---
title: "Battery backup design with inverter at 48 V"
subject: "Industrial Power Distribution and Illumination"
chapter: 9
key: "semester_6/industrial_power_distribution_and_illumination:battery-backup-design-48v"
source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/battery-backup-design-48v
site: BCT Engineering
---

# Battery backup design with inverter at 48 V

Industrial Power Distribution and Illumination · Chapter 9: Emergency and Backup Supply Systems

- Asked in: 2080 Chaitra, 8 marks
- 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/battery-backup-design-48v
- Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-9

## Question

Design a Battery Backup system with an inverter for the following loads for 1 day, if DC bus voltage is 48V.

| Devices | Qty. | Wattage | Backup Time |
|---|---|---|---|
| Computer | 6 | 100 | 4 |
| Tube light | 24 | 10 | 6 |
| Camera | 1 | 60 | 24 |
| Backup Supply | 3 | 400 | 6 |
| Induction Cooker | 6 | 550 | 5 |
| Water Pump | 2 | 2000 | 6 |

Following are the battery available:

| Option | Voltage | Capacity | DoD |
|---|---|---|---|
| Option 1 | 12V | 200Ah | 80% DoD |
| Option 2 | 2V | 1500Ah | 80% DoD |

## Short answer

Daily energy requirement is 52.98 kWh. A 15 kVA inverter is selected. Battery bank needs 1533 Ah at 48 V; using 12 V 200 Ah batteries at 80% DoD and 90% inverter efficiency, 32 batteries (4 series × 8 parallel) providing 1600 Ah are recommended.

## Answer

**Assumptions:** inverter efficiency 90%, battery depth of discharge 80% (given), all loads may run at the same time, DC bus 48 V, back-up for 1 day as listed.

#### Step 1: Load and daily energy

| Device | Qty | W | Total W | h | Wh/day |
|---|---|---|---|---|---|
| Computer | 6 | 100 | 600 | 4 | 2400 |
| Tube light | 24 | 10 | 240 | 6 | 1440 |
| Camera | 1 | 60 | 60 | 24 | 1440 |
| Backup supply | 3 | 400 | 1200 | 6 | 7200 |
| Induction cooker | 6 | 550 | 3300 | 5 | 16 500 |
| Water pump | 2 | 2000 | 4000 | 6 | 24 000 |
| **Total** | | | **9400** | | **52 980** |

#### Step 2: Inverter rating
```text
Connected load        = 9400 W
With 25 % margin      = 11 750 W
At pf 0.8 → kVA       = 11 750/0.8 = 14.7 kVA
```
Choose a **15 kVA, 48 V DC input, pure sine-wave inverter** (pumps need surge capability of about 3 × running current; a soft starter/VFD for pumps is advised).

#### Step 3: Battery energy and Ah
```text
Energy from battery = 52 980 / (η_inv × DoD)
                    = 52 980 / (0.9 × 0.8)
                    = 73 583 Wh
Ah at 48 V          = 73 583 / 48 = 1533 Ah
```

#### Step 4: Battery bank options
**Option 1: 12 V, 200 Ah**
```text
Series  = 48 / 12     = 4
Parallel = 1533 / 200 = 7.67 → 8 strings
Total   = 4 × 8       = 32 batteries
Bank    = 48 V, 1600 Ah
Usable  = 1600 × 48 × 0.8 × 0.9 = 55 296 Wh ≥ 52 980 ✓
```

**Option 2: 2 V, 1500 Ah**
```text
Series   = 48 / 2      = 24 cells
Parallel = 1533 / 1500 = 1.02 → 2 strings
Total    = 48 cells, 48 V, 3000 Ah
```
One string (24 cells, 1500 Ah) gives 1500 × 48 × 0.8 × 0.9 = 51 840 Wh, about 2% short (enough only if inverter losses are ignored: 52 980/(0.8 × 48) = 1380 Ah).

#### Selection
**Option 1 (32 × 12 V, 200 Ah, 4 series × 8 parallel)** matches the need closely (1600 Ah vs 1533 Ah). Option 2 with one string is cheaper in wiring and 2 V tubular cells last longer, but slightly undersized; with two strings it is oversized (3000 Ah). Option 2 (single string) can be chosen only if the load hours are reduced a little.

#### Step 5: Charger
To recharge 1600 Ah in about 10 h: charging current ≈ 0.1 C = 160 A at 48 V (≈ 8–9 kW charger, or the inverter's built-in charger / solar charge controller).

```text
AC ─►[Charger 48V,160A]─┬─►[15 kVA Inverter]─► Loads
                        │
              [48 V battery bank]
```

**Answer: Daily energy 52.98 kWh; 15 kVA inverter; battery 1533 Ah at 48 V → 32 nos. 12 V 200 Ah (4S × 8P) recommended.**

## Check yourself

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

1. What is the total daily energy of the loads?
   A. 52 980 Wh (about 52.98 kWh)
   B. 9400 Wh
   C. 73 583 Wh
   D. 11 750 Wh
   Answer: A. Sum of load x hours: 2400+1440+1440+7200+16500+24000 = 52 980 Wh. (Key point: Daily energy)

2. How is energy from the battery calculated?
   A. Daily energy x 0.9 x 0.8
   B. Daily energy / (inverter efficiency x DoD) = 52 980 / (0.9 x 0.8) = 73 583 Wh
   C. Daily energy / 0.9 only
   D. Daily energy x 48 V
   Answer: B. Inverter losses and limited depth of discharge both increase the required stored energy. (Key point: Battery energy)

3. What Ah is required at 48 V?
   A. About 52 980 Ah
   B. About 195 Ah
   C. About 1533 Ah
   D. About 3000 Ah
   Answer: C. 73 583 Wh / 48 V = 1533 Ah. (Key point: Ah at DC bus)

4. How many 12 V, 200 Ah batteries are needed, and how are they connected?
   A. 8 batteries: 4 series x 2 parallel
   B. 24 batteries: 24 in series
   C. 48 batteries: 4 series x 12 parallel
   D. 32 batteries: 4 in series x 8 strings in parallel
   Answer: D. 48/12 = 4 in series; 1533/200 = 7.67, so 8 parallel strings; 4 x 8 = 32. (Key point: Bank layout)

## Related in Industrial Power Distribution and Illumination

- [Types of battery charging system](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Abattery-charging-system-types) — Industrial Power Distribution and Illumination, chapter 9, asked 3×
- [Power/light sub-circuits and cable for 50 kW, 60 kW](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Asub-circuits-cable-50kw-60kw) — Industrial Power Distribution and Illumination, chapter 5, asked 4×
- [Single-phase Cu cable for three loads](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Asingle-phase-cable-three-loads) — Industrial Power Distribution and Illumination, chapter 5
- [Battery charger circuit and charging methods](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Abattery-charger-two-methods) — Industrial Power Distribution and Illumination, chapter 9
- [Waste light factor and floodlight calculation](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Awaste-light-factor-floodlight) — Industrial Power Distribution and Illumination, chapter 8
- [Light sources and specular reflection principle](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Alight-sources-specular-reflection) — Industrial Power Distribution and Illumination, chapter 6

## Related in other subjects

- [Square-wave inverter ±48 V: fundamental and third harmonic](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Ainverter-48v-rl-fundamental-third) — Power Electronics, chapter 5
- [EVA; 10 days, 100,000, day 6, spi and cpi](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Aeva-numerical-day6-spi-cpi-b80) — Project Engineering, chapter 4
- [Design DC voltage regulator for 6 V](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Azener-regulator-design-6v) — Electronics Devices and Circuit, chapter 6, asked 4×
- [Solar panel and battery for a house (7 h sun)](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Apv-sizing-12v-7h-sun) — Power Plant Design, chapter 1, asked 2×