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Cylindrical resistor current density and p.d.

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.

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?

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

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

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

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

✓ Answer checked by Claude Opus 5.5 (high effort)

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