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There's quite a few places online where you can find out "how to add an indicator light into an existing circuit," but I can't find any that tell WHY the light needs to be added in a certain place, while other locations will fail.

enter image description here

From: Furnace wiring

Connect the light:

  1. Between E and F. (Won't that always use the path of least resistance? Current will flow through the light, and not the load, or the current will flow through the load, and not the light.)
  2. Between B and C. (Won't that ALWAYS complete the circuit, even when the switch is off/open? When the switch is on/closed, the light is basically no longer in the circuit at all.)
  3. Between A and C. (Won't that ALWAYS complete the circuit, even when the switch is off? The switch is basically no longer in the circuit.)
  4. Cut the wire at D and put the light inline.
  5. Cut the wire at G and put the light inline.
  6. Between A and B. When the switch is off/open, the light is just mounted to the switching arm. When the switch is on/closed, current flows through the light (but also the switch.)

Notes:

A. The blue shaded area is located behind walls, a furnace, and permanent ceiling. Very difficult to access. Can this light be added in the non-blue area only? Choices 1, 4, 5 would be tricky to implement.

B. I'd like the light to be mounted as close to the switch as possible, for best visibility

C. The "load" is a small solenoid that opens/closes a water valve.

D. The light is already a 24 V version, so no resistor should be needed. (Or will it?)

E. I have no idea if the power is AC or DC. (Or will it not matter here?)

F. The custom-designed switch used must remain and can't be replaced with a 'lighted one.'

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    \$\begingroup\$ Whoiever first said "always use the path of least resistance" should be shot!! An electric current follows all possible paths, in inverse proportion to the path's resistances. The path with the least resistance will carry more current than a path with a higher resistance, but both paths will carry current. \$\endgroup\$ Commented Jan 15, 2022 at 0:43
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    \$\begingroup\$ so, what exactly are you trying to do? \$\endgroup\$
    – jsotola
    Commented Jan 15, 2022 at 1:29
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    \$\begingroup\$ "But won't that always use the path of least resistance? Current will flow through the light, and not the load. Or the current will flow through the load, and not the light." If this was true then if you had two lights in your house of different powers (watts) then switching on one would switch off the other. You would have to cook in the dark. \$\endgroup\$
    – Transistor
    Commented Jan 15, 2022 at 11:48

3 Answers 3

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Electrically, you only have three points or nodes in your circuit.

A, B, G, and the positive terminal of the supply are one node. Whether you connect something to A, B, G or the positive terminal of the supply makes no difference - they are electrically the same point.

C, D, and E are another node

F and the negative terminal of the supply are the third node.

Where you connect an indicator light depends on what you want it to indicate.

Connecting the indicator light between node 1 (A, B, G, +) and node 2 (C, D and E) will make the light shine when the switch is open - you might do this with the indicator light inside the light switch, so you can find the switch in the dark. This assumes that the light requires much less current than the load, so that the load won't have enough power to operate when the switch is off.

If you connect the light between node 2 (C, D, E) and node 3 (F and -) it will light when the load has power.

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In answer to your questions:

  1. You would connect the light between E & F. Note that C,D & E are essentially the same wire, so between C & F, D & F and E & F are electrically the same. As for 'path of least resistance' - the houses in your street are connected to the same utility supply. If Joe down the street turns on his air conditioner, does your TV stop working? No. There are two current paths.

  2. This won't work. It is putting the lamp in series.

  3. Same as 2.

  4. Same as 2.

  5. Same as 2.

  6. Assuming the switch is changeover, A & B makes no sense as in the AB position, the switch sorts the lamp, in the AC position, B is open circuit.

The lamp must be suitable for the power that is supplied. For an incandescent lamp, it doesn't care others may differ. If the lamp spec says 24V AC/DC, you're home free.

Basically, you'll need to run another wire from the lamp to F. Can we do an inline trick? Possible, but you need to design it specifically for the load.

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    \$\begingroup\$ Actually, connecting the light between A & C (#3) is often used to light the indicator when the circuit is off. However, this works only if the indicator takes much lass current than the load, typically an neon or LED indicator and an incandescent light as a load. This works by trickling a small current through the load, too small to active the load. Unfortunately, if you replace the incandescent light with a large LED light, the current through the indicator may be enough to make the large light glow dimly but visibly. \$\endgroup\$
    – DoxyLover
    Commented Jan 15, 2022 at 6:05
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It is presumed that the indicator light is to come on when the load is switched on.

Since the load and the light have the same voltage rating, the right position for the light would be across the load.

The light could be conveniently located close to the switch.

enter image description here

Should it be located in any of the lines, it would be in series with the load. It would light up but subject the load to a voltage far below its rating.

Should the light be across the switch, it would light up when the switch is off but subject the load to under-voltage as above. When the switch is on the light would be off with the load operating at its rated voltage.

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