Variable Voltage and Resistance Question

[Jellyfish]

First, I understand that the wattage is the driving force and is calculated by: W = V^2/Ohms. Also, from what I've seen, the watts value is a personal preference and can vary with the juice being vaped.

So my basic question is, is there an advantage to firing a higher voltage with a higher resistance head over using a lower voltage and lower resistance head if they both produce the same watts?

 

[never_enough_juice]

It seems like higher resistance and higher voltage is better to me. . .maybe it's just in my head but i do feel like it's a better vape. . .i was using an ego twist with low res cartos but now i got a five volt box mod with 3ohm cartos and it's better to me .

 

[Black Strat]

Less current is required. Another way to express power is current squared times resistance.

Assume you prefer to vape at 8 Watts:

Current required for a 1.8 ohm Atty is 2.1 amps.

Current required for 2.8 ohm Atty is 1.7 amps.

Of course the voltage boost/regulation also uses some power to accomplish.

 

[gardnerd4me]

Yeah, watts being the driving factor, your higher res. atties are going to give you more mileage. They draw less amps so in turn your battery has more amps for longer.

 

[tj99959]

Which is more advantagious for bringing the teapot to a boil, the large burner or the small burner, their both heated to the same temperature? While wattage is important, it's not the only factor involved.

 

[Aoi~]

Lower current draw = saves battery life. At least, to my understanding anyway.

 

[enoryt]

Is the general consensus that warmer/higher watts are used flavors such as coffee/cream to bring out the flavor and lower for fruity ones?

 

[enoryt]

Sorry I realize its a bit off topic, for some reason reading your replies lead me to that train of thought.

 

[Black Strat]

Which is more advantagious for bringing the teapot to a boil, the large burner or the small burner, their both heated to the same temperature? While wattage is important, it's not the only factor involved.

I am not understanding your point.

 

[digunderground]

If you follow ohms law (you have to its physics law lol), then the following is true:

3ohms at 10 watts = 1.82 amps draw on source (battery)
1ohm at 10 watts = 3.16 amps draw on source (battery)

The difference is, on a regulated mod (ProVari/zMax) the 3ohm/10 watts is possible. So for unregulated mods (cube, don, GGTS etc) the 1ohm/10 watts is much more attainable. It depends on what you are using to power your coil. On my unregulated mods I use 30g kanthal which is thick as hell and will last damn near forever..

 

[tj99959]

I am not understanding your point.

What is the surface aria of the two coils at the same wattage? How fast does each come up to temperature? Why does a 306 taste different than a 510?

As a generalized statement, you will find that the LR tastes hotter, but produces less volume than the SR at the same wattage.

 

[gthompson]

The same same as the surface area at different wattages......

 

[Black Strat]

And a little I squared R loss from the boost circuit needs to come from the battery as well. But there is a net gain.

Electric utilities also prefer higher transmission voltage to lower current. The losses transforming down to end users are smaller than the heat losses due to lower voltage//higher current.

 

[Jellyfish]

Electric utilities also prefer higher transmission voltage to lower current. The losses transforming down to end users are smaller than the heat losses due to lower voltage//higher current.

Great analogy...and thanks to all that have contributed and helped me understand things better.

 

[steved5600]

Here are some things to help you calc wattage.
\ View attachment 144615

 

[dillydobbs]

Here are some things to help you calc wattage.
\View attachment 144614View attachment 144615

Do you have that pic in a larger resolution ??? Its hard to see its so small

Thanks

 

[gthompson]

http://www.vaporrater.com/images/Ohmschart.jpg

 

[dBm0]

Agree, I use higher res atties with higher voltage settings to hit my sweet-spot wattage as well. Pulling less amps while still hitting the same target wattage would extend charge times and overall battery longevity.

 

[Black Strat]

I am not understanding your point.

What is the surface aria of the two coils at the same wattage? How fast does each come up to temperature?

Oh, OK. Now I see what you are getting at.

Power is power, but- Each liquid requires a given amount of energy to vaporize. From a saturated liquid to a saturated vapor is the "latent heat of vaporization" another way to express power is "unit of energy per unit time". If you transfer 1 BTU of energy to 1pound mass of water it will heat 1 degree Fahrenheit.

Here are some calculations based on 8 Watts. You can look up all of the conversions.

8 Watts - 27.3 BTU/HR
27.3 BTU/HR - .0076 BTU/sec
8 Watts/7 second draw-. 053 BTU
VG latent heat of vaporization- 419 BTU/lbm
.053 BTU can vaporize. 00013 lbm
1ml of VG-. 00278 lbm
Now, if all the energy was transferred directly to VG, you could vape 1ml in 21 7second drags at 8 Watts.

We know it takes three times as many drags. Why? Efficiency. Some of the heat is transferred to the VG which has already been vaporized, warming it instead of liquid for one. Also sensible heat must be added as well as the latent heat of vaporization. So it also depends on the starting temperature of the liquid too.

By the way, power plants which rely on vaporizing liquid has an efficiency of.... 33%.

 

[Rader2146]

As long as watts is unchanged the battery drain current will (ideally) be the same no matter the voltage or resistance. Law of Conservation applies: power in must equal power out. The only variable is the efficiancy of the device.