Why are CFL wattage ratings wrong as the light may be rated for xx watts but when checked with a wattage meter the actual usage is significantly higher?
Usually a CFL has two ratings. The first is wattage and the second is in mA I suspect the wattage is only the wattage of the bulb itself and the second is the actual current draw of both the light bulb and the ballast. They are not compatible. If you figure the mA and multiply times the voltage you will obtain close to the actual wattage of the combination of the bulb and the ballast. I have a 100 watt Feit BPESL25T which indicates it draws 25 Watts when it actually draws 47.5 Watts. Very close to the mA (400 x 120 VAC =48 Watts) indicated on the base. Almost all CFLs I've tested with my very accurate Fluke RMS meter draw twice the wattage they indicate. Not such a good deal.
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Actually the NEC says that the load on a branch circuit can only be80% of the breaker size since Some appliances use a more energy tostart up. so to be legal, you could only have a 1440W load on a 15amp circuit, and 2880W with a 240 v 15 amp breaker. Another Answer You really should be asking how… you can estimate ENERGY (not'power') consumption, based upon a wattage rating. The term,'wattage', is a slang term for 'power', which is expressed inwatts. Power is simply a rate, the rate at which energy is beingconsumed, and energy (for the purpose of billing, at least) isexpressed in kilowatt hours.. So if you know the power ('wattage') of your load, then you mustensure that it is expressed in kilowatts and, then, multiply it bythe number of hours that the load is operating. This will then tellyou how much energy has been consumed over that period, expressedin kilowatt hours.. ( Full Answer )
Yes, Of Course The lifetime depends on the wattage. The wattage is the amount of Power the light bulb uses, and the life time depends on the power. Thank you
Answer . \nYes it should, however most flourescent ballast can serve different wattages check the label on the ballast case.
\n. \n Answer \n. \nDon't - \nYou replace the wires just as you would for the original wiring but you must use a lower guage wire size - meaning a thicker wire.\nThe AC cord for lamps in the past 30 years was at least good for 500 watts.\n. \n. \nDon't forget that the wattage rating my not …be limited by electrical considerations but by thermal limits. A larger bulb may melt or burn the fixture itself, not just the socket. ( Full Answer )
Answer . Power = Voltage x Current P=V.I Power (in Watts) = 110V x 8.70A = 957W (Appx. 1kW). - Neeraj Sharma
The higher wattage rating means that your computer will run more efficiently with more components (hard drives, DVD burners, high end video and sound cards etc.) without robbing power from the processor and mother board.
\nWattage = voltage x amperage. Every appliance in North America is built to work at 120 V, so you have the voltage. The amperage rating is probably written in the microwaves manual if the wattage is not. Remember the amperage you put on a circuit can't be more that 80% of what the wiring is rate…d for in the branch circuit. ( Full Answer )
How is the wattage of a circuit divided amongst resistors in series of varying ohmic and wattage ratings?
The voltage supplying the circuit will be divided across the series resistors in proportion to their resistance. The wattage of the resistors has no effect on the distribution, but if you put an under rated resistor in the circuit, it will fail.. For example, if you have a 10v source, and a 1 ohm r…esistor in series with a 3 ohm resistor, the 1 ohm resistor, being only a quarter of the total resistance, will see a quarter of the voltage, or 2.5 volts. The other 7.5 volts will seen across the 3 ohm resistor.. The total power consumed by the circuit is given by P = VI or V 2 /R or I 2 R, so for this circuit, the resistors will consume 25 watts (current is 10/4 = 2.5 amps according to Ohms Law), and 10 x 12.5 gives 25 watts.. Hope that helps. ItAintMe ( Full Answer )
The wattage is stamped on the bulb. The rating might be hard toread, because of the heat dimming the stamping over time.
Watts can be checked and measured by using an instrument called amultimeter. The multimeter will measure different details of anelectrical circuit.
You should not use a power adapter with a higher voltage output than the input rating on the computer. It can cause damage to the computer and the battery. You can use a lower voltage. If you do and you are using your computer, it might not charge but it will prolong the time you have before it r…uns out. If you do not use it, then it will charge but slower than with the correctly rated power adapter. ( Full Answer )
Fluorescent lamps can output a total of 100 lumens per watt.Fluorescent lamps were invented by Peter Cooper Hewitt in 1901.
The wattage for your kitchen table light will depend on severalfactors. The size of the room, the type of lighting, and theceiling height will all need to be considered. You will probablyneed between 200 and 400 watts of light in your kitchen.
How do you arrive at choke losses in street light fittings what is total wattage what is ballast wattage?
Assuming the question is about simple reactive control gear, e.g. a choke or inductor in series with the discharge lamp, then the following answer applies. In an ideal situation, an ideal inductor, that is an inductor exhibiting pure inductance by virtue of its construction and the use of wire wit…h no resistance, then the losses would be zero and the power consumed by the circuit would be V.I.Cos Phi, where Phi is the phase angle between the voltage and current. In the real world electronic components are rarely ever perfect, and although a component may be called an inductor or choke, it will possess some parasitic resistance and possibly some stray capacitance too. The main source of the parasitic resistance is due to the resistance of the wire used to wind it, the diameter and length of the wire required to obtain a particular inductance and current rating and any eddy-current losses in an iron core if one is used. In the case of street lighting control gear designed to function at 50 or 60Hz, laminated iron cores are almost always used to obtain the relatively large inductances required. (At 240v R.M.S., 50Hz approximately 1 Henry of inductance is required to regulate a 20 Watt discharge lamp and 100mH for 200Watt lamp and 10mH for a 2Kw lamp, particular lamp types such as fluorescent, high pressure sodium or metal halide of the same power rating requiring slightly differing inductance values dependent on their discharge voltages at their specified operating currents.) The laminated iron cores are often constructed with an air gap to control the current value at which the core saturates. In a real choke power is wasted in I squared R losses in the parasitic resistance of the wire, which appears as a low value, (hundreds of ohms for low current, high inductance chokes down to a an ohm or less for high current low inductance chokes used for large H.I.D. lamps) in series with the actual reactive inductance obtained by the number of turns, thickness of the wire and the nature of the core. Eddy current losses appear as a parallel resistance across the series combination of the actual inductance and resistance of the wire. Both of these resistive components add a small in-phase current to the large lagging current of the inductive component. (An ideal inductor exhibits a pure quadrature lagging current.) The series resistive component due to the wire is easy to measure and can simply be measured with a multi-meter on the appropriate resistance range for relatively high inductance chokes used with lamps up to about 400w rating. A four wire milli-ohm meter may be needed to measure the lower resistances of "larger" , lower inductance chokes with fewer turns of thicker wire, used with lamps rated above 400w. Simply measuring the operating current, squaring it and multiplying this by the previously measured resistance will give the loss in the choke due to its wire resistance, but this will not include any eddy current losses in the core. Another method would be to connect the choke directly across a low source impedance A.C. voltage of the intended operating frequency. Generally most series regulating discharge lamp chokes can be connected directly across the supply voltage they are intended to be used from, (lamp shorted out), although some may saturate and in this case the voltage would need to be reduced. Under these conditions one could use a low value shunt or a suitably terminated current transformer in series with the choke under test to obtain the current waveform on an oscilloscope. The voltage waveform across the choke could also be displayed on the second channel of the oscilloscope to give both the amplitudes and phase angle between the voltage an current. Rembering to take into account the ratios of the current and voltage transformers used and to convert the peak values to R.M.S., the in-phase component should be able to be extracted by resolving the phase ange and magnitudes from the oscilloscope into two orthogonal components. The magnitude of this in-phase component of choke current can be used to calculate all the lossy resistive components, Rt, of a real choke and thus using "Iop" squared "Rt" to get the trus losses. ("Iop", being the operating current through the choke when used with a fully run-up lamp in series.) Some notes, low inductance, high current chokes wound on large cores with few turns of thick wire and used for H.I.D. lamps with power ratings of aroud 1Kw or more are a lot less lossy than high inductance, low current chokes used for small fluorescent tubes which are wound with many thousands of turns of fine wire with high resistance. High frequency control gear, such as that used in electronic compact fluorescent lamps. operating at frequencies of around 45-50KHz make use of the fact that inductive reactence = 2pifL, and by increasing the frequency, f, the inductance, L, can be reduced by an inverse amount and since the inductance, L, is roughly proportional to the number of turns, these can be reduced requiring a lot less wire and thus possessing a lot less resistance and thus are a lot less lossy. The losses in the rectifier and oscillator used to raise the frequency are more than compensated for by the large increase in inductance. Measuring omic "I squared R" losses in control gear such as leak transformers and "constant wattage", CWA gear is more demanding but not impossible. ( Full Answer )
In a PCB its very hard to find out. the best way is to look at a component catalogue and check there.
How does a 3-phase meter handle using on 220 volt single phase and does the meter measure the wattage usage accurately?
If you mean 220 single phase as in two hots and a neutral 110V-0-110V (in the US this would be 240V 120V-0-120V), then the answer is most 3-phase meters cannot be used.. A few meters can be setup for either configuration, check the manual for the specific one you want to use.. If you mean 220V sin…gle phase with just one hot wire and a neutral, then a larger percentage of 3-phase meters can be used, and will meter accurately. But again, the meter has have this as one of the setup choices. Check the manual.. One possibility would be if you have a meter that can be setup to use 3 CT's and tell it that it is hooked to a wye system with neutral. If it sees current on just one phase to neutral, this is a valid condition, since the system could be feeding unbalanced loads (remember the meter thinks this is a 3-phase system). This would only work if the meters wiring diagram calls for CT's on all 3 phases. Many meters call for only two CT's, calculating the 3rd leg by using the two known values. This type of setup would not meter accurately for your single phase, I believe. And this will not work on a 110V-0-110V system, because a 3-phase meter expects 120 degrees between phases, not 180.. Oh, I didn't mention the case where you might have two hots and no neutral. The 3-phase meter would have to be set up for a delta connection as far as the meter was concerned, since there is no neutral. It could work, not sure on this one. ( Full Answer )
The wattage rating of a bulb is in direct relation to its energyusage per unit time. A 100 watt bulb consumes twice as much energyas a 50 watt bulb over the same period. Keep in mind that thelatest generation of energy saving bulbs (Compact Fluorescent Lamp)can be labeled both how much energy they a…ctually consume, and alsoby how much light they produce as compared to an incandescent bulb.For instance, a CFL rated at 60 watts equivalent consumes only 14watts. In other words, it only takes 14 watts from a CFL to producethe light equal to a 60 watt incandescent bulb. ( Full Answer )
I have 14 recessed lights in your tv room what wattage should you use it is quite bright with 65 watts?
Use 40 watt. You do not have to use all the lights just because they are there. Unscrew every other one to bring down the brightness.
it would depend on what brand and what size sub it is i wouldrecommend 250 watts
The wattage rating of a light bulb is a measure of how much power is dissipated to create the light and heat that radiates from the bulb. Higher wattage means more power is used and that means more energy is radiated as light and heat.
The short answer: they are identical: 50VA (through a resistive load) = 50Watts The long answer: Power can be divided into real power and reactive power. Real power is used to do real work. reactive power is stored and released in reactive elements (capacitors and reactors). A purely resistor is ha…rd to get, having absolutely no reactance or capacitance. Because of this, watts defines the amount of real power available to do work; volt-amps defines the maximum current drawn at a specified voltage. ( Full Answer )
Practically its 15 watts. One can see the power ratings on the bulb package also.
A higher wattage HPS bulb may work with a ballast that is rated for a lower wattage bulb , but may appear to be dim and will not produce the rated light output. It is best to match ballast and bulb accordingly.
Some light fixture are configured for multi wattage bulbs but as a general rule of thumb only replace a bulb with the fixtures recommended bulb other wise there are serious safety issues including bulb explosion and fire.
That depends on how much resistance is called for through the engineering, there is tiny resistance (mf) microfared on up huge resistance seen on electrical poles. Your circuit engineering formula dictates the needed resistance or it will just burn or not flow at all, you cannot answer the question …without first asking a question..How much resistance do you need. They can be produce for virtually any tolerance. ( Full Answer )
No, each manufacturer has their own method, use the specs. for a ballpark figure. You can't hear the difference between 100 watts and 120 watts.
No. All light from light bulbs (incandescent) are equally bright. Higher wattage bulbs simply produce a higher quantity of light measured in lumens.
The manufacturers of light bulbs want to give their customers light. This requires the use of energy (Watts) which cost money for their use. Manufacturers design their light sources so that users get as much light as possible from the bulb for a given wattage while maintaining a reasonable lifetime …(otherwise we would just get flash bulbs). ( Full Answer )
PSU wattage is determined before the unit arrives on the shelf at the store. Wattage rating is easy to understand. Your power supply will provide 600w, but the power being drawn from the wall might be 680w. It's important to understand that more electricity enters the PSU than what comes out to you…r components. In layman's terms: More power in, less power out. The higher the rating, the more efficient that PSU is. These ratings can also be associated with other electrical units such as light bulbs. ( Full Answer )
So to replace a traditional 60 -watt bulb, buy a 15 -watt CFL: 60 -watt incandescent / 4 = 15 watts . Note: Some brands of 60 -watt equivalent CFLs still do not seem to give off as muchlight as a 60watt incandescent bulb.
60 is a standard house bulb; 40 for a dull one, 100 for a bright one
The rating is about 1500W. This is for both the input and theoutput. Output voltage is usually 2,000 volts. Divide watts byinput volts to get input current. And divide watts by outputvoltage to get output current. -Joe
What is the minimum wattage rating for a motor that is required to lift a 1000 N load a vertical distance of 4 meters in 2 seconds or less?
1 watt = 1 joule/sec; 1 joule = 1 N*m (Newton * meter) 1000 N * 4 m / 2 sec = 2000 N*M/sec = 2000 watts But this is just the power to lift the load and does not take into account any friction or the power required to create the magnetic field in the windings.
Resistance wise, the rated wattages are immaterial. When calculating the amount of power the two of them would be able to dissipate, (or the maximum current) you would have to use V=IR and P=VI to work out which resistor would run into it's wattage rating first.
It stated the amount of energy that the bulb will consume when the bulb is in operation.
Reasonobaly low. maximum 7 for Ds. just fiddle around with differend bulbs and batteries unless the battery is lead-acid
Yes, and it's proportional. V * I = P Where V is the voltage in Volts, I is the current in Amperes, and P is the power in Watts. So we get: I = P / V For example, with a 240 Volt supply, a 12 Watt lamp would draw: 12/240 1/20 0.05 Amperes, or 50 milliamps
each resistance drops the same current so the wattage disipated would be 2times the single one.
The wattage will depend upon the type of bulb and the number of lights on the string. It will tell you on the package what each bulb requires in wattage, so you only need to multiply that by the number of bulbs in the string. so, number of bulbs x wattage requirement for each bulb = wattage requirem…ent. ( Full Answer )
I am not sure how you would accomplish that in a home environment? You would have to increase the Voltage somehow. CFL bulbs have electronic circuitry in their bases. That circuitry would be more sensitive to Voltage changes than a simple filament bulb. You could expect a shorter 'Life' of the bulb …at a higher voltage. You can exceed the equivalent light output, but not the actual maximum wattage for a fixture. If the fixture is rated for 75 watts maximum, you can install a 23 watt CFL, even though the "equivalent" light output is equal to a 100 watt incandescent. The light fixtures are rated based on their ability to dissipate heat or the power handling capacity you will not exceed either as long as the actual wattage is below the rated capacity. ( Full Answer )
No, wattage is the product of amps times volts. You are charged by a utility company for the amount of watts that are used in a hour. These watts are totaled up over a month and multiplied by a specific kw/h (kilowatt per hour) that is set by the utility company. In my area I am charged .08 cents fo…r every kilowatt (1000) watts used per hour. ( Full Answer )
A high wattage light bulb heat up faster then a lower power light bulb, Also since a low watt light bulb is producing less heat then a high wattage light bulb. More heat, faster the water will evaporate.
How does a 3 phase meter handle using on 220 volt single phase and does the meter measure the wattage usage accurately?
A 3-phase meter measures the sum of the power in the three phases, so if power is drawn on one phase only it will still be correctly measured.
The typical wattage for closet lights depends on the person. Some want their closet lights to not be very bright, so they will choose something that has 20-30 watts. Some other people prefer something a little brighter, so they will choose 50 watts.
No. Just make sure that the ballast you use is rated for at least as many watts as the lighting you plan to operate with it.
Answer #1: 800 watt is the wattage in that case. That refers to how much power a device uses. ==== Answer #2: When you see an ad for an "800-watt microwave oven", that number is invariably the "cooking power" ... the 2.5 GHz RF power that'sgenerated by the magnetron and radiated into the cha…mber to heat themeatloaf. Your question suggests that you understand that the appliance usesmore power than that from the outlet while it's cooking. The relationship between cooking power and total operating power isnot necessarily the same for all brands, sizes, and styles. We'rereally talking about efficiency here ... the fraction of the input power that'sultimately presented to the meatloaf ... and that can certainly vary accordingto the design of the machine. The microwave oven that hangs from the bottom of the cabinet overmy stove is advertised as "1200 watts" of cooking power, and thelittle plate on the back says that it needs 1.58 KW of electric power tooperate. That means that it's nominally 76% efficient, and it uses 1.32 asmuch power as it delivers to the cooking chamber. There's no way to know whether your 800-watt unit is similar. IF itis, then it would need about 1,050 watts from the wall while it's running... about 8.8 amps on a nominal 120VAC circuit. That's not a horrendous intermittent load for a modern kitchenappliance. Still, if your house was wired some years ago, AND you have somelights on AND the refrigerator is running AND you're making toast at thesame time that the microwave is cooking, the combination could trip abreaker or blow a fuse. ( Full Answer )
A reptile light is used to generate heat for the reptile, so youmust use a bulb that uses 100 watts, and an incandescent bulb iswhat you need.
The ohms tells you the resistance; the wattage tells you how much heat energy it can handle. Not that any resistor ever lived up to its specifications -- I always needed some type of heat sink. Bottom line -- yes.
Optimal wattage depends on several factors including the size ofthe growing area and the type of plants. Plants that require morelight (i.e. tomatoes), need a higher wattage system.
Also CFL and LED light bulbs use much less actual wattage thantheir equivalent wattage. It is best to look at the datasheet for each individual type oflight bulb. They can vary from milliwatts to kilowatts, dependingon the purpose the light bulb was intended for.