---
title: "Solar panel and battery for a house (7 h sun)"
subject: "Power Plant Design"
chapter: 1
key: "semester_8/power_plant_design:pv-sizing-12v-7h-sun"
source: https://www.bctengineer.com/q/semester_8/power_plant_design/pv-sizing-12v-7h-sun
site: BCT Engineering
---

# Solar panel and battery for a house (7 h sun)

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

- Asked in: 2071 Bhadra, 8 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-7h-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 system) for the house.

| Load type | Numbers | Wattage | Operating hours (per day) |
|---|---|---|---|
| Lamps | 5 | Each 10 W | 4 |
| TV | 1 | 80 W | 5 |
| Computer | 1 | 85 W | 3 |

Assume equal efficiency for both inverter and battery as 80%. Consider sun shining duration 7 hours a day.

## Short answer

Daily load is 855 Wh. With inverter and battery each 80% efficient, required PV energy is about 1336 Wh/day; with 7 sunshine hours, panel size is about 191 Wp, choose 200 Wp. Battery capacity with DoD 0.6 is about 186 Ah, choose 12 V, 200 Ah; use a 20 A charge controller.

## Answer

The solar panel is sized from the daily energy demand divided by the combined efficiency and the sunshine hours; the battery is sized 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.

#### 1. Daily energy demand

| Load | No. | W each | Total W | h/day | Wh/day |
|---|---|---|---|---|---|
| Lamps | 5 | 10 | 50 | 4 | 200 |
| TV | 1 | 80 | 80 | 5 | 400 |
| Computer | 1 | 85 | 85 | 3 | 255 |
| **Total** | | | **215** | | **855** |

#### 2. Solar panel size

$$
\begin{aligned}
E_{PV} &= \frac{855}{\eta_{inv}\,\eta_{b}} = \frac{855}{0.8 \times 0.8} = 1335.9\ \text{Wh/day} \\
P_{PV} &= \frac{1335.9}{7\ \text{h}} = 190.8\ \text{W}_p
\end{aligned}
$$

Select **200 Wp** (two 100 Wp, 12 V panels in parallel). Array 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:

$$
\begin{aligned}
E_{b} &= \frac{855}{\eta_{inv}\,\eta_{b}} = \frac{855}{0.8 \times 0.8} = 1335.9\ \text{Wh} \\
\text{Ah (usable)} &= \frac{1335.9}{12} = 111.3\ \text{Ah} \\
C &= \frac{111.3}{0.6} = 185.5\ \text{Ah}
\end{aligned}
$$

Select a **12 V, 200 Ah** deep-cycle battery.

#### 4. Inverter (extra)
Connected load 215 W × 1.25 ≈ 269 W → **12 V, 300 VA** inverter.

**Answer:** Solar panel ≈ 191 Wp → **200 Wp**; battery ≈ 186 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 demand of the house?
   A. 215 Wh
   B. 855 Wh
   C. 1335.9 Wh
   D. 1068.75 Wh
   Answer: B. Sum of load x hours: 200 + 400 + 255 = 855 Wh/day. 215 W is only the connected power. (Key point: Daily energy demand)

2. Why is the daily load of 855 Wh divided by 0.8 x 0.8 when sizing the panel?
   A. Because the panel is 80% efficient
   B. Because inverter and battery losses mean the array must produce more than the load uses
   C. To convert Wh to Ah
   D. To allow for depth of discharge
   Answer: B. Energy is lost in the inverter and battery, so the array must supply 855/0.64 = 1335.9 Wh/day. (Key point: PV energy requirement)

3. How is the panel rating in Wp obtained from 1335.9 Wh/day?
   A. Multiply by 7 h
   B. Divide by 12 V
   C. Divide by 7 sun hours to get about 191 Wp
   D. Divide by 0.6
   Answer: C. Panel power = energy / sunshine hours = 1335.9/7 = 190.8 Wp, so 200 Wp is chosen. (Key point: Panel peak power)

4. How is the 150 Ah battery size reached?
   A. 855/12 = 71 Ah directly
   B. 855/0.8 = 1068.75 Wh, /12 V = 89.1 Ah usable, /0.6 DoD = 148.4 Ah
   C. 1335.9/12 = 111 Ah
   D. 215 W x 7 h / 12 V
   Answer: B. Battery energy is corrected for efficiency, converted to Ah at 12 V, then divided by the allowed depth of discharge 0.6. (Key point: Battery capacity and DoD)

## Close variants set in other exams

- [Solar panel and battery for a house (6 h sun)](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Apv-sizing-12v-6h-sun) — Power Plant Design, chapter 1, asked 2×