Voltage or Amps

[Youngvapor]

Ok guys, so i know how steam engine works. All i wonder is, will more voltage take more power of the battery or amps? My build right now is .18 ohm and it sucks out the batteries. Its a 4 battery mod guys, so i have a capacity of atleast 16 volts and 80 amps.

 

[Sljuka381]

More voltage will draw more amps,more amps will draw more battery juice...Sorry for asking,but do you know ohms law?

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[Youngvapor]

More voltage will draw more amps,more amps will draw more battery juice...Sorry for asking,but do you know ohms law?

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Yes nevermind. I went lower in ohms to stabilze the amps and voltage

 

[BoomStick]

What type of mod? Variable wattage?

 

[Youngvapor]

What type of mod? Variable wattage?

Yup, it is variable wattage. Wismec RX 300 )

 

[David Wolf]

Ok guys, so i know how steam engine works. All i wonder is, will more voltage take more power of the battery or amps? My build right now is .18 ohm and it sucks out the batteries. Its a 4 battery mod guys, so i have a capacity of atleast 16 volts and 80 amps.

Um no you don't have 16V and 80 amps you can't just add up battery voltage and current. That would be 1280 watts haha. Here are your specs and you should only set anywhere near maximum power with at least 25A cells:
http://www.wismec.com/product/reuleaux-rx300/

 

[BoomStick]

Watt setting divided by total battery voltage equals battery current in a variable wattage mod. Build resistance has nothing to do with how fast batteries drain when using vw. Only watt setting.

 

[MWorthington]

More voltage will draw more amps,more amps will draw more battery juice...Sorry for asking,but do you know ohms law?

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It's actually the other way around. Less voltage will draw more amps with a given resistance. On a regulated mod resistance doesn't matter, though. Only the watt level selected.

 

[Sljuka381]

Nope,think again...3V on 1oHm will take 3A,4V on 1oHm will take 4A...simple ohms law...A increases with W/V.On a meh,since V is,lets imagine, constant value,resistance determines amp draw,lower the resistance,higher amp draw.On a regulated device resistance is constant,increasing power invreases A,also increasing power (W),increases V

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[The Cromwell]

It's actually the other way around. Less voltage will draw more amps with a given resistance. On a regulated mod resistance doesn't matter, though. Only the watt level selected.

You are thinking of the battery drain in a regulated mod.
Yes the mod will draw more amps as the battery voltage declines to maintain the output voltage required for the desired wattage on the applied coil resistance.

But that is not really ohms law it is the functioning of a voltage converter circuit.

 

[MWorthington]

Yeah but the discussion wasn't about mech mods. It was about the RX300 which is a regulated mod.

 

[Angrygod50]

A variation of Watts law is used to find the amp draw on a regulated mod. Resistance doesn't apply directly to a regulated mod because you need to factor in what goes on with the computer(chip). This is the equation for battery side amp draw.
Wattage / # of cells / voltage at cutoff / efficiency(%) = amps.
But to answer your question; for a set power (10 watts) it can be 10V(1A)=10 watts or 1V(10A)=10 watts.
Batteries in series add voltage but amps stay the same 2 18650 = 8.4V and 20A
Batteries in parallel the amps add and the voltage is the same 2 18650 = 4.2V and 40A
Ohms law only applies to the atty side of a regulated mod, but it doesn't tell you what the chip or batteries are doing.
Ohms law; Volts = Amps x Resistance
Watts law; Watts = Voltage x Amps
Confusing isn't it...

 

[The Cromwell]

Watts in is watts out plus any losses due to voltage conversion inefficiencies.

 

[David Wolf]

These power volts amps threads are always fun and educational really good work guys.

 

[Sljuka381]

As battery charge decreases amp draw will increase to compensate V lack.
I=P/V
Lets take 100W on freshly charged batt
I=100/(4.2*number of batteries in a mod,this case 3)
I=100/12.6
I=7.93A
Drained batteries at 100W
I=100/(3.4*3)
I=9.80A
On a sereies mod we get higher nominal V and they are not "amp hungry" like parallel batteries mod.
OP sorry,i didnt understand your question at first.I think it doesnt matter,discharge is going to be the same

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[fq06]

As battery charge decreases amp draw will increase to compensate V lack.
I=P/V
Lets take 100W on freshly charged batt
I=100/(4.2*number of batteries in a mod,this case 3)
I=100/12.6
I=7.93A
Drained batteries at 100W
I=100/(3.4*3)
I=9.80A
On a sereies mod we get higher nominal V and they are not "amp hungry" like parallel batteries mod.
OP sorry,i didnt understand your question at first.I think it doesnt matter,discharge is going to be the same

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That's the opposite of what you said but I agree, as a cells voltage declines, the amp load on it increases pulling the same watt output.

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And thread, correct me if I'm wrong but the OP's question is on the cells, not what the board is doing right?

And this is always a fun discussion

 

[Sljuka381]

Its not oposite,i just didnt understand OP's question than i saw a post stating that he ment batteries A draw.My post (#9) was answer to post #8,where was stated that the lower V draws hihger A,thats not true,decreasing V decreases A and oposite.Thats the coil current.There is A on the coil and A drawn from battery,and formulas are diferent.If you wonna go in to details add chip efficiency to I=P/V its ussualy anywhere from 70% to 90%,i think there is no mod with 100% efficiency...

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