---
title: "Cylindrical resistor current density and p.d."
subject: "Engineering Physics"
chapter: 8
key: "semester_1/engineering_physics:cylinder-resistor-current-density-power"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/cylinder-resistor-current-density-power
site: BCT Engineering
---

# Cylindrical resistor current density and p.d.

Engineering Physics · Chapter 8: Electromagnetism

- Asked in: 2070 Asar, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/cylinder-resistor-current-density-power
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-8

## Question

A cylindrical resistor of radius 6 mm and length 2.5 cm is made of material that has a resistivity of 4×10⁻⁵ Ω.m. What are (i) the magnitude of the current density and (ii) the potential difference when the energy dissipation rate in the resistor is 2 Watt?

## Short answer

Resistance R = ρL/(πr²) = 8.84×10⁻³ Ω. Current I = √(P/R) = 15.04 A gives current density J = I/A = 1.33×10⁵ A/m². Potential difference V = P/I = 0.133 V.

## Answer

**Given:** radius r = 6 mm, length L = 2.5 cm, ρ = 4×10⁻⁵ Ω·m, power P = 2 W.

```text
Area   A = π r² = π (6×10⁻³)² = 1.131×10⁻⁴ m²
Resistance R = ρ L / A = (4×10⁻⁵ × 2.5×10⁻²) / 1.131×10⁻⁴
             = 8.84×10⁻³ Ω
```

#### (i) Current density
```text
P = I² R  →  I = √(P/R) = √(2 / 8.84×10⁻³) = 15.04 A
J = I / A = 15.04 / 1.131×10⁻⁴
  = 1.33×10⁵ A/m²
```

#### (ii) Potential difference
```text
V = P / I = 2 / 15.04 = 0.133 V
(check: V = IR = 15.04 × 8.84×10⁻³ = 0.133 V)
```

**Answer: (i) J ≈ 1.33×10⁵ A/m²; (ii) V ≈ 0.133 V**

## Check yourself

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

1. The resistance of the cylindrical resistor is found from:
   A. R = ρL/A with A = πr²
   B. R = ρA/L
   C. R = ρ/(LA)
   D. R = ρL/(2πr)
   Answer: A. Resistance = resistivity × length / cross-sectional area. (Key point: Resistance)

2. Given P = 2 W and R = 8.84 mΩ, the current is found from:
   A. P = I R, so I = P/R ≈ 226 A
   B. P = I²R, so I = √(P/R) ≈ 15 A
   C. P = I/R, so I = PR
   D. I = P R² ≈ 0.0002 A
   Answer: B. Dissipation rate is I²R. (Key point: Current from power)

3. The current density is:
   A. J = I A ≈ 1.7×10⁻³ A/m²
   B. J = I/L ≈ 600 A/m²
   C. J = I/A ≈ 1.33×10⁵ A/m²
   D. J = P/A ≈ 1.8×10⁴ A/m²
   Answer: C. J = 15.04 A / 1.131×10⁻⁴ m². (Key point: Current density)

4. The potential difference follows from:
   A. V = P I ≈ 30 V
   B. V = I/R ≈ 1700 V
   C. V = P R ≈ 0.018 V
   D. V = P/I = IR ≈ 0.133 V
   Answer: D. Both P/I and IR give 0.133 V. (Key point: Potential difference)

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