Ohms and Voltage - if the watts are the same why does it matter?

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Oktyabr

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That's not really correct... Because you won't get a 10 watt differential in similar setups.... And that goes to prove what I was saying.

A large gauge wire will have lower resistance and not get as hot at the same wattage as a thinner wire...

See your 20 watts will pop the 1/4" of 40 gauge wire you're pumping 6 volts through...

My 10 watts will heat my 2" of 30 gauge nicely.


But since the question was same watts..

17 watts

0.8 ohms
2" of 28 gauge kanthal
3.7 volts
Cool vape, big flavor, huge plumes of vapor

Or

2.1 ohms
3/4" of 36 gauge kanthal
6 volts
HOT vape, burnt flavor, low vapor

But one thing is missing from this equation and that is current, measured in amps.

A few simple formulas where V=voltage, R=resistance, W=watts, A=amps (sqrt = square root):

W = V x A
W = (V x V)/R
V = sqrt (W x R)
A = V/R

So if we look at the two examples, which are both producing (roughly) the same 17 watts:

3.7 volts / 0.8 ohms (R) = 4.625 amps of current

and

6 volts / 2.1 ohms (R) = 2.857 amps of current

Since the resistance itself consumes power and the rate of consumption increases as current (amps) is increased there is a benefit in using higher voltage to achieve the same wattage because the overall system uses less amperage and is therefore more efficient, which should translate into a longer battery life....

Just to make things EXTRA confusing! :p
 
Clear as mud... Just kidding, there have been a lot of helpful things in this thread so thanks. I did not realize my 1.8ohm heads are probably not 1.8ohm - and it's fascinating to me the way amperage plays into this with battery life.

I now realize why the best answer is just mess around with different ones and see what floats my boat, there are too many things that can effect it outside of the simple equation of ohms law.

Cheers -

b
 
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