explain Y the initial current for the lamp is likely to be higher than the normal working current Read 4 more.
814978
The resistance of the filament will increase as it heats up. So you're going to get something called a "transient", which is when the initial value will stabilize (to the working current in this case), after a period of time.\nA lower starting resistance equals a higher starting current.
695376
The resistance of the filament of the lamp varies with temperature. When the filament is at room temperature the resistance is much lower than when it is white hot. \n\nCurrent = V/R. If the voltage is constant and the resistance increases, the current decreases.
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.
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.
613982
Old fluorescent tubes, which required a ballast to start, used a bunch of electricity to turn on - but a small amount of electricity to keep them on. So it was more economical to leave them on rather than turn them off for a few hours.\nImprovements in design have eliminated that anomaly and there is now little difference in the power needs to start and maintain the bulb.
657073
electricity passes through the wire in the light bulb. as it goes through the curved part, it heats up because it has to push its way through ( meatphorically speaking). By heating up it emits thermal energy (heat) and light energy, and thanks to this, you can read this answer at night, all for this moment. ;)
729732
that's right..............BUT, the conventional current flows from positive to the negative
975896
Energy is measured in watts, so I assume you mean a voltage source of 200v.\n\n(1)\nThe voltage source of 200v applied to the light bulb (laminar) will be 200v.  Since the light bulb is designed for 130v, the supply voltage will likely blow the bulb.\n\n(2)\nHere you have two possibilities, depending on your circuit.\n(a) If you have a parallel circuit, then each light bulb will have 200v across it, and again the bulbs will likely blow because the voltage is to high.\n(b) If you have a series circuit, the voltage across the light bulbs will be subject to their resistance.  Assuming the resistance is the same, then the voltage will be the same, that is, 100v each.\nThe voltage across each light bulb will be equally proportional to it's resistance ratio.
1009589
The current rating on the battery is higher than the solar cell.  You need to add solar cells in parallel in order to boost you amp load.
882455
Not sure what you are trying to do but it kind of sounds like you are attempting to replace an arc-discharge light with an incandescent.\n\nThey have fundamentally different drive requirements. The arc-discharge has a negative resistance and it will not operate without a ballast designed specifically for that lamp. Such ballasts sometimes include active power-monitoring. The lamp and the ballast need to be paired with each other. Using a lamp and ballast not designed to work together will result in one of: won't work at all; will have very short life; will have unstable operation.\n\nThe incandescent lamp has a resistance that changes dramatically as the filament heats up and it is therefore self-ballasting. It works best when operated from an approximately constant-voltage source, not a constant-power source.
698625
Maybe because it is depicting Current flow NOT Electron flow??
216568
We'd need to either see the circuit or have it described to us in order to answer.  If there are just two lamps and a voltage source and nothing else then the voltage across the second lamp would be 6V-2V= 4V\n\nSo 4V
775688
Hmmm, curious.\n\nIf the bulb has a filament you can see, then yet, it will almost certain be dimmable.\n\nI cannot see why there would need to be a trasformer in the box, if you try and wore the lamp in series with a 400Watt rated dimmer, th e worst that is likely to happen is you kill the dimmer.\n\nTry and find a bit more about this (make/model/type) and pop up another question, we can then give better feedback
609858
Its is the amount of resistance.  Same reason the wire in a lightbulb grows.  There is some level of electrical resistance, even in conductive materials.  The resistance creates heat and eventual combustion.
1001343
There is an evident discrepancy between what you claim is actual use as to what the electric company has actually read.  Something is not right.  When you first moved in, there should have been an accurate starting point reading.  If not, then that is one issue.  Another, sometimes a person uses enormous amounts of electricity (growing pot for example).  Another possibility is that the meter itself is defective and so you could ask that it be checked out.  You would need to rule out that you have actually used an excessive amout each month.  There is a reason the company came up with its figure.  Find out exactly what the rest of the story is.
371630
the first 1000 blocks 85% of the light and passes 15%.\n\nThat means only 15% of the original light "makes it" to the second 1000.  The second 1000 ALSO blocks 85% of the light and pssses 15%; BUT those numbers apply to the light that "made it", not to the "original" light source.\n\nand so on.\n\nThe total light passing through, converting 15% to decimals (0.15) is:\n\n(0.15) * (0.15) * (0.15).
808047
By touching the lamp you are changing the capacitance of the lamp and a circuit inside the lamp detects the change and turns on the light. \n\nIn order to offset sensitivity to temperature and other objects near the lamp it is designed to be triggered by a sudden change in capacitance - so when you slowly touch it, or tap it with an object with insufficient cpacitance then the circuit does not detect enough of a change to trigger.