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Q39PE

Expert-verifiedFound in: Page 734

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**(a) To what temperature must you raise a resistor made of constantan to double its resistance, assuming a constant temperature coefficient of resistivity? (b) To cut it in half?**

**(c) What is unreasonable about these results?**

**(d) Which assumptions are unreasonable, or which premises are inconsistent?**

(a)The temperature to which the resistor must be raised to double its resistance is obtained as: \(\Delta T{\rm{ }} = {\rm{ }}5.00{\rm{ }} \times {\rm{ }}{10^{5{\rm{ }}o}}C\).

(b)The temperature at which the resistor must be to cut its resistance in half is obtained as: \(\Delta T{\rm{ }} = {\rm{ }} - 2.50{\rm{ }} \times {\rm{ }}{10^{5{\rm{ }}o}}C\).

(c)The unreasonable about the results of parts (a) and (b) is the temperature.

(d)It is wrong to assume that the temperature coefficient of resistivity is constant.

**In an electrical circuit, resistance is a measure of the resistance to current flow. The Greek letter omega \({\rm{(\Omega )}}\) is used to represent resistance in ohms.**

**The resistance of a conductor varies approximately linearly with temperature according to the expression given as:**

\(R{\rm{ }} = {\rm{ }}{R_0}{\rm{ }}(1{\rm{ }} + {\rm{ }}\alpha {\rm{ }}(T{\rm{ }} - {\rm{ }}{T_0}))\)

**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 constantan is:**\(\alpha {\rm{ }} = {\rm{ }}0.002{\rm{ }}\times{\rm{ }}{10^{ - 3}}^\circ {C^{ - 1}}\).**The value \({\rm{\alpha }}\) is assumed to be constant.**

**(a)Resistance of the resistor is expressed as a function of temperature change by the above equation as:**

**\(R{\rm{ }} = {\rm{ }}{R_0}{\rm{ }}(1{\rm{ }} + {\rm{ }}\alpha \Delta T)\)**

**As, we are interested in the temperature at which the resistance of the resistor doubles, we substitute \({\rm{2}}{{\rm{R}}_{\rm{0}}}\) for \({\rm{R}}\):**

**Dividing both sides by \({R_0}\), we obtain:**

**\(2 = {\rm{ }}(1{\rm{ }} + {\rm{ }}\alpha \Delta T)\)**

**Subtracting one from both sides:**

**\(\alpha \Delta T{\rm{ }} = {\rm{ }}1\)**

**Divide both sides by \(\alpha \):**

**\(\Delta T{\rm{ }} = {\rm{ }}\frac{1}{\alpha }\)**

**Substituting the numerical values:**

**\(\begin{aligned}{c}\Delta T{\rm{ }} = {\rm{ }}\frac{1}{{0.002{\rm{ }}x{\rm{ }}{{10}^{ - 3}}^\circ {\rm{ }}{C^{ - 1}}}}\\ = {\rm{ }}5.00 \times {10^{5{\rm{ }}o}}{\rm{ }}C\end{aligned}\)**

**Therefore, temperature to which the resistor must be raised to double its resistance is: \(\Delta T{\rm{ }} = {\rm{ }}5.00{\rm{ }} \times {\rm{ }}{10^{5{\rm{ }}o}}C\).**

**(b)Resistance of the resistor is expressed as a function of temperature change with the help of the above equation as:**

**\(R{\rm{ }} = {\rm{ }}{R_0}{\rm{ }}(1{\rm{ }} + {\rm{ }}\alpha \Delta T)\)**

**As, we are interested in the temperature at which the resistance of the resistor cuts in half, we substitute \(\frac{{{R_0}}}{2}\) for \(R\):**

**\(\frac{{{R_0}}}{2}{\rm{ }} = {\rm{ }}{R_0}{\rm{ }}(1{\rm{ }} + {\rm{ }}\alpha \Delta T)\)**

**Dividing both sides by \({R_0}\), we obtain:**

**\(\frac{1}{2}{\rm{ }} = {\rm{ }}1{\rm{ }} + {\rm{ }}\alpha \Delta T\)**

**Subtracting one from both sides:**

**\(\alpha \Delta T{\rm{ }} = {\rm{ }} - \frac{1}{2}\)**

**Divide both sides by \(\alpha \):**

**\(\Delta T{\rm{ }} = {\rm{ }} - \frac{1}{{2\alpha }}\)**

**Substituting the numerical values:**

**\(\begin{aligned}{c}\Delta T{\rm{ }} = {\rm{ }} - \frac{1}{{2({\rm{ }}0.002{\rm{ }}x{\rm{ }}{{10}^{ - 3}}^\circ {C^{ - 1}})}}\\{\rm{ }} = {\rm{ }} - 2.50 \times {10^{5{\rm{ }}o}}C\end{aligned}\)**

**Therefore, temperature at which the resistor must be to cut its resistance in half is: \(\Delta T{\rm{ }} = {\rm{ }} - 2.50{\rm{ }} \times {\rm{ }}{10^{5{\rm{ }}o}}C\).**

**(c)The changes in temperature in parts (a) and (b) are not reasonable.**

** **

**To put forward this into perceptive, one should know that the sun's surface is about \(6000^\circ {\rm{ }}C\) and the absolute zero (nothing can go below absolute zero) which is \( - 273^\circ {\rm{ }}C\). Also, the constantan would change phase long before reaching these temperatures.**

**(d)It will be very wrong to assume that the temperature coefficient of resistivity is constant because of the huge temperature range. It is constant only over a small temperature range (mainly up to \(100^\circ {\rm{ }}C\) or so).**

**Therefore, we get:**

**(a)The temperature to which the resistor must be raised to double its resistance is obtained as: \(\Delta T{\rm{ }} = {\rm{ }}5.00{\rm{ }} \times {\rm{ }}{10^{5{\rm{ }}o}}C\).**

**(b)The temperature at which the resistor must be to cut its resistance in half is obtained as: \(\Delta T{\rm{ }} = {\rm{ }} - 2.50{\rm{ }} \times {\rm{ }}{10^{5{\rm{ }}o}}C\).**

**(c)The temperature is unreasonable about the results of parts (a) and (b).**

**(d)It is wrongly assumed that the temperature coefficient of resistivity is constant.**

** **

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