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Length of heater wire from conductivity

Short answer

R = V²/P = 48.4 Ω. Resistivity ρ = 1/σ = 6×10⁻⁷ Ω·m. Area A = π(0.5×10⁻³)²/4 = 1.9635×10⁻⁷ m². Length l = RA/ρ = 15.84 m ≈ 15.8 m.

You are using a heater 220V/1KW made of a material of conductivity 16666.67 mho/cm. The wire has thickness of 0.5mm. What could have been the length of the given wire used in building heating element?

The heater rating gives its resistance; the wire dimensions and conductivity then give the length.

Step 1: Resistance of the element

R = V²/P = 220²/1000 = 48.4 Ω

Step 2: Resistivity

σ = 16666.67 mho/cm = 16666.67 × 100 S/m = 1.6667 × 10⁶ S/m

ρ = 1/σ = 6 × 10⁻⁷ Ω·m (that is 60 μΩ·cm)

Step 3: Cross-sectional area

d = 0.5 mm, so A = π d²/4 = π (0.5 × 10⁻³)²/4 = 1.9635 × 10⁻⁷ m²

Step 4: Length

R = ρ l/A, so

l = R A/ρ = 48.4 × 1.9635 × 10⁻⁷/(6 × 10⁻⁷) = 15.84 m

Answer: Length of wire ≈ 15.8 m (assuming the wire is round with 0.5 mm diameter)

✓ Answer checked by Claude Opus 5.5 (high effort)

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