---
title: "Selecting main panel board and DB rating"
subject: "Industrial Power Distribution and Illumination"
chapter: 5
key: "semester_6/industrial_power_distribution_and_illumination:panel-board-db-rating"
source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/panel-board-db-rating
site: BCT Engineering
---

# Selecting main panel board and DB rating

Industrial Power Distribution and Illumination · Chapter 5: Electrification of Industrial Building

- Asked in: 2076 Bhadra, 8 marks; 2070 Bhadra, 8 marks (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/panel-board-db-rating
- Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-5

## Question

Explain the procedure for selection of rating of main panel board and distribution board with suitable example.

## Short answer

The rating of an MDB/SDB is based on its incomer MCCB/MCB. List loads, calculate circuit currents, apply simultaneity factors for demand, and select incomer rating ≥1.25× demand; e.g., a 56.82 A demand leads to an 80 A TPN MCCB. Cables are chosen to match or exceed the incomer rating.

## Answer

The rating of a panel board (MDB) or distribution board (SDB) means the rating of its incomer (MCCB/MCB), busbar and incoming cable. It is fixed from the design current of the loads it feeds, after applying utilization and simultaneity (diversity) factors and a safety margin.

#### Procedure
1. **List the loads** on each DB: kW, pf, efficiency, single or three phase. Balance single-phase loads on R, Y, B.
2. **Find the current of each outgoing circuit**: I = P/(√3·V·η·cosφ) for 3-phase, I = P/(V·cosφ) for single phase.
3. **Select outgoing MCBs**: In ≥ 1.25 × Ib (next standard size: 6, 10, 16, 20, 25, 32, 40, 50, 63 A); C curve for general loads, D curve for motors.
4. **SDB demand**: add circuit currents and multiply by the simultaneity factor Ks (0.9 for 2–3 circuits, 0.8 for 4–5, 0.7 for 6–9, 0.6 for 10 or more).
5. **SDB incomer**: In ≥ 1.25 × SDB demand; MCB up to 63/100 A, MCCB above.
6. **MDB demand**: add the SDB demands, apply Ks of the MDB, add spare for future.
7. **MDB incomer**: MCCB/ACB ≥ 1.25 × MDB demand; check that it does not exceed the transformer LT full-load current, and its breaking capacity ≥ prospective fault current.
8. **Busbar**: rating ≥ incomer rating (copper about 1.2–1.6 A/mm²). Keep 20 % spare ways.
9. **Coordination**: Ib ≤ In ≤ Iz (cable rating), and upstream device larger than downstream for discrimination.

#### Example
A factory, 400 V, 3-phase: SDB-1 feeds 3 motors of 7.5 kW (η = 0.88, pf = 0.85); SDB-2 feeds 15 kW balanced lighting and sockets at pf 0.9.

```text
Motor current = 7500 / (√3 × 400 × 0.88 × 0.85)
              = 14.47 A
Motor MCB     = 1.25 × 14.47 = 18.09 A → 20 A (D curve)

SDB-1 demand  = 3 × 14.47 × Ks 0.9 = 39.08 A
SDB-1 incomer = 1.25 × 39.08 = 48.84 A → 50 A TP MCB

SDB-2 current = 15000 / (√3 × 400 × 0.9) = 24.06 A
SDB-2 incomer = 1.25 × 24.06 = 30.07 A → 32 A TPN MCB

MDB demand = (39.08 + 24.06) × Ks 0.9 = 56.82 A
MDB incomer = 1.25 × 56.82 = 71.02 A → 80 A TPN MCCB
```

| Board | Demand (A) | Incomer | Outgoings |
|---|---|---|---|
| SDB-1 | 39.08 | 50 A TP MCB | 3 × 20 A TP MCB |
| SDB-2 | 24.06 | 32 A TPN MCB | SP MCBs 6–16 A + RCCB |
| MDB | 56.82 | 80 A TPN MCCB | 63 A, 40 A MCCB/MCB + spares |

Cables are then chosen so that their rating is above the protective device rating (e.g. MDB–SDB-1 with 50 A MCB → 10 mm² Cu), and checked for voltage drop.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is an outgoing MCB rating chosen from the circuit current Ib?
   A. In = Ib exactly
   B. In ≥ 1.25 x Ib, then the next standard size
   C. In ≤ 0.8 x Ib
   D. In = 2 x Ib
   Answer: B. A 1.25 safety margin is applied and the next standard size is taken; e.g. 14.47 A gives 18.09 A, so 20 A. (Key point: MCB rating rule)

2. How is the demand of an SDB found from its circuit currents?
   A. Add the currents and multiply by a simultaneity factor Ks (e.g. 0.9 for 2-3 circuits)
   B. Take the largest circuit current only
   C. Add the currents with no reduction
   D. Divide the sum by the number of circuits
   Answer: A. All loads do not run together, so the sum is reduced by Ks: 3 x 14.47 x 0.9 = 39.08 A. (Key point: Simultaneity factor)

3. Why are motor circuits given D-curve MCBs while general loads use C curve?
   A. D curve trips faster at low current
   B. D curve trips at a lower fault current
   C. D curve gives earth-fault protection
   D. Motor starting surges are tolerated without nuisance tripping
   Answer: D. D-curve MCBs allow the high starting current of motors without tripping. (Key point: D curve)

4. What condition links design current, protective device rating and cable rating?
   A. Iz ≤ In ≤ Ib
   B. In ≤ Ib ≤ Iz
   C. Ib ≤ In ≤ Iz
   D. Ib = In = Iz
   Answer: C. Design current must not exceed the device rating, and the device rating must not exceed the cable's current capacity. (Key point: Coordination)

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