First of all, I've actually been vaping for a total of more than 12 months. However it was only a few days ago that I switched to RBA, I'm using a Kayfun Lite+. Previously I've been vaping on the basic ego clearomizers. And therefore I am newbie in the world of true vaping.
I've been doing a lot of readings on how to rebuild an atomizer and there are a few things that always come across in all the discussions I've read... that is Ohms / voltage / woltage / low and high resistance.
Now can someone kindly explain to me in layman terms what are all these? And how it will affect my vaping experience?
I am really a newbie here... I've haven't even really started rebuilding my Kayfun yet! I scare I'll mess it up! Haha.
Lastly, I loving my Kayfun!
Welcome to the Forums
I'll give this a shot to try to help you. If I get too simplistic, I'll apologize up front.
Ohms (Ω): resistance to the flow of current (typically symbolized by "R")
Voltage: the amount of "potential" drop (think of it as stored energy?) between the positive and negative end of a battery (I decline to use the word "power" because that has its own electrical meaning) typically symbolized by "V"
Current: the measure of the flow of electric charge, typically symbolized by "I"
Ohms Law: V=IR (voltage = current times resistance) This rule is the most simple definition because I'm not going into the differences attributed to efficiency of all the components in question.
Watts: power as calculated by V²/R (voltage squared divided by Resistance), or VI (volts times current)
So, the way an e-cig device works (in the way my simple mind understands) is that the battery (source of electric charge) passes current through the heating coil (resistance) and vaporizes the e-juice. Playing the numbers game now, if I have a 2Ω coil and a 4V battery, I can pass 2Amps of current through the coil to heat the juice. (V=IR, therefore 4V = 2Ω * 2A) The Power is then calculated by 4V * 4V/2Ω = 16/2 = 8 Watts.
In all this, only one factor can vary during any given draw on the device. So, if I decide to use a 4V battery, but lower my ohms to 1.5, I will now be drawing 2.667A, which means a HOTTER draw (which means I'm consuming the electric charge of the battery more quickly and have the potential to burn juices or wicks) On the flip side, if I use a 2.5Ω coil on the same 4V battery, I'm only consuming 1.6A current, which means a COOLER draw (which then means that my battery can last a little bit longer and possibly prolongs the life of the wick, but may not be sufficient to vaporize the juice properly.)
Long-winded reply sort of shorter is that if you vary two elements of the equation, the third MUST calculate itself. You have control over the coil resistance (low = warmer vape, more current draw on battery; high = cooler vape, less current draw on battery) and possibly either the voltage level or the wattage level. You can't control BOTH volt and watts.
So, with your RBA, you must DILIGENTLY measure the resistance of the coil, and based on your device (knowing if it's regulated or unregulated is only a start) so that you don't accidentally "overdrive" the battery and hurt yourself. If you have a mech mod (likely unregulated), your batteries probably won't shut down on their own like a regulated device will.
Your vaping experience becomes experimenting with different coil resistances and power settings to determine what works best for you.
Hope this helps