Calculation questions — short-circuit current & voltage drop

Two types of calculation questions recur on almost every NEN 1010/3140-related theory or practical exam: short-circuit current (needed to check whether a distribution board/circuit breaker can handle it) and voltage drop (needed to check whether equipment still functions within the permitted margin). This article covers both with a worked example.

Short-circuit current at the origin (Icc)

The maximum short-circuit current a transformer can deliver in the event of a direct short circuit on the secondary terminals:

Icc = In / %Ucc

• In — rated secondary current of the transformer.
• %Ucc — short-circuit voltage of the transformer, as a percentage of the rated voltage (typically 4–6% for distribution transformers, stated on the nameplate).

Worked example

Transformer 1000 kVA, 400 V, Ucc = 6%:

In = 1 000 000 / (1.732 × 400) ≈ 1 443 A
Icc = 1 443 / 0.06 ≈ 24 057 A ≈ 24 kA

This figure determines the minimum breaking capacity (kA) that the main circuit breaker or main distribution board must have — a circuit breaker with too low a breaking capacity can itself be damaged by this short-circuit current instead of switching off safely.

Ikmax versus Ikmin

Two different short-circuit currents are relevant for two different purposes:

An installation checked only against Ikmax can still have a fault: at the end of a long, thin cable, the short-circuit current may be too low to make the circuit breaker trip within the required time (see §411, automatic disconnection of supply).

Voltage drop — the limits

NEN 1010 §525 (table 52.G.1) gives the maximum permitted voltage drop between the origin of the installation and any point of use:

• 3% for lighting installations.
• 5% for other uses (socket outlets, motors, general loads).

Note: older sources and rules of thumb sometimes mention higher percentages — for exams, use the official table values 3% / 5%.

Calculation formula

ΔU = (2 × ρ × L × I) / A

• ρ — resistivity of copper (≈ 0.0175 Ω·mm²/m at 20°C).
• L — length of the cable in metres (single direction).
• I — load current in amperes.
• A — cross-sectional area of the conductor in mm².
• The factor 2 accounts for both the outgoing and return conductor (phase + neutral) — a common exam mistake is forgetting this factor.

Worked example

A socket outlet circuit, 25 m single length, 2.5 mm², load 10 A, 230 V:

ΔU = (2 × 0.0175 × 25 × 10) / 2.5 = 3.5 V
% = 3.5 / 230 × 100 ≈ 1.5%

This stays well within the 5% limit for other uses.

Common mistakes

1. Forgetting the factor 2 in the voltage-drop formula — this gives a result that is half the actual voltage drop.
2. Confusing Icc and Ikmin — Icc (at the origin) is almost always the highest current in the installation, Ikmin at the end of a long circuit the lowest; both are needed, for different checks.
3. Using outdated voltage-drop percentages (e.g. 6%/8% from non-official rules of thumb) instead of the table values 3%/5% from §525.
4. Confusing the %Ucc from the nameplate with the load factor — %Ucc is a fixed transformer property, not a percentage of the actual load.

Related

• §531 — Residual Current Devices (RCD)
• §523 — Cable cross-section & current-carrying capacity
• §536 — Distribution boards & selectivity
• Theory exam questions