Question

Light bulb 1 operates with a filament temperature of 2800 K, whereas light bulb 2 has a filament temperature of 1700 K. Both filaments have the same emissivity, and both bulbs radiate the same power. Find the ratio A1/A2 of the filament areas of the bulbs. A1/A2

Answer 1

Answer:

A₁/A₂ = 0.136

Explanation:

The power radiated by a filament bulb is given by the following formula:

E = σεAT⁴

where,

E = Emissive Power

σ = Stephen Boltzman Constant

ε = emissivity

A = Area

T = Absolute Temperature

Therefore, for bulb 1:

E₁ = σε₁A₁T₁⁴

And for bulb 2:

E₂ = σε₂A₂T₂⁴

Dividing both the equations:

E₁/E₂ = σε₁A₁T₁⁴/σε₂A₂T₂⁴

According to given condition, the emissive power and the emissivity is same for both the bulbs. Therefore,

E/E = σεA₁T₁⁴/σεA₂T₂⁴

1 = A₁T₁⁴/A₂T₂⁴

A₁/A₂ = (T₂/T₁)⁴

where,

T₁ = 2800 K

T₂ = 1700 K

Therefore,

A₁/A₂ = (1700 K/2800 K)⁴

A₁/A₂ = 0.136