A cylindrical tungsten filament 15.0 cm long with a diameter of 1.00 mm is to be used in a machine for which the temperature will range from room temperature 120°C2 up to 120°C. It will carry a current of 12.5 A at all temperatures (consult Tables 25.1 and 25.2). (a) What will be the maximum electric field in this filament, and (b) what will be its resistance with that field? (c) What will be the maximum potential drop over the full length of the filament?

Given data:

Length of the cylinder of tungsten$L=15.0cm=0.15m$

And the radius is$r=1.0mm/2=0.50mm=0.00050m$

Temperature range$T_0=20.0C°uptoT=120.0C°$

Current$l=12.5A$

$E=\rho J=\rho \frac{l}{A}=\frac{\rho l}{{\mathrm{\pi r}}^2}$

Here,is the resistivity of the tungsten at a temperature$T=120.0C°$

As the increase in temperature, the resistivity of metal increases at a temperature $T_0=20.0C°$and resistivity ${\rho }_o=5.25\times {10}^{-8}\Omega .m$

So $\rho$is;

$$\begin{gathered} \rho ={\rho }_o\left[1+\alpha \left(T-T_o\right)\right] \\ =5.25\times {10}^{-8}\Omega .m\left[1+0.0045{\left(C°\right)}^{-1}\left(120C°-20C°\right)\right] \\ =7.6125\times {10}^{-8}\Omega .m \end{gathered}$$

Here $\alpha$is the temperature coefficient and$\alpha =0.0045{\left(C°\right)}^{-1}$

Putting the values of$\rho ,landr$ :

$$\begin{gathered} E=\frac{\rho l}{{\mathrm{\pi r}}^2} \\ =\frac{\left(7.6125\times {10}^{-8}\Omega .m\right)\left(12.5A\right)}{\mathrm{\pi }{\left(0.0005\mathrm{m}\right)}^2} \\ =1.21V/m \end{gathered}$$

Hence, the maximum electric field in the filament is$E=1.21V/m$

$R=\frac{\rho L}{A}$

Putting the values:

$$\begin{gathered} R=\frac{\rho L}{{\mathrm{\pi r}}^2} \\ =\frac{\left(7.6125\times {10}^{-8}\Omega .m\right)\left(0.15m\right)}{\mathrm{\pi }{\left(0.0005\mathrm{m}\right)}^2} \\ =0.0145\Omega \end{gathered}$$

Hence, the resistance is$R=0.0145\Omega$

$$\begin{gathered} V=IR \\ =\left(12.5A\right)\left(0.0145\Omega \right) \\ =0.181V \end{gathered}$$

Therefore, the maximum potential drop is$V=0.181V$