Q. Two wires of resistance $R_{1}$ and $R_{2}$ have temperature coefficient of resistances $\alpha_{1}$ and $\alpha_{1}$ 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) $\frac{\alpha_{1}R_{1} + \alpha_{2}R_{2}}{R_{1} + R_{2}}$
D) $\frac{\sqrt{R_{1}R_{2}\alpha_{1}\alpha_{2}}}{\sqrt{R_{1}^{2} + R_{2}^{2}}}$
Explanation
(c) $R_{t_{1}} = R_{1}\left( 1 + \alpha_{1}t \right)$ and $R_{t_{2}} = R_{2}\left( 1 + \alpha_{2}t \right)$
Also $R_{eq} = R_{t_{1}} + R_{t_{2}} \Rightarrow R_{eq} = R_{1} + R_{2} + (R_{1}\alpha_{1} + R_{2}\alpha_{2})t$
$\Rightarrow R_{eq} = \left( R_{1} + R_{2} \right)\left\lbrack 1 + \left( \frac{R_{1}\alpha_{1} + R_{2}\alpha_{2}}{R_{1} + R_{2}} \right).t \right\rbrack$
So $\alpha_{eff} = \frac{R_{1}\alpha_{1} + R_{2}\alpha_{2}}{R_{1} + R_{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.