4.4 Cable Short-Circuit Capacity

Fault heat is a different problem from continuous load heat

Current-carrying capacity (Sections 4.1–4.3) limits steady temperature under continuous load. A short-circuit dumps far higher current for a short time. Insulation and conductor metal must not reach destructive temperatures before the fuse or circuit-breaker clears.

AS/NZS 3008 provides data and methods for the short-circuit capacity of conductors — commonly taught through the adiabatic assumption: for brief faults, heat stays in the conductor (negligible heat loss during the event), so temperature rise tracks the electrical energy input.

Queensland licence and the capstone assessments expect you to recognise when a cable that "passes CCC" still fails fault withstand, especially on submains close to a MEN switchboard where prospective fault current is high.

The adiabatic relationship (teaching form)

A widely used adiabatic expression relating minimum conductor size to fault duty is:

S = √(I²t) / k

or equivalently I²t = S²k²

Where:

k is not a universal constant. Copper with a given insulation temperature limit has a different k from aluminium, and initial/final temperatures assumed in the Standard matter. Always take k from the AS/NZS 3008 tables matching the cable.

Rearrangements you will use

• Minimum S for a known I and t: S ≥ √(I²t) / k
• Maximum withstand time for known S, I, k: t ≤ (S²k²) / I²
• Maximum I for known S, t, k: I ≤ S k / √t

Worked example 1 — minimum size for a bolted fault duty

Given (illustrative): Prospective short-circuit current I = 6 000 A, protective device clears in t = 0.1 s, copper conductor with k = 115 (illustrative — verify in AS/NZS 3008 for the insulation/temperature pair).

I²t = 6000² × 0.1 = 36 000 000 × 0.1 = 3 600 000 A²s

√(I²t) = √3 600 000 ≈ 1 897

Smin = 1897 / 115 ≈ 16.5 mm²

So a 16 mm² copper conductor is borderline/inadequate on this illustrative arithmetic; 25 mm² would be the next common size that satisfies S ≥ 16.5 mm². Even if CCC derating only required 16 mm² for load, fault energy forces 25 mm².

That is the exam insight: two independent checks.

Worked example 2 — checking an existing submain

Given: Cu S = 25 mm², k = 115 (illustrative), fault I = 10 000 A.

Maximum adiabatic time:

t_max = (S²k²) / I² = (25² × 115²) / 10 000²

25² = 625; 115² = 13 225; product = 8 265 625

10 000² = 100 000 000

t_max ≈ 0.083 s

If the protective device's clearing time at 10 kA is 0.2 s, the cable fails the adiabatic check — change protection (faster device / better current-limiting), reduce fault level (design change), or increase S.

If the device is current-limiting and its published let-through I²t is used instead of the unbounded I²t from prospective current × full time, use the manufacturer's let-through data with the same S²k² ≥ I²t_let-through inequality. Licence answers should state whether prospective I with clearing time or device let-through energy was used.

Overload versus short-circuit versus earth fault

Protective earthing conductors also need fault withstand — undersized earths that melt or vaporise leave exposed conductive parts live. Do not celebrate a fat active/neutral pair with a fragile earth.

Coordination with protective devices

Circuit-breakers and fuses differ in let-through energy. A current-limiting HRC fuse may allow a smaller conductor for the same prospective fault level than a non-limiting breaker that holds longer. Conversely, a slow device on a high fault level near the supply transformer is hostile to small cables.

Also confirm the device breaking capacity (Icu/Icn) equals or exceeds prospective fault current at its point of installation — that is device selection, but it pairs with cable withstand in any serious fault study.

Voltage drop reminder (linked constraint)

Although this section focuses on short-circuit capacity, remember a cable enlarged for fault withstand often improves voltage drop automatically. On long Queensland rural runs, voltage drop may govern before fault energy; on short submains at a CBD main board, fault energy may govern before voltage drop. Run all checks.

Practical selection sequence (complete)

1. Estimate Ib, select In.
2. Choose installation method → base CCC → apply multiplied derating → verify Iz ≥ In.
3. Check voltage drop against AS/NZS 3000 limits for the circuit type.
4. Determine prospective fault current and protective clearing time / let-through.
5. Verify adiabatic withstand with correct k and S.
6. Verify earth conductor fault capacity and disconnection requirements.
7. Document assumptions (method, factors, fault level, k source).

Exam traps for Section 4.4

• Stopping after CCC and declaring the cable "compliant".
• Using aluminium k for a copper cable (or the reverse).
• Taking t from a random guess instead of device curves or stated exam data.
• Applying adiabatic heating to justify exceeding insulation temperature continuously (adiabatic is for short faults, not steady overload).
• Ignoring that reduced cross-section control cores or tap-offs may be the weak link on a faulted submain tee-off.

Capstone narrative example

"Consumer mains: 25 mm² Cu V-90, buried method per AS/NZS 3008, Iz after factors = 98 A, In = 80 A, OK for load. Prospective fault 12 kA; main fuse clears in 0.05 s; k from AS/NZS 3008 = …; S required = … mm²; 25 mm² adequate / inadequate. Earth conductor checked separately for the same fault duty."

Candidates who speak that language show they understand AS/NZS 3008 as a selection system, not a single ampacity poster on the workshop wall.

Master continuous rating, temperature class, derating multiplication and short-circuit adiabatic checks together, and cable selection questions on the Queensland electrical licence pathway become structured calculations instead of memorised folklore.