Question

The temperature coefficient of resistance of a wire is 0.00125 per degree Celsius. At 300K its resistance is 1 ohm. The resistance of the wire will be 2 ohms at the following temperature: 

A) 1127 K

B) 400K

C) 600K

D) 1400K

Answer 1

Hint: Resistance of any substance is linearly proportional to the absolute temperature of that substance. The proportionality coefficient is known as the temperature coefficient of that particular substance and the value of that coefficient varies from substance to substance.

Formula used:

The relationship between resistance and its absolute temperature is:

${R_T} = {R_0}\left( {1 + \alpha \Delta T} \right)$ ……………..(1)

Where,

${R_T}$ is the resistance at absolute temperature $T$,

${R_0}$ is the resistance at reference temperature,

$\alpha $ is the temperature coefficient of resistance for the substance,

$\Delta T$ is the temperature difference between reference and T.

Complete step by step solution:

Given:

The temperature coefficient of the wire is $\alpha  = 0.00125{/^ \circ }C$

Temperature of the wire is \[{T_1} = 300K = 27 ^oC\].

Resistance of the wire at $300K$ is \[{R_{{T_1}}} = 1\Omega \].

Later, resistance of the wire at ${T_2}$ will be ${R_{{T_2}}} = 2\Omega $.

To find: The temperature of ${T_2}$ when the resistance of the wire will be 2 Ohm.

Step 1:

Using eq.(1) take the ratio of two resistance at two different temperatures \[{T_1}\] and ${T_2}$, then find an expression to get the value of ${T_2}$ as:

$0 ^oC$ is the reference temperature.

$\Delta T\;for\;1\Omega = (T_1 - 0) = T_1 $

$\Delta T\;for\;2\Omega = (T_2 - 0) = T_2 $

$  \dfrac{{{R_{{T_2}}}}}{{{R_{{T_1}}}}} = \dfrac{{{R_0}\left( {1 + \alpha {T_2}} \right)}}{{{R_0}\left( {1 + \alpha {T_1}} \right)}} $

$   \Rightarrow {R_{{T_2}}} + \alpha {T_1}{R_{{T_2}}} = {R_{{T_1}}} + \alpha {T_2}{R_{{T_1}}} $

$   \Rightarrow \alpha {T_2}{R_{{T_1}}} = ({R_{{T_2}}} - {R_{{T_1}}}) + \alpha {T_1}{R_{{T_2}}} $

$  \therefore {T_2} = \dfrac{{({R_{{T_2}}} - {R_{{T_1}}})}}{{\alpha {R_{{T_1}}}}} + \dfrac{{{T_1}{R_{{T_2}}}}}{{{R_{{T_1}}}}} $………………. (2)

Step 2:

Now, substitute the values of $\alpha $, ${T_1}$, ${R_{{T_1}}}$, ${R_{{T_2}}}$ in eq.(2) to get the value of ${T_2}$ as:

$  {T_2} = \dfrac{{\left( {2 - 1} \right)\Omega }}{{0.00125{C^{ - 1}} \times 1\Omega }} + \dfrac{{27 \times 2\Omega }}{{1\Omega }} $

$  \therefore {T_2} = \dfrac{1}{{0.00125}} + 54 = 800 + 54 = 854 ^oC = 854 + 273 = 1127K $

The temperature when the resistance will be 2 Ohm is 1127 K. So, option (A) is correct.

Note: There may be a chance of making mistakes by simply dividing the resistance difference ($1\Omega $ here) by the temperature coefficient ($\alpha $) and adding that to the existing temperature (300 K). But that can’t be done here because the ${R_0}$ term is multiplied with $\alpha $ also (see eq.(1)). So be careful not to make this common mistake.