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Discussion Starter #1 (Edited)
Hello - I've done some simple wiring in the past and thought this would be fairly straightforward, but I'm now a bit confused. I'm looking to replace a light switch (in a closet, controlling a single overhead light) with a Leviton switch/plug combo (model 5225-I). I need the new switch to continue controlling the light, while the new plug remains constantly powered.

When I pulled out the current switch I discovered there are 3 romex cables running to the box (labeled in the image below as X, Y, & Z). All three white wires are together under a single cap (X-w, Y-w, and Z-w). All three black wires are connected to the current switch. Y-b runs to it's own brass screw. X-b and Z-b are each connected to the other brass screw (one via screw and the other plugs in the back). (The 3 grounds are all also together and connected to the ground screw on the switch).

Aside from moving the ground wire to the ground screw on the new switch/plug combo, any guidance for where I should run the other screws? There are two black screws, one silver, and one brass on the new switch/plug combo. Thanks for any advice you can offer.

Images at:
ibb.co/FzPS8Tz
ibb.co/KLfrPFJ


 

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First of all, you cannot run a screw. They are physically attached to the device. Find the line feed and simply connect that to both lines on the device. There will be no load for the outlet. There will be a load for the switch. Determine the load for the current switch, or use an ohm meter. You are not allowed to post a graphic here. You simply split a parallel circuit to have two devices. You have only one line, that will feed both line of the device. A neutral will go on the same side of the outlet. The load going to the light will be on the load side of the switch. Neutral coming back from the light will connect to the neutral coming into the box.


Draw it out on paper, a simple parallel circuit with each device.
 
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