Question

The filament of a lightbulb has a resistance of R0=12Ω at 20 ∘C and 140 Ω when hot. Part APart complete Part B In this temperature range, what is the percentage change in resistance due to thermal expansion? Express your answer using two significant figures. Rexpansion−R0R0R e x p a n s i o n − R 0 R 0 = 1100 %

Answer 1

Answer:

T = 2390 degree C

Explanation:

Given data:

Resistance 12 ohm

Temperature 20 degree C

$Resistance _{HOT} = 140 ohm$

we know that linear relation between the resistance and temperature is given as$R = R_O[ 1+ \alpha ( T - T_o)]$

where $R_O$ is resistance at $_o,$ and $\alpha$ = coefficient of resistivity

$140 = 12 \times ( 1 + 0.0045 \Delta T}$

SOLVING FOR $\Delta T$

$\Delta T = 2370$ degree C

WE KNOW

$\Delta T = T - T_o$

T = 2370 + 20

T = 2390 degree C