Question

A wire of cross-section area $A$, length $L_1$, resistivity ${\rho }_1$ and temperature coefficient of resistivity ${\alpha }_1$ is connected in series to a second wire of length $L_2$, resistivity ${\rho }_2$, temperature coefficient of resistivity ${\alpha }_2$ and the same area $A$, so that wires carry same current. Total resistance $R$ is independent of temperature for small temperature change if (Thermal expansion effect is negligible)

• A. None
  
• B. ${\alpha }_1=-{\alpha }_2$
  
• C. ${\rho }_1{L}_1{\alpha }_1+{\rho }_2{L}_2{\alpha }_2=0$
  
• D. $L_1{\alpha }_1+L_2{\alpha }_2=0$
  

Answer 1

The correct option is C ${\rho }_1{L}_1{\alpha }_1+{\rho }_2{L}_2{\alpha }_2=0$


Let initial resistances of the wire be $R_1$ and $R_2$ ,

Now after increase in temperature by $t$, the final resistances are $R_1$′ and $R_2$′ respectively.

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

$R_2\text{'}=R_2[1+{\alpha }_{2}t]$


Since the wires are joined in series,

$\therefore R_{eq}=R_1^{\prime }+R_2^{\prime }\to \left(1\right)$

$\Rightarrow R_{eq}=R_1+R_2+(R_1{\alpha }_1+R_2{\alpha }_2)t\to \left(2\right)$

Now,

$R_{eq}=R$

But it is given that $R_{eq}$ is independent of temperature $t$

$\Rightarrow R_1^{\prime }=R_1\text{and}{R}_2\text{'}=R_2\to \left(3\right)$


From equation (1) , (2) and (3), we conclude that

$R_1{\alpha }_1+R_2{\alpha }_2=0\to \left(4\right)$

We know that,

$R_1=\frac{{\rho }_1{L}_1}{A}\text{and}{R}_2=\frac{{\rho }_2{L}_2}{A}\to \left(5\right)$

Since area of cross-section is same for both wires,

$A_1=A_2=A$

Substituting equation (5) in equation (4), we get

$\Rightarrow \frac{{\rho }_1{L}_1{\alpha }_1}{A}+\frac{{\rho }_2{L}_2{\alpha }_2}{A}=0$

$\Rightarrow {\rho }_1{L}_1{\alpha }_1+{\rho }_2{L}_2{\alpha }_2=0$

Hence, option (b) is correct.