NationStates Jolt Archive

Is there an actual relationship between voltage and amperage? I ask this because part of my job as a DJ is to check the signal towers, see how much electricity they are using etcetera. I don't know much about electricity other than the basics like don't stick you fingers in the light socket, and electrons and yadda yadda... but I have never been sure on that question. So now I turn to you NSers for the answer!

The amperage is the actual current, that is the flow of electrons. Voltage measures electric potential; there is no such flow if there isn't a potential difference.

Ask me if you need clarification, or something more specfic ... you can also check wikipedia, though the physics articles are usually difficult to understand if you don't have some knowledge of it.

Is there an actual relationship between voltage and amperage?

Q=W/V

Is there an actual relationship between voltage and amperage?

Civil partners.

Civil partners.

I knew it!

While we are at it... is there an actual relationship to wattage in the whole thing?

While we are at it... is there an actual relationship to wattage in the whole thing?

W=QV

Is there an actual relationship between voltage and amperage? I ask this because part of my job as a DJ is to check the signal towers, see how much electricity they are using etcetera. I don't know much about electricity other than the basics like don't stick you fingers in the light socket, and electrons and yadda yadda... but I have never been sure on that question. So now I turn to you NSers for the answer!
Being an ex-physics student, I could nitpick all day, but I must resist the temptation.

The relationship between potential difference (Voltage) and current (Amperage) is V=IR, where V is Potential Difference (Volts), I is Current (Amps) and R is Resistance (Ohms).

If you want to see how much electricity you're using you should do W=IV, where W is the rating of the equipment you're using in Watts.

Civil partners.

Last I heard, it was friends with benefits ...

I knew it!

While we are at it... is there an actual relationship to wattage in the whole thing?
The Wattage is the amount of energy you're using.:)

I knew it!

While we are at it... is there an actual relationship to wattage in the whole thing?

Yeah; wattage is power, which is the rate of work done by the electric current. Watts = Voltage x Current, as so noted by others in this thread.

Being an ex-physics student, I could nitpick all day, but I must resist the temptation.

The relationship between potential difference (Voltage) and current (Amperage) is V=IR, where V is Potential Difference (Volts), I is Current (Amps) and R is Resistance (Ohms).

If you want to see how much electricity you're using you should do W=IV, where W is the rating of the equipment you're using in Watts.

I studied Industrial Art (website design, archetectural drawing, stuff like that) history, and business.

The only way electricity figures into those is... the bill!

How muchdo you know and how much do you need to know? I know Standard Grade physics so I could tell you V=IR which means the voltage is equal to the amps times by the resistance so if you know two of the three you can figure out the last one.

Honestly Undogs, i was just curious about the relationship between the two. (or three). Because honestly when I'm taking the readings on the towers sometimes I have no idea what the hell that means.

Just as it has been said, Voltage is directly related to Amperage.

V=I.R

Voltage (d.d.p) is the cause, resistance (ohmn) makes it harder, and current is the consequence (ampere).

If you increase the voltage (like from 110 to 220) and dont increase the resistance, it will directly increase the amperage, which is the number of coulombs ("electrons") by second.

For example, if you take an equipment made for 110 volts and plug it to a 220 volts in the light socket, it will receive twice as much amperage, which will probabily fry its circuits. You would need an adapter to use it, or set the equipment to 220 volts.

Honestly Undogs, i was just curious about the relationship between the two. (or three). Because honestly when I'm taking the readings on the towers sometimes I have no idea what the hell that means.
Basically, the relationship is a defined constant (sometimes partially imaginary!) for each body. We call it the Resistance or Impedance.

I studied Industrial Art (website design, archetectural drawing, stuff like that) history, and business.

The only way electricity figures into those is... the bill!
Damn paying for electricity.:p

*laughs at the 'merkins and their 110v domestic supply*

240, baby, 240.

*laughs at the 'merkins and their 110v domestic supply*

240, baby, 240.
Is that 240dc, 240ac, or 240rms?

Just out of interest...;)

Is that 240dc, 240ac, or 240rms?

Just out of interest...;)

AC, but technically anything between 215-250V in the real world.

Where do you get 240?

That can easily kill someone lol.

Being an ex-physics student, I could nitpick all day, but I must resist the temptation.

The relationship between potential difference (Voltage) and current (Amperage) is V=IR, where V is Potential Difference (Volts), I is Current (Amps) and R is Resistance (Ohms).

If you want to see how much electricity you're using you should do W=IV, where W is the rating of the equipment you're using in Watts.
Is that how they teach it now? We always learned E=IR and P=IE, where E meant electromotive force and P was just power -- you supply the units.

Where do you get 240?

That can easily kill someone lol.
Most of the world that uses meters also uses 220-240 VAC. Edison and Tesla had it out about whether AC or DC was the best means to distribute power and Tesla won.

Where do you get 240?

That can easily kill someone lol.
When you increase the voltage you can decrease the amps and still get the same watts out of it. So it isn't more dangerous as it balances out when you increase one but decrease the other.

The difference as far as I know is that transfer is more efficient with more vatts and less amps so you can use thinner and therefore cheaper wires.

Here is linky (http://science.howstuffworks.com/question501.htm) to splain clearily ;)

Here is linky (http://science.howstuffworks.com/question501.htm) to splain clearily ;)

now that is what was needed.

i think this thread demonstrates perfectly why i could never get to grips with physics at school. ask a physicist a simple question and all they do is rearrange it and put it into a formula, like formulas actually mean anything to the person asking the question. my gcse physics teacher was a cambridge graduate with a doctorate in some mad physics thing, but he couldn't teach for shit. it's a testament to how much of a pisstake gcses are that i got an A.

Where do you get 240?

That can easily kill someone lol.

UK - volts give you jolts, but its the amps that kill you.

Is that how they teach it now? We always learned E=IR and P=IE, where E meant electromotive force and P was just power -- you supply the units.
I have a British education.;)

Out of interest, do you guys do physics in Imperial or Metric?

I have a British education.;)

Out of interest, do you guys do physics in Imperial or Metric?

Both. I think the grade school students are largely taught in metric. College students too -- by and large. Engineering students typically get a larger dose of customary units because they are more likely to deal with them in practice.

My field is high power microwaves and that's just about all metric by default.

AC, but technically anything between 215-250V in the real world.

That is the peak voltage isn't it?

When you increase the voltage you can decrease the amps and still get the same watts out of it. So it isn't more dangerous as it balances out when you increase one but decrease the other.

The difference as far as I know is that transfer is more efficient with more vatts and less amps so you can use thinner and therefore cheaper wires.

You try to transfer with more amps. There is a phenomenon called the 'skin effect'. What happens is that if you transfer, on average, over 500 volts on a copper power line you end up having the electric flow only on a small, closer to surface section of the wire. This massively increases the resistance and ends up causing power losses to rise.