---
title: "Solar panel and battery for a house (6 h sun)"
subject: "Power Plant Design"
chapter: 1
key: "semester_8/power_plant_design:pv-sizing-12v-6h-sun"
source: https://www.bctengineer.com/q/semester_8/power_plant_design/pv-sizing-12v-6h-sun
site: BCT Engineering
---

# Solar panel and battery for a house (6 h sun)

Power Plant Design · Chapter 1: Energy Sources and Electric Power Generation

- Asked in: 2074 Bhadra, 10 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/power_plant_design/pv-sizing-12v-6h-sun
- Whole chapter: https://www.bctengineer.com/md/semester_8/power_plant_design/chapter-1

## Question

You have planned to electrify a house with electric load as below in table. Make a suitable selection for Solar Panel and battery size (of 12 V systems) for the house.

| Load type | Numbers | Wattage | Operating hours (per day) |
|---|---|---|---|
| Lamps | 5 | Each 10 W | 4 |
| TV | 1 | 60 W | 5 |
| Computer | 1 | 65 W | 3 |

Assume equal efficiency for both inverter and battery as 80%. Consider sun shining duration 6 hours a day.

## Short answer

Daily load is 695 Wh. With inverter and battery each 80% efficient, required PV energy is about 1086 Wh/day; with 6 sunshine hours, panel size is about 181 Wp, choose 200 Wp. Battery capacity with DoD 0.6 is about 151 Ah, choose 12 V, 200 Ah; use a 20 A charge controller.

## Answer

A stand-alone solar home system is sized by finding the daily energy demand, adding inverter and battery losses to get the energy the panel must produce, and sizing the battery to supply one day's load within an allowed depth of discharge.

**Assumptions:** all loads are AC through the inverter; one day of autonomy; depth of discharge (DoD) = 0.6; no extra wiring loss beyond the given efficiencies.

#### 1. Daily energy demand

| Load | No. | W each | Total W | h/day | Wh/day |
|---|---|---|---|---|---|
| Lamps | 5 | 10 | 50 | 4 | 200 |
| TV | 1 | 60 | 60 | 5 | 300 |
| Computer | 1 | 65 | 65 | 3 | 195 |
| **Total** | | | **175** | | **695** |

#### 2. Solar panel size
Energy passes through the battery and the inverter, each 80% efficient:

$$
\begin{aligned}
E_{PV} &= \frac{695}{0.8 \times 0.8} = 1085.9\ \text{Wh/day} \\
P_{PV} &= \frac{1085.9}{6\ \text{h}} = 181.0\ \text{W}_p
\end{aligned}
$$

Select **200 Wp** (two 100 Wp, 12 V panels in parallel). Panel current $\approx 200/12 = 16.7$ A, so use a **12 V, 20 A** charge controller.

#### 3. Battery size
Taking both the inverter and battery efficiencies into account:

$$
\begin{aligned}
E_{b} &= \frac{695}{\eta_{inv}\,\eta_{b}} = \frac{695}{0.8 \times 0.8} = 1085.9\ \text{Wh} \\
\text{Ah (usable)} &= \frac{1085.9}{12} = 90.5\ \text{Ah} \\
C &= \frac{90.5}{DoD} = \frac{90.5}{0.6} = 150.8\ \text{Ah}
\end{aligned}
$$

A 150 Ah battery would be just short with no margin, so select a **12 V, 200 Ah** deep-cycle (tubular lead-acid) battery, which also gives some margin for cloudy days.

#### 4. Inverter (extra)
Connected load 175 W × 1.25 ≈ 219 W → **12 V, 250–300 VA** inverter.

**Answer:** Solar panel ≈ 181 Wp → **200 Wp**; battery ≈ 151 Ah → **12 V, 200 Ah**.

## Check yourself

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

1. What is the total daily energy?
   A. 175 Wh
   B. 500 Wh
   C. 695 Wh
   D. 1086 Wh
   Answer: C. 200 + 300 + 195 = 695 Wh. (Key point: Daily energy)

2. How is the panel size found?
   A. 695/0.8/6 = 145 Wp
   B. 695/(0.8 x 0.8)/6 = 181 Wp, choose 200 Wp
   C. 695/6 = 116 Wp
   D. 695 x 0.64/6
   Answer: B. Both inverter and battery are 80% efficient: 695/0.64 = 1085.9 Wh, divided by 6 h gives 181 Wp. (Key point: Panel size)

3. How is the battery capacity found?
   A. 868.75/12 = 72.4 Ah, divided by DoD 0.6 gives 120.7 Ah
   B. 695/12 = 58 Ah
   C. 868.75 x 0.6/12 = 43 Ah
   D. 1085.9/12 = 90 Ah
   Answer: A. Battery supplies 695/0.8 = 868.75 Wh, giving 72.4 Ah usable and 120.7 Ah total, so 12 V, 150 Ah. (Key point: Battery size)

4. How is the charge controller current chosen?
   A. From battery capacity
   B. From panel current about 181/12 = 15.1 A, so 20 A
   C. From inverter VA
   D. From daily Wh
   Answer: B. Panel current 181 W/12 V = 15.1 A, so a 12 V, 20 A controller. (Key point: Charge controller)

## Close variants set in other exams

- [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×