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Why does the electricity does not take the shortest path and just skip past all the bulbs?

Here is a parallel circuit with two sets of 3 bulbs in series. Notice at the bottom of the circuit there is a negative wire and a positive wire that goes all the way across.

Why does the electricity still enter the parallel circuits? Also is that power wire that goes all the way across really necessary?

See my illustrations below:

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I understand now from the answerer that circuit 1 it is not connected to anything. But how about now?

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2 Answers 2

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I'll try to answer each part piece by piece.

1st circuit: The power doesn't "skip" the circuit because it reaches live out, and has no where to go. The live - neutral in "power out" is an open circuit, until something is connected externally.

Of course, if you were to short live-out to neutral-out, then you have a short circuit and you could consider the power to "skip" the parallel strings of lights.

2nd circuit: This path is not needed in the circuit, it just allows additional components to be plugged in downstream, and essentially acts like an extension cord in parallel with the light strings.

3rd circuit: No, here the live-out voltage is only 2/3 of the live-in voltage, when no load is connected to power-out, and will change depending on the load connected from live out to neutral out.

Edit: I think it is easier to understand if you envision this parallel circuit like the one below. The switches shown represent the connection to a 2nd, identical circuit. Here you can see that regardless if a 2nd load is connected or not, the "easiest" path from live to neutral will require going through 3 light bulbs:

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Great, just one more thing, in the first circuit, I understand what you said about it's not connected to anything. But then say you plug in the second string of identical lights, why does it not skip the first string and only light the second string? \$\endgroup\$
    – hbsrnbnt
    Commented Oct 28, 2016 at 20:35
  • \$\begingroup\$ See edit, I think it will help. \$\endgroup\$
    – Jim
    Commented Oct 28, 2016 at 20:41
  • \$\begingroup\$ Excellent, thank you! So I see that no matter how many equal strings, the path to the ground will still be the same across all of them, just divided among the number of strings. \$\endgroup\$
    – hbsrnbnt
    Commented Oct 28, 2016 at 20:48
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The circuit in your first picture is effectively a current divider. Part of the current will go to the bulbs, and part of the current will go to the Power Out device.

Think of this like a power strip. One input is divided into several outputs. In this case both the lamps and a fan can operate at the same time from the same power source.

It is better to say that electricity takes all the paths to ground, it is that some paths are more favored than others.

If you want the Power Out device to operate properly then the path is needed.

In your third picture the problem with hooking up the Power Out device as pictured is that the bulbs will induce a voltage drop before the device. For instance if you're using 120V, then each bulb in each string would drop 40V. If the power out device was designed to operate from 120v, then 40V would not be enough. This is ignoring the issues with current that will also arise.

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  • \$\begingroup\$ Thank you, this is very helpful! Upvoted, already accepted an answer though. \$\endgroup\$
    – hbsrnbnt
    Commented Oct 28, 2016 at 20:43

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