0

I want to wire a circuit that has 3 switches and 3 lights. S1 controls L1 and S2 and S3 control all three lights so basically when I turn on S1 it only turns on L1 and when I turn on either s2 or s3 all three lights turn on.

Here's a table of how I'd like this to work:

S1=ON  S2=Off S3=off == L1=on L2=off L2=off
S1=off S2=On  S3=off == L1=on L2=on  L3=on 
S1=off S2=off S3=On  == L1=on L2=on  L3=on
S1=on S2=on S3=on == L1 =on L2= on L3=on
S1=on S2=on S3=off== L1 =on L2=on L3=on
S1=on S2=off S3=on== L1=on L2=on L3=on
7
  • Yeah it should stay on if S2 was off – Carlos Nov 11 '20 at 16:08
  • Is there anyway you can make a wiring diagram to explain how what your saying – Carlos Nov 11 '20 at 16:11
  • 1
    I started reading the "Is there any way you can make a wiring diagram..." comment and thought, "Yeah, that's a good question to ask the OP for some clarification." Then I saw that was the OP asking it. Maybe it would be good for you to draw up what you're after, @Carlos. Maybe instead of a wiring diagram, you could show a truth table S1 = On, S2 = Off, S3 = Off and show what bulbs you expect to be on, then S1=On, S2=On, S3=Off and show what bulbs you expect to be on, etc. Showing that for all conditions will help people understand what you're after. – FreeMan Nov 11 '20 at 16:26
  • When you draw a diagram, consider: if the lead from S1 is hot, then it will make the wire from L1 back to S2 hot, and that wire could be connected to the leads to L2 and L3. You need to figure out how to isolate the two "hot" feeds to L1 . I'm not sure you can. – Carl Witthoft Nov 11 '20 at 16:38
  • 5
    I'm still sensing an X-Y problem here. What is the ultimate goal/purpose? The definition of an X problem is fixating on a particular method to solve a problem and excluding from scope any other method, which also excludes any normal methods others use to solve the same problem. – Harper - Reinstate Monica Nov 11 '20 at 20:24
2

Steam wiring is not for this

By "steam wiring" I mean all the doddy old-fashioned wiring we do with 1880-era SCADA tech (i.e. brains made of wire and nuts, snap switches, etc).

What you want is smart switches in a hub system. So you have smart modules (or simply smart bulbs), and then smart switches in the locations of interest.

At that point you can simply define events in the system:

 S1 turnon: turn on L1
 S1 turnoff and S2=off and S3=off: turn off L1, L2, L3
 S2 turnon: turn on L1, L2, L3
 S2 turnoff and S1=off: turn off L1, L2, L3
 S2 turnoff and S1=on: turn off L2, L3 

or whatever the heck you choose to define.

1

This seems to be a bit of an X-Y problem. Specifically, I suspect the issue isn't so much that "all 3 switches control L1" but rather "all 3 switch locations control L1". In which case:

  • Box 1: S1a
  • Box 2: S1b, S2a
  • Box 3: S1c, S2b

S1a, S1b, S1c are all one circuit. Two of these are 3-way switches and one is a 4-way switch. Assuming typical wiring (but could be arranged differently depending on the physical room layout):

Panel /2 -> S1a (3-way) -> /3 -> S1b (4-way) -> /3 -> S1c (3-way) -> /2 -> Light L1

S2a and S2b are 3-way switches:

Panel /2 -> S2a -> /3 -> S2b -> /2 -> Lights L2 & L3

Assuming typical cables, /2 means a black/white (black = hot or switched hot, white = neutral) and /3 means a black/red/white (black and red = travellers, white = neutral).

The end result is that from any location if you want L1 on or off, you can do that. If you want L2 & L3 on or off, you can do that from the 2 designated locations.

5
  • I’m a little confused so is there going to be 5 switches in total – Carlos Nov 12 '20 at 15:18
  • Correct. 3 switches for L1, 2 switches for L2+L3. Based on my (possibly incorrect) understanding of the goal: One location to control just L1, two locations to control L1/L2/L3. – manassehkatz-Moving 2 Codidact Nov 12 '20 at 15:20
  • The XY Problem defined. – FreeMan Nov 12 '20 at 15:30
  • If I just want one switch for L1 is that possible – Carlos Nov 12 '20 at 15:50
  • Yes, that is possible. – manassehkatz-Moving 2 Codidact Nov 12 '20 at 15:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.