Question

Two resistances $R_1$ and $R_2$ are made of different materials. The temperature coefficient of the material of $R_1$ is $\alpha$ and of the material of $R_2$ is $-\beta$. The resistance of the series combination of $R_1$ and $R_2$ will not change with temperature, if $R_1/R_2$ equals

• A. $\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }$
  
• B. $\frac{\beta }{\alpha }$
  
• C. $\frac{\alpha +\beta }{\alpha -\beta }$
  
• D. $\frac{\alpha }{\beta }$
  

Answer 1

The correct option is B $\frac{\beta }{\alpha }$


Step 1: Variation of resistance with temperature

After increasing temperature by $t$, let the final resistance of $R_1$ and $R_2$ be $R_1^{\prime }$ and $R_2^{\prime }$ respectively

Then,

$R_1\text{'}=R_1[1+\alpha t]$

$R_2\text{'}=R_2[1-\beta t]$


Step 2: Series combination of resistances

It is given that $R_1$ and $R_2$ are joined in series then,

$R_{eq}=R_1^{\prime }+R_2^{\prime }$

$\Rightarrow R_{eq}=R_1+R_2+(R_1\alpha -R_2\beta )t$

But it is given that $R_{eq}$ will not vary with temperature

$\therefore R_1\alpha -R_2\beta =0$

$\Rightarrow R_1\alpha =R_2\beta$

$\Rightarrow \frac{R_1}{R_2}=\frac{\beta }{\alpha }$

Hence, option (d) is correct.