Electric Efficiency

  • Thread starter Bullkank
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  • #1
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Homework Statement



A Student Performs an experiment in which an electric motor is used to lift a 200g weight through 2m, thus increasing its potential energy by 4j. From measurements of the rate at which the weight is lifted the efficiency of the motor is to be determined. Two different voltages were used an the current was measured.




Homework Equations



A/

In the first experiment at 6V, a current of 0.25A was measured and the weight took 5 seconds to rise the 2 metres. What was the efficiency of the motor?

B/

In the Next experiment the voltage was increased to 8V, the current was found to be 0.30A and the efficiency worked out to be 60%. How long did the motor take to lift the weight the 2m this time?


The Attempt at a Solution




I have no idea even how to work out efficiency yet so my attempts were quite pathetic


Answers for reference

a) 53%


b) 2.8 seconds
 
Last edited:

Answers and Replies

  • #2
Filip Larsen
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Considering that your results are correct you must have understood something correct. Feel free to ask specific questions if you are unsure what exactly it is you have understood.
 
  • #3
Lok
555
23
Considering that your results are correct you must have understood something correct. Feel free to ask specific questions if you are unsure what exactly it is you have understood.

there are no results there, maaaaan.

Anyway, you should find the energy given off by the motor at 6V with 0.25 A for 5 s. (multiply the power with time and you should get energy). You will get a value in J. Efficency is found by Energy needed/Energy spent, and for expressing it in % you just multiply the result by 100.

So Eff= 100*4J/xJ %

Have fun!!!
 
  • #4
Filip Larsen
Gold Member
1,360
276
there are no results there, maaaaan.

Ah, "answers for reference ". I did not get that.
 
  • #5
Lok
555
23
Ah, "answers for reference ". I did not get that.

Somebody had to put you on the right path ... :P
 

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