Question

Two metallic wires $A$ and $B$ of equal dimensions but made of different materials, having resistivities $\rho$ and $2\rho$, temperature coefficients of resistivities $2\alpha$ and $\alpha$ are connected in series. The temperature coefficient of resistance of the composite wire equals:

• A. ${\rho }_{eq}=\frac{3}{2}\rho$
• B. ${\alpha }_{eq}=\frac{3}{2}\alpha$
• C. ${\rho }_{eq}=3\rho$
• D. ${\alpha }_{eq}=\frac{4}{3}\alpha$

Answer 1

The correct option is D ${\alpha }_{eq}=\frac{4}{3}\alpha$

Let the length of the two wires be $\ell$

And cross section area$=A$

Let the resistance of composite wore be $R_{eq}$

$R_{eq}=\frac{\rho \ell }{A}+\frac{2\rho \ell }{A}$

$\frac{{\rho }_{eq}2\ell }{A}=\frac{3\rho \ell }{A}$

${\rho }_{eq}=\frac{3\rho }{2}$

$R=R_1+R_2$

$R_{eq}^0(1+{\alpha }_{eq}\Delta T)=R_1^0(1+2\alpha \Delta T)+R_2^0(1+\alpha \Delta T)$

Also,

$R_{eq}^0=R_1^0+R_2^0$

And,

$R_{eq}^0=3R_1^0$

$R_2^0=2R_1^0$

Or,

$3R_1^0{\alpha }_{eq}\Delta T=R_1^{0}2\alpha \Delta T+2R_1^0\alpha \Delta T$

${\alpha }_{eq}=\frac{4\alpha }{3}$

Options $\text{D}$ is the correct answer.