Can someone explain breifly - the relationshsip between battery size and cart Ohms?

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CBB

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The lower the resistance the hotter your vape will be. it takes a 450mAh or larger battery to handle the extra effort required to run a low resistance atomizer. You can test the voltage of a battery with a meter designed to measure voltage, there are different types of meters for different applications for different meters so most people use a multimeter that can be used for several applications. A multimeter can be a bit tricky to learn to use if you don't have any experience with them.
 

Dalton63841

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mAh will tell you NOTHING about the voltage of the battery. Rather, it is a relative measurement of how long a battery will last before it needs to be recharged. As a general rule e-cig batteries, and the li-ions used in mods, are all 3.7v nominal, but will put out 4.2v fully charged. There are a few exceptions, but they are few.

Lower ohm carto's allow more power draw, so they discharge the battery a bit quicker. For instance, a 2.5 ohm carto with a 3.7v 650mAh battery might last 1.5 days, while a 1.5ohm carto with the exact same battery might only last 0.75 days. On the same token, a 1.5ohm carto on a 3.7v 650mAh battery might only last 0.75 days, while the exact same carto on a 3.7v 1100mAh battery may last 1.5 days. Of course these are just examples, but in these examples the voltage was always the same. The only difference was the power draw of the carto, or the charge life of the battery.
 

Iffy

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If you're asking about concept, here is a elementary analogy:

Resistance = water pipe diameter
Voltage = water pressure at supply
Amperage (current flow) = water volume through pipe
Battery = supply water tank

If you enlarge the pipe diameter (R), the volume (A) will increase and empty the tank (B) faster while maintaining the pressure (V) until the supply tank (B) can no longer keep the pipe full.

If you're looking for actual values, they're on the ECF somewheres... haven't found 'em yet... :blush:

Looky what I FOUND!
 
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dormouse

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Low Resistance (low ohms) atomizers and cartomizers suck power too hard for the electronics in a slim battery. That is why those items should only be used on batteries 450mah or larger charge because those batteries usually have beefier electronics inside that can deal with the heat and the current flow. And even then, some people will not use anything lower than 2.0 ohms on batteries like Ego and Riva etc and will only use lower on mods that have removable industrial type protected batteries (like aa-size but strong) because those can be replaced and are charged in an external charger where any damage or leaking can be seen.

Other than that mah has nothing whatsoever to do with what carto you should get. MAH is just the length of the charge. The VOLTAGE of the battery combined with the OHMS of of the attachment, in an equation, determine how much current (or wattage? I do not know) is produced = heat and also how fast the battery is being used.

An Ego battery is about 3.4v (more when freshly charged)

I do not know the equation

AHHH - Here is the big chart!! Click on the picture to zoom it when you bring it up.
http://img97.imageshack.us/img97/46...tbnw=158&start=16&ndsp=17&ved=1t:429,r:6,s:16
 
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zoiDman

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I do not know the equation
...

I'm about the Last person to answer an electrical question but if my Algebra is right,

Watts = Volts x Amps

and

Ohms = Volts / Amps, => Amps = Volts / Ohms

Sub in Volts / Ohms into Watts = Volts x Amps => Watts = Volts x (Volts / Ohms)

So...

Watts = (Volts x Volts) / Ohms

"Watts is equal to Volts squared divided by ohms."


Seems to jive with the chart you posted.
 

AttyPops

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I am currently enjoying a 1.5 dual cart on a 1000mah eGoT battery. But the more I read, the more confused I have become. How does one arrive at which strength(?) carto to use, how do I find the voltage in any given battery, what exactly is low resistance and is it a good thing,etc...

You're using low resistance (LR) sort of.... but it's really two 3.0 ohm coils (hence 1.5 ohms). Explaining dual coils makes my head hurt. Think of it as two 3.0 ohm atties (like a double barreled mod). It takes more amps tho, since it's two coils.

Lithium Ion batteries are 3.7 volts... ish.... it's actually a curve. They start out at 4.2 volts (max charge), go down to 3.6/3.7 fairly quickly, and stay there for most of the operating time. Then dwindle to 2.8 (?) volts and cut out.

Ohms determines amps needed/drawn from the battery. Other than the smallest batteries (like in a cig-sized super-mini), size of the battery doesn't matter. It matters for the super-mini because they are amp restricted due to size. They can only produce so many electrons per second. This means that the battery could be AAA sized, AA sized, C-Cell, D-Cell, house-sized.... doesn't matter (a generalization). Still 3.7 volts. Amps still determined by atty ohms.

The device matters... eGo's, for example, are modulated... so they net out to about 3.4 volts rather than 3.7. Many models are true 3.7. Others still, have voltage regulators and multiple cells in series (like a flashlight-- end to end) and are higher volts. This is more "oomph".

Resistance = ohms = Think of resistance like "electrical friction". Higher = more friction, less electricity per second.

mAH = milliamp hours. Battery capacity ... like size of a gas tank. Not the gas pump volume per second (amps) nor the line PSI (volts) nor the %clogging of the fuel filter (resistance), but rather capacity of storage space (xx gallon gas tank). (like iffy said in #5)

Watts = an instantaneous measure of work being performed = volt * amps. This is what you pay the electrical company for over time (Kilowatts used per hour).

The VV/R calc above gives you watts. Watts is a good ballpark for vape heat and oomph. A good number is 8 watts for a single coil.
 
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Starspecks

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If you're asking about concept, here is a elementary analogy:

Resistance = water pipe diameter
Voltage = water pressure at supply
Amperage (current flow) = water volume through pipe
Battery = supply water tank

If you enlarge the pipe diameter (R), the volume (A) will increase and empty the tank (B) faster while maintaining the pressure (V) until the supply tank (B) can no longer keep the pipe full.

If you're looking for actual values, they're on the ECF somewheres... haven't found 'em yet... :blush:

Looky what I FOUND!

Thank you!

Sent from my Evo
 

Stosh

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I do wonder if batteries with more mah have lower internal resistance. If that *is* the case, then you would get more of a voltage drop across your atty with those, and thus more current.

In a series DC circuit, the current / amperage is the same anywhere it is measured. The electricity goes from the battery to the switch to the atty and back to the battery, only one path, current will not vary. Voltage drop will vary across each component in the circuit and having the most V drop accross the atty is :thumb: way to go.
 

scratch777

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True, but the voltage drop / internal resistance of the battery will indeed affect the overall current, i.e. given a 1 ohm atty, a battery with an internal resistance of one ohm will give you 25% less current than battery with an internal resistance of .5 ohm. Again, I have no idea if a larger capacity e cig battery has a lower internal resistance, but it would not surprise me seeing that, say, a big honking car battery has a lower internal resistance than a small 12v battery.
 
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