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 DIY Chatroom Home Improvement Forum Open Hot, Still Use Power?

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01-06-2016, 08:14 PM   #1
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## Open Hot, Still Use Power?

Hi,

First, thanks in advance for your assistance in helping me understand the following question that has me perplexed.

Suppose there are two wires, A and B, a DC power source, and a light bulb. Suppose both wires A and B have zero (0) voltage difference between them and the ground of the power source. Suppose one end of each wire is connected to the light bulb. Then, lastly, suppose only one of the two wires, say wire A, is connected to the hot side of the power source. One end of wire B remains open (not connected to anything), so the entire "circuit" is open.

Will the light bulb consume some power during the moment wire A is initially connected to the power source?

The reason I think it will is that wire A and wire B need to change electric potential energy (relative to the power source's ground) from zero (0) to the voltage of the hot side of the power source. In doing so, electrons will ultimately pass through wire A, through the light bulb's filament, and into wire B forming a higher electron density than before wire A was connected. That said, this electron motion against the resistance of the wires and light bulb consumes some, albeit, small amount of power (and maybe not even enough power to cause the filament to emit light, but still some amount of power that is greater than zero).

Is it true that there would be some current while the voltage initially equalizes.

Again, thanks in advance for you help.

 01-06-2016, 08:35 PM #2 Engineer   Join Date: Nov 2012 Location: Chicago Posts: 191 Rewards Points: 161 A complete circuit is necessary for current to flow.

 01-07-2016, 10:54 AM #3 Member   Join Date: Sep 2011 Location: western Ny Posts: 1,662 Rewards Points: 1,336 You are thinking about this as if it were a pipe to fill with water. It is not. There is no physical substance to electricity in the way we think of it. In other words, unlike water which has mass, electricity does not and does not "fill" a wire

 01-07-2016, 11:50 AM #4 Semi-Pro Electro-Geek   Join Date: Jul 2009 Location: Arizona, USA Posts: 3,376 Rewards Points: 3,652 As a practical matter, there is no significant current flow to "charge" the wire without a complete circuit. Nothing you could measure with normal equipment (or even fairly advanced equipment), and certainly not even close to enough to light a lamp. The normal basic models for electrical circuits will tell you there is no current flow in this scenario. There is actually a tiny bit of current flow to charge the wires when the power source is first connected though. It's extremely small, and brief. So small and brief that very specialized lab equipment would be needed to sense it. Whenever you create a difference in potential between two objects, a small amount of current must flow in order to create that voltage difference by charging the objects. Electrons flow off of the positive object, and onto the negative object. The amount of charge which much flow (i.e. the number of electrons) to achieve a certain voltage difference depends on the size and shape of the objects, how far apart they are, and what's between them. The measure of how much charge must flow per volt is called "capacitance", and it can be calculated using some deceptively simple-looking equations. Because the capacitance depends on the exact shape, size, and spacing of the objects, the calculation gets complicated in real life. With DC circuits and normal voltages, the amount of charge stored on open-ended wires is so completely insignificant that it's irrelevant and practically unmeasurable. But in very high voltage situations and with long wires, this is not always the case. Buried high-voltage cables can store a whole lot of charge. The wire itself (with nothing connected at either end) can store enough energy when charged to several thousand volts to be dangerous. The other situation where open-ended wires matter is in high-frequency AC circuits. The amount of charge that flows into an open-ended wire may be tiny, but if the charging and discharging process is repeated millions of times per second, the amount of current can add up to quite a bit. This is how radio antennas work, essentially. A transmitting antenna is a wire which is connected only at one end, and has charge repeatedly forced into and out of it very rapidly. Of course there is a lot more to antennas and radio electronics than just this, but it is one of the interesting examples of how high-frequency circuits act differently from DC and regular AC power wiring. It is common to have current flows of several amps in an antenna - a wire connected only at one end. __________________ I am a lawyer, but not your lawyer. And who cares anyways? We're here to talk construction. This is DIY advice, not legal advice.
 The Following User Says Thank You to mpoulton For This Useful Post: Oso954 (01-07-2016)

 Tags basics , current , open circuit , theory , understanding

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