
#1
Feb1414, 10:45 PM

P: 67

Hello, I'm building a generator and have completed initial phase. I made coil with 300 turns, which gives me 12v when rotor is turned at 2000rpm, if I short the coil it gives me 16 amp.
1] Now I'm wondering what is the output wattage of the generator I built / described? 2] If I use 18, similar coils and I rotate the rotor at 2000 rpm as before, what could be the output/wattage I may get from the generator in ideal conditions? Any help? Thank you. 



#2
Feb1514, 01:27 AM

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It will depend on the load  simplest is to measure it.




#3
Feb1514, 01:42 AM

P: 285

If you want the maximum wattage, you will have to multiply 12v ( if you are sure it is turned at 2000 rpm only) by the maximum current through it. But shorting will not give you the maximum wattage. Because by doing so, the voltage appears on the coil of the generator. So first measure it's internal resistance, which is equal to the load you will have to use in order to measure the max watts or current. (Just applying max power transfer theorem). Hence by that theorem if R is the int. resistance of the generator coil, the max power is v^{2}/4R which is 36/R in your case.




#4
Feb1514, 02:17 AM

P: 285

Generator wattage calculationIf you are connecting 18 coils parallel, the internal resistance decreases (which is favourable) and I think voltage remains same but power again increases, as is obvious from V^{2}/4R. 



#5
Feb1514, 11:53 AM

P: 67

Thank you Simon and Raj.
I wanted to get an Idea of the output before I make another coils, as it is very hard to wind them by hand. Took almost 45 hours for one coil and my hands were cursing me. The resistance for 300 turns coil shows me 0.8 ohm. Also does the output wattage depend on the way I connect them [parallel or series]? 



#6
Feb1514, 10:26 PM

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#7
Feb1514, 11:10 PM

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You can  but the results will likely be misleading.
Much the same way as you get a misleading idea of output power from a battery from the shortcircuit current. The shortcircuit current depends on the internal impedance from all that wire in coils, ... you could work out what that power is based on ##P=I^2R_{int}## or something but that would be the power dissipated as internal losses. You need to know the power that can be delivered to a load. Try it and see  use different loads and use ##P=VI## and see what happens. You could get a ballpark figure off faraday's law with the rpm of what the generator could, in principle, supply ... but it really depends on the power that is supplied to whatever turns the handle. 



#8
Feb1614, 01:13 AM

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#9
Feb1614, 01:14 AM

P: 285

P_{m}=V^{2}/4R =(12)^{2}/4*0.8 =45W. You also told about connecting 18 similar coils. Then if in series, resistance will be 18 times, that is 18*0.8=14.4Ω but o/p voltage will change, Iam not sure if that will also become 18 times so if you connect 18 coils measure the voltage. If they are connected parallel, R=0.8/18=0.0444..Ω. But then Iam not sure about the voltage which you will have to measure. Any way the formula V^{2}/4R where R is the effective internal resistance will serve you through out. 



#10
Feb1614, 01:19 AM

P: 285

So you will have to be sure for what purpose you are using the gen. at 12v..? 



#11
Feb1614, 01:23 AM

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#12
Feb1614, 06:49 PM

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You can buy generators that are rated at a particular power output  the rating will be assuming some standard use for that kind of generator. The manual will warn you not to expect the rated power for all uses. When you build your own generator by putting bits together to see what happens, the question becomes quite open ended. You can use Faraday's Law to get you a ##V_{peak}## and thus a ##V_{rms}## for the open circuit case  then figure out the current for different loads/usages. That will give you an idea of what to expect from your generator. At some point you will need to measure things to work out the detailed performance. 



#13
Feb1614, 10:51 PM

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#14
Feb1614, 11:07 PM

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<Puzzled> these are odd questions for an engineer to ask in a general community.
Just to be clear: in NZ the word "engineer" is legally reserved. An engineer of the sort who would reasonably be commissioned to build a generator would be a specialist electrical engineer... I think it would help us to help you if we knew your education better, just so we don't short change you. 



#15
Feb1614, 11:15 PM

P: 67

Now internal Power dissipated = 16*16*14.4 = 3686.4 watt when the generator coil is shorted. [Am I right?] Now let's do parallel arrangement of all coils. V = 12, R = 0.044 Ohm I = 16 Amp in each coil. Which makes total current of 16*18 = 288 amp if shorted without load. This gives 288*288*0.044 = 3649.536 watt [again is this right?] I'm taking current in calculation, as I know what is the amount of current when I short it. Both calculations are little off from each other. 3686.4 watt != 3649.536. But I think that is the power generated by either Parallel or series arrangement of the coils. Can we consider this as the max power output this generator can deliver? 



#16
Feb1614, 11:23 PM

P: 67

I'm Software Engineer, but not Electrical Engineer, though I think I've more curiosity of Electrical Engineering [The Technology, which runs the world]. 



#17
Feb1714, 12:50 AM

P: 285

The best thing is to first get an idea for what use you will put the generator to (lighting or heating etc.) and decide the output voltage. With that done you can check the wattage needs of your load and design the generator according to that. If you go the other way around, it will be trickier. 



#18
Feb1714, 03:06 AM

P: 67




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