How to calculate cable size
The cable must carry the current without overheating (current-carrying capacity) and keep voltage drop low enough. We calculate the minimum area for each check and pick the next standard size that satisfies both.
Min. area (1φ) = 2 × L × I × ρ ÷ ΔV
Min. area (3φ) = 1.732 × L × I × ρ ÷ ΔV
ρ copper ≈ 0.0225 Ω·mm²/m · ρ aluminum ≈ 0.036 Ω·mm²/m
Example: 32 A, 30 m, 230 V single-phase, copper
Allowed drop = 3% × 230 = 6.9 V. Area = 2 × 30 × 32 × 0.0225 ÷ 6.9 = 6.26 mm². Current capacity alone would allow 6 mm² (46 A), but voltage drop needs more, so the answer is 10 mm².
Current capacity, copper PVC (IEC 60364-5-52, method C)
| Size | 2 loaded conductors | 3 loaded conductors |
|---|---|---|
| 1.5 mm² | 19.5 A | 17.5 A |
| 2.5 mm² | 27 A | 24 A |
| 4 mm² | 36 A | 32 A |
| 6 mm² | 46 A | 41 A |
| 10 mm² | 63 A | 57 A |
| 16 mm² | 85 A | 76 A |
| 25 mm² | 112 A | 96 A |
| 35 mm² | 138 A | 119 A |
| 50 mm² | 168 A | 144 A |
| 70 mm² | 213 A | 184 A |
Frequently asked questions
What size cable for 32 amps?
6 mm² copper carries 32 A in most installations, but runs over about 20 m at 230 V usually need 10 mm² for voltage drop.
How do I convert mm² to AWG?
Approximate matches: 1.5 mm² ≈ 16 AWG, 2.5 mm² ≈ 14 AWG, 4 mm² ≈ 12 AWG, 6 mm² ≈ 10 AWG, 10 mm² ≈ 8 AWG, 16 mm² ≈ 6 AWG, 25 mm² ≈ 4 AWG.
Why does the calculator suggest a bigger cable than the current table?
On long runs voltage drop, not heating, decides the size.