Question

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

• A. 40Ω
• B. 100Ω
• C. $\frac{5}{2}$Ω
• D. $\frac{40}{3}$Ω

Answer 1

Answer: (A)

40Ω

Answer 2

$\frac{R_{1}}{R_{2}} = \frac{A_{3}}{A_{1}} = \frac{1}{3}$As ρ and l is same

R₁ = 10Ω (As A₁ > A₃) ∴ R₂ = 3R₁ = 30Ω

R_eq = R₁ + R₂ = 30 + 10 = 40Ω