Chapter 25 Problem 28

Problem

The filament of a light bulb has a resistance of 12 $\Omega$ at 20°C and 140 $\Omega$ when hot.

(a) Calculate the temperature of the filament when it is hot, and take into account the change in length and area of the filament due to thermal expansion (assume tungsten for which the thermal expansion coefficient is $\approx 5.5\times 10^{-6}$ C°$^{-1}$ ).

(b) In this temperature range, what is the percentage change in resistance due to thermal expansion, and what is the percentage change in resistance due solely to the change in $\rho$

use

 $\rho _{T}=\rho _{0}\left[1+\alpha (T-T_{0})\right]$

Solution

(a)

Lower temperature resistance:

$R_{0}=\rho _{0}{\frac {l_{0}}{A_{0}}}=\rho _{0}{\frac {4l_{0}}{\pi d_{0}^{2}}}$

(remember we use diameter instead of radius for standard compliance)

at higher temperature ($\alpha$ being the temperature coefficient and $\alpha _{T}$ being the therma expansion coefficient)

$R=\rho {\frac {l}{A}}=\rho _{0}{\frac {4l}{\pi d^{2}}}=\rho _{0}\left[1+\alpha (T-T_{0})\right]{\frac {4l_{0}\left[1+\alpha _{T}(T-T_{0})\right]}{\pi \left(d_{0}\left[1+\alpha _{T}(T-T_{0})\right]\right)^{2}}}$ $=\rho _{0}{\frac {4l_{0}\left[1+\alpha (T-T_{0})\right]}{\pi d_{0}^{2}\left[1+\alpha _{T}(T-T_{0})\right]}}$ $=R_{0}{\frac {\left[1+\alpha (T-T_{0})\right]}{\left[1+\alpha _{T}(T-T_{0})\right]}}$

$\rightarrow R\left[1+\alpha _{T}(T-T_{0})\right]=R_{0}\left[1+\alpha (T-T_{0})\right]$

$T=T_{0}+{\frac {(R-R_{0})}{(R_{0}\alpha -R\alpha _{T})}}=20{\textrm {C}}^{\circ }+{\frac {(140-12)\Omega }{(12\Omega )(0.0045{\textrm {C}}^{\circ -1})-(140\Omega )(5.5\times 10^{-6}{\textrm {C}}^{\circ -1})}}=2425{\textrm {C}}^{\circ }$

(b)

The net effect of thermal expansion is that both the length and diameter increase, which lowers the resistance.

${\frac {R}{R_{0}}}={\frac {\rho _{0}{\frac {4l_{0}}{\pi d_{0}^{2}}}}{\rho {\frac {4l}{\pi d^{2}}}}}={\frac {ld_{0}^{2}}{l_{0}d^{2}}}={\frac {l_{0}\left[1+\alpha _{T}(T-T_{0})\right]}{l_{0}}}{\frac {d_{0}^{2}}{(d_{0}\left[1+\alpha _{T}(T-T_{0})\right])^{2}}}={\frac {1}{\left[1+\alpha _{T}(T-T_{0})\right]}}$

${\frac {1}{\left[1+(5.5\times 10^{-6}{\textrm {C}}^{\circ -1})(2405{\textrm {C}}^{\circ })\right]}}=0.9869\approx -\%1.3$

The net effect of resistivity change is that the resistance increases.

${\frac {R}{R_{0}}}={\frac {\rho {\frac {4l_{0}}{\pi d_{0}^{2}}}}{\rho _{0}{\frac {4l_{0}}{\pi d_{0}^{2}}}}}={\frac {\rho }{/rho_{0}}}=\left[1+\alpha (T-T_{0})\right]=\left[1+0.0045{\textrm {C}}^{\circ -1}(2405{\textrm {C}}^{\circ })\right]=11.82\approx +\%1082$