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The formula you are looking for is, R = Volts (squared)/Watts.
Wiki User
∙ 13y ago
Wiki User
∙ 6y ago121 ohms. P = VI. I=0.909 Amps. P = I x I x R. R= P/(I x I) R = 100/(0.826)
Wiki User
∙ 6y agoVW as in Volkswagen I suppose...
The resistance of an incandescent(glow wire) bulb depends on the power - the wattage.
Wiki User
∙ 12y ago8.26 ohms
Wiki User
∙ 10y ago960 ohms
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What is the current being drawn by a 100W light bulb and what is the resistance of a 100W light bulb?
Assume the rating of 100W refers to operation on a supply of 117 volts.Power = (voltage) x (current)Current = (power) / (voltage) = 100/117 = 0.855 ampere (rounded)Power = (voltage)2 / (resistance)Resistance = (voltage)2 / (power) = (117)2 / 100 = 136.89 ohms
A 2.5 Amperes current flows through a 100W lamp what is the voltage?
The formula you are looking for is E = W/I.
Why it is not possible to determine the value of low resistance using neon flash lamp experiment?
In the neon flash lamp experiment, the resistance of the lamp is very low, often in the range of a few ohms. When measuring low resistance, the resistance of the connecting wires and contacts becomes significant and can interfere with the accuracy of the measurement. These additional resistances create uncertainties in determining the exact value of the low resistance of the lamp. Additionally, the behaviour of the neon lamp itself, which is a non-linear device, further complicated the measurement process. Therefore, it is challenging and difficult to accurately determine the precise value of low resistance using this method.
What happens to the ammeter reading when a lamp is added to a circuit which contains two cells and a lamp in series?
Assuming the new lamp is in series, the ammeter reading falls because the total resistance has increased. By how much depends on how the lamp resistance depends on voltage. If the lamp is added in parallel to the first, then the ammeter reading doubles.
Two lamps A and B are rated 60W 220V and 100W 220V respectively so why does the answer say that when connected in series bulb A is brighter?
This seeming contradiction occurs on most physics tests. But it's true, and here's why: Lamp A by itself has power of P=IxE ; 60=IxE ; so I=60/220=0.27 amperes. Thus the lamp has a resistance E=IxR ; R=220/0.27 = 807 ohms. Lamp B by itself has a power of P=IxE ; so I=100/220 = 0.45 amperes. Thus the lamp has a resistance E=IxR ; R=220/0.45 = 489 ohms. Since the resistance in series is added up, the total resistance is 807+489 =1296 ohms and the current I= E/R = 220/1296 = 0.17..... Phew... Since P= IxIxR For 60W lamp A, P= 0.17 x 0.17 x 807 = 23.3 watts. For 100W lamp B, P=0.17 x 0.17 x 489 = 14.1 watts. So, when they are wired in series, the bright one and the dim one appear to be reversed as to their respective "normal rated powers". Their normal rated powers are only correct when both of them are supplied with their normal rated voltage of 220 volts, i.e. when they are wired in parallel. This makes a lot of sense if you think about it.
Why would a high watt lamp draw more current than a low watt lamp?
Power is measured in Watts, power (Watts) = E (volts) x I (current - amps) current is determined by the internal resistance (R) of the lightbulb, the lower the resistance the more current will flow. 120v x 0.5a = 60W 120V x 0.83a = 100W the 100W lightbulb will draw more current We also have Ohm's law: E(volts) = I (amps) x R (ohms) Household voltage stays the same at 120v we have for a 100w lamp: 120v = I x R R = 120v/0.83 amps R = 144.6 ohms for a 60w lamp: 120v = I x R R = 120v/0.5 amps R = 240 ohms The higher watt lamp has lower resistance.
The electrical resistance of a 100w 240v bulb will be?
42 ohm
Does a 100w lamp use more energy than two 60w lamps?
NO. Two 60W lamps will use 120W - which is more than 100W !!
What is the VA of a 100W metal halide lamp?
For general calculations VA is the same as Watts.
What is the current being drawn by a 100W light bulb and what is the resistance of a 100W light bulb?
Assume the rating of 100W refers to operation on a supply of 117 volts.Power = (voltage) x (current)Current = (power) / (voltage) = 100/117 = 0.855 ampere (rounded)Power = (voltage)2 / (resistance)Resistance = (voltage)2 / (power) = (117)2 / 100 = 136.89 ohms
A 2.5 Amperes current flows through a 100W lamp what is the voltage?
The formula you are looking for is E = W/I.
Can you use a led lamp in a 100w sodium fixture?
Probably not. Sodium light fixtures generally have a "ballast".
110v and the current is 10 a what is the resistance?
Resistance is Volts over Current 11 Ohm = 110Volt / 10 Amp
If a light bulb is a hundred watts but puts out 23 watts can it be used in a smaller light fixture?
A 100W incandescent lightbulb is rated this way because it consumes 100W of power to produce a given amount of lumens. The wattage is a power number derived from the voltage supplied multiplied by the current that will flow through the filament. Consequently, the filament acts as a resistive component and will always draw the rated current when supplied with the rated voltage. Now, lets say you have a light bulb rated at 100W for 430V and you want to use it in a 110V 40W fixture. This would be an example of a 100W bulb drawing only 23W. Assuming the screw base is the same, there is no electrical reason it couldn't be used because the voltage rating of the bulb exceeds the supply and the current draw is lower than the maximum rating of the lamp. But what of fluorescents? This is a no go. You cannot run a 430V ballast on 110V and expect it to work correctly.
Can you rewire a 240v lamp light fitting or small appliance to 110v and if so how?
Yes, you need a transformer
How many lumens in a 125 watt CFL?
Depending on the brand, it will give about 18.000 Lumens and equals a 100W HPS lamp
Which has a higher resistance a 5w lamp bulb or a 25w lamp bulb and how many times?
As per the formula for power (Power (Watt) = Voltage (V) x Current (i) & Resistance (R) = V / i), 25w lamp bulb would have higher resistance compared to that of 5w lamp bulb.
