---
title: "Lead and iron wire cross-section ratio"
subject: "Basic Electrical Engineering"
chapter: 1
key: "semester_1/basic_electrical_engineering:lead-iron-wire-parallel-area-ratio"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/lead-iron-wire-parallel-area-ratio
site: BCT Engineering
---

# Lead and iron wire cross-section ratio

Basic Electrical Engineering · Chapter 1: General Electric System

- Asked in: 2069 Asar, 4 marks
- Video: [A lead  and an iron wire are connected in parallel. specific resistances  ratio 49 : 24 BE ENGINEER](https://www.youtube.com/watch?v=PcIuDzKrGSA) — Target Nasa
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/lead-iron-wire-parallel-area-ratio
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-1

## Question

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.

## 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.

## Answer

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)**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For two wires in parallel, which relation holds?
   A. I1 R1 = I2 R2 since they have the same voltage
   B. I1 = I2
   C. R1 = R2 always
   D. I1 / R1 = I2 / R2
   Answer: A. Same voltage across both means I1 R1 = I2 R2. (Key point: Parallel relation)

2. If lead carries 80 % more current than iron, R_lead/R_iron equals:
   A. 1.8
   B. 1/1.8
   C. 0.8
   D. 1/0.8
   Answer: B. R is inversely proportional to current in parallel: I_iron/I_lead = 1/1.8. (Key point: Resistance ratio)

3. Using R = rho l / A with rho ratio 49:24 and l_iron = 1.47 l_lead, A_lead : A_iron is:
   A. 0.4 : 1
   B. 1 : 2.5
   C. 1.8 : 1
   D. 2.5 : 1
   Answer: D. A_iron/A_lead = (1/1.8)(24/49)(1.47) = 0.40, so A_lead/A_iron = 2.5. (Key point: Area ratio)

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