91Ó°ÊÓ

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.

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

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

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

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

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.

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.