Question

The tungsten filament of an electric lamp has a surface area A and a power rating P. If the emissivity of the filament is $ϵ$ and $\sigma$ is Stefan's constant, the steady temperature of the filament will be

• A. $T={\left(\frac{P}{Aϵ\sigma }\right)}^2$
• B. $T=\frac{P}{Aϵ\sigma }$
• C. $T={\left(\frac{P}{Aϵ\sigma }\right)}^{\frac{1}{2}}$
• D. $T={\left(\frac{P}{Aϵ\sigma }\right)}^{\frac{1}{4}}$

Answer 1

The correct option is D

$T={\left(\frac{P}{Aϵ\sigma }\right)}^{\frac{1}{4}}$

The energy radiated per second per unit area at temperature T is $\sigma ϵT^4$.
Thus, the energy radiated per second (or power radiated) from the filament of area A will be given by $P=A\sigma ϵT^4\Rightarrow T={\left(\frac{P}{A\sigma ϵ}\right)}^{\frac{1}{4}}$
Hence, the correct choice is (d).