Two wires of equal length and equal diameter and having resistivities ρ₁ and ρ₂ are connected in series. The equivalent resistivity of the combination is

Options

(a) ρ₁+ρ₂/2
(b) ρ₁+ρ₂
(c) ρ₁ρ₂/ρ₁+ρ₂
(d) √ρ₁ρ₂

Correct Answer:

ρ₁+ρ₂/2

Explanation:

The resistance of two wires are

R₁ = ρ₁ (l / A) and R₂ = ρ₂ (l / A)

Now, equivalent resistance of series connection of wire

Req = R₁ + R₂ = ρ₁ (l / A) + ρ₂ (l / A) = (l / A) (ρ₁ + ρ₂)

The equivalent resistance Req can be given by Req = ρ(eq) (l / A)

⇒ ρ(eq) (l / A) = (l / A) (ρ₁ + ρ₂)

Hence, ρ(eq) = (ρ₁ + ρ₂) / 2