Question

Two wires of same length and same area of cross-section but made of different material of resistivities ρ₁ and ρ₂ are connected in series. The equivalent resistivity of the combination is ______.

पर्याय

• `1/2 (rho_1 + rho_2)`

• ρ₁ + ρ₂

• `sqrt(rho_1 rho_2)`

• 2(ρ₁ + ρ₂)

Answer 1

Two wires of same length and same area of cross-section but made of different material of resistivities ρ₁ and ρ₂ are connected in series. The equivalent resistivity of the combination is `bbunderline(1/2 (rho_1 + rho_2))`.

Explanation:

We are given two wires of the same length l and the same area of cross-section A, but different resistivities ρ₁ and ρ₂. They are connected in series.

The resistance of a wire is given by:

R = `(rho l)/A`    ...(i)

For the two wires:

R₁ = `(rho_1 l)/A`

R₂ = `(rho_2 l)/A`

For series combination:

R_eq = R₁ + R₂

= `(rho_1 l)/A + (rho_2 l)/A`

= `((rho_1 + rho_2)l)/A`    (ii)

For the combination, the total length becomes 2 l, and the cross-sectional area remains A.

If we consider the combination as a single wire of length 2l, area A, and equivalent resistivity ρ_eq, then:

R_eq = `(rho_(eq)(2 l))/A`

Equating with the expression from equation (ii):

`((rho_1 + rho_2)l)/A = (rho_(eq)(2 l))/A`

Cancelling `l/A` from both sides:

ρ₁ + ρ₂ = 2ρ_eq

ρ_eq = `1/2 (rho_1 + rho_2)`

∴ The equivalent resistivity is the average of the two resistivities.