A common flashlight bulb is rated at 0.30 Aand 2.9 V(the values of the current and voltage under operating conditions). If the resistance of the tungsten bulb filament at room temperature ($\mathbf{20}\mathbf{°}\mathbf{C}$) is$\mathbf{1}\mathbf{.}\mathbf{1}\mathbf{\Omega }$, what is the temperature of the filament when the bulb is on?

$$\begin{gathered} \mathrm{a})\mathrm{Current}\mathrm{in}\mathrm{the}\mathrm{bulb},\mathrm{l}=0.30\mathrm{A} \\ \mathrm{b})\mathrm{Voltage}\mathrm{of}\mathrm{the}\mathrm{bulb},\mathrm{V}=2.9\mathrm{V} \\ \mathrm{c})\mathrm{Initial}\mathrm{resistance},{\mathrm{R}}_0=1.1\mathrm{\Omega } \\ \mathrm{d})\mathrm{Initial}\mathrm{temperature},{\mathrm{T}}_0={20}^0\mathrm{C} \end{gathered}$$

The resistance of all the materials changes with temperature. The relation between resistance at different temperatures is given by the equation,

$\mathit{R}\mathbf{-}{\mathit{R}}_{\mathbf{0}}\mathbf{=}{\mathit{R}}_{\mathbf{0}}\mathit{\alpha }\left(T-T_0\right)$

Here, R is resistance at temperature T,${\mathit{R}}_{\mathbf{0}}$is resistance at temperature ${\mathit{T}}_{\mathbf{0}}$and a is called as temperature coefficient of resistance.

We have to use the concept of temperature coefficient of resistance to find the temperature of the filament when the bulb is on.

Formulae:

The resistance formula using Ohm’s law, $R=\frac{V}{l}$ …(i)

Here, R is resistance, is voltage, V and l is current.

The new resistance due to change in temperature, $R=R_0\left(1+\alpha \left(T-T_0\right)\right)$ …(ii)

Using the given data in equation (i), we can get the final resistance of the bulb as follows:

$$\begin{gathered} R=\frac{2.9\mathrm{V}}{0.30\mathrm{A}} \\ =9.67\mathrm{\Omega } \end{gathered}$$

Using the given data in the rearranged equation of resistance of equation (ii), we can get the final temperature of the filament when the bulb is on as follows:

(The value of thetemperature coefficient of resistance for tungsten,$\alpha =\left(4.5\times {10}^{-3}\right)°$C)

$$\begin{gathered} T=T_0+\frac{1}{\alpha }\left(\frac{R}{R_0-1}\right) \\ =20°C=\frac{1}{\left(4.5\times {10}^{-3}\right)°\mathrm{C}}\left(\frac{9.67\mathrm{\Omega }}{1.1\mathrm{\Omega }}\right) \\ =\left(1.751\times {10}^3\right)°\mathrm{C} \\ \approx \left(1.8\times {10}^3\right)°\mathrm{C} \end{gathered}$$

Hence, the value of the final temperature is $\left(1.8\times {10}^3\right)°\mathrm{C}$.