(a) Redo Exercise\({\rm{20}}{\rm{.25}}\)taking into account the thermal expansion of the tungsten filament. You may assume a thermal expansion coefficient of\({\text{12}} \times {\text{1}}{{\text{0}}^{{\text{ - 6}}}}{\text{ / C}}\).

(b) By what percentage does your answer differ from that in the example?

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 temperature of an object is changed by an amount\(\Delta T\), its length changes by an amount\(\Delta L\)that is proportional to\(\Delta T\)and its initial length\({{\rm{L}}_{\rm{i}}}\):

\(\Delta L{\rm{ }} = {\rm{ }}\alpha {L_i}\Delta T\)…………………..(I)

Here the constant\({\rm{\alpha }}\)is the average coefficient of linear expansion.

The temperature of an object is changed by an amount\(\Delta T\), its area changes by an amount\(\Delta A\)that is proportional to\(\Delta T\)and its initial area\({A_i}\):

\(\Delta A{\rm{ }} = {\rm{ }}\alpha {A_i}\Delta T\)………………(II)

Here the constant\(\alpha \)is the average coefficient of linear expansion.

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

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

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 electrical resistance of the value\(R\)of a resistive circuit element depends on its geometric structure (its length symbolized as\(L\)and cross-sectional area symbolized as\(A\)) and on the resistivity\(\rho \)of the material of which it is made:

\(R{\rm{ }} = {\rm{ }}\rho \frac{L}{A}\)…………..(IV)

• Resistance of the filament at the reference temperature\({T_0}{\rm{ }} = {\rm{ }}{20.0^o}{\rm{ }}C\), it is found in the Example\(20.25\)to be:\({R_0}{\rm{ }} = {\rm{ }}0.350{\rm{ }}\Omega \)
• The initial length of the filament is:\(\begin{aligned}{L_i}{\rm{ }} = {\rm{ }}(4.00{\rm{ }}cm)(\frac{{1{\rm{ }}m}}{{100{\rm{ }}cm}})\\ = {\rm{ }}0.0400{\rm{ }}m\end{aligned}\)

• The operating temperature of the tungsten filament is:\(T{\rm{ }} = {\rm{ }}{2850^o}{\rm{ }}C\).
• The initial diameter of the tungsten filament is:\({D_i}{\rm{ }} = {\rm{ }}9.0{\rm{ }} \times {\rm{ }}{10^{ - 5}}{\rm{ }}m\).
• The resistivity of tungsten at temperature\({T_0}{\rm{ }} = {\rm{ }}{20.0^o}{\rm{ }}C\)is:\({\rho _0}{\rm{ }} = {\rm{ }}5.6{\rm{ }}x{\rm{ }}{10^{ - 8}}{\rm{ }}\Omega .m.\)

\(\begin{aligned}\Delta L{\rm{ }} = {\rm{ }}\alpha {L_i}\Delta T\\{L_f} - {L_i}{\rm{ }} = {\rm{ }}\alpha {L_i}(T - {T_0})\\{L_f}{\rm{ }} = {\rm{ }}{L_i} + \alpha {L_i}(T - {T_0})\\{L_f}{\rm{ }} = {\rm{ }}{L_i}(1 + \alpha {L_i}(T - {T_0}))\end{aligned}\)

Putting the numerical values and we get:

\(\begin{aligned}{L_f}{\rm{ }} = {\rm{ }}0.0400{\rm{ }}m{\rm{ }}(1 + (12{\rm{ }} \times {\rm{ }}{10^{ - 6}}{\rm{ }}^\circ {C^{ - 1}})({\rm{ }}2850^\circ {\rm{ }}C{\rm{ }} - {\rm{ }}20.0^\circ {\rm{ }}C))\\ = {\rm{ }}0.0413584{\rm{ }}m\end{aligned}\)

Similarly, the final cross-sectional area of the filament is found from the second equation as:

\(\begin{aligned}\Delta A{\rm{ }} = {\rm{ }}2\alpha {A_i}\Delta T\\{A_f} - {A_i}{\rm{ }} = {\rm{ }}2\alpha {A_i}(T - {T_0})\\{A_f}{\rm{ }} = {\rm{ }}A({1_i} + 2\alpha (T - {T_0}))\end{aligned}\)

The value of\({A_f}{\rm{ }} = {\rm{ }}\pi \frac{{D_i^2}}{4}\)which is the initial cross-sectional area of the filament:\({A_f}{\rm{ }} = {\rm{ }}\pi \frac{{D_i^2}}{4}{\rm{ }}({1_i}{\rm{ }} + {\rm{ }}2\alpha (T{\rm{ }} - {\rm{ }}{T_0}))\)

Putting the numerical values and we get:\(\begin{aligned}{A_f}{\rm{ }} = {\rm{ }}\frac{{\pi (9.0{\rm{ }} \times {\rm{ }}{{10}^{ - 5}}{\rm{ }}m)}}{4}({1_i}{\rm{ }} + {\rm{ }}2(12{\rm{ }} \times {\rm{ }}{10^{ - 6}}{\rm{ }}^\circ {\rm{ }}{C^{ - 1}})(2850^\circ {\rm{ }}C{\rm{ }} - {\rm{ }}20.0^\circ {\rm{ }}C))\\ = {\rm{ }}6.793813494{\rm{ }} \times {\rm{ }}{10^{ - 9}}{\rm{ }}{m^2}\end{aligned}\)

The resistivity of the tungsten is expressed as a function of temperature with the help of the third equation as:

\(\rho {\rm{ }} = {\rm{ }}{\rho _0}{\rm{ }}(1{\rm{ }} + {\rm{ }}{\alpha _r}(T{\rm{ }} - {\rm{ }}{T_0}))\)

The value of\({\alpha _r}\)is the temperature coefficient of resistivity of the tungsten.

Entering numerical values and we obtain:

\(\begin{aligned}\rho {\rm{ }} = {\rm{ }}(5.6{\rm{ }}x{\rm{ }}{10^{ - 8}}{\rm{ }}\Omega .m)(1 + (4.5{\rm{ }} \times {\rm{ }}{10^{ - 3}}^\circ {\rm{ }}{C^{ - 1}})(2850^\circ {\rm{ }}C{\rm{ }} - {\rm{ }}20.0^\circ {\rm{ }}C))\\ = {\rm{ }}7.6916{\rm{ }} \times {\rm{ }}{10^{ - 7}}{\rm{ }}\Omega .m\end{aligned}\)

Therefore, the resistance of the tungsten filament, taking into account its thermal expansion, is: \({R_f}{\rm{ }} = {\rm{ }}4.7{\rm{ }}\Omega \).

\(\begin{aligned}Percentage{\rm{ }} = {\rm{ }}(\frac{{{R_f} - {R_0}}}{{{R_0}}})(100\% )\\ = {\rm{ }}(\frac{{4.7{\rm{ }}\Omega {\rm{ }} - {\rm{ }}0.350{\rm{ }}\Omega }}{{0.350{\rm{ }}\Omega }})(100\% )\\ = {\rm{ }}1200{\rm{ }}\% \end{aligned}\)

Therefore, the Percentage by which the answer of part (a) differs from the Example\(20.25\)is:\(1200{\rm{ }}\% \).