Power/light sub-circuits and cable for 50 kW, 60 kW

Light and power sub-circuits

A sub-circuit is a branch circuit taken from a distribution board (DB) through its own fuse/MCB to feed a group of points. In building wiring the points are split into two separate kinds of sub-circuit:

• Light sub-circuit: feeds lamps, fans and 6 A (5 A) socket points. As per IS 732 / NEA practice it carries at most 10 points or 800 W, is wired in 1.5 mm² Cu and is protected by a 6 A (or 10 A) MCB.
• Power sub-circuit: feeds 16 A (15 A) power sockets for heaters, irons, AC, machines, etc. It carries at most 2 power sockets (about 3000 W), is wired in 2.5–4 mm² Cu and protected by a 16–20 A MCB.
• The two are kept on separate circuits (often separate DBs) so that a fault or overload on a power point does not plunge the area into darkness, and so that each can be sized for its own current.

Cable size from MDB to SDB1 and SDB2

Assumptions: power factor 0.8 lagging; Cu resistivity 0.0178 Ω·mm²/m; permissible voltage drop 2.5 % of 400 V (= 10 V) on a sub-main.

Current: I = P / (√3 × V_L × cos φ)

SDB1: I1 = 50000 / (√3 × 400 × 0.8) = 90.21 A
SDB2: I2 = 60000 / (√3 × 400 × 0.8) = 108.25 A

From the table (3.5-core Cu, laid in ground):

• SDB1: 90.21 A → 25 mm² (rating 99 A)
• SDB2: 108.25 A → 35 mm² (rating 120 A)

Voltage drop check: ΔV = √3 × I × R × cos φ, with R = ρL/A (reactance neglected).

SDB1: R  = 0.0178 × 60 / 25  = 0.0427 Ω
      ΔV = √3 × 90.21 × 0.0427 × 0.8 = 5.34 V
         = 5.34/400 = 1.33 %   < 2.5 %  OK
SDB2: R  = 0.0178 × 100 / 35 = 0.0509 Ω
      ΔV = √3 × 108.25 × 0.0509 × 0.8 = 7.63 V
         = 7.63/400 = 1.91 %   < 2.5 %  OK

Answer: MDB → SDB1: 3.5-core 25 mm² Cu cable; MDB → SDB2: 3.5-core 35 mm² Cu cable.

Note: if a 25 % margin for future load is added (1.25 × I = 112.76 A and 135.32 A), the sizes become 35 mm² and 50 mm² respectively.