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Voltage VS Amps

This may have been answered before, but I have searched. Throughout all my reading of High Voltage devices, I have noticed that some have high amps and some don't. For instance a Van De Graff generator creates very nice HV but little amps. Well, to be to the point, my question is: What determines a charge having a large amount of amps or not?

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olddawg9 years ago
My question is a bit different. I am attempting to calculate my amp need for my garage in order to set up a 12v battery system. Using the formula; Watts / voltage = amps drawn, I have hit a question. If using a 12v battery system and converting to 110v, should I be using 12v or 110v in the formula?

Example; 100 watt light bulb / 110 volts = .91 amps. Nice!
But if I use the 12v; 100watt light / 12v = 8.3 amps.

I'm believing I would be drawing 8.3 amps from the battery to power this light.
Plus the converter draw...

Please tell me which is correct?
gmoon olddawg9 years ago
Both.

Your figures illustrate quite nicely the reciprocal relationship between volts and amps. For the same amount of power, insert a smaller voltage figure, and the amperage rises, and vice-versa.

For most devices, using a higher voltage also drives more current through. For this example, however, we'll say the bulbs are appropriate -- there are certainly 100W 12V lights, so we'll assume the bulbs in each equation are correctly voltage-rated (the light bulbs were 12V and 110V, respectively.)

So, both examples are correct. In each case the overall power consumption is the same.

But you're probably wondering about battery capacity, right? The 12V example will be closest to the estimated draw on the batteries. As you know, that doesn't include the power required to run the inverter...the 12V calculation assumes a "lossless" conversion.

(ever thought about using 12V lights?)
gmoon gmoon9 years ago
oooops. I see you've since added a separate topic on this discussion, and others have responded already.... Just disregard.
olddawg gmoon9 years ago
I will never disregard anyone willing to me. Thank you! I have considered a 12v lighting system, just not sure if I should truly go that way. I have items (vacuum, power tools, garage type items) that will be used. If I need to run them with 110v, then I 'assume' it is preferable to run them all this way. So far a 10% converter loss doesn't seem that bad... Wiring for both, even if possible is beyond me right now.
yourcat olddawg9 years ago
What do you want to light?
gmoon olddawg9 years ago
Sure, just a thought about 12V lighting. There's a lot of RV and boat lighting available. You've been getting good response from the other thread--I didn't want to split efforts, hence the "just disregard."
tech-king9 years ago
its also worth adding that volts jolt, but milliamps will kill you. a cattle fence transformer, for instance, will give upwards of 1kv. a lethal current is 50ma. 20ma causes permanent damage, and you fell pain at 5 ma.
PKM9 years ago
This is partly, I believe, due to a slight misunderstanding about what current actually is. Current is not a property of a power supply the way voltage is (actually that's a slight lie) - the current of your mains supply depends what you plug into it. A current is, in a traditional circuit, defined by the supply voltage and resistance or power consumption of the circuit. 3000W draw / 240V = 12.5 amps, 60W draw / 240V = 0.25 amps. A power supply may have a maximum current it can provide before the voltage drops or the magic smoke comes out, but within this limit it will provide any amount of current that may be drawn from it.

The reason I say "is a slight lie" is that actually current, not voltage, defines a Van De Graaf generator and similar electrostatic devices- they are actually a class called "constant current source". In this case, the voltage reached by the charge is the voltage at which the leakage current to the air is equal to the supply current. If a lower-resistance circuit is provided the dome voltage will drop accordingly.

Current * voltage is also limited by power- a VDG using 50 watts clearly won't be able to provide a high current at 100,000V, but a high-powered NST or flyback at several hundred watts will provide more current. Also note that a lot of AC high voltage sources (like ignition coils) supply intermittent current rather than constant, which increases the instantaneous current that can be supplied from a constant power consumption (rather like the millions of amps that flow for a tiny fraction of a second in a lightning strike from a much smaller constant charge build up)

I hope this answers some of your questions- you will find most of the answers you are looking for in a physics textbook, this is just a very brief summary.
westfw10 years ago
From a physics standpoint, voltage is the energy of each electron, while current is related to the number of electrons involved. Something like a van de Graff generator produces high energy electrons, but it doesn't produce very many of them, so the maximum current available is very low (in fact, in electrostatics, it seems that the quantities of electrons are so low that traditional electronics concepts like ohm's laws lose whatever intuitiveness they might have had at one time, and you end up dealing in other concepts like the actual amount of charge. "Joules" replaces "Watts" as the relevant measurement, and all "normal" resistances are "approximately zero" compared to the voltages present...)
Cryptonat (author) 10 years ago
Sorry to double post but I had another sudden question. I'm sorta grasping that electricity page Curve12 posted (thanks Curve). But anyways, what I am seeing is that you increase the so called flow to raise amps, which is an effect I am not wishing to utilize. How do you reduce flow to reduce amps? Is it just as simple as creating resistance? If you increase resistance does this also lower the volts? I should have paid more attention in physics. :P
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