Question:
A solar cell has a light gathering area of \( 10\,\text{cm}^{2} \) and produces 0.2 A at 0.8 V (D.C.) when illuminated with sunlight of intensity \( 1000\,\text{Wm}^{-2} \). The efficiency of the solar cell is:
A solar cell has a light gathering area of \( 10\,\text{cm}^{2} \) and produces 0.2 A at 0.8 V (D.C.) when illuminated with sunlight of intensity \( 1000\,\text{Wm}^{-2} \). The efficiency of the solar cell is:
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Be sure to convert the area from square centimeters (\( \text{cm}^2 \)) into standard square meters (\( \text{m}^2 \)) by multiplying by \( 10^{-4} \). Skipping this conversion step will cause your calculations to be off by several orders of magnitude.
Updated On: Jun 8, 2026
- 16%
- 12%
- 8%
- 20%
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The Correct Option is A
Solution and Explanation
Concept:
The efficiency \( \eta \) of a solar cell is defined as the ratio of the output electrical power generated to the total input solar radiant power captured across its surface area:
\[
\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \frac{V \cdot I}{\text{Intensity} \cdot A} \times 100\%
\]
Step 1: Calculating the generated output electrical power \( P_{\text{out}} \).
Given parameters: Voltage \( V = 0.8\,\text{V} \), Current \( I = 0.2\,\text{A} \). \[ P_{\text{out}} = V \cdot I = 0.8 \times 0.2 = 0.16~\text{W} \]
Step 2: Calculating the captured input solar power \( P_{\text{in}} \).
Given parameters: Intensity \( = 1000\,\text{Wm}^{-2} \), Area \( A = 10\,\text{cm}^2 = 10\times10^{-4}\,\text{m}^2 = 10^{-3}\,\text{m}^2 \). \[ P_{\text{in}} = \text{Intensity} \cdot A = 1000 \times 10^{-3} = 1~\text{W} \]
Step 3: Finding the percentage efficiency fraction.
\[ \eta = \frac{0.16}{1} \times 100\% = 16\% \] This matches option (A) perfectly.
Step 1: Calculating the generated output electrical power \( P_{\text{out}} \).
Given parameters: Voltage \( V = 0.8\,\text{V} \), Current \( I = 0.2\,\text{A} \). \[ P_{\text{out}} = V \cdot I = 0.8 \times 0.2 = 0.16~\text{W} \]
Step 2: Calculating the captured input solar power \( P_{\text{in}} \).
Given parameters: Intensity \( = 1000\,\text{Wm}^{-2} \), Area \( A = 10\,\text{cm}^2 = 10\times10^{-4}\,\text{m}^2 = 10^{-3}\,\text{m}^2 \). \[ P_{\text{in}} = \text{Intensity} \cdot A = 1000 \times 10^{-3} = 1~\text{W} \]
Step 3: Finding the percentage efficiency fraction.
\[ \eta = \frac{0.16}{1} \times 100\% = 16\% \] This matches option (A) perfectly.
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