Open Hot, Still Use Power?

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  #1  
Old 01-06-16, 06:46 PM
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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.
 
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Old 01-06-16, 07:05 PM
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Why are you asking this? That answer may help us help you better.
 
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Old 01-06-16, 07:42 PM
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There are a whole lot of "conditions" that you have not fully explained.

You mention a "DC power source" but give no particulars on what this "power source" is made up of. Is it a solar voltaic array? Or maybe a switching power supply with an AC input? Simplest would be a battery.

...say wire A, is connected to the hot side of the power source.
Again, what kind of power source and why would it have a "hot" (or cold) side?

(relative to the power source's ground)
What is this "ground" connection? Where does it originate and where does it end? What is its purpose?

And assuming that you choose not to answer the questions, no, there is NO electron flow UNTIL the circuit is completed.
 
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Old 01-06-16, 07:49 PM
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Ultimately, this question came from me trying to understand why phantom voltage (a.k.a. ghost voltage, stray voltage, inductive coupling) goes "away" when the open wire is closed.

In particular, I am trying to understand the following sentence from Fluke support paper titled, "Ghost voltages – phantom readings can lead to the wrong diagnosis".
http://support.fluke.com/find-sales/...105317_A_w.pdf

"So, when the leads are placed on the open circuit that contains a ghost voltage, the low input impedance will cause the ghost voltage to dissipate..."

So I tried to break a problem that I couldn't solve down into smaller problems that I might have a chance at solving.

I imagined a light bulb connected to an open wire that was being inductively coupled from an AC source. I imagined that the open wire will still have the free electrons in the wire gyrating back and forth due to the AC phantom voltage. Furthermore, the jostling of the electrons is 'current' and therefore would in fact be using power (which would not be desired). To make matters simpler yet, I removed the inductive coupling element with a direct power source and instead of AC I imagined a DC source first energizing a system.

At this time, I don't want to engage in a conversation about inductive coupling, AC vs. DC, or some other ancillary topic, so I asked the simplest question I could imagine that drives at the heart of my confusion.
 
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Old 01-06-16, 07:54 PM
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Imagine a capacitor. Unless it is has a very high capacitance such as supper capacitors used for back up in some electronic devices when a load is attached there will be a current flow but only for a few milliseconds till the capacitor is drained.
 
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Old 01-06-16, 07:59 PM
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Suppose it is a battery. By the "hot" side, I mean the cathode of the battery. By the "ground" of the battery, I mean the anode of the battery.

The purpose of the ground is to set a reference point for zero electric potential for the discussion.

If there is no electron flow, at least initially, how could wire A and wire B come to be at the same electric potential as the cathode when they were originally at the same electric potential as the anode?
 
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Old 01-06-16, 08:13 PM
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Thank you ray2047. Yes, thinking of wire A, the bulb's filament, and wire B as one (poor) capacitor explains the situation that 'yes, current will flow, but only very briefly'.
 
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