Cable selection is not just dependent on the load. There is also a requirement that the cable withstand all the short circuit current that the source might be able to deliver. It has to carry this large current, without significant damage, for the duration that it takes for the circuit protection to operate. This method applies to cables rated up to 600 volts phase to neutral (or 1kV phase to phase) and has some basis in IEC 60287.
The first step is to determine the maximum prospective short circuit (PSC) current that could flow in any part of the cable. If you don't know how to do this using the transformer impedance and supply cable impedance etc, then it is safe to use the potential current at the transformer secondary terminals. For a transformer with 4% impedance, then the short circuit current could be 25 times the nominal rated secondary current. So a 200 kVA 3 phase transformer, if rated at 240 volts phase to neutral, will have a rated secondary current of 200,000/3/240 = 277.7 amps and a PSC current of 277.7 * 25= 6,945 amps (say 7,000 amps) under short circuit conditions. Check with your supply authority for them to tell you the potential fault current at your premises.
The main circuit breaker after the transformer might be rated at 250 amps per phase but there will be circuit breakers protecting lights or socket outlets rated at 20 amps per phase. The PSC at the 20 amp circuit breaker is still 7kA. If you know the length and impedance of the consumer mains and any cables from the main circuit breaker to the 20 amp breaker, then you can calculate the slight reduction in PSC at the 20 amp breaker.
Check the breaker manufacturer's data to determine the circuit breaker operating time to clear a current equal to the PSC. If you do not have this data, it is usual to assume a one second operating time.
Then apply the formula:
I²t = K²S²
I= short circuit current in amperes
t= duration of short circuit in seconds
K= a constant depending on the material of the conductor, its initial temperature and its final temperature (from a table in the standard)
S= cross sectional area of the current carrying conductor.
If you do not have access to the Standard (IEC 60287), then for a copper cable with PVC insulation, you can assume a value of K = 111.
So for an installation with a 200 kVA transformer close to the circuit breaker:
I=7,000 and t=1 and K= 111, find S.
7000² x 1²= 111² x S²
49,000,000 = 12,321 x S²
So the minimum cable size has to be at least 63 square millimetres (nearest commercially available size is 70 mm².)
All cables, even for 20 amp load, that originate from the main switchboard near a 200 kVA transformer have to be at least 70 mm² because of the potential short circuit current.. If you use a small cable like 2.5 mm², then a short circuit would totally destroy the cable before the circuit breaker trips. If somebody operates the (20A) breaker after it has tripped due to the short circuit. It possible that the metalwork of the switchboard would be live if the damaged cable is now in contact with any metal parts.
Many people do not understand the importance of selecting the cable to not just carry the load current continuously, but it must also carry the short circuit current until the circuit breaker or fuse operates. Some people have the mistaken idea that the maximum current is 7.5 times the normal current. That is the point where the breaker will trip instantly on magnetic operation.
The cable has to be selected to satisfy all the following conditions:
- continuous current carrying capacity
- voltage drop
- prospective short circuit withstand
- earth fault loop impedance
The notes in this comment only describe the prospective short circuit withstand. If you are working in a country that uses NEC, the follow your local wiring rules.