Check logic for Watts calculation

  • #1

Main Question or Discussion Point

Thanks to anyone kind enough to help check my logic. Using 13 gauge copper wire and a magnet I can generate .05 watt every time the magnet passes the coil (according to the voltmeter). Am I correct in assuming that if the magnet passes the coil 60 times a minute, then (.05 watts X 60rpm X 60 min/hr) = 180 watts per hour?

Much appreciate your help.
 

Answers and Replies

  • #2
russ_watters
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Hi.
Watts is already a rate: J/sec. There is no such unit as "watts per hour". You need to multiply by the duration of the passes to get energy in joules or watt-hours.
 
  • #3
berkeman
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Thanks to anyone kind enough to help check my logic. Using 13 gauge copper wire and a magnet I can generate .05 watt every time the magnet passes the coil (according to the voltmeter). Am I correct in assuming that if the magnet passes the coil 60 times a minute, then (.05 watts X 60rpm X 60 min/hr) = 180 watts per hour?

Much appreciate your help.
What voltmeter do you have that measures power (Watts)? Can you post a picture or say what the model number is?
 
  • #4
CWatters
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Using 13 gauge copper wire and a magnet I can generate .05 watt every time the magnet passes the coil (according to the voltmeter).
Did you mean "0.05 Volts" ? In which case what you probably got was some sort of pulse that might be displayed as 0.05 Volts. That could be the peak voltage or some sort of average voltage depending on the meter. Either way you can't calculate the power without knowing what the load is. If the load was just the volt meter then it's likely the power was very very low because the resistance of a voltmeter is designed to be high. Connect a known load resistor and repeat the experiment.

If you really did mean "0.05 Watts" then you need to tell us how long it was at 0.05 Watts? If each pass generated 0.05W for half a second and you repeat that once per second (60 times per min) then the average power will be 0.05/2 = 0.025W. That's because half the time it's generating 0.05W and half the time nothing.
 
  • #5
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If you actually did the “experiment” let’s say you have a good permanent magnet of 2”*2” area and you moved it
with 1m/sec.
E.M.F.=Br*L*v [see: http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/genwir2.html]
E.M.F.=1[Wb/m^2]*[2*2.54/100]*1=0.0508 V
60 RPM=60/60=1 sec/turn. If the length of the wire loop is only 2”[0.0508 m] then the velocity v=0.0508 m/sec, only.
You’ll need 1200 rpm in order to achieve 0.05 V.
 

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