Two wires of equal lengths and equal diameters having resistivity P1 and P2 are connected in series. The equivalent resistivity of the combination is

R=R1+R2→1R=ρ.2lA R1=ρ1lA R2=ρ2lASubstitute in (1)∴ρ.2lA=ρ1lA+ρ2lA=2ρlA=lAρ1+ρ2∴ρ=ρ1+ρ22

Two wires of equal lengths and equal diameters having resistivity P1 and P2 are connected in series. The equivalent resistivity of the combination is

R=R1+R2→1R=ρ.2lA R1=ρ1lA R2=ρ2lASubstitute in (1)∴ρ.2lA=ρ1lA+ρ2lA=2ρlA=lAρ1+ρ2∴ρ=ρ1+ρ22

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Two wires of equal lengths and equal diameters having resistivity P₁ and P₂ are connected in series. The equivalent resistivity of the combination is

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P₁P₂

$\sqrt{ρ_{1} ρ_{2}}$

P₁ + P₂

$\frac{ρ_{1} + ρ_{2}}{2}$

answer is B.

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Detailed Solution

$R = R_{1} + R_{2} \to (1)$
$R = \frac{ρ . 2 l}{A} R_{1} = \frac{ρ_{1} l}{A} R_{2} = \frac{ρ_{2} l}{A}$
Substitute in (1)
$∴ \frac{ρ . 2 l}{A} = \frac{ρ_{1} l}{A} + \frac{ρ_{2} l}{A}$
$\begin{array}{l} = 2 ρ \frac{l}{A} = \frac{l}{A} (ρ_{1} + ρ_{2}) \\ ∴ ρ = \frac{ρ_{1} + ρ_{2}}{2} \end{array}$