}\left( {1 + \alpha \left( {{T_{lower}} - {T_0}} \right)} \right)$$ ………………(I)

The value we take here is: $${T_0}{\rm{ }} = {\rm{ }}{20^\circ }\;{\rm{C}}$$ which has to be the reference temperature.

Resistance of the carbon resistors corresponding to the upper limit of the temperature range is found from the above equation as:

$${R_{Upper}} = {R_0}\left( {1 + \alpha \left( {{T_{Upper}} - {T_0}} \right)} \right)$$ …………….(II)

## Step 4: Calculation of the carbon resistance, which increases over the temperature

Dividing the first equation with the second equation, we get:

\begin{align}\frac{{{R_{lower}}}}{{{R_{upper}}}}{\rm{ }} &= {\rm{ }}\frac{{{R_0}\left( {1 + \alpha \left( {{T_{lower}} - {T_0}} \right)} \right)}}{{{R_0}\left( {1 + \alpha \left( {{T_{Upper}} - {T_0}} \right)} \right)}}\\ &= {\rm{ }}\frac{{1 + \alpha \left( {{T_{lower}} - {T_0}} \right)}}{{1 + \alpha \left( {{T_{Upper}} - {T_0}} \right)}}\end{align}

Entering the values and we obtain is:

\begin{align}\frac{{{R_{lower}}}}{{{R_{upper}}}}{\rm{ }} &= {\rm{ }}\frac{{1 + ( - 0.5 \times {{10}^{ - 3}}^ \circ \;{{\rm{C}}^{ - 1}})( - {{10.0}^ \circ }\;{\rm{C}} - {{20.0}^ \circ }\;{\rm{C}})}}{{1 + ( - 0.5 \times {{10}^{ - 3}}^ \circ \;{{\rm{C}}^{ - 1}})({{55.0}^ \circ }\;{\rm{C}} - {{20.0}^ \circ }\;{\rm{C}})}}\\ &= \frac{{1.015}}{{0.9835}}\\ &= {\rm{ }}1.03\end{align}

Therefore, the factor by which the resistance increases is: $$1.03$$. ### Want to see more solutions like these? 