I read a article in a paper,that says a tube light consumes 75 W. Please explain?
830736
It means a tube-light consumes 75 joules of energy per second.But usually a tube light conumes still less power. \n           1watt(W)=1joule/second.\nThe power of the tube... is only dependent on the materials used inside it and it does not depend on voltage applied,current supplied etc.Whatever may be the current supply,the power remains the same.only reqirement is we have to apply adequate voltage and current depending on the power of the tube light.
507766
depends on the type of bulb, For incadesant bulbs thee is NO extra power required, they get hot, they glow. For florescent tubes it can be quite long, certainly of the orer of hours. once up and running they use very little
416072
It depends on the wavelength of the light and its energy.
958515
You can try the following web sites:\n\nhttp://www.glenbrook.k12.il.us/GBSSCI/PHYS/CLASS/light/u12l1e.html\n\nhttp://www.uwgb.edu/dutchs/petrolgy/genlight.htm
20622
You'll need to know what the electricity costs per kilowatt hour. They you need to know how many watts each bulb uses per hour. There are many different kinds of flood lights, the standard incandescent fllod light, mercury vapor, halogen, high pressure sodium, low pressure sodium. Each of these has an advantage in light and uses energy differently in producing light. You can determine the light output by the number of lumens the bulb produces.
948698
Power = E x I, where "E" is the voltage accelerating the electrons toward the screen and "I" is the electron current hitting the screen.\n\nIf the accelerating voltage is typically regulated at 20,000 volts or so, and a typical electron beam current might be 1 to 10 milliamperes (0.001 to 0.010 amperes), then the power would be in the range of 20 to 200 watts. Except for a very small amount that is converted to visible light, this power appears as heat dissipated in the phosphor face plate.\n\nBear in mind that the image on the picture tube screen is formed not only by scanning (a raster) but also by modulating the current of the electrons impacting the screen. In general, more current means more light until the phosphor saturates and cannot produce any more light.\n\nA lot of the power in the electron beam of a color television is wasted because some of the electrons never reach the phosphorescent screen. Instead they are intercepted by a shadow mask or grill that is only there to make sure each of three electron guns illuminates its own color and not its two adjacent neighboring colors.
1132158
Those references are fine, but you can both lose amps and volts.  I understnd that if you have a constant current source driving the wire then you don't lose amps, but that is not the case.\n\nYOu normally have a good voltage source like a utility feed driving the wire and load (say a light bulb).  If you put the bulb at the end of 2000 ft of wire, then yes you will have lower voltage at the bulb compared to putting it right at the source.  But all the resistance in teh wire and bulb add up - so you current is lower too!\n\nFor example - 120 VAC driving a load of 10 ohms would give 12 A.  Now, say you have 2000 ft of wire that has 2 ohms of resistance.  The current would now by only 10 A.  The voltage at the load would be (10/12)*120 VAC = 100 VAC.  SO you lose both amps and volts - don't know where these other guys got their EE degrees but I hope it wasn't anywhere I went.
114097
casey's last sentence is wrong:\nthe more resistance the slower the flow rate \npressure is determined by voltage\n\njdub says:\n"for example...the higher the voltage, the greater the resistance, and the higher the current, the lower the resistance...etc.."\nshould say_ that a a higher voltage will need a greater resistance to oppose flow in order to keep the current the same, or if you want to raise current you would either raise voltage across given resistance, or lower the resistance at a given voltage.\n\nbrmick says:\nthat air is a better conductor than concrete,\nThis largely depends on the frequency of the source, gases are not generally considered conductive, while concrete which is largely water is a moderate conductor especially if wet or wetted.\nDO this take two light bulbs and wire one of each lamps leads to the other then tie it to its intended source (series) using the remaining leads. check the voltage across the source and between the lamps. Assuming you have two equal size lamps you have {nearly-ignoring impedance} doubled the resistance. and so the current and wattage are cut.
272513
rate; how is energy "radiated" from the following- a heated piece of platinum metal; a fluorescent light; an incadescent light; the human body
739842
To work this out you must find out the power rating (in watts). Most of the time this information is written on the appliance. For example a standard electric kettle may use 1300 watts. \n\nElectricity is sold in units of 1000 watts per hour. For example if you use 10 100w light bulbs for an hour it will cost you 1000watts or 1 unit of Electricity\n\nYou will have to find your power rating on your appliance then compare an hour of usage of that appliance with the price of 1 unit of power (1 Kilowatt per hour)
275900
Thomas Edison explored this very same issue. He experimented many different components and found that a steel wire in a vacuum lasted longest.\n\nSo yes the light would work, but for a short period of time when the wire, exposed to air, would oxidize then burn up.
393843
b/c a constant amt. of electricity is going into it, and it will only get brighter if you add more electricity to it, or get a brighter bulb!
798595
The light-bulb has a tungsten filament, which has electrical resistance.  The voltage applied across the resistance causes a current to flow.  The value of the current is: I = V/R, where I is the current in amps, V is the voltage in volts, and R is the resistance in ohms.  The current flowing through the resistance causes it to heat up--it dissipates power.  The power is: P = I*V, or P = R*I^2, where the power is in watts.  MOST of that power produces heat; in an incandescent bulb only a little goes into light.  Power, incidentally, is energy per unit time.  Watts are Joules/sec, where the joule is the SI standard unit of energy.\n\nAs for the motors, connect them in parallel.  They will each receive the same voltage, and each will draw the amount of current it needs.
1040071
100 W == 100 joules per second, or 86.4e7 joules (roughly) per day.\n\nAssume all the energy here is transferred to heat (not true, as the bulb also gives out light)\n\nYou would need to know the heat capacity of the material in the closet to know exactly, but for air it's approx 1 joule per gram per deg K (http://en.wikipedia.org/wiki/Specific_heat_capacity)\n\nA 6 x 8 x 9 foot closet has volume 432 cubic feet, or ~ 12000 cubic decimetres ~ 500 moles ~ 14000 g which makes an increase of ~ 600 degrees K.\n\nSeems a bit high, probably made some errors somewhere....\n\nBut to compare this to a 14W lamp all you need to observe that this is 14 joules per second compared to 100, so the heat given out is about a factor of a seventh as much, so the increase would be around 100 K
131801
When electricity is given, the bulb glows because of the power dissipated from the tungsten. This is because of the heat produced when current flows through the filament. So when the current is switched off, no current flows through the filament causing the heat not to be created. So the bulb stops glowing and light energy is not given out.\n\nSince no electric energy is given, no light energy is given out. The already produced light energy (before shutting off) i.e. the photons, are absorbed by surrounding objects so that it no longer produces light.\n\nThe photons are the smallest unit of light like cells in biology. Photons have energy in them. These energies are absorbed by surroundings.
35547
By definition, a watt is "the SI unit of power. Power is the rate at which work is done, or (equivalently) the rate at which energy is expended."\n\nA 50-watt bulb is expending energy at 1/2 the rate of a 100-watt bulb.\n\nSo - if both are shining (not "shinning") for the same amount of time, the 50-watt bulb would use 1/2 the energy of the 100-watt bulb.
336571
The ad is little misleading. It is true that light sources are evaluated and quantified in terms of candles, but there are two different terms and they mean different things. The first term is "candela" which can also be referred to as "candlepower" and it refers to the amount of light given out by a source IN ONE PARTICULAR DIRECTION actually decribed at one angle. So to know how bright your source was in candlepower, you would have to know something about its beamspread. But- making a bunch of assumptions here- IF the light was more or less shaped as a floodlight (all the light being emitted in one direction like a wide spot), and if it had 1,000,000 candlepower at 0°, then your light could give out roughly 150 times as much light as a 100-watt incandescent yard light like you'd get at Home Depot.\nSecond term: Lumen. A lumen is the term for the amount of light coming out of the source in ANY direction. The lumen is also tied the the candle so there's a relationship between the candela and the lumen. But the thing to remember is that it's non-directional. Knowing how many lumens a source has doesn't help you to know what it does because it could be very intense like a spotlight or very diffuse like an ordinary light bulb - both can give out the same amount of total light.\nSo if your emergency light has 1,000,000 lumens, then it would have 100 times as much light as a 500-watt incandescent, or 300 times as much as a standard fluorescent tube, or 40 times as much as a tall streetlight.\nNow- I have to say that both of these numbers are really really really high. I mean we're talking tanning bed here. I would be very skeptical of this myself, and I would ask some hard questions of the seller. In lighting, as in life, if it seems too good to be true it probably is!