Skip to main content

Lead and iron wire cross-section ratio

Short answer

I_lead = 1.8 I_iron, so R_lead/R_iron = 1/1.8. Using R=ρl/A and length ratio, A_lead/A_iron = (49/24)×(1/1.47)×1.8 = 2.5.

A lead wire and an iron wire are connected in parallel. Their specific resistances are in the ratio 49:24. The former carries 80% more current than the latter and the latter is 47% longer than the former. Determine the ratio of their cross-sectional areas.

Two wires in parallel have the same voltage, so I₁R₁ = I₂R₂.

Data

  • ρ_lead : ρ_iron = 49 : 24
  • I_lead = 1.8 × I_iron (80 % more)
  • l_iron = 1.47 × l_lead (47 % longer)

Step 1: Resistance ratio

R_lead/R_iron = I_iron/I_lead = 1/1.8

Step 2: Write resistance in terms of ρ, l and A

R = ρ l/A

R_lead/R_iron = (ρ_l/ρ_i) × (l_l/l_i) × (A_i/A_l)

1/1.8 = (49/24) × (1/1.47) × (A_i/A_l)

Step 3: Solve

A_i/A_l = (1/1.8) × (24/49) × 1.47 = 0.40

A_lead/A_iron = 1/0.40 = 2.5

Answer: A_lead : A_iron = 2.5 : 1 (that is, 5 : 2)

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Basic Electrical Engineering chapter 1 questions →