OK I have been debating this with some friends and they all disagree with me. Here is the question. If you have a simple circuit with a source switch and a load (lamp), you turn the switch on and then back off before the light comes on will the light still "blink"? From the source if there is 186,000 miles of wire it take exactly one second for the light to come on after closing the circuit. If you close it and open it before one second will the light come on for the amount of time you have it closed. I say that there would be no more emf to push the current so therefore it wouldn't light up. Do you know the answer? Please let me know what you think and why.

Thanks,

 Capuchin Posts: 5,319, Reputation: 3601 Uber Member #11 Jun 9, 2007, 08:00 AM
Sorry strat?
 worthbeads Posts: 569, Reputation: 280 Senior Member #12 Jun 13, 2007, 08:25 AM
The bulb would not light because the circuit would be broken and the flow of electrons would stop.
 Capuchin Posts: 5,319, Reputation: 3601 Uber Member #13 Jun 13, 2007, 08:49 AM
Worth, if you imagine the water analogy, in your case you would end up with more water in one side of the circuit than the other, the water would still flow to even out the pressure.
 worthbeads Posts: 569, Reputation: 280 Senior Member #14 Jun 21, 2007, 05:16 PM
Electricity is not water.
 caibuadday Posts: 450, Reputation: 61 Full Member #15 Jun 21, 2007, 05:17 PM
Quote:
 Originally Posted by Capuchin worth, if you imagine the water analogy, in your case you would end up with more water in one side of the circuit than the other, the water would still flow to even out the pressure.
can't use the water analogy, because the pipe is full of water and the wire donot have electricity store in it, and if the pipe is full of water and it is vertical or incline water may not come out the other end or it even flow backward.. I think the lite won't blink, because the energy would dissipate as heat
 ebaines Posts: 10,033, Reputation: 5529 Expert #16 Jun 22, 2007, 11:00 AM
The light will blink. While the circuit is closed electrons start moving, pushed along by the EMF of the power source. But because of the length of this circuit, the electrons don't all start moving at once: the EMF can only propagate through the wire at a speed a bit less than the speed of light (in typical cables the propagation of the electrical signal is around 2/3 the speed of light). So in effect there's a shockwave (as capuchin describes it) which is the front of the of electrons that are being pushed by the EMF. Now, when you disconnect the power source, the "news" that the power is no longer connected also travels down the wire at about 2/3 the speed of light. So the electrons in the vicinity of the light continue to feel the EMF until over a second after the power is turned off, and during that period of time electrons continue to move through the light filament and the bulb turns on.

It all comes down to the fact that nothing can travel faster than the speed of light, and in this case that includes the news that the power source is no longer connected.
 sparkyibew32 Posts: 3, Reputation: -1 New Member #17 Jun 22, 2007, 10:40 PM
If this is purely hypothetical , not taking into account for voltage drop and all. The light will light instantly when the switch is made. Electrons are like ping pong balls in a tube. If you push one in one end one will be pushed out the other end at the exact same time.
Capuchin (Jun 22, 2007 11:43 PM): Ouch, nothing can travel faster than the speed of light. Ever.   Source:
 ebaines Posts: 10,033, Reputation: 5529 Expert #18 Jun 23, 2007, 07:28 AM
Regarding ping pong balls in a tube - the speed that the pressure waves moves down the tube to push those ping pong balls along is the speed of sound (through the ping pong balls) - substantially slower than the speed of light, and certainly not instantaneous!

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