Q. Two wires of equal diameters, of resistivities $\rho_{1}$ and $\rho_{2}$ and lengths $l_{1}$ and $l_{2}$, respectively, are joined in series. The equivalent resistivity of the combination is
A) $\frac{\rho_{1}l_{1} + \rho_{2}l_{2}}{l_{1} + l_{2}}$
B) $\frac{\rho_{1}l_{2} + \rho_{2}l_{1}}{l_{1} - l_{2}}$
C) $\frac{\rho_{1}l_{2} + \rho_{2}l_{1}}{l_{1} + l_{2}}$
D) $\frac{\rho_{1}l_{1} - \rho_{2}l_{2}}{l_{1} - l_{2}}$
Explanation
(a) $R_{1} = \frac{\rho_{1}l_{1}}{A}$ and $R_{2} = \frac{\rho_{2}l_{2}}{A}$. In series $R_{eq} = R_{1} + R_{2}$
$\frac{\rho_{eq.}(l_{1} + l_{2})}{A} = \frac{\rho_{1}l_{1}}{A} + \frac{\rho_{2}l_{2}}{A} \Rightarrow \rho_{eq} = \frac{\rho_{1}l_{1} + \rho_{2}l_{2}}{l_{1} + l_{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.