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-   -   Connecting Power Source to 3 Single Pole switch(s) (https://www.askmehelpdesk.com/showthread.php?t=342914)

  • Apr 17, 2009, 08:32 PM
    funbanker
    Connecting Power Source to 3 Single Pole switch(s)
    :confused:I am building a stand alone garage and I am doing the basic wiring myself. The 3 switch(s) are beside a walk through door on the other side of the building from the panel box. How can I limit the number of source feeds from across the room? One switch controls the two fluorescent ceiling lights inside the garage, one controls two outside lights on each side of the garage door, and one switch controls an outside motion light beside the walk in door. I prefer not to run 3 separate power source cables across the garage. I am thinking I can use a box and get two sources but unsure on the third. Also what type of box and where is it placed (should it be accessible after the garage is sheetrocked?). Thanks!! :):confused:
  • Apr 17, 2009, 09:01 PM
    Missouri Bound
    Help me understand. You have 3 switches controlling lighting loads, correct? These require only one circuit. You only need one feeder to all three switches, then each switch will feed it's respective circuit. Am I missing something?
  • Apr 18, 2009, 05:22 AM
    stanfortyman
    Fun, I agree with Missouri, and I am also not sure what the problem is. Why do you need another box. Just run one feed to the box with the three switches.
    You could double what you described and still only need one circuit.

    Is there a problem we are not seeing? :confused:
  • Apr 18, 2009, 07:23 AM
    funbanker
    Quote:

    Originally Posted by Missouri Bound View Post
    Help me understand. You have 3 switches controlling lighting loads, correct? These require only one circuit. You only need one feeder to all three switches, then each switch will feed it's respective circuit. Am I missing something?

    I understand that one circuit is enough, I am trying to figure out how to run the hot wire to all three switches. My understanding was that I could only have a maximin of 5 wires in the box. I am using a single box with mountings for three switches. Do I use 3 pigtails (one hot to each switch) or what? Sorry I was unclear. Thanks again!
  • Apr 18, 2009, 07:32 AM
    stanfortyman
    Quote:

    Originally Posted by funbanker View Post
    My understanding was that I could only have a maximin of 5 wires in the box.

    Where did you find this? It is certainly not true.


    What is the cu/in capacity of the box? It should imprinted in the back of the box.
  • Apr 18, 2009, 08:02 AM
    Stratmando

    Use a 3 gang box and it should handle what you are trying to do.
    If the box can only handle 5 wires, it is too small.
    If connections come from junction or lighting, you could do with 5 wires(Ground, Hot and 3 switchlegs.
    I would use a box that would allow 3 dimmers in the future if desired.
  • Apr 19, 2009, 05:15 AM
    tkrussell
    2 Attachment(s)
    And, if the box is plastic, not only will the cubic inch be stamped inside, the total quantity of the various size wires allowed will also be stamped inside.

    Remember to deduct two wires for each device from the total, so, three switches will deduct 6 wires from the total overall amount allowed.
  • Apr 19, 2009, 08:18 AM
    Tev

    Also when calculating your box fill, regardless of how many you actually have, the grounding wires count as a total of 1. The pigtails you won't count at all. Devices count as 2 for each yoke.

    As an example, the diagram TK provided requires a box suitable for 15 conductors.

    Here's how we can get that number

    1- all grounding
    4- neutrals
    4- hots (1 constant, 3 switched)
    6- 3 switches each with it's own yoke.
    15 total

    Now assuming you are using 12AWG wire, you need 2.25 cubic inches per wire so 15 X 2.25 will give you the minimum box size in cubic inches. If you use 14AWG you need 2 cu in per wire.
  • Apr 19, 2009, 08:26 AM
    Stratmando

    Also if a wire enters and exits the box and not cut, it counts as 1, cut it, and it becomes 2.

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