Question

Two wires of equal diameters of resistivity’* 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}]