Is it possible add an outlet to an existing light switch where the outlet is switched along with the light? The switch is not at the end of the run. So far I've only been able to find answers where the outlet has constant power.
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1Have a look at the maximum current rating of the switch and select fuses accordingly. Note that a switch increases the resistivity of the complete circuit. A 10 A fuse will likely trigger too slow for a 10 A switch. See en.wikipedia.org/wiki/Circuit_breaker#Standard_current_ratings– Jonas SteinCommented Aug 3, 2017 at 20:36
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3 Answers
Yes, that can be done.
- Connect the neutral from the light, and the neutral from the receptacle to the feeder neutral.
- Connect the feeder hot to one side of the switch.
- Connect together a wire from the other side of the switch, the hot to the light, and a wire to the hot side of the receptacle.
- Connect all grounding conductors.
If you're a more visual learner, search google for "switched receptacle wiring".
answered Aug 3, 2017 at 11:22
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Thanks. I was thinking that the outlet could also come off of the switch, but I found this, which has the light coming off of the outlet. do-it-yourself-help.com/images/…. My outlet will be directly below the switch, so that shouldn't be difficult.– JNoCommented Aug 3, 2017 at 17:37
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1The light and receptacle can both come off the switch. You'd just have to use a pigtail, since you can't connect two wires to a single screw terminal on the switch. Commented Aug 3, 2017 at 17:45
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@tester101 (Resident electrical guy) Just looking at the images above. Is it correct that both the box and the outlet need grounding? Don't the wings ground the box? Commented Aug 3, 2017 at 18:59
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1@ChrisCudmore: If a device is self-grounding (and mounted to a metal junction box) then you don't need to jump the EGC to the device. But you always need to bond the EGC to a metal junction box. That is, you can ground devices via the box, but not the other way around. I assume the reason for the asymmetry is that if the device were to be removed and/or not be self-grounding, then you'd have no bonding between the box and the EGC. Commented Aug 3, 2017 at 22:28
Thanks Tester101. For the visually oriented:
Adding the outlet between the switch and the light. Outlet is switched with the light.
Outlet and light come off of the switch. Outlet has constant power (Thanks brichins). To make it switch with the light, follow the instructions in Tester101's answer above.
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I think your 2nd image is wrong for what you're asking - tracing the wire back from the outlet, it appears that the outlet would have constant power instead of being switched. Moving the outlet's black lead wire from the "hot source" to the other side of the switch (where the light's black is connected) would make it switched with the light.– brichinsCommented Aug 3, 2017 at 20:45
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For comparison / illustrating the difference, I would note another option I personally prefer much more - a partially-switched outlet, allowing flexibility of use. In either arrangement shown above, this would be done by using a 3-wire cable from the switch to the outlet. At the switch, one lead is connected directly to the source and the other to the disconnected side of the switch. At the outlet, the jumper on the hot side is removed, and the 2 leads connected separately to top and bottom receptacles. In the top diagram, the light would also be connected to the switched leg.– brichinsCommented Aug 3, 2017 at 20:53
As long as the outlet is after the switch then it will work when the switch is energized. I have a 4 outlet at the end of my garage light wiring. If the lights are on it works. If it overloads the circuit it's on you'll know when the breaker pops but there generally aren't limits on outlets per circuit just be aware the amps being pulled
