Ask Question
30 March, 14:45

A common flashlight bulb is rated at 0.32 A and 4.3 V (the values of the current and voltage under operating conditions). If the resistance of the bulb filament at room temperature (20°C) is 1.6 Ω, what is the temperature of the filament when the bulb is on? The temperature coefficient of resistivity is 6.4 * 10-3 K-1 for the filament material.

+4
Answers (2)
  1. 30 March, 15:10
    0
    Answer: 1.16*10^3°C

    Explanation:

    It is known that resistance depends on temperature

    Recalling ohms law of v = ir

    R = V/I

    R = 4.3/0.32

    R = 13.44

    R = R• (1 + α (T - T•))

    13.44 = 1.6 (1 + 6.5*10^-3 (T - 20))

    13.44/1.6 = 1 + 6.5*10^-3T - 0.13

    8.4 = 0.87 + 0.0065T

    7.53 = 0.0065T

    T = 1158.46°C

    T = 1.16*10^3°C
  2. 30 March, 15:10
    0
    1176.01 °C

    Explanation:

    Using Ohm's law,

    V = IR ... Equation 1

    Where V = Voltage, I = current, R = Resistance when the bulb is on

    make R the subject of the equation

    R = V/I ... Equation 2

    R = 4.3/0.32

    R = 13.4375 Ω

    Using

    R = R' (1+αΔθ) ... Equation 3

    Where R' = Resistance of the bulb at 20°, α = Temperature coefficient of resistivity, Δθ = change in temperature

    make Δθ the subject of the equation

    Δθ = (R-R') / αR' ... Equation 4

    Given: R = 13.4375 Ω, R' = 1.6 Ω, α = 6.4*10⁻³ K⁻¹

    Substitute into equation 4

    Δθ = (13.4375-1.6) / (1.6*0.0064)

    Δθ = 11.8375/0.01024

    Δθ = 1156.01 °C

    But,

    Δθ = T₂-T₁

    T₂ = T₁+Δθ

    Where T₂ and T₁ = Final and initial temperature respectively.

    T₂ = 20+1156.01

    T₂ = 1176.01 °C
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “A common flashlight bulb is rated at 0.32 A and 4.3 V (the values of the current and voltage under operating conditions). If the resistance ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers