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Q39PE

Expert-verified
Found in: Page 734

### College Physics (Urone)

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.

See the step by step solution

## Step 1: Define Resistance

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.

## Step 2: Concepts and Principles

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.

## Step 3: The given data

• 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.

## Step 4: Evaluating the temperature of the resistor must be raised and doubled its resistance

(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: $$\Delta T{\rm{ }} = {\rm{ }}5.00{\rm{ }} \times {\rm{ }}{10^{5{\rm{ }}o}}C$$.

## Step 5: Evaluating the temperature at which the resistor is half

(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: $$\Delta T{\rm{ }} = {\rm{ }} - 2.50{\rm{ }} \times {\rm{ }}{10^{5{\rm{ }}o}}C$$.

## Step 6: Determining unreasonable results for parts (a) and (b)

(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.

## Step 7: Determine which assumptions or premises are unreasonable

(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.