Question

5. It is found that the resistance of a coil of wire increases from 50 Ω at 15 0

C to 58 Ω at 55 0

C. Calculate

the temperature co-efficient of resistance of the material.​

Answer 1

Answer:

Temperature coefficient = 4 × 10⁻⁴ C⁻¹

$\huge \mathfrak \colorbox{purple}{explaination࿐}$

Given:

Initial resistance = 50 Ω

Final resistiance = 58 Ω

Initial temperature = 150° C

Final temperature = 550° C

To Find:

Temperature coefficient of resistance of the material

Solution:

The temperature coefficient of a material is given by the equation,

$\sf R=R_0(1+\alpha(T-T_0))R=R$

where R is the final resistance, R₀ is the initial resistance, α is the temperature coefficient of the material, T is the final temperature, T₀ is the initial temperature.

Substitute the given data,

$\sf 58=50\times (1+\alpha(550-150)58=50×(1+α(550−150)$

$\sf \\58=50\times (1+400\:\alpha)58=50×(1+400α)$

$\sf 58=50+20000\: \alpha58=50+20000α$

$\sf 20000\: \alpha=820000α=8$

$\sf \alpha=4\times 10^{-4}\: C^{-1}α=4×10$

Therefore the temperature coefficient of the material is 4 × 10⁻⁴ C⁻¹