QuestionDecember 13, 2025

On clear day, sunlight delivers approximately 1000 J each second to a 1m^2 surface; smaller areas receive proportionally less-half the area receives half the energy. A 12% efficient solar cell, a square 14 cm on a side, is in bright sunlight. How much electric power does it produce? 0.8 W 1.6 W 2.4 W 3.2 W

On clear day, sunlight delivers approximately 1000 J each second to a 1m^2 surface; smaller areas receive proportionally less-half the area receives half the energy. A 12% efficient solar cell, a square 14 cm on a side, is in bright sunlight. How much electric power does it produce? 0.8 W 1.6 W 2.4 W 3.2 W
On clear day, sunlight delivers approximately 1000 J each second to a 1m^2 surface; smaller areas receive proportionally less-half the area receives half the energy. A 12% efficient solar cell, a square 14 cm on a side, is in bright sunlight. How much electric power does it produce? 0.8 W 1.6 W 2.4 W 3.2 W

Solution
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Answer

2.4\,\text{W} The correct option is **2.4 W**. Explanation 1. Calculate the Area of the Solar Cell The side length is 14\,\text{cm} = 0.14\,\text{m}, so the area is A = (0.14\,\text{m})^2 = 0.0196\,\text{m}^2. 2. Find the Total Sunlight Power Incident on the Cell Sunlight delivers 1000\,\text{J/s} = 1000\,\text{W} per 1\,\text{m}^2. For 0.0196\,\text{m}^2, the incident power is P_{\text{in}} = 1000 \times 0.0196 = 19.6\,\text{W}. 3. Calculate the Electric Power Output Using Efficiency The cell is 12\% efficient, so output power is P_{\text{out}} = 0.12 \times 19.6 = 2.352\,\text{W}.

Explanation

1. Calculate the Area of the Solar Cell<br /> The side length is $14\,\text{cm} = 0.14\,\text{m}$, so the area is $A = (0.14\,\text{m})^2 = 0.0196\,\text{m}^2$.<br />2. Find the Total Sunlight Power Incident on the Cell<br /> Sunlight delivers $1000\,\text{J/s} = 1000\,\text{W}$ per $1\,\text{m}^2$. For $0.0196\,\text{m}^2$, the incident power is $P_{\text{in}} = 1000 \times 0.0196 = 19.6\,\text{W}$.<br />3. Calculate the Electric Power Output Using Efficiency<br /> The cell is $12\%$ efficient, so output power is $P_{\text{out}} = 0.12 \times 19.6 = 2.352\,\text{W}$.
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