Discussion
Two devices in series achieve \( 90 \) and \( 95 \) per cent efficiency. The overall efficiency is
Answer: B
The penetrations multiply, so \( 0.10 \times 0.05 = 0.005 \) passes and 99.5 per cent is collected. Working in penetration rather than efficiency is much the easier way to handle devices in series, and it makes the diminishing returns obvious.
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