Question

The filament of a certain lamp has a resistance that increases linearly with temperature. When a constant voltage is switched on, the initial current decreases until the filament reaches its steady-state temperature. The temperature coefficient of resistivity of the filament is 4 times 10-3 K-1. The final current through the filament is one- eighth the initial current. What is the change in temperature of the filament

Answer 1

Answer:

The change in temperature is $\Delta T = 1795 K$

Explanation:

From the question we  are told that

   The temperature coefficient is  $\alpha = 4 * 10^{-3 }\ k^{-1 }$

The resistance of the filament is mathematically represented as

           $R = R_o [1 + \alpha \Delta T]$

Where $R_o$ is the initial resistance

Making the change in temperature the subject of the formula

     $\Delta T = \frac{1}{\alpha } [\frac{R}{R_o} - 1 ]$

Now from ohm law

           $I = \frac{V}{R}$

This implies that current varies inversely with current so

           $\frac{R}{R_o} = \frac{I_o}{I}$

Substituting this we have

       $\Delta T = \frac{1}{\alpha } [\frac{I_o}{I} - 1 ]$

From the question we are told that

    $I = \frac{I_o}{8}$

Substituting this we have

   $\Delta T = \frac{1}{\alpha } [\frac{I_o}{\frac{I_o}{8} } - 1 ]$

=>     $\Delta T = \frac{1}{3.9 * 10^{-3}} (8 -1 )$

        $\Delta T = 1795 K$