Sup,
I got this Samsung ICR18650-26F.
Can i use it for my aspire Pegasus 70w + triton sub ohm tank?
Thanks in advance
I got this Samsung ICR18650-26F.
Can i use it for my aspire Pegasus 70w + triton sub ohm tank?
Thanks in advance
Sup,
I got this Samsung ICR18650-26F.
Can i use it for my aspire Pegasus 70w + triton sub ohm tank?
Thanks in advance
It's even easier than that, if you accept that the battery must be able to meet the mod's maximum demand capability of 70W.How many Ohm's is your coil for your tank this is easy to pop into Ohm's law to see why this battery is not a good choice..
Tha is incorrect that Mod itself does not put out that Amperage rate you surely do have the formulas kinda correct but your values are incorrectIt's even easier than that, if you accept that the battery must be able to meet the mod's maximum demand capability of 70W.
What the unit requires: Watts = Volts * Amps. Thus Amps = Watts / Volts. Nominal voltage for an 18650 is 3.7V. 70W / 3.7V = 19A. Thus the battery needs to be able to supply at least 19A of current intermittently.
What that battery can supply: The max discharge current of the Samsung ICR18650-26F is 2C (2 x Capacity). It's capacity is 2600mAh. So it's maximum discharge rate is 2 x 2600mA = 5200mA or 5.2A.
The most the OP's battery can do is 3.7V x 5.2A = 19W. Less, actually, due to various losses. I wouldn't try to drive more than, say, 15W (?), with that battery.
Jim
(looks at MrScaryZ weakly...) I have the formulas "kinda correct?" No, MrScaryZ, I have the formulas correct, period.Tha is incorrect that Mod itself does not put out that Amperage rate you surely do have the formulas kinda correct but your values are incorrect
That will come as quite a revelation to no end of Electrical Engineers, Physicists and others.You cannot convert Watts into Amps directly without a resistance.
Whatever you are thinking you are Engineering please make sure to keep me the fuck away from it..(looks at MrScaryZ weakly...) I have the formulas "kinda correct?" No, MrScaryZ, I have the formulas correct, period.
That will come as quite a revelation to no end of Electrical Engineers, Physicists and others.
Now it is likely that, due to inefficiencies in design: Current draw from the battery may be higher than that suggested by the wattage setting, but it can never be lower. You cannot make something from nothing. Exception: A freshly-charged 18650 may be a tenth of a volt or two above nominal, but the difference in current draw will be fairly insignificant (less than two or three percent).
Jim
I'm not engineering anything, so you can relaxWhatever you are thinking you are Engineering please make sure to keep me the fuck away from it..
That's what I was doing.You can calculate amps from watts and volts.
Never said you could "convert" watts to amps. I said you could calculate current knowing power and voltage. You can also calculate maximum wattage knowing voltage and maximum current.You can't convert watts to amps since watts and amps units do not measure the same quantity.
Context is the issue for the OP you just confused the whole issue.. they surely have no clue WTF we are talking aboutI'm not engineering anything, so you can relax
That's what I was doing.
Never said you could "convert" watts to amps. I said you could calculate current knowing power and voltage. You can also calculate maximum wattage knowing voltage and maximum current.
You asserted "You cannot convert Watts into Amps directly without a resistance." That is not correct. You can derive amps (current) from watts if you know either resistance (I = (P / R)½) or voltage (I = P / V).
(The above applies only to DC circuits or AC circuits with purely resistive loads.)
Jim
Why did you bring coil resistance into a post about amp draw at the battery of a regulated mod (Pegasus 70w)? It's irrelevant and confuses the issue.