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I have a set of christmas lights I have shortened. I don't understand what is happening here.

Both wires work to light the second half, but when I connect it all together (1), the light on the first half does not work??

In (2) and (3) I showed that power is still passing through both wires, except that it does not light the first half. So the problem is not a short in the wires themselves.

In (4) I show that the bulb lights up when I put them in series. So I also know the problem is not the bulb.

What is going on here? I don't understand what I am missing.

Here is a picture showing illustrations:

enter image description here

UPDATE:

Ok, so by looking at a new uncut set, I have learned that the original was NOT in fact wired like above. Thanks to a commentor who pointed out that it is short-circuited, I went to a new set and learned that it was not wired as it originally looked. I also cracked open one of those lights with three wires to see what it looked like inside.

Here is a modified diagram of the actual wiring of the original. I popped off the bulb so you can see how it is attached:

enter image description here

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  • \$\begingroup\$ A riddle? When you short-circuit a light bulb, the voltage across it is zero. And a light-bulb which has zero voltage across its terminals will light up? \$\endgroup\$
    – Janka
    Commented Oct 28, 2016 at 1:05
  • \$\begingroup\$ Please note that the black line is the negative. The green line is the positive. The light is not short-circuited. In (1), all four lights are connected to the power source. \$\endgroup\$
    – hbsrnbnt
    Commented Oct 28, 2016 at 1:06
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    \$\begingroup\$ Yes it is. You've short-circuited it with the green. \$\endgroup\$
    – Ignacio Vazquez-Abrams
    Commented Oct 28, 2016 at 1:07
  • \$\begingroup\$ The same thing happens on the other half though, look, why does the other half light up? \$\endgroup\$
    – hbsrnbnt
    Commented Oct 28, 2016 at 1:08
  • \$\begingroup\$ These pictures are misleading. The green line below the upper middle candle should be another conductor, right? That's how it is used in #2. \$\endgroup\$
    – Janka
    Commented Oct 28, 2016 at 1:12

1 Answer 1

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In #1 you are giving a (near) zero-resistance path alternative to the first light bulb, meaning it will have very, very little current flowing through it.

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  • \$\begingroup\$ I accepted this answer, because you are right, in fact the first original would have been impossible, because that is not how it was wired. I have gone back and fixed it after carefully studying a new set. I accepted your answer because although it is unclear, it is actually the solution to the actual question. \$\endgroup\$
    – hbsrnbnt
    Commented Oct 28, 2016 at 1:58
  • \$\begingroup\$ I am curious though how the bulb is preventing it from short circuiting in the correct original diagram? \$\endgroup\$
    – hbsrnbnt
    Commented Oct 28, 2016 at 2:00
  • \$\begingroup\$ I have been thinking about this a lot more, and I think I understand better now. So basically because there is no resistance on the path without the resistor (the bulb is the resistor in this case), and since electricity finds the shortest and easiest path, it won't bother wasting its time going through the resistor and instead will just skip the resistor and head down the easy path with no resistor. Is this correct? If so I think I understand why it didn't light up when I closed the circuit! \$\endgroup\$
    – hbsrnbnt
    Commented Oct 28, 2016 at 2:12
  • \$\begingroup\$ Precisely! Technically speaking, a very small amount of current will flow through the resistor, relative to the ratio of the resistance of the resistor and the resistance of the alternate path (the wire, which will be on the order of milliohms). So for example the amount of current flowing through the resistor may be something like 0.00001% of that flowing through the "no resistance" path, but for all intents and purposes you can consider there to be no current. \$\endgroup\$
    – Benjamin Wharton
    Commented Oct 28, 2016 at 4:02

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