Question

Two wires of resistance $R_1$ and $R_2$ have temperature coefficient of resistance ${\alpha }_1$ and ${\alpha }_2$, respectively. These are joined in series. The effective temperature coefficient of resistance is

• A. $\frac{{\alpha }_1+{\alpha }_2}{2}$
  
• B. $\sqrt{{\alpha }_1{\alpha }_2}$
• C. $\frac{{\alpha }_1{R}_1+{\alpha }_1{R}_2}{R_1+R_2}$
  
• D. $\frac{\sqrt{R_1{R}_2{\alpha }_1{\alpha }_2}}{\sqrt{R_1^2+R_2^2}}$
  

Answer 1

The correct option is C $\frac{{\alpha }_1{R}_1+{\alpha }_1{R}_2}{R_1+R_2}$


Step 1: Resistance of wire $1$ and $2$

Resistance of wire $1$,

$R_1^{\prime }=R_1[1+{\alpha }_{1}t]$

Resistance of wire $2$,

$R_2^{\prime }=R_2[1+{\alpha }_{2}t]$


Step 2: Calculate equivalent resistance

Now, since the wires are joined in series,

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

$\Rightarrow R_{eq}^{\prime }=R_1[1+{\alpha }_{1}t]+R_2[1+{\alpha }_{2}t]$

$\Rightarrow R_{eq}^{\prime }=R_1+R_2+\left(R_1{\alpha }_1\right)t+\left(R_2{\alpha }_2\right)t$

$R_{eq}^{\prime }=(R_1+R_2)[1+\left(\frac{R_1{\alpha }_1+R_2{\alpha }_2}{R_1+R_2}\right)].\left(i\right)$

Also,

$R_{eq}\text{'}=R_{eq}[1+{\alpha }_{eff}t]\dots \left(ii\right)$

Comparing equation (𝑖) and equation (𝑖𝑖)

We get,

$\Rightarrow {\alpha }_{eff}=\frac{R_1{\alpha }_1+R_2{\alpha }_2}{R_1+R_2}$

Hence, option (c) is correct.