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
Cross-section and breaker chosen together — Iz from Annex 5.10, In ≤ corrected Iz (Art. 4.3.2.1.3 I7-2011)
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 only K1. The Iz table stays the PVC one, so below 30 °C XLPE can give a larger cross-section than PVC
Factor 0.86 — Annex 5.29, for cables with 4 or 5 conductors: applied to three-phase circuits only (LED, switched-mode supplies)
Conservative derating ~0.93 (dry/sandy soil)
Current-carrying capacity table Iz (A) — I7-2011 Annex 5.10, method B2, PVC, 30 °C
| Cross-sectional area (mm²) | 1.5 | 2.5 | 4 | 6 | 10 | 16 | 25 | 35 |
|---|---|---|---|---|---|---|---|---|
| I max Cu (A) · B2, 2 loaded cond. | 16.5 | 23 | 30 | 38 | 52 | 69 | 90 | 111 |
| I max Al (A) · B2, 2 loaded cond. | — | — | — | — | 41 | 54 | 71 | 86 |
Voltage drop calculator
Check against the 3% limit (lighting) / 5% (sockets) — I7-2011 Art. 5.2.5.1
Conductivity used in the calculation: Cu = 56 m/(Ω·mm²), Al = 35 m/(Ω·mm²). These are the app’s working values, close to a cold conductor; the standard does not fix a conductivity, and a hot conductor has higher resistance, so the real drop can be up to ~20–25 % higher than the one calculated here — a result close to the limit must be rechecked.
Calculation assumptions and scope
- Included: Cu/Al material; single-phase (2 loaded conductors) or three-phase (3); installation method B2 or C, with the current-carrying capacities from Annex 5.10; the breaker chosen together with the cross-section (In ≤ corrected Iz, Art. 4.3.2.1.3), plus the app’s 1.25 design reserve; PVC/XLPE insulation, K1 (temperature, Annex 5.18), K2 (grouping, Annex 5.19), 3rd-order harmonics (0.86, Annex 5.29, three-phase only), dry-soil derating. For aluminium, cross-sections from 10 mm² upwards — a convention of the app’s validation engine, not a limit from the standard (Annex 5.10 also covers aluminium from 4 mm²); minimums per circuit type are in Annex 5.32
- The „dry soil” factor is a conservative estimate of the app (~0.93), not a factor from the standard, and it only makes sense for laying in the ground; for buried cables use the in-ground laying table (method D, Annexes 5.22–5.28)
- The current-carrying capacities in the table are those for PVC insulation; with XLPE only K1 changes, the Iz table stays the PVC one (a conservative result). The voltage drop is calculated with the conductivity stated below the calculator, not with a value imposed by the standard
- 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 % design reserve (the app’s own choice, not a code requirement)
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 and ≥ In
Iz_table × K ≥ max(I_design; In) (equivalent: Iz_required = max(I_design; In) / K)
Ideal case (K=1), single-phase — 2 loaded conductors: 20 A → 2.5 mm² (23 A). Real case (K=0.609): 20 / 0.609 = 32.8 A → 6 mm² (38 A). Three-phase (3 loaded conductors) gives the same cross-sections: 2.5 mm² (20 A) in the ideal case, where Iz meets the design current exactly — equality with the 1.25 reserve is accepted, because Iz stays above the 16 A In —, and 6 mm² (34 A) in the real case. Here the breaker is 16 A (the first rating ≥ Ib = 16 A), so the design current of 20 A exceeds it and sets the cross-section. When the rating is the larger of the two — the rating series has steps of more than 25 % — the cross-section follows In, because Art. 4.3.2.1.3 requires In ≤ the corrected Iz; at exact equality In = corrected Iz, the app goes up one size.
| Cross-sectional area (mm²) | 1.5 | 2.5 | 4 | 6 | 10 | 16 |
|---|---|---|---|---|---|---|
| Iz B2 Cu, 2 loaded cond. (A) | 16.5 | 23 | 30 | 38 | 52 | 69 |
| Iz B2 Cu, 3 loaded cond. (A) | 15 | 20 | 27 | 34 | 46 | 62 |
| Al cross-sectional area (mm²) | 10 | 16 | 25 | 35 |
|---|---|---|---|---|
| Iz B2 Al, 2 loaded cond. (A) | 41 | 54 | 71 | 86 |
| Iz B2 Al, 3 loaded cond. (A) | 36 | 48 | 62 | 77 |
Current-carrying capacities from I7-2011 Annex 5.10, method B2 (cable in conduit, on or in the wall), PVC insulation, 30 °C: 2 loaded conductors for single-phase (L + N), 3 for three-phase (Art. 5.2.3.1.10 b) and Art. 5.2.3.0.7.1). They are the same values the app’s validation (V06) uses. For aluminium the app’s table starts at 10 mm² — an internal convention; Annex 5.10 also covers aluminium from 4 mm², and the minimums per circuit type are in Annex 5.32.
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 (the smallest cross-section with a tabulated Iz in the app — not a minimum from the standard; the minimums are in Annex 5.32) |
| 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 (2 loaded) | R + S + T + N + PE (3 loaded) |
| 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
In three-phase, the calculation gives 1.5 mm², which the code allows (Annex 5.32: at least 1.5 mm² for power circuits). The app’s rule V20, however, asks by convention for at least 2.5 mm² on three-phase circuits and flags a smaller cross-section in the editor; in practice 2.5 mm² is chosen.
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), hard-wired without a socket, 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
A 3000 W load (single-phase, cos φ = 1), on a power circuit, with aluminium cable. What is the minimum cross-section (mm²)? Note: the app’s Iz table starts at 10 mm².
Discussion
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