Physics Help Forum Wind power experiment questions

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Nov 16th 2015, 10:27 AM   #11
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Been watching Youtube videos to try and get my head around current and voltage and resistance. Semi-confused. Here is what I understand:

 Originally Posted by studiot The source produces a voltage, whether there is a load there or not. But voltage is not power.
A battery provides potential to do work. If there is a high potential difference (PD) between the two terminals of the battery (called voltage) then lots of current can flow.

Power is just a way of assigning a value to the battery by saying P = IV

If battery A is assigned a value of 100 W then this means that the chemicals inside it allow it to provide 20 Amps of current when its current is passing through a LOAD that causes a potential drop of 5 V

If battery B is assigned a value of 100 W then this can mean that the chemicals inside it allow it to provide 50 Amps of current when its current is passing through a LOAD that causes a potential drop of 2 V

 Originally Posted by studiot If you connect a load current will flow and the source will attempt to maintain its voltage across the load. As you reduce the resistance of the load you increase the load. This seems odd but this is correct.
I'm losing the plot
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Nov 16th 2015, 10:31 AM   #12
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 Originally Posted by Hero The turbine is effectively a battery so I would find out the voltage output of the turbine by connecting the voltmeter in parallel with it + connect an ammeter in series to find the current supplied. With voltage and current know I will know the power generated by the turbine from the formula P=IV.
So as a after all this discussion I've concluded that I need to place a LOAD in the circuit. But what size of load (wattage of bulb)? How do i work that out.

How do i measure the power output of the turbine?
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Nov 16th 2015, 11:52 AM   #13
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 I'm losing the plot
I'm not suprised.

I have been carefully restricting things I introduce to present them in a logical order and not to introduce inappropriate or downright incorrect material.

 A battery provides potential to do work. If there is a high potential difference (PD) between the two terminals of the battery (called voltage) then lots of current can flow. Power is just a way of assigning a value to the battery by saying P = IV If battery A is assigned a value of 100 W then this means that the chemicals inside it allow it to provide 20 Amps of current when its current is passing through a LOAD that causes a potential drop of 5 V If battery B is assigned a value of 100 W then this can mean that the chemicals inside it allow it to provide 50 Amps of current when its current is passing through a LOAD that causes a potential drop of 2 V
These misconceptions are most unfortunate and will hinder you until you can dump them.

If you want to put them right tell me what you think a watt is.

Nov 16th 2015, 12:46 PM   #14
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 Originally Posted by studiot If you prefer youtube, then by all means go ahead.
Was trying to get a quick fix.

 Originally Posted by studiot These misconceptions are most unfortunate and will hinder you until you can dump them. If you want to put them right tell me what you think a watt is.

WATT gives a measurement of the rate of consumption of energy.

One WATT is equivalent to one joule of energy consumed in one second.
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Last edited by Hero; Nov 16th 2015 at 12:52 PM.

Nov 16th 2015, 01:28 PM   #15
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So how do you reconcile this

 WATT gives a measurement of the rate of consumption of energy. One WATT is equivalent to one joule of energy consumed in one second.
with this?

 Power is just a way of assigning a value to the battery by saying P = IV
What does it say on the case of a battery about the electrical characteristics?

Here are some facts.

You can flatten a 12 volt car battery in 45 seconds cranking the engine.

It takes 45 minutes to flatten the same battery leaving the headlights on.

After six hours of leaving the doors open and the interior lights on the battery is flat.

What physics can you say about all these facts?

Last edited by studiot; Nov 16th 2015 at 01:33 PM.

 Nov 16th 2015, 01:55 PM #16 Senior Member     Join Date: Apr 2008 Location: Bedford, England Posts: 668 Experimental practice (iii) Explain the number and range of readings that you will take. No matter how carefully an experiment is constructed, there will be some unavoidable variability (the exact position of the hair dryer, random breezes in the room, mains voltage fluctuations, etc, etc). Repeating the readings several times allows you to determine the variability of your experiment. Ideally you should be doing enough repeats to allow basic statistical analysis, to get the average and the standard deviation. The average is an estimate of what the value "really" is, the standard deviation is an estimate of how good an estimate the average is. If all the results (for the same experimental conditions) are pretty much identical, then you just need a handful of repeats to prove this. If the results are more variable, you need enough repeats to give an acceptably well defined average and standard deviation. Wildly varying results could point to an experimental problem. However some experiments are unavoidably variable, CERN for example will repeat billions of particle smashups to achieve the required statistically viable result. __________________ You have GOT to Laugh !
Nov 16th 2015, 02:29 PM   #17
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 Originally Posted by MBW (iii) Explain the number and range of readings that you will take. No matter how carefully an experiment is constructed, there will be some unavoidable variability (the exact position of the hair dryer, random breezes in the room, mains voltage fluctuations, etc, etc). Repeating the readings several times allows you to determine the variability of your experiment. Ideally you should be doing enough repeats to allow basic statistical analysis, to get the average and the standard deviation. The average is an estimate of what the value "really" is, the standard deviation is an estimate of how good an estimate the average is. If all the results (for the same experimental conditions) are pretty much identical, then you just need a handful of repeats to prove this. If the results are more variable, you need enough repeats to give an acceptably well defined average and standard deviation. Wildly varying results could point to an experimental problem. However some experiments are unavoidably variable, CERN for example will repeat billions of particle smashups to achieve the required statistically viable result.
Thanks MBW. Very useful
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Nov 16th 2015, 02:45 PM   #18
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 Originally Posted by studiot You can flatten a 12 volt car battery in 45 seconds cranking the engine. It takes 45 minutes to flatten the same battery leaving the headlights on. After six hours of leaving the doors open and the interior lights on the battery is flat. What physics can you say about all these facts?
Volt = joule per coulomb
Current = coulombs per second
Power = joule per second

The car battery is the LOAD that draws a high amount of current (coulombs per second) from the battery that only provides on a small number of joules per coulomb (12 V or energy per coulomb of charge).

The battery needs a high value of joules per second (energy per second) to operate and it depletes the current in the battery quickly.

Dear Studiot,
My answer may make no sense and you may think I'll never get it BUT I'm happy to work through your instruction. I won't give up if you don't.
Thanks
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