Electro Questions

[Vego]

Hello folks. Wonder if some electrical gurus might could help me here.

Of the 4 measurements referenced in the ohms law calculator (Volts, Amps, Resistance and Watts) does just one of them determine the rate of drain on the charge level of your battery, or is it a combination of 2 or more... or perhaps all of them?

Secondly, can anybody explain to me, say using an analogy of water flowing through a tube, what attributes of this model would describe the 4 measurements? Like say the diameter of the tube, or the volocity of the water through the tube, etc... If this is a ridiculous analogy for this purpose, is it possible to use another equally simple one? I'm usually lost without simple analogys.

 

[LucentShadow]

Hi. The electrical current (amperage) is the only thing that directly affects the drainage of charge in the battery.

You could easily compare electrical current to water current. Electrical current (through wires) is the flow of electrons through it. More electrons means more amperage. Equate that to volume of water flowing through a pipe.

Voltage is just a difference in potential between two points. One point is more positively charged, and one point is more negatively charged. When a circuit is made between them, they will naturally seek balance. Imagine an empty tank and a tank full of water, waiting for a pipe to be opened between them.

Resistance is anything that impedes the flow of current. An electrical resistor limits the amount of current that can pass through a circuit, converting some of the current into heat. This is how an atomizer works as well. Imagine putting some sort of impeller in the water pipe, with the impeller attached to a wheel that turns a load. It will lower the water flow, but also convert the lost flow into some other power (or work.)

Wattage is just voltage multiplied by amperage. The important thing to understand about wattage is that it represents the total work done. This is heat in an atomizer. With two devices of different voltages set up for the same current drain to the battery, the higher voltage device will create more heat, as a tank with more water would get more work done in the above example.

EDIT: I should say, that while similar, water and electricity work a bit differently overall, so the above examples are a bit oversimplified.

And it's been about 20 years since I worked in electronics, so my electronics theory is a bit rusty. I hope this makes it a bit more understandable, though.

 

[johnnymaestro]

moving this to modder's forum. I think you may get some more info there.

 

[Vego]

Good idea, Dav... I mean johnnymaestro. Thank you

 

[Vego]

Hi. The electrical current (amperage) is the only thing that directly affects the drainage of charge in the battery.

You could easily compare electrical current to water current. Electrical current (through wires) is the flow of electrons through it. More electrons means more amperage. Equate that to volume of water flowing through a pipe.

Voltage is just a difference in potential between two points. One point is more positively charged, and one point is more negatively charged. When a circuit is made between them, they will naturally seek balance. Imagine an empty tank and a tank full of water, waiting for a pipe to be opened between them.

Resistance is anything that impedes the flow of current. An electrical resistor limits the amount of current that can pass through a circuit, converting some of the current into heat. This is how an atomizer works as well. Imagine putting some sort of impeller in the water pipe, with the impeller attached to a wheel that turns a load. It will lower the water flow, but also convert the lost flow into some other power (or work.)

Wattage is just voltage multiplied by amperage. The important thing to understand about wattage is that it represents the total work done. This is heat in an atomizer. With two devices of different voltages set up for the same current drain to the battery, the higher voltage device will create more heat, as a tank with more water would get more work done in the above example.

EDIT: I should say, that while similar, water and electricity work a bit differently overall, so the above examples are a bit oversimplified.

And it's been about 20 years since I worked in electronics, so my electronics theory is a bit rusty. I hope this makes it a bit more understandable, though.

Awesome information, Lucent... That helps me greatly! Well, except for Voltage maybe... that's still a little fuzzy. I understand potential energy, and what I get from your description seems like you're talking about the charge of the battery. But I'm not sure I've understood that correctly. I'm not catching that bit about the full and empty tank... especially how it relates to an electic circuit. Hmmm, I'm sorry. Let me dwell on it for awhile, maybe it'll come to me. Thank you so much for helping me understand.

Edit: btw, your answer to the first question really helps me. I've been trying to determin, if I have a vv pv, and I want to vape at around 10watts, which ohm atty I should use to get the most use from my batterys before they need a charge. It seems that as you increase the ohms, and are able to raise the voltage to maintain wattage, you automatically use less amps. This was my speculation and you have confirmed it. So now I know it's best to use high ohm attys, and why that is. Thanks again.

 

[LucentShadow]

Awesome information, Lucent... That helps me greatly! Well, except for Voltage maybe... that's still a little fuzzy. I understand potential energy, and what I get from your description seems like you're talking about the charge of the battery. But I'm not sure I've understood that correctly. I'm not catching that bit about the full and empty tank... especially how it relates to an electic circuit. Hmmm, I'm sorry. Let me dwell on it for awhile, maybe it'll come to me. Thank you so much for helping me understand.

Edit: btw, your answer to the first question really helps me. I've been trying to determin, if I have a vv pv, and I want to vape at around 10watts, which ohm atty I should use to get the most use from my batterys before they need a charge. It seems that as you increase the ohms, and are able to raise the voltage to maintain wattage, you automatically use less amps. This was my speculation and you have confirmed it. So now I know it's best to use high ohm attys, and why that is. Thanks again.

The correlation isn't perfect, but imagine a full tank and an empty tank, with a pipe that runs between the tanks, at the bottom of the tanks. In the middle of the pipe is the impeller that turns a work load (think water wheel that drives a grain mill.)

That's similar to a battery with atomizer attached. The two tanks are the battery. I'd call the full tank the negative part of the battery (full of excess electrons,) and the empty tank the positive (receptive to those electrons,) but most would think of it the other way around because of convention. It doesn't really matter.

When you open the valve in the pipe (press the switch in the circuit,) the water will naturally flow to the side with no water, as electrons will flow from the part of the battery with excess electrons, to the part lacking electrons. At some point, the tanks will equalize, just as the two parts of the battery will. This is when the difference in potential ceases to exist, and the battery is depleted.

Meanwhile, the water flowing through the pipe correlates to the amperage of an electrical circuit. A bigger pipe would represent a lower resistance circuit, and result in higher volume of water flow in a given time than a smaller pipe would.

I can't think of a good analogy for the atomizer coil, with water, but the impeller works, I guess. With more volume of water flowing, it will do more work, faster, just as more amperage through the atomizer will create more heat.

I also can't think of a way to apply the voltage amount to a water analogy very well. I guess if you thought of the tank with water being pressurized to some degree, that would represent the voltage. More pressure would represent more voltage, and cause the water to flow faster through a fixed-size pipe than a lower pressure would. Higher votage will cause more current through a given resistor than a lower voltage would.

With that idea, the amount of difference in the two tanks would be the amount of charge left in the battery.

I think that makes more sense to me, this time.

Anyway, you're quite correct about the benefit of higher voltage devices. I posted the results of a simple comparison of the math in a thread a while ago, here:

http://www.e-cigarette-forum.com/forum/new-members-forum/276590-3-7v-vs-5v-vaping.html#post5522901

Hope that helps.

 

[bstedh]

You are still doing the same amount of work i.e. Wattage.... So you are still using the same amount of energy out of your battery. You are just doing it differently.

This is why some times it is better to list battery capacity in Watt hours instead of amp hours. A lot of 12V car batteries do this.

Sent from my NookColor using Tapatalk

 

[Vego]

Yes, now pressurising the water to 'push' it faster makes much more sense to me for representing voltage. I can understand that. I think I got it.

Hey, thanks again for extrapolating on that, Lucent, and to bstedh for chiming in. I really appreciate it

 

[asdaq]

I just wish we were using the toilet for the analogy as we have the switch, the size of the pipes and two tanks reaching equilibrium. There's resistors and work too.

 

[LucentShadow]

No problem, Vego.

Now, if I understand bstedh's post correctly, then you don't gain reduced drain from the battery by lowering your current demand. I guess that actually makes sense to me, instinctively, even if it is a bummer. I've only recently started researching how batteries work, since I've taken up vaping, so I'm a bit fuzzy on how they work, exactly. I've always worked with electronics that ran on mains power sources.

It's kind of hard to compare in practice, I'd guess, because you just can't use a battery at a voltage other than it's own voltage without introducing some sort of regulator that will waste efficiency. I'd venture to say that most any two battery VV device will waste energy when it's operating at less than the voltage supplied by the batteries, so you're better off running at the higher potential, anyway. I'm not sure how the efficiency is on the one battery VV devices.

Also, from what I've read on batteries, running them with high current drain applications, like LR atties, can be detrimental to the chemistry in them, causing them less ability to take as many charges in the future. Seems to me that running higher voltage, less amperage, should help with that.