Question

Two wires of the same metal have same length, but their cross-sections are in the ratio 3 : 1. They are joined in series. The resistance of the thicker wire is 10 Ω. The total resistance of the combination will be:

Options

• 10 Ω

• 20 Ω

• 40 Ω

• 100 Ω

Answer 1

40 Ω

Explanation:

Length of each wire = ℓ; Area of thick wire (A₁) = 3A; Area of thin wire (A₂) = A and resistance of thick wire (R₁) = 10 Ω. Resistance (R) = `ρ ℓ/"A" ∝ 1/"A"` (if ℓ is constant)

∴ `"R"_1/"R"_2 = "A"_2/"A"_1 = "A"/(3"A") = 1/3`

or, R₂ = 3R₁ = 3 × 10 = 30 Ω

The equivalent resistance of these two resistors in series = R₁ + R₂ = 30 + 10 = 40 Ω.