Two wires of equal diameters of resistivities ρ1 and ρ2 and lengths l1 and l2 respectively are joined in series. The equivalent resistivity of the combination is :
Answer & explanation
Correct answer: option 1
Resistance of a wire, $R=\frac{4 \rho l}{\pi D^2}$
Where $l$ is the length, D is the diameter and $\rho$ is the resistivity of the material of the wire.
As the wires are connected in series, then
$\frac{4 \rho_{S}\left(l_1+l_2\right)}{\pi D^2}=\frac{4 \rho_1 l_1}{\pi D^2}+\frac{4 \rho_2 l_2}{\pi D^2}$ [∵ Since the wires of same diameter]
Where $\rho_{S}$ is the equivalent resistivity
$\rho_{S}\left(l_1+l_2\right)=\rho_1 l_1+\rho_2 l_2$ or $\rho_{S}=\frac{\rho_1 l_1+\rho_2 l_2}{l_1+l_2}$