Question

The temperature coefficient of resistance of a wire is $0.00125{/}^{\circ }\text{C}$. At $300\text{K}$ its resistance is $1\text{ohm}$. The temperature at which the resistance becomes $2\text{ohm}$ is:

• A. $1154\text{K}$
• B. $1100\text{K}$
• C. $1400\text{K}$
• D. $1127\text{K}$

Answer 1

The correct option is D $1127\text{K}$

Given:
The temperature coefficient, $\alpha =0.00125{/}^{\circ }\text{C}$
Temperature, $T_1=300\text{K}=27{}^{\circ }\text{C}$
Resistance of the wire at $300\text{K}$ is $R_{T_1}=1\Omega$

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

Let $0{}^{\circ }\text{C}$ is the reference temperature.
$\Delta T\text{for}1\Omega =(T_1-0)=T_1$
$\Delta T\text{for}2\Omega =(T_2-0)=T_2$

Now,
$\frac{R_{T_2}}{R_{T_1}}=\frac{R_0(1+\alpha T_2)}{R_0(1+\alpha T_1)}$

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

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

$\therefore T_2=\frac{(R_{T_2}-R_{T_1})}{\alpha R_{T_1}}+\frac{T_1{R}_{T_2}}{R_{T_1}}$

Now, substitute the values of $\alpha ,T_1,R_{T_1},R_{T_2}$ in above equation. We get value of $T_2$ as:

$T_2=\frac{(2-1)}{0.00125\times 1}+\frac{27\times 2}{1}$

$\therefore T_2=\frac{1}{0.00125}+54=800+54=854{}^{\circ }\text{C}=854+273=1127\text{K}$