Question

The filament of a light bulb has a resistance of 20.0 Ω at 20oC and 160 Ω when the light is on. Find the temperature of the filament when the light is on. (The temperature coefficient of resistivity is 3.50x10-3 oC-1.)Group of answer choices939oC1120oC1440oC2020oC2310oC

Answer 1

2020°C

Step-by-step explanation

Given:

• The initial resistance of the light bulb, R₀ = 20.0 Ω

,

• The initial temperature of the filament of the light bulb, T₀ = 20°C

,

• The final resistance of the fi light bulb when it is on, R = 160 Ω

,

• The temperature coefficient of resistivity, α = 3.50 * 10⁻³ °C⁻¹

Unknown:

• The final temperature of the filament of the light bulb when it is on, T

The resistance and the change in temperature of a resistive element is given by,

`R=R_o\lbrack1+\alpha(T-T_o)_{}\rbrack`

We have to solve this equation for T. First, divide both sides by R₀,

`(R)/(R_o)=1+\alpha(T-T_o)`

Subtract 1 from both sides,

`(R)/(R_o)-1=\alpha(T-T_o)`

Divide both sides by α,

`((R)/(R_o)-1)/(\alpha)=T-T_o`

And add T₀ to both sides,

`T=((R)/(R_o)-1)/(\alpha)+T_o`

Replace with the known values and solve,

`T=\frac{\frac{160\Omega}{20\Omega_{}}-1}{3.5\cdot10^(-3)(1)/(\degree C)}+20\degree C=(8-1)/(3.5\cdot10^(-3)(1)/(\degree C))+20\degree C=(7)/(3.5\cdot10^(-3)(1)/(\degree C))+20\degree C=2000\degree C+20\degree C=2020\degree C_{}`

Hence, the temperature of the filament of the light bulb when it is on is 2020°C.