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22 March, 18:41

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

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  1. 22 March, 19:05
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    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
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