How cable size is chosen in an installation

A cable is sized so that it carries the design current without overheating, keeps voltage drop within limits and withstands short-circuit current. The same method is used for each stage, but the governing condition changes from stage to stage.

Transformer to MDB

• Design current = full-load LT current of the transformer: I = kVA × 1000 / (√3 × 400). Example: 250 kVA → 360.8 A.
• Cable (or parallel runs/bus duct) must carry this current after derating for grouping and temperature; it must also withstand the transformer LT fault current for the clearing time (A ≥ I_sc·√t / k).
• Run is short, so current rating and short-circuit withstand govern; voltage drop is usually small (aim < 1 %).

MDB to SDB

• Design current = maximum demand of the SDB (connected load × Ku × Ks), plus 20–25 % future margin.
• Cable rating ≥ rating of the protective MCCB/MCB at the MDB.
• Feeder lengths are moderate to long, so check voltage drop (about 1.5–2.5 %); long feeders often need a larger size than the current alone requires.

SDB to end use

• Design current of each final circuit: motor full-load current (I = P/(√3·V·η·pf)), lighting or socket circuit current.
• Cable rating ≥ MCB rating; minimum 1.5 mm² Cu for lighting, 2.5–4 mm² for power sockets.
• For motors, check voltage drop during starting (5–7 × FLC) and running.
• Total drop from transformer to the farthest load should be within about 5 %.

Factors on which selection depends

1. Load current (design current) and type of load.
2. Current-carrying capacity from the manufacturer's table.
3. Derating factors: ambient temperature, grouping, method of laying (air, duct, ground), depth and soil resistivity.
4. Permissible voltage drop and cable length.
5. Short-circuit current and fault clearing time.
6. Rating of the protective device (coordination Ib ≤ In ≤ Iz).
7. Conductor material (Cu/Al), insulation (PVC/XLPE) and armouring.
8. Future load growth and economics (cost of losses).