Question

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?

Short answer

(a) Maximum electric field in the filament is$E=1.21V/m$

(b) Resistance is$R=0.0145\Omega$

(c) Maximum potential drop is$V=0.181V$

Step-by-step solution

Step 1:

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$l=12.5A$

Step 2:

(a) To determine the electric field passing through the filament Where the electric field is connected to the resistivity, the greater the resistivity, the greater the electric field required to induce a particular current.

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

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$E=1.21V/m$

Step 3:

(b) For determining the resistance R of the filament, where the resistance of the cylinder is determined by its area, resistivity, and length,

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

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$R=0.0145\Omega$

Step 4:

(c) For the voltage of the rod, putting the value of R in equation V=IR;

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

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