SuperNetGo Ninja,Go Ninja Go.. Premium Member join:2002-10-08 Hoffman Estates, IL |
SuperNet
Premium Member
2015-Jan-21 10:33 am
[Electrical] How many watts are in an amp?How many watts are in an amp? |
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rjackal Premium Member join:2002-07-09 Plymouth, MI |
rjackal
Premium Member
2015-Jan-21 10:41 am
Depends on the volts. |
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HarryH3 Premium Member join:2005-02-21 |
to SuperNet
There must be voltage AND current to have watts.
Current * voltage = watts |
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rjackal Premium Member join:2002-07-09 Plymouth, MI |
to SuperNet
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1 recommendation |
to SuperNet
said by SuperNet:How many watts are in an amp? How many seconds are in a yard? Different, independent units of measurement. |
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garys_2k Premium Member join:2004-05-07 Farmington, MI
1 recommendation |
to SuperNet
power = current X volts, true for DC and AC where there is little phase shift (resistive loads like incandescent lamps, heaters, etc.). |
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pende_tim Premium Member join:2004-01-04 Selbyville, DE
1 recommendation |
to SuperNet
For DC watts=amps*Volts For most AC the same as DC. In 99% of the cases with AC you can simply use volts*amps and get very close However if there is a significant powerfactor involved ( such as motor under light loads ) it is watts=volts*amps*powerfactor |
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SuperNetGo Ninja,Go Ninja Go.. Premium Member join:2002-10-08 Hoffman Estates, IL |
SuperNet
Premium Member
2015-Jan-21 10:54 am
What I am trying to do is figure out how many amps my family room is using. I have 6 T8 Bulbs for lights, 52in Sony LCD TV, Bathroom fan, very old stereo(1999) receiver,Wii, and a Bluray player. |
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to HarryH3
said by HarryH3:There must be voltage AND current to have watts. Current * voltage = watts Not necessarily. You can have Newtons and meters/sec A Watt is a unit of measurement for POWER (not necessarily electricity!!!). It is [1 Joule] / [1 second] An Ampere is a unit of measurement for electric current intensity. It is [1 Coulomb] / [1 second] The Volt is derived from the two... the potential difference between which a current of 1A generates 1W of power. |
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cowboyro |
to SuperNet
said by SuperNet:What I am trying to do is figure out how many amps my family room is using. I have 6 T8 Bulbs for lights, 52in Sony LCD TV, Bathroom fan, very old stereo(1999) receiver,Wii, and a Bluray player. 6-7A ballpark. |
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DarkLogixTexan and Proud Premium Member join:2008-10-23 Baytown, TX 1 edit |
to cowboyro
said by cowboyro:A Watt is a unit of measurement for POWER You sure its not a measue of work? w=mass*acceleration*distance |
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cowboyro
Premium Member
2015-Jan-21 11:01 am
said by DarkLogix:You sure its not a measue of work? w=mass*acceleration*mass Power=Work/time |
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SuperNetGo Ninja,Go Ninja Go.. Premium Member join:2002-10-08 Hoffman Estates, IL |
to cowboyro
said by cowboyro:said by SuperNet:What I am trying to do is figure out how many amps my family room is using. I have 6 T8 Bulbs for lights, 52in Sony LCD TV, Bathroom fan, very old stereo(1999) receiver,Wii, and a Bluray player. 6-7A ballpark. That many?? |
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unless he knows the wattage of the each device its a complete guess |
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DarkLogixTexan and Proud Premium Member join:2008-10-23 Baytown, TX |
to cowboyro
said by cowboyro:said by DarkLogix:You sure its not a measue of work? w=mass*acceleration*mass Power=Work/time Ok guess I forgot a bit. So power=MAD/T |
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to SuperNet
You can calculate it if you get the power consumption from every device. On the stickers, they'll note either the maximum amps used, or the wattage. If it gives Watts, then divide by 120 to get the Amps.
And note that those are maximums, and are usually higher than actual usage (except for the bulbs).
Make sure you get the wattage for every bulb too. Though the ballast may come into play on consumption. But probably doesn't matter unless you want to be super-accurate. |
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DarkLogixTexan and Proud Premium Member join:2008-10-23 Baytown, TX
3 recommendations |
DarkLogix
Premium Member
2015-Jan-21 11:10 am
Just get a kill-o-watt |
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1 recommendation |
Word. As long as you can route everything through one outlet. Or get your hands on a clamping ammeter. Turn everything on, open up your electrical panel, and clamp the ammeter around the hot wire for that circuit. |
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to DarkLogix
Technically no. You can drag a box on the floor at constant speed (no acceleration). Clearly there is mechanical work involved... |
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mackey Premium Member join:2007-08-20
3 recommendations |
mackey
Premium Member
2015-Jan-21 11:23 am
said by cowboyro:Technically no. You can drag a box on the floor at constant speed (no acceleration). Clearly there is mechanical work involved... Actually if it were along a frictionless surface then it would not require any work to maintain a constant speed once you got it moving. You actually are accelerating it when pushing a real life box at a constant speed; the acceleration is equal to the rolling resistance (friction). |
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SuperNetGo Ninja,Go Ninja Go.. Premium Member join:2002-10-08 Hoffman Estates, IL |
to cybersaga
said by cybersaga:You can calculate it if you get the power consumption from every device. On the stickers, they'll note either the maximum amps used, or the wattage. If it gives Watts, then divide by 120 to get the Amps.
And note that those are maximums, and are usually higher than actual usage (except for the bulbs).
Make sure you get the wattage for every bulb too. Though the ballast may come into play on consumption. But probably doesn't matter unless you want to be super-accurate. » www.homedepot.com/p/GE-1 ··· 68676700That is the ballast i bought, there are 3 of them. |
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to mackey
said by mackey:You actually are accelerating it when pushing a real life box at a constant speed Ummm.. nope. Someone skipped the classes... tsik tsik tsik... Acceleration is the rate of change of velocity. |
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Boooost to SuperNet
Anon
2015-Jan-21 12:03 pm
to SuperNet
said by SuperNet:What I am trying to do is figure out how many amps my family room is using. I have 6 T8 Bulbs for lights, 52in Sony LCD TV, Bathroom fan, very old stereo(1999) receiver,Wii, and a Bluray player. If you total up the wattages, take the total and divide by 120. That's the approximate number of amps. e.g., 800 Watts / 120 V = 6.7 Amps |
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DarkLogixTexan and Proud Premium Member join:2008-10-23 Baytown, TX |
to cowboyro
said by cowboyro:Technically no. You can drag a box on the floor at constant speed (no acceleration). Clearly there is mechanical work involved... You forget about friction and air resistance and the accel and deccel of your legs as you move them back and forth? |
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DarkLogix
1 recommendation |
to cowboyro
said by cowboyro:said by mackey:You actually are accelerating it when pushing a real life box at a constant speed Ummm.. nope. Someone skipped the classes... tsik tsik tsik... Acceleration is the rate of change of velocity. Force is F=MA and if you're pushing against a force (friction and air resistance) then you are applying a force if that force is divided by the mass the force is acting on then you have the acceleration. And A=D/T/T so F can also be expressed as F=MD/T^2 |
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DarkLogix |
to cybersaga
said by cybersaga:Word. As long as you can route everything through one outlet. Or get your hands on a clamping ammeter. Turn everything on, open up your electrical panel, and clamp the ammeter around the hot wire for that circuit. Just us it on one at a time and total them up. |
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DavesnothereChange is NOT Necessarily Progress Premium Member join:2009-06-15 Canada 3 edits
1 recommendation |
to cybersaga
said by cybersaga:Word. As long as you can route everything through one outlet. Or get your hands on a clamping ammeter. Turn everything on, open up your electrical panel, and clamp the ammeter around the hot wire for that circuit. A clamping ammeter is best used at the power panel, where it can measure the current of each hot wire, and then use that number as a basis to calculate the watts. A kill-o-watt (a brand of energy meter) can be used to figure out amps, or watts, or kilowatt-hours (the last one being the energy unit by which we are charged on our electric bills, and prob'ly the reason why this thread began). But be aware that consumer energy meters can be fooled, as can all other types of consumer meters and consequent calculations. If the load is not purely resistive (a traditional toaster or an incandescent light bulb are purely resistive), then the numbers which you measure will be off by enough to likely make the reading useless. Computers and today's consumer electronics equipment seem to be common examples of loads which are NOT purely resistive (IOW, they are at least partially inductive or capacitive), and readings of these, performed on commonly available consumer plug-in energy meters, will usually end up estimating a much lower consumption than what the professional energy meter at your power panel will count for your electricity bill. My own tests with a consumer plug-in energy meter have shown that my computer gear is estimated to use substantially less power than it really does. Compact flourescent light bulbs are not purely resistive and will be off too, in the same direction. Incandescent bulbs will use pretty much what they say (not much less, if your line voltage is on-spec), and so readings will be close to true on those. = = = = = P = E x I (Power in Watts = EMF in Volts x Current in Amperes) EMF is ElectroMotive Force, which is why voltage is represented by 'E' rather than 'V' in the above formula. I forget what 'I' stands for - possibly Induced Current ? = = = = = Energy in Kilowatt Hours = P (from the above formula) multiplied by the Time during which it has been drawn, and divided by 1000, as a Kilowatt is 1000 watts. This is what shows up on the energy meter at your power panel. |
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to SuperNet
If I live to be 200 I will never forget this one:
P=IE
Pie. Power (Watts) equals Current(Amps) times Energy (volts)
It's come in handy so many times I wish I could remember who taught me this!
My favorite pie is probably Pecan (since moving to TX) though when I lived in WI is was Rhubarb. Mmmm. |
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cowboyro
Premium Member
2015-Jan-21 12:56 pm
said by tomupnorth:Pie. Power (Watts) equals Current(Amps) times Energy (volts) You already forgot. Energy is expressed in Joules and/or derived units (such as watt-hour) |
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cowboyro |
to DarkLogix
said by DarkLogix:Force is F=MA and if you're pushing against a force (friction and air resistance) then you are applying a force if that force is divided by the mass the force is acting on then you have the acceleration. Dude... you are clearly mixing things... You have to specify that the force is actually the net force (sum of all force vectors) acting on an object - not just one of the forces. An object moving at constant velocity has no acceleration. PERIOD. No, it's not. It's expressed in units of [distance]/[time]^2 - as it is the change of velocity in respect to time (dv/dt) and the velocity is the change in position rate in respect to time. An object moving 4m in 2s does NOT have an acceleration of 1m/s^2 as you are implying. |
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