Interactive compliance calculators
Calculation tools per I7-2011 and IEC 60364: cable cross-sectional area sizing by current and installation method, voltage drop check per circuit. The formulas are shown step by step.
Cable sizing calculator
Minimum cross-sectional area per I7-2011 Table 5.1 + IEC 60364-5-52
0.85 typical residential, 1.0 purely resistive
K1 — Annex 5.18 (reference 30°C, PVC insulation)
K2 — Annex 5.19 (1 = a single circuit)
Affects K1 (XLPE withstands heat better)
Factor 0.86 — Annex 5.29 (LED, switched-mode supplies)
Conservative derating ~0.93 (dry/sandy soil)
Current-carrying capacity table (Cu + Al, concealed B2) — I7-2011 reference
| Cross-sectional area (mm²) | 1.5 | 2.5 | 4 | 6 | 10 | 16 | 25 | 35 |
|---|---|---|---|---|---|---|---|---|
| I max Cu (A) · B2 | 15 | 21 | 28 | 36 | 50 | 66 | 84 | 104 |
| I max Al (A) · B2 | — | — | 22 | 28 | 39 | 52 | 66 | 82 |
Voltage drop calculator
Check against the 3% limit (lighting) / 5% (sockets) — I7-2011 Art. 5.2.5.1
Conductivity: Cu = 56 m/(Ω·mm²), Al = 35 m/(Ω·mm²) at 70°C per IEC 60228.
Calculation assumptions and scope
- Included: Cu/Al material, single-phase/three-phase, B2/F installation, PVC/XLPE insulation, K1 (temperature, Annex 5.18), K2 (grouping, Annex 5.19), 3rd-order harmonics (0.86, Annex 5.29), dry-soil derating and the 10 mm² minimum cross-sectional area for Al (Art. 5.2.4.6.1(d))
- The "dry soil" factor is a conservative estimate (~0.93) applied to table B2; for accurate sizing of buried cables use the in-ground installation table (method D, Annexes 5.22–5.28)
- Tabulated current-carrying capacities for PVC insulation at 70°C (XLPE is adjusted via K1); voltage drop calculated using the resistivity at 70°C
- For complex industrial installations, validate the result with a full calculation + short-circuit check (breaking capacity + thermal stability)
How it’s calculated — step by step
The full method per I7-2011 / IEC 60364, from power to cross-sectional area, length and voltage drop.
A. The operating current — the starting point
Work out the current drawn from the power
Single-phase: I = P / (U × cos φ), U = 230 V
Three-phase: I = P / (√3 × U × cos φ), U = 400 V
E.g.: a 2000 W socket, cos φ = 0.9 → I = 2000 / (230 × 0.9) = 9.66 A.
B. Choosing the cable cross-sectional area (with K correction factors)
Start from the operating current I (step A)
E.g.: I = 16 A.
Apply the 25% safety margin (IEC 60364-5-52 Art. 523)
I_design = I × 1.25
16 × 1.25 = 20 A.
Apply the K correction factors (real-world conditions)
K = K1 (temperature, Annex 5.18) × K2 (grouping, Annex 5.19) × K_harmonics × K_soil
At 30 °C and a single circuit, K = 1. At 40 °C + 3 grouped circuits: K = 0.87 × 0.70 = 0.609.
Additional factors (all in the calculator above): 3rd-order harmonics → 0.86 (Annex 5.29); dry soil / in-ground installation → derating; XLPE insulation adjusts K1.
Choose the cross-sectional area: the first in the table with Iz × K ≥ I_design
Iz_table × K ≥ I_design (equivalent: Iz_required = I_design / K)
Ideal case (K=1): 20 A → 2.5 mm² (21 A). Real case (K=0.609): 20 / 0.609 = 32.8 A → 6 mm² (36 A).
| Cross-sectional area (mm²) | 1.5 | 2.5 | 4 | 6 | 10 | 16 |
|---|---|---|---|---|---|---|
| Iz B2 Cu (A) | 15 | 21 | 28 | 36 | 50 | 66 |
| Al cross-sectional area (mm²) | 10 | 16 | 25 | 35 |
|---|---|---|---|---|
| Iz B2 Al (A) | 39 | 52 | 66 | 82 |
Iz tables for B2 installation, PVC insulation (Cu and Al conductors): I7 Annex 5.10 / NTE 007/08/00 (XLPE: Annex 5.13). Aluminium — minimum cross-sectional area 10 mm² (Art. 5.2.4.6.1(d)).
C. The real cable length
Measure the horizontal route on the plan
The sum of the horizontal segments between consumer unit → junction boxes → equipment.
Add the vertical drops/rises (mounting heights)
junction box 2.0 m · socket 0.3 m · switch 1.3 m · light/ceiling 2.6 m
E.g.: from the junction box (2.0 m) to the socket (0.3 m) → drop = 2.0 − 0.3 = 1.7 m.
Add the floor-to-floor crossings (slab height)
~2.6–3 m per floor crossed.
Add it all up
E.g.: 8 m horizontal + 1.7 m socket drop = 9.7 m.
D. Voltage drop over the length
Apply the voltage drop formula
ΔU% = (factor × L × P) / (k × S × U²) × 100
factor = 2 (single-phase, out and back) or 1 (three-phase); k = 56 (Cu) / 35 (Al); S = cross-sectional area (mm²).
Quick rule (single-phase, copper)
ΔU% ≈ (L × P) / (14800 × S)
E.g.: L = 20 m, P = 2000 W, S = 2.5 → 20 × 2000 / (14800 × 2.5) ≈ 1.08 %.
Compare with the regulatory limit (I7-2011 Art. 5.2.5.1)
3 % for lighting, 5 % for sockets/power. If you exceed it → increase the cross-sectional area or shorten the route.
Conductivity k per IEC 60228 (Cu 56, Al 35 m/Ω·mm²). Voltage drop limits: I7-2011 Art. 5.2.5.1.
E. Copper (Cu) vs. Aluminium (Al)
| Aspect | Copper (Cu) | Aluminium (Al) |
|---|---|---|
| Conductivity k | 56 | 35 → ~60% more voltage drop |
| Current-carrying capacity | reference | ~78% of Cu → larger cross-sectional area |
| Minimum cross-sectional area | 1.5 mm² | 10 mm² (Art. 5.2.4.6.1(d)) |
| Typical use | indoor circuits | service connection / distribution circuit |
In the formulas, the only change is k (56 → 35) and the current-carrying capacity table. The Cu/Al toggle in the comparator below applies both automatically.
K1 — temperature (PVC, Annex 5.18)
| °C | 30 | 35 | 40 | 45 | 50 |
|---|---|---|---|---|---|
| K1 | 1 | 0.94 | 0.87 | 0.79 | 0.71 |
K2 — circuit grouping (Annex 5.19)
| circ. | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| K2 | 1 | 0.8 | 0.7 | 0.65 |
Single-phase vs. three-phase
The same calculations, but 3 differences that matter. Enter a power and see both options at once.
| Quantity | Single-phase (230 V) | Three-phase (400 V) |
|---|---|---|
| Current | I = P/(230·cosφ) | I = P/(√3·400·cosφ) |
| Voltage drop | factor 2 · U = 230 | factor 1 · U = 400 |
| Conductors | L + N + PE | R + S + T + N + PE |
| Protection | 1P+N | 3P / 4P · R/S/T balancing |
At the same power, three-phase "draws" ~1/3 of the current (→ smaller cross-sectional area) and has ~6× less voltage drop.
Single-phase · 230 V
Three-phase · 400 V
Three-phase / single-phase current = 230 / (√3 × 400) = 0.33 (≈ 1/3)
Cross-sectional area: 1.5 mm² (three-phase) vs 10 mm² (single-phase), same power
At the same cross-sectional area, the three-phase drop would be 6.0× smaller (factor 1 vs 2 and 400² vs 230²)
Exercises — pick the correct option
You go through each step by choosing the correct answer. Wrong? Try again.
Cross-sectional area — kitchen socket
A socket circuit feeds a 3000 W appliance (single-phase, cos φ = 1), B2 installation, at 30 °C, a single circuit.
Step 1: What is the operating current I = P / (U × cos φ)?
Length — route to the socket
The consumer unit → socket route is 6 m horizontal on the plan. The junction box is at 2.0 m, the socket at 0.3 m.
Step 1: What is the vertical drop from the junction box to the socket?
Voltage drop — socket circuit
Socket circuit, L = 25 m, P = 2200 W, 2.5 mm² Cu cross-sectional area, single-phase.
Step 1: What factor do you use in the formula (single-phase)?
Cross-sectional area with K factors — water heater in a hot loft
2000 W water heater (single-phase, cos φ = 1), mounted in a loft at 45 °C, grouped with 2 other circuits (3 in total).
Step 1: The operating current?
Exercises — you calculate each step
Now you enter the value for each step yourself. You get hints if you go wrong.
Cross-sectional area — air conditioning (you calculate)
1800 W AC unit (single-phase, cos φ = 0.9), at 35 °C, grouped with one more circuit (2 in total). Use the tables above.
Step 1: The operating current I = P / (U × cos φ) = ?
Length — two components
Consumer unit → junction box route = 9 m horizontal. The junction box at 2.0 m, the socket right below it at 0.3 m.
Step 1: Vertical drop junction box → socket = ?
Voltage drop — the full formula
L = 30 m, P = 3000 W, 4 mm² Cu cross-sectional area, single-phase. Formula: ΔU% = (2 × L × P) / (56 × S × 230²) × 100.
Step 1: ΔU% = ?
Cross-sectional area — three-phase hob
7200 W three-phase hob (cos φ = 1), 30 °C, one circuit. Three-phase: I = P / (√3 × 400 × cos φ).
Step 1: The operating current = ?
Voltage drop — ALUMINIUM distribution circuit
Aluminium distribution circuit: L = 30 m, P = 6000 W, 16 mm² cross-sectional area, single-phase. For aluminium k = 35 (not 56 as for copper).
Step 1: ΔU% on aluminium = ?
Exercises — direct answer
You solve it on your own and give the final answer. The step-by-step solution is available if needed.
Cross-sectional area — oven
2500 W oven (single-phase, cos φ = 1), 30 °C, one circuit. Which minimum cross-sectional area do you choose (mm²)?
Total length
15 m horizontal route + drop from the junction box (2.0 m) to the socket (0.3 m). What is the cable length (m)?
Voltage drop
L = 40 m, P = 2000 W, 2.5 mm² Cu cross-sectional area, single-phase. What is ΔU% (percentage)?
Cross-sectional area with K factors
4000 W load (single-phase, cos φ = 1), at 40 °C, grouped with 2 other circuits (3 total). Which cross-sectional area (mm²)?
Cross-sectional area — ALUMINIUM cable
3000 W load (single-phase, cos φ = 1), but an aluminium cable. Which minimum cross-sectional area (mm²)? Watch the Al minimum.
Discussion
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