Questions about resistance, wattage, and voltage (and heat)

[BARFREFRE]

Hey, I've been reading threads here and even about light bulbs and electricity on How Stuff Works, but I have a few questions and think it might be best to ask them directly here.

First of all, what I am using: Hana dna 30 with sony vtc5, 4nine with sony vtc5, Patriot and Gaia RDAs, 28 awg kanthal, and an ohm meter.

1) If I have the gaia built with dual coils, each about 1 ohm, together reading .53 ohms, is the power also split across the two loads? For example, if the Gaia is on my 4nine and I am guessing the not fully charged vtc5 is putting out 3.7v, so using Ohm's Law it is running at 25.83 W, then is 12.915 W being delivered to each load/coil?

2) This question has to do with resistance and heat and wattage. Ok, so I have my patriot (single coil) at 1.30 ohms on the Hana and I get decent heat between 15-20W. If I put the .53 ohm Gaia (dual coil) on the Hana, it bearly fires until it gets to at least 25 W. So my question is why does it seem that higher resistance builds need less wattage than lower resistance builds? Even when I build a higher resistance dual coil, I run it at less wattage than a lower resistance dual coil. The lower ohm coil takes longer to heat up than the higher ohm coil at the same wattage, why?. Thank you, to whomever can answer this, thank you.

With light bulbs in the US, we are generally working with a constant 120 V, so we use a lower resistance to get more Watts, and thus a brighter bulb (analogous to a mechanical mod). So question (2) has to do with choosing the wattage (using variable wattage) and why higher resistance gives more heat than lower resistance at the exact same wattage. Also using known values, is there a way to calculate for heat production? It would seem as though we caculate Watts for light bulb brightness, but as this questions suggests, I don't think we can use Watts for attomizer heat...

3) Lets say I build an rda at 1ohm and another at 2 ohms and I decide to vape either one at exactly 20W on the Hana. Using Ohms law, I will be pulling 4.47 amps on the 1 ohm and 3.16 amps on the 2 ohm, so I will get better battery life with the higher ohm rda than the lower ohm, right?

Thank you for your help with any of these questions. Equations, web links, or metaphors to illustrate your explanation would be appreciated love the metaphors, thanks!

 

[Alien Traveler]

Since nobody answered so far, I’ll try my best, though I never was in sub-ohm vaping myself.

1. Yes.

2. I - current
R – resistance
V - voltage
W – wattage
W=V*I=V*V/R
So, if you decrease R in half (say, from 2 to 1 ohm), at the same voltage you’ll have twice as high wattage. That’s the idea of sub-ohm coils. If you use same wattage for low and high resistance coils low resistance one will not be working properly (too low heating).
You see, constant wattage in VW devices is not working exactly as it was describe in a sales pitch. Actually for best results you need to change wattage when you place coil with different resistance.

3. As was already discussed, you need to change wattage, you cannot use the same setting for so different resistances. However, higher resistance coils do need lower voltage and battery life with them will be better.

 

[BARFREFRE]

Thank you for your answer to (2) Alien Traveler. It seems to correspond with my experience. Your example with contant wattage (using a VW device) using different resistances seems to imply that resistance is the factor that affects heat (under constant wattage), and therefor you should dial wattage up with lower resistance and dial wattage down with higher resistance to control for heat production.

So you and I are both saying the same Wattage on different resistance builds does not = the same heat production on different resistance builds.

I have read other people saying that with a VW device your resistance doesn't matter, but in my experience ohms do matter (in regards to heat production) with a VW device.

Thank you also for your answer (to question 1). I was thinking about that and that is quite interesting. I'll need to study that a bit more and experiment.
Using W=V*V/R, if I am vaping 24 W in a dual coil .5 ohm, each parallel coil having 1.0 ohms, 12 W are delivered to each 1.0 ohm coil. If it was a single coil built at .5 ohms, then 24 W would be delivered to the single .5 ohm coil. The wattage is split on a dual coil, but the resistance is doubled across each parallel coil. I think this works out well, since as you have stated Alien Traveler (for question 2) and my experience suggests, we want higher wattage for lower resistance and lower wattage for higher resistance (in regards to heat production)

(for question 1) I will try to break the equations apart for each parallel coil, but I don't know if it works this way exactly.
Dual coil (whole build):
24W=V*V/.5 --> V=6.928
Dual coil splitting the Wattage (and doubling the resistance) to each parallel coil:
12W(coil 1)=V*V/1 and 12W(coil 2)=V*V/1 V(coil 1)=3.464 V(coil 2)=3.464 added together V(total)=6.928
Cool. I hope that is right.
Thanks Alien Traveler