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.

Note — assumption: Icc = In / %Ucc assumes an infinitely strong upstream network (zero network impedance). In reality the short-circuit current is somewhat lower, so the formula gives a safe, high upper bound for choosing the breaking capacity. For an exact value add the network impedance (from the network operator) to that of the transformer.

Ikmax versus Ikmin

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

	Where calculated	What it is used for
Ikmax	At minimum cable impedance (short, thick, close to the source)	Checking whether the breaking capacity of protective devices is sufficient.
Ikmin	At maximum cable impedance (long, thin, far from the source — often the end of the longest final circuit)	Checking whether the protective device actually trips fast enough for a fault far away.

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

The limits for the maximum voltage drop between the origin of the installation and any point of utilisation come from IEC 60364-5-52 (Annex G, Table G.52.1), which also underlies the Dutch rule (NEN 1010 §525 — check the exact value in your edition):

Supply	Lighting	Other uses
Installation directly on the public low-voltage network	3 %	5 %
Installation on its own transformer (MV/LV, e.g. large businesses, greenhouses)	6 %	8 %

With a private transformer there is more headroom because the voltage drop in the public service connection does not count. In a standard exam situation use 3 % / 5 % unless the question mentions a private transformer.

Formula (single-phase)

ΔU = (2 × ρ × L × I) / A

• ρ — resistivity of copper. Two values are in use:
  • 0.0175 Ω·mm²/m — copper at 20 °C. This is usually what the exam gives; the result is the voltage drop with cold conductors.
  • 0.0225 Ω·mm²/m (= 1.25 × ρ₂₀) — copper at operating temperature (about 70 °C for PVC). IEC 60364-5-52 (Annex G) uses this because a loaded cable is warm. It gives a 29 % higher voltage drop and is the more realistic design value.
• L — one-way length in metres, I — current in A, A — cross-section in mm².
• The factor 2 accounts for outgoing and return conductor (phase + neutral). Forgetting it is a classic exam error.

Worked example (both values of ρ)

Socket-outlet circuit, 25 m, 2.5 mm², 10 A, 230 V:

ρ = 0.0175:  ΔU = (2 × 0.0175 × 25 × 10) / 2.5 = 3.5 V   → 1.52 %
ρ = 0.0225:  ΔU = (2 × 0.0225 × 25 × 10) / 2.5 = 4.5 V   → 1.96 %

Both stay well within 5 %; the 29 % difference can decide on longer circuits or with little margin. Read the question: if it gives a ρ value, use it.

Three-phase: power, current and voltage drop

P = √3 × U × I × cos φ     →     I = P / (√3 × U × cos φ)
ΔU(3~) = √3 × I × L × (ρ / A) × cos φ     (resistive part; reactance neglected)

Example: motor 11 kW, 400 V, cos φ = 0.85, cable 30 m, 4 mm² copper:

I  = 11 000 / (1.732 × 400 × 0.85) ≈ 18.7 A
ΔU = 1.732 × 18.7 × 30 × (0.0225 / 4) × 0.85 ≈ 4.6 V   → 1.16 % of 400 V
     (with ρ = 0.0175: ≈ 3.6 V → 0.90 %)

For large cross-sections (from about 50 mm²) reactance can no longer be neglected.

Choosing the cable cross-section

Steps:

1. Ib — design current of the load (18.7 A above).
2. In — choose a protective device with Ib ≤ In: here a 20 A breaker.
3. Iz — the cable must carry In: In ≤ Iz, where Iz is the table value after correction factors (ambient temperature, grouping, installation method). With a total factor of 0.8 a table value of ≥ 20 / 0.8 = 25 A is needed. In the table for your installation method pick the smallest cross-section with ≥ 25 A.
4. Then check voltage drop (above) and automatic disconnection (next section). The thinnest cable that satisfies step 3 often fails step 4.

Fault loop Zs and short-circuit current at the end of the cable

For automatic disconnection (TN system):

Zs × Ia ≤ U₀

• Zs — fault-loop impedance (source + line conductor + PE).
• Ia — current that trips the device within the required time (for a B-curve breaker: the upper limit of magnetic tripping, 5 × In).
• U₀ — line-to-earth voltage (230 V).

Example: B16 breaker, final circuit ≤ 32 A (required disconnection time 0.4 s):

Ia = 5 × 16 = 80 A
Zs(max) = 230 / 80 ≈ 2.9 Ω

Measured Zs = 1.9 Ω → passes. Measured Zs = 3.2 Ω → fails: shorter or thicker cable, smaller breaker, or an RCD as an additional measure.

Ik at the end of the cable (single-phase, line–neutral):

Zcable = 2 × ρ × L / A = 2 × 0.0225 × 25 / 2.5 = 0.45 Ω   (hot cable)
Ik,min ≈ 230 / (Zsource + Zcable) = 230 / (0.05 + 0.45) = 460 A

(assuming a source impedance of 0.05 Ω). 460 A ≥ 80 A → the B16 trips magnetically. For Ik,min use the hot resistance (higher ρ), for Ik,max the cold one (lower ρ) — the opposite of intuition. See also protective conductor continuity.

Common mistakes

1. Forgetting the factor 2 in single-phase voltage drop — the result is half the real value.
2. Mixing up Icc and Ikmin — Icc (at the origin) is almost always the highest current, Ikmin at the end of a long circuit the lowest; both are needed, for different checks.
3. Always applying 3 %/5 % — with a private transformer 6 %/8 % apply (IEC 60364-5-52, G.52.1); conversely do not use 6 %/8 % for an ordinary connection to the public network.
4. Using ρ = 0.0175 when the question means operating temperature (or vice versa) — the difference is 29 %.
5. Confusing %Ucc on the nameplate with the load factor — %Ucc is a fixed transformer property.
6. Selecting on Ib and current-carrying capacity only and skipping voltage drop and disconnection time.

Related

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