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yan [13]
3 years ago
12

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 temperatu

re 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 %
Physics
1 answer:
solong [7]3 years ago
4 0

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 asR = 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

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