19/04/2026
How to do Cable Sizing – Step by Step
As per IEC 60364 / IS 732
Step 1: Calculate Load Current (I)
Single Phase: I = P / (V x PF)
Three Phase: I = P / (1.732 x V x PF)
Where P = Load in Watts, V = Voltage, PF = Power Factor 0.8 to 0.9
Example: 15 kW motor, 415V, 3-phase, 0.85 PF
I = 15000 / (1.732 x 415 x 0.85) = 24.5 A
Step 2:
Select Base Cable from Current Table
Pick rating ≥ Load Current.
From table: 24.5 A → 4 sqmm Cu = 25 A. Don’t stop here.
Step 3:
Apply Derating Factors – IEC 60364-5-52
Required Cable Capacity = Load Current / Total Derating Factor
Common factors:
1. Ambient Temp >30°C: 0.91 to 0.71.
If the surrounding temperature of equipment and cable is greater than 30°,the correction factor need.
Ambiant temperature is one iff Tempreature is30°.
Ex: 45°C = 0.79
2. Grouping of cables: 0.8 to 0.5.
The same size cable are grouped together,they impart heat to each other.
Ex: 6 cables in tray = 0.69
3. Installation Method:
Conduit, buried, tray – check table
Total Derating = Temp x Grouping x Installation
Example: 0.79 x 0.69 = 0.545
New required capacity = 24.5 / 0.545 = 44.9 A
Now from table: 44.9 A → 10 sqmm Cu = 45 A
Step 4: Check Voltage Drop
Max allowed: 3% lighting, 5% power
Single Phase: VD% = (2 x L x I x mV) / (V x 10)
Three Phase: VD% = (1.732 x L x I x mV) / (V x 10)
L = length meters, mV/A/m from cable datasheet, V = system voltage
If VD > allowed %, increase cable size.
Step 5: Check Short Circuit Withstand*
Min size: S = (Isc x √t) / k
Isc = fault current, t = breaker time sec, k = 115 for Cu-PVC, 143 for Cu-XLPE
Result: Final cable = largest size from Step 3, 4, and 5.
*Common Mistake*
Picking cable only by load current. Do all 5 steps.
Site example:
25 A load, 80m, grouped, 40°C → 4 sqmm fails. Correct = 16 sqmm.