Two wires of equal diameters, of resistivities ρ 1 and ρ 2 and length l 1 and l 2 , respectively are joined in series. The equivalent resistivity of the combination is

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Wires having frame diameter, hence area of cross-section will be equal,

i.e. $A_{1} = A_{2} = A$

$R_{1} = ρ_{1} . \frac{l_{1}}{A}$

and $R_{2} = ρ_{2} . \frac{l_{2}}{A}$

Since, both resistance are joined in series.

Hence, $R_{e q} = R_{1} + R_{2}$

$= ρ_{1} . \frac{l_{1}}{A} + ρ_{2} . \frac{l_{2}}{A}$

$R_{e q} = \frac{1}{A} (ρ_{1} l_{1} + ρ_{2} l_{2})$

$ρ_{e q} . \frac{(l_{1} + l_{2})}{A} = \frac{1}{A} (ρ_{1} l_{1} + ρ_{2} l_{2})$

$\Rightarrow ρ_{e q} = \frac{ρ_{1} l_{1} + ρ_{2} l_{2}}{l_{1} + l_{2}}$

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