Can someone please explain ohms to me?

[Amiles]

O.K., I'm not new to whole e cig world, but there are still a few things that I don't quite understand. One of them being ohms. I'm using a K-go, which I believe to be 3.7v. I have used different ohms with them and I don't want to ruin a battery or waste juice either. Also, whats a lr atty and a sr atty? I know that lr is low resistance, but what does that even mean? what devices can it be used on and what should it not be used on? And whats the difference between bridged and non bridged attys? If anyone could dumb these down for me I would be eternally greatful

 

[Amiles]

I think I put this in the wrong place, mods, could you move it for me pretty please?

 

[yo vapo]

OK, in generic terms resistance is just that -- resistance to the flow of electricity. Resistance is measured in units called ohms.
So less resistance (lower number of ohms) = more electricity flows. More resistance (more ohms) = less electricity flows.
That's ohms & resistance in general.

So, all else being equal (i.e. using your 3.7 volt battery), a low resistance (LR) carto or atty will give you a hotter vape, because more electricity is heating the coil. The opposite is true of a standard resistance (SR) carto / atty, and even more so with a high resistance.

So LR = more vapor which is warmer, while SR = less, cooler vapor (all else being equal -- your 3.7 volt battery).

HTH

EDIT: As far as specifics concerning various batteries, I'm afraid I'm new to vaping. I've read that if your battery is 450 mah (milli amp-hours) or more, then you can use a LR atty / carto without hurting your battery, but the battery will run down quicker than with a SR. Hopefully someone who knows more about the specifics of vaping can be more help with the finer points.

 

[Todd Mulske]

In electricity flow there are 4 things to look at.

1. Volts
2. Amps
3. Resistance
4. Watts

Volts(E) x amps(I) = watts(P)
Volts(E)=watts(P)/Amps(I)
Amps(I) x resistance(R) = volts(E)
Resistance(R)= Volts(E)/Amps(I)
Amps(I)=Volts(E)/Resistance(R)

The biggest thing you need to be concerned with as far as your batter is how many amps you pull from it. Your battery has a switch on it and current goes through the switch with many of the batteries on the market. Usually they should be good for 2 amps. Your battery is just discharging voltage and the resistance of your atty determines how many amps are being pulled through the circuit.

The two things that you need to know the most is the voltage of the battery and the resistance of your atty because this will determine the amperage or current flowing through you system

Amps(I)=Volts(E)/Resistance(R)

With an ecig you will always know your voltage and your resistance
If a battery is rated at 3.7 volts and you use a 2.3 ohm atty using the above formula you will pull amps(I)=3.7(E)/2.3(R) or 1.61 amps

If you use a low resistance atty like a 1.7 ohm you would pull (I)=3.7(E)/1.7(R) or 2.18 amps

I have an ego-t 1000mah battery that is rated at 3.7 volts but off the charger it only tests at 3.3 volts and stays that way to the last puff before it turns off and I am using a 1.7 ohm atty on it right now and not having any problems as of yet.

A bridged atty has a triangular piece of metal with mesh wire going around it and there is a wick that is in the mesh that feed juice to the heating coil. A non bridged atty does not have this mesh or triangular piece of metal in it. The non bridged attys are usually used for dripping or a few of them have a wick that the coil wraps around, but I believe they can only be used for dripping as well except for the bulli a2 tank that has a non bridged coil with a wick inside of it and another wick that feed the coiled wick through a tube.

Hope this helps

 

[aut0tek]

You could use the old plumbing analogy they teach in Tech Schools:

Amps = How much water is moving through the pipes.
Resistance = How big the pipe is. (Inversely. Higher resistance = smaller pipe)
Volts = How hard the water is pushing. i.e. "Water Pressure"

 

[Tanis143]

Todd, you forgot one formula that helps figure out the wattage (a lot of vapers look at the watts to judge how warm their vapor is going to be) which is wattage (w) = v^2/r or given the above parameters: (3.7 * 3.7)/1.7 = 8.052 watts. The reason this is important is when people look at either raising their voltage and raising their resistance or vice versa. In essense, for me 7-9 watts is my "sweet spot", meaning thats gives me the amount of vapor and warmth that I like. Now, say I started with a basic kit, a 650mah Ego system. The Ego batts generally run around 3.1 to 3.3 volts due to the regulator board in them. The standard JoyeTech atties are 3.2 ohms. So using this basic setup I would get 3.00 to 3.40 watts. Now, I can raise the wattage two ways: increase the voltage or lower the resistance. If I go to a 6volt battery mod, that would give me 11.2 watts. Oops, too high. Ok, so using the same batt I can lower the resistance to a lr dual coil carto (mine are 1.5 ohm), that would give me 6.40 to 7.26. Ok, thats closer. So, I go to a 3.7 volt mod with the lr dual coil (or commonly called lr dc cartos), that gives me 9.12 watts.

The amperage being used isn't as big of a deal as most mods that do not have a regulator board use 3 amp rated switches. If you do something to exceed 3 amps, well your just wasting juice and/or boiling your lungs. Most mods that have a regulator board have amperage overdraw protection and will shut off if too much amperage is pulled. That is why most people look at the watts and not the amperage.

Now, when you look at atties and carto's you have to check what the manufacturer states is their resistence, or buy a multimeter that goes down to 100 ohm in measurement and test them. I've had a few carto's that say they are 1.5 but in reality they are either much less or slightly higher (I've measured them from 1.1 ohm to 1.7 where they state they are 1.5). I hope that helps some.

 

[Liv2Ski]

Amiles - If you would like to play around with the variables in ohms law here is a nice calculator

Ohm's Law Calculators

 

[Todd Mulske]

Todd, you forgot one formula that helps figure out the wattage (a lot of vapers look at the watts to judge how warm their vapor is going to be) which is wattage (w) = v^2/r or given the above parameters: (3.7 * 3.7)/1.7 = 8.052 watts. The reason this is important is when people look at either raising their voltage and raising their resistance or vice versa. In essense, for me 7-9 watts is my "sweet spot", meaning thats gives me the amount of vapor and warmth that I like.

Your right, the formula you posted above volts squared divided by resistance does give Watts. Watts is the end product of the combination of voltage, amperes and resistance. Watts is power. I just tried to answer the question being asked which was what kills batteries or actually switches in batteries. Amperage is what will destroy a switch. A battery is rated in MAH (miliiamp hours) If you take a 450 MAH battery and connect it to a atomizer say at 3.7 volts with a 1.7 ohm resister you would be pulling 2.18 amp or 2180 Milliamps per hour. If your battery is rated at 3000 MAH and you still were at 3.7 volts and a 1.7 ohm resistor you still would be pulling the same or 2.18 amps or 2180 milliamps per hour.

The 450 MAH battery could maintain a continuous output for 450/2180=..21 hours x 60 miinutes=12.8 minutes. The 3000 MAH battery would maintain a continuous output for 3000/2180=1.38 hours x 60 minutes=82.8 minutes.

I used a standard 3.7 volt battery and a low resistance 1.7 atomizer to give Amiles an idea of how long one could vape with the two different batteries and a 1.7 atty with a standard kit. A 450 MAH would only last 12.8 minutes; that would be about 2 hours actual time for me as I vape about 6 or 7 minutes per hour.

Back to the switches, most switches are rated at 12 volts for the most part. the amperage rating is what you need to look at. If a 12 volt switch rated at 2 amps were pushed to 3 amps at 3 volts it would still fry. Amps or current is what destroys them. Although you never want to exceed the volt rating either.

Little technical, but hey it works when you need to size your battery for your vaping requirements, By the way my sweet spot depends on what I'm looking at as far as watts go. If I want to taste the fullness of the flavor 4.7 watts is perfect. If I am after a better throat hit 6.4 watts is right where I like to be. The hotter or more watts you use the less the flavor comes out; that's what I have found anyway.

Rather than trying to figure out ohms law on paper or a calculator this is a good application that even includes a DIY recipe software program. Just click on tools; pv tuning and enter any two known variable and enter 0 for the other two and hit calculate and wholla you have einstein.

http://ejuice.breaktru.com/

 

[Amiles]

Hey guys, thanks for all the awesome info!!! those calculators will def. come in handy! And thank you for explaining to me...I think I might finally understand!!!