Two wires of equal diameters of resistivities ρ1 and ρ2 and length X1 and X2 respectively are joined in series. The equivalent resistivity of the combination is

ρx1+x2A=ρ1x1A+ρ2x2A⇒ρ=ρ1x1+ρ2x2x1+x2

Two wires of equal diameters of resistivities ρ1 and ρ2 and length X1 and X2 respectively are joined in series. The equivalent resistivity of the combination is

ρx1+x2A=ρ1x1A+ρ2x2A⇒ρ=ρ1x1+ρ2x2x1+x2

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Two wires of equal diameters of resistivities $ρ_{1} and ρ_{2}$ and length X₁ and X₂ respectively are joined in series. The equivalent resistivity of the combination is

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$\frac{ρ_{1} X_{1} + ρ_{2} X_{2}}{X_{1} + X_{2}}$

$\frac{ρ_{1} X_{2} + ρ_{2} X_{1}}{X_{1} - X_{2}}$

$\frac{ρ_{1} X_{2} + ρ_{2} X_{1}}{X_{1} + X_{2}}$

$\frac{ρ_{1} X_{1} + ρ_{2} X_{2}}{X_{1} - X_{2}}$

answer is A.

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$\frac{ρ (x_{1} + x_{2})}{A} = \frac{ρ_{1} x_{1}}{A} + \frac{ρ_{2} x_{2}}{A} \Rightarrow ρ = \frac{ρ_{1} x_{1} + ρ_{2} x_{2}}{x_{1} + x_{2}}$

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Two wires of equal diameters of resistivities ρ1 and ρ2 and length X1 and X2 respectively are joined in series. The equivalent resistivity of the combination is

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Two wires of equal diameters of resistivities $ρ_{1} and ρ_{2}$ and length X₁ and X₂ respectively are joined in series. The equivalent resistivity of the combination is