Question

The filament of the light bulb is made of tungsten. The resistance of the light bulb at room temperature (20∘C), measured by an ohm-meter, is about 3.0 Ω. Based on this information, estimate the temperature of the light bulb at 12.0 V. Neglect the thermal expansion of the filament.

Answer 1

Explanation:

Formula to showing relation between resistance and temperature is as follows.

           R = $R_{o} + \alpha [T_{2} - T_{1}]$

where,   R = final of resistance

        $R_{o}$ = initial of resistance

        $\alpha$ = temperature coefficient of resistivity

        $T_{2}$ = final temperature    

        $T_{1}$ = initial temperature

As the given data is as follows.

      $T_{1} = (20 + 273) K = 293 K$

      $T_{2}$ = ?

          R = 36 ohm,   $R_{o}$ = 3 ohm

         $\alpha$ = 0.0045

Putting the given values into the above formula as follows.

        R = $R_{o} + \alpha [T_{2} - T_{1}]$

        36 = $3 + 0.0045 \times [T_{2} - 293]$

      $T_{2}$ = $\frac{34.3185}{0.0045}$

                 = 7626.33 K

Thus, we can conclude that the temperature of the light bulb at 12.0 V is 7626.33 K.