Posted by CB on Sunday, January 24, 2010 at 3:12pm.
Efficiency = Power out/ Power in
= (900/14) / (1100/12)
= 70%
This assumes that the power is delivered continously and the stated measurements are typical of the average power in and out.
If the work were done only during a 12 second interval and then recovered or "delivered" more slowly in a following 14 sec period, the efficiency would be (work out/(work in) = 900/1100 = 81.8%
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