91影视

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$T_0=20.0C掳uptoT=120.0C掳$

Current$l=12.5A$

$E=蚁J=蚁\frac{l}{A}=\frac{蚁l}{{\mathrm{蟺谤}}^2}$

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

As the increase in temperature, the resistivity of metal increases at a temperature $T_0=20.0C掳$and resistivity ${蚁}_o=5.25脳{10}^{-8}惟.m$

So $蚁$is;

$$\begin{gathered} 蚁={蚁}_o\left[1+伪\left(T-T_o\right)\right] \\ =5.25脳{10}^{-8}惟.m\left[1+0.0045{\left(C掳\right)}^{-1}\left(120C掳-20C掳\right)\right] \\ =7.6125脳{10}^{-8}惟.m \end{gathered}$$

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

Putting the values of$蚁,landr$ :

$$\begin{gathered} E=\frac{蚁l}{{\mathrm{蟺谤}}^2} \\ =\frac{\left(7.6125脳{10}^{-8}惟.m\right)\left(12.5A\right)}{\mathrm{蟺}{\left(0.0005\mathrm{m}\right)}^2} \\ =1.21V/m \end{gathered}$$

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

$R=\frac{蚁L}{A}$

Putting the values:

$$\begin{gathered} R=\frac{蚁L}{{\mathrm{蟺谤}}^2} \\ =\frac{\left(7.6125脳{10}^{-8}惟.m\right)\left(0.15m\right)}{\mathrm{蟺}{\left(0.0005\mathrm{m}\right)}^2} \\ =0.0145惟 \end{gathered}$$

Hence, the resistance is$R=0.0145惟$

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

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