Ohm's Law and Vaping
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Thread: Ohm's Law and Vaping

  1. #1
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    Default Ohm's Law and Vaping

    It has been a looooong time since I studied E = I over R but now that I am a vaper, perhaps I need to hit the books again.

    I cannot understand why a low resistance atty would burn hotter than a higher resistance one. Resistance is, after all, resistance, and the more of it there is, the more current has to fight to get through. Or is my ohms law bass ackwards?

    I hope this will be answered by someone who knows the basics of ohms law.

    It also seems to mean that what we are taking about is the production of energy. The more of it, the more the vape - well at least until it starts frying. I would think that it isn't so much about voltage and resistance but the relation between the two. In theory, at least, I would think the same thing could be accomplished with different atties of different voltages and resistances as long as the relationship between the two produce the same amount of energy.

    However, I could understand that the amount of coil exposure to the juice and the size and shape of the atty also come into play.

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    You are correct. It is about the relationship between the voltage and resistance, and power (heat) created. That's why the term 'high voltage vaping' is so misleading. We really should be talking about 'high wattage vaping'. The formula for wattage is voltage squared divided by resistance. And yes, you can get the same wattage from different voltages by just using different resistance attys.

    Kent explains it very well here:

    Carto Master Thread - Just the Facts Ma'am - No Chit Chat Please

    And here's a handy little calculator. Just plug in the numbers.

    Online Conversion - Ohm's Law Calculator

    Hope this helps.
    Last edited by Katya; 11-29-2010 at 11:34 AM.
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    Hmm, according to Kent, Mr. Ohm, and that online converter, wattage (power) goes up when resistance goes down. But I would have thought it would be the other way around. If you force the same amount of voltage through a smaller wire, it should produce more heat. I would think, smaller, meant more resistance. Hence more chance to fry an atty or carto.

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    so, the lower resistance carto can hit more vapor ? how to get more vapor if the resistance should keep normal for protecting the carto get very hot ? higher voltage ?

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    Well, I usually do the math and try to figure out what wattage (power) satisfies me the most. If your vape is too cool, your juice won't vaporize properly and you won't get much vapor or nicotine in your bloodstream. If your vape is too hot, you're runnig a risk of frying your juice rather than vaporizing it; not too mention that you're very likely to burn your cartomizer also.

    Power (wattage) = V/R x V

    The good old trial and error method works for me. I try different combinations of voltages and resistances to find my sweet spot. For example, a 3.0Ω carto on a 3.7v battery (like most kr8 batteries) will give you 4.56 Watts [3.7/3.0 x 3.7 = 4.56]. That's a recommended safe and effective wattage. Now the same 3.0Ω carto on a 3.2v battery (like an eGo) will result in 3.4 Watts--a much cooler vape, but still perfectly acceptable. However, when you put a 2.0 Ω carto on your eGo, your wattage will go up to 5.1W, much warmer than the 3.0Ω carto on a 3.7v battery. Generally, 3.5-5.5W range is where most vapers like to be.
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    The less resistance there is, the more current gets passed through. less RESISTANCE. This is because all our batteries are Voltage sources. They maintain a constant voltage, so they want to stay at that voltage. If your resistance is lowered by half, twice as much current is needed to get the desired voltage. Hence, less resistance, more current and consequently more power.
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    Think of it like this. If -0- resistance = a short circuit, and Infinity resistance = no flow, which one is likely to produce a high current spike and trip your protection on the battery.

    You have a 100w light bulb. you have a 120v service. power=current times voltage P=IxE 100w/120v=.83amps
    voltage=current times resistance E=IxR 120v/.83amps=24.1ohms

    now you have a 60w bulb with 120v service 60w/120v= .5amps
    120v/.5amps= 240ohms

    the brighter 100w bulb is pulling .83amps with 24.1 ohms of resistance verses the dimmer 60w bulb pulling 1/2 amp at 240 ohms

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    I just like expressing Ohm's law differently...to me it makes more sense in a vaping context...

    Wattage= Voltage (squared)/ resistance
    and
    Amps= Voltage/ resistance

    along with the problem for max drain rates for batteries which is
    mAh rating (in Amps) * C rate

    so that tells me that say for example using a 1.5Ω LR atty versus a joye 2.2Ω atty on a 3.7V battery

    well the 1.5Ω atty is going to put out 9.13W and draw 2.47A
    and the 2.2Ω atty is going to put out 6.22W and draw 1.68A

    however depending on the battery it may not be capable of delivering the Amperage because of it's max drain rate so we experience what we would call "voltage sag"
    like on a ultrafire 900mAh 14500 battery (which we know has a C rate of 1.5C)
    that would have a max drain rate of 1.3A

    so it can't deliver the 2.47A that the high drain is asking for
    and since the resistance is fixed the only way to adjust the amp draw to align with the max drain rate is by dropping the voltage (hence the "sag")

    meanwhile since a 600mAh high drain IMR 14500 has a max drain rate of 5A it can deliver the amperage of the LR atty without having to sag so it can deliver the wattage in full...
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    Current PVs: Precise Plus 18650 10/7/11, Super6 3/13/10, eGo 1/23/10 (shelved)...

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