Discussion
A solar panel of 2 m² in full sunlight of 1 kW/m² and 20 per cent efficiency delivers
Answer: D
Correct answer: 400 W
\[ 2\ \mathrm{m^2} \times 1000\ \mathrm{W/m^2} \times 0.20 = 400\ \mathrm{W}. \] The power density is multiplied by the area to give the power intercepted, 2 kW, and then by the efficiency to give what is delivered. Applying the efficiency without the area, or the area without the efficiency, produces two of the distractors offered.
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