Question

The resistance of a wire at room temperature ${30}^{o}C$ is found to be $10\Omega$. Now to increase the resistance by 10%, the temperature of the wire must be: [The temperature coefficient of resistance of the material of the wire is 0.002 per ${}^{o}C$]

• A. ${36}^{o}C$
• B. ${83}^{o}C$
• C. ${63}^{o}C$
• D. ${33}^{o}C$

Answer 1

The correct option is B ${83}^{o}C$

Given: Temperature coefficient of resistance $=\alpha =0.002{/}^{o}C$
We know that, $R_T=R_0(1+\alpha T)$
Initially, $10=R_0(1+\alpha .30)...\left(1\right)$
After $10\%$ increase the resistance becomes $=10+10\times \frac{10}{100}=11\Omega$ and corresponding temp is $T$.
Now, $11=R_0(1+\alpha T)...\left(2\right)$

$\left(2\right)/\left(1\right)\Rightarrow \frac{11}{10}=\frac{R_0(1+\alpha T)}{R_0(1+30\alpha )}$

$1+\alpha T=1.1+33\alpha$

$T=\frac{1.1+33\alpha -1}{\alpha }=\frac{0.1+33\times 0.002}{0.002}={83}^{o}C$