Question

Two wires of equal diameters of resistivity’s ρ₁ and ρ₂ are joined in series. The equivalent resistivity of the combination is....

(A) {(ρ₁ℓ₁ + ρ₂ℓ₂)/(ℓ₁ + ℓ₂)}

(B) {(ρ₁ℓ₂ + ρ₂ℓ₁)/(ℓ₁ – ℓ₂)}

(C) {(ρ₁ℓ₂ + ρ₂ℓ₁)/(ℓ₁ + ℓ₂)}

(D) {(ρ₁ℓ₁ + ρ₂ℓ₂)/(ℓ₁ – ℓ₂)}

Answer 1

The correct option (A) {(ρ₁ℓ₁ + ρ₂ℓ₂)/(ℓ₁ + ℓ₂)}

Explanation:

wires are connected in series (1)

hence   R = R₁ + R₂

Let length be ℓ₁ & ℓ₂.

resistivity given ρ₁ & ρ₂.

Area is same

∴ R₁ = {(ρ₁ℓ₁)/A}, R₂ = {(ρ₂ℓ₂)/A}

R = [{ρ(ℓ₁ + ℓ₂)}/A] ---- series

combination hence lengths get added

∴ from (1)

[{ρ(ℓ₁ + ℓ₂)}/A] = {(ρ₁ℓ₁)/A} + {(ρ₂ℓ₂)/A}

∴ ρ = [{ρ₁ℓ₁ + ρ₂ℓ₂}/{ℓ₁ + ℓ₂}]