Ohms and watts?!?

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sigtausteve

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Low ohms (resistance) = less volta needed to fire the coil. The opposite goes for higher ohm. What it translates to is wattage. Lower watts is a cooler vape with less vapor. Higher watts is a hotter vape with more vapor production. The formula for that is (volts x volts/ohms). So if you are running say a 2.0ohm atty or carto at 3.7volts, you would be pulling around 7watts. Which is a nice cool vape for most people. I like to sit around 7.5-9 myself.






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Wuzznt Me

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Camden

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Here's a handy-dandy chart that may help you out (not made by me, and I forget the member who deserves all the credit). It was immensely helpful for me; hopefully it can help clarify things for you as well

Ig0a3Xk.png
 

DKP#

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Hi,

Can somebody explain to me how to play around with the voltage of my ego c twist?
What I do is to go as high as I can before the burned taste arrives and then go back a bit. It kind of works but I'd like to know a bit more about it. Also, how do I choose the resistance? What do I get with lower, higher ohms?

Thanks

That's the way I do it too.
The twist goes from 3.2 to 4.8 volts. If you find you can't go over 4V without getting a burnt taste, higher resistance (ohms) will give you more adjustment range.

For Those that like math:
Voltage is electrical pressure measured in Volts
Current is electrical flow measured in Amps
Resistance is opposition to current flow measured in Ohms
Power is the rate of work done measured in Watts (heat)

Ohms law:
Voltage = Current x Resistance (V = I*R)
Power = Current x Voltage (P = I*V)

Example:
Say you have a 1.8 ohm head and 4V gives you a burnt taste.
How much power is that?
V=I*R
I = V/R = 4V/1.8ohms = 2.22Amps

P = I*V = 2.22Amps*4V = 8.88 watts

What resistance would I need to not get that power until 4.8 Volts?
P = I*V
I = P/V = 8.88watts/4.8Volts = 1.85Amps

V = I*R
R = V/I = 4.8volts/1.85amps = 2.59ohms


Here's the 2 formulas (V = I*R) and (P = I*V) arranged in every way
Ohms law.jpg
 
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