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. 16%
• B. 12%
• C. 8%
• D. 20%

Hint:

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.

Answer 1

The Correct Option is A

Step 1: Define efficiency.
The efficiency of a solar cell is the useful electrical power it gives out divided by the solar power it takes in, written as a percentage: \[ \eta = \frac{P_{out}}{P_{in}}\times100\% \]
Step 2: Find the electrical output power.
Output power is voltage times current. With $V = 0.8$ V and $I = 0.2$ A: \[ P_{out} = VI = 0.8\times0.2 = 0.16\ \text{W} \]
Step 3: Convert the area to square metres.
\[ A = 10\ \text{cm}^{2} = 10\times10^{-4}\ \text{m}^{2} = 10^{-3}\ \text{m}^{2} \]
Step 4: Find the solar input power.
Input power is intensity times area. With intensity $1000\ \text{Wm}^{-2}$: \[ P_{in} = 1000\times10^{-3} = 1\ \text{W} \]
Step 5: Divide output by input.
\[ \eta = \frac{0.16}{1}\times100\% \]
Step 6: Get the efficiency.
\[ \eta = 16\% \] \[ \boxed{\eta = 16\%} \]