An experiment is set up to determine the efficiency of a 60-W light bulb. The light output is measured as 6 watts.
(CONTINUATION) The power used is assumed to be 60 watts. The experimenter calculates the efficiency using this equation. Why is this experiment flawed?
Efficiency = Measured light output (watts)/ Measured power used (watts)
A- The efficiency cannot be measured.
B- The efficiency of the light bulb can only be compared to a second light bulb.
C- The light output needs to be measured more precisely.
D- The actual power used was not measured.
2 Answers | Add Yours
The definition of efficiency in physics is how well a device or machine is doing what it is supposed to do. Given the formula, "measured light output/given power input", one has but to put the 6 watts the light is giving off on top of the equation and divide it by the power being input, which would be 60 watts. That would give .1, which we would multiply by 100%, because efficiency is always expressed as a percentage. The answer would be 10%, meaning the light bulb is not doing a very good job using the energy coming in and converting it to light. 90% of the energy is being dissipated, or wasted, in some form or fashion. In terms of picking a flaw, I would tend to select answer "B". Every good experiment needs a control to compare to. If we are comparing the amount of watts being given off, we should have another light bulb that is giving off the correct amount of wattage to compare the first light bulb to. Having control factors in an experiment insures the variable we are examining is likely the cause of the outcome of the experiment. Without them, our experiment is little more than a guessing game.
My answer would be 'D' because the power consumed will be equql to the power rating of the bulb only when it is connected exactly at the voltage rated on the bulb. if the voltage is higher then the bulb consumes more power and if the voltage is low it consumes less power.
We’ve answered 317,824 questions. We can answer yours, too.Ask a question