Suggested languages for you:

Americas

Europe

Q32PE

Expert-verifiedFound in: Page 734

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 \(1.03\).

**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.**

- 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 \({55.0^o}\;{\rm{C}}\).

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)

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: \(1.03\).

94% of StudySmarter users get better grades.

Sign up for free