Mech Mod & Ohms Law Question

[DDV]

Hi all,

I'm new here so please be gentle lol.

I have a decent understanding of Ohms law, I own and have used a mech mod (furyan squonk).

I have only ever used dual coil rda's (bonza, reload x, oumier VLS) I am looking at going onto a single coil build as i want to use the recurve rda on my mech and i tend to vape at under 80w on a regulated device anyway.

I have a Samsung 30T 21700 battery which is believe has a genuine 35A discharge. My question is...

Could I use say a 0.23 resistance build with my 35A battery on a mech mod without it melting the coil lol. I had used a 0.12 ohm build dual coil and know that it would require approximately 30A battery discharge so i know thats suitable for my battery, but a 0.23 ohm would only require a 16A (approx) battery. So is my 35A battery on a build that only requires a 16A going to fry it haha. From what ive researched, it wouldnt. It would just use less of the batteries power; meaning the life of the battery would decrease slower. But i thought that unregulated devices will always output the maximum power available in the battery - this is why i am now confused lol.

Please help me guys!!

 

[Chapperz]

The ohm`s represent how much of the batteries 'power' is allowed to flow/be released. Lower resistance equals more power, higher amps, more watts and heat all round (wire, battery, mod).

Your single coil will be fine if the ohms are what you say

 

[Simon G]

It will always output the voltage of the cell, that can't be changed ..... but the wattage (and the amperage) will be determined by the resistance of your coil.

 

[domejunky]

The battery can supply 'up to' 35amps. The actual current drawn is governed by the resistance of the load.

 

[Simon G]

I thought you said you had "a decent understanding of Ohms law"

 

[oldhippydude]

You could probably make better use of your "decent understanding of Ohms law" if you realised that the reason you were doing the calculations was not to avoid melting your coil but to avoid overloading you battery by drawing too many amps. The more you are under the the max continuous discharge rating (max safe amps) of your battery the safer your build is.

 

[allen]

4.2v at 0.23 ohms = 76.7 watts

That's the only answer I can give.
Will 76.7w fry your coil? You decide.

If your decent understanding of Ohms law leads you to think that unregulated devices will always output the maximum power available in the battery I would politely suggest you start learning it again.
I suspect you're confused because you've completely missed the key points of ohms law.

As your battery discharges, the wattage will decrease. That's ohms law.
If you thought anything else was true please take the time to understand it better before something goes wrong for you.

 

[allen]

Attached is a really boring table of numbers. It'll show you what happens to wattage and amps drawn as your battery discharges. Maybe it'll be of some use to you

 

[Mawsley]

And here a picture of a really interesting table

 

[DMAN]

And here a picture of a really interesting table

View attachment 168121

That table will definitely fry your coil if you drop the mod down one of those cracks, and the fire button gets stuck on fire.