Explore wasted energy and efficiency in GCSE Physics, learning how machines lose energy, how to reduce losses, and how to calculate efficiency using simple examples and equations.
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You can find more about this topic by visiting BBC Bitesize - Changes in energy stores - AQA
The key useful energy transfer is to produce kinetic energy. The noise of the train and friction waste energy in the form of heat and sound
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These crop up almost every year in the exams
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Efficiency is always a percentage worked out by dividing the output energy by the input energy and multipying by 100
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Answer four would be what you got if you did the division the wrong way round. The majority of the electrical energy is wasted as heat
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Not only does LED lighting consume a fraction of the power of a filament lamp, it is also much more efficient at transferring the electrical energy to useful light energy
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Since this is a useful energy transfer, it increases the overall efficiency by using some of the heat energy that would otherwise be wasted
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It is important to remember that efficiency is a measure of the useful energy transfer. The way the question is worded could make you forget that the calculation gives you the percentage of energy that has been wasted and needs to subtracted from 100% to find the efficiency. Watch out for this sort of question in your exams
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The efficiency is 25 ÷ 29 x 100 = 86.2% efficient
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The total energy transfer losses are 1.87% so to the nearest whole number that is 2% wastage, leaving 98% efficiency. Efficiency is worse in the lower voltage sections of the National Grid because the higher currents cause more energy loss as heat
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This question could look very complicated at first. The knack is to break down complicated questions into smaller parts which makes them more manageable. So, first you need to work out the area of the solar cell which is 0.1 m2. That means that it receives 100 W of the insolation (incoming solar energy). Finally, rearranging the equation for efficiency provides the answer:
(efficiency x input wattage) ÷ 100 = output wattage |