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.

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

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

the ima (higher than 1m) will be greater than the ima with the height of 1m.