How  much light  and how much heat do you get out of a light bulb if input power is 50 watts?
137865
Deppends on the bulb, and it's efficiency, total energy in = total out. 50W at 10 min is 30000 Jouls of energy.\nThe efficency of an average incad. bulb is about 15% efficient, so I would say you get about 4500J of light, and the rest would mostly be heat, about 25500J of heat.\n\nI am not sure how to convert the Jouls into Lumens, and Calories/BTUs, but it can be done.
1023774
450 watts min or 600 watts to max it out
943748
use voltage dividers with a transistorized switch.  this method lights  a number of lights matching the input voltage in this case, all the way up the tree to it, rather than just one, so note that.  (7V in gives 7 lamps on).\n\nthe first transistor comes on if the base reaches .7V (2N2222 or so will work).  the next transistor is attached to the base of the first transistor through a voltage divider, the third ....  From the collector stick in a 1V Zener for regulation (make sure you get one that withstands 12v in), and the lights then go from there straight to ground (include a resistor here to limit current, try 15 ohms to start, experiment!).\n\ndecide how many lights you want, divide your final voltage by that number, and that is the voltage per step (here, 1V).  the voltage divider then comes from the voltage per step versus the transistor base ON voltage, across a good sized resistor (say 470 ohms or 1k, whatever you want (watch the current and the power drain here).  in your case, if you use a 470 load resistor on the voltage divider, then the second leg of the divider is (470 / (1V/1V-.7V)) = 141 is about a 130 (less is more on the base!).\n\nrecap:\n\nso the voltage divider goes from the input into the 130, into both the base and the 470, the 470 goes to ground.  the emitter can tie to the input voltage, the collector goes to a 1V Zener, to a 1V lamp, a small resistor, and ground.\n\nthe other stages stack right onto this.  the input still goes to the 130, then to the base and the 470, the 470 ties to the base of the previous stage.  the emitter again goes to the input voltage, and the collector goes to a 1V Zener, a 1V lamp, and a small resistor to ground.  sorry, i can't attach a picture here.
1110126
I assume that a power factor is zero and current and voltage are given in  RMS values then \nPin=IV= 4x120=480 Watts\nPout=.8 (Pin)= 384 Watts\nHorsepower = Pout/ 746W  = .515 HP or ~1/2 HP
951221
Floodlights are anywhere from 150 to 500 watts, while indoor lights range from 25 - 100 watts.\n\nAlso flood lights are designed with a reflector that throws the light in one direction, while indoor bulbs give off light in every direction.
688307
Well, 36 Watts tube means they can handle a maximum of 36 watts. If the circuit provides 40 Watts, it will most likely fry. However, just because it says 40W doesn't necessary mean the circuit will be shelling out 40W, but I'd say go and get a 40W tube just to be safe.
530475
It depends on the rating of the breaker.  Common values for the maximum current in houshold lighting circuits are 15 and 20 amps.\n\nCurrent = power/voltage, so each 60 W bulb running on 110V draws 60W/110V = 0.545 amps.  You could run 27 of these on a 15 amp circuit and not blow the breaker (15 * 0.545 = 14.73 amps).  The 28th light would exceed allowable current.\n\nFor a 20 amp circuit, you could run 36 60W lights. (36*0.545amp = 19.64 amps)
165350
wattage of the bulb is important...prob not generating enough heat to make the 'lava' flow (wax).\n\nIt should show signs of flow within an hour...at least!
1001733
It sounds like you are creating a basic electrical circuit. with the batteries, wires, bulbs and paper clip you can make either a series or parallel circuit. The light bulbs will light up when you have a complete single loop for series. When you measure the current it should be the same anywhere on the circuit. In a parallel circuit you are splitting the circuits into three or more branches, with one bulb on each side. When measured each branch should be using half the current. You shouldn't have any problem finding a site online that shows this experiment. use keywords, including science experiments for kids. I found the experiment in Reader's Digest,"How Science Works"
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.
1043980
You can, but there is a danger of overheating and fire. Here's why.\n\nThe power of a light bulb in watts is determined by the equation P=V^2/R where V = voltage and R = resistance. The higher power (higher wattage) light bulbs actually have lower resistance, so they draw more current.\n\nThat same current must pass through the wires, socket, etc. These also have resistance. The greater the current that passes through the wire, socket, etc. generates heat from the equation P = I^2 * R.  So, if the current is doubled, there is 4 times as much heat that is generated by the wires, socket, etc.\n\nSockets and wires have different ratings depending on their ability to handle high current and heat. You may have seen ceramic light sockets - they provide better heat resistance that lightweight sockets with plastic that can melt.
728334
might not have enough watts!!!
1130014
V=IR\n\nWire 1:\n50 = 2 * R1 => R1 is 25 ohms\nCurrent = 2 amps\npower = V*I = 2*50 = 100 watts\n\nWire 2: \n100= 2* R2 => R2 = 50 ohms\nCurrent=2amps\npower = V*I = 100*2 = 200 watts\n\nare you looking for ratios when you say "compare" ?\nif yes: ratio of resistances is  1:2\nratio of currents is 1:1\nand ratio of power is 1:2
38597
P= work/time and work is = to force x distance so its 500x4 =2000joules\nand 2000/4 =500 watts
711657
If this is 230 V circuit\n\nwatts = 4600 w\n\nIf this is 110 V circuit\n\nwatts = 2200 w
932209
Basic-   15 amp circuuit should not hold more than 1440.00watts, or 12 amps ( 1440watts / 120v= 12 amp).  you should load the circuit for lighting or continual use more than 75%.  the same for 20 amp circuits. (1800 watts/120v=15amps).   Check the back of the space heater. It should the watts listed.
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.