Question

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

Explanation:

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,

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

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,

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

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

$\sf 58=50+20000\: \alpha$

$\sf 20000\: \alpha=8$

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

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

Answer 2

Given that :

• Initial temperature = 150° C
• Final temperature = 550° C
• Initial resistance = 50 Ω
• Final resistiance = 58 Ω

Need to find :

• The temperature coefficient of resistance of the material.

SolutioN :

We know that,

The temperature coefficient of a material is shown by:

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

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.

Substituting,

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

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

$\sf 58=50+20000\: \alpha$

$\sf 20000\: \alpha=820000$

$\sf \red{ \alpha=4\times 10^{-4}\: C^{-1}} \bigstar$

Therefore,

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