Running electrical to a shed

Question

I have a shed in my backyard about 200ft from the house. I would like to run electrical to the shed to install 2 or 3 power outlets, and two LED lights for the inside.

Currently from the house to about 50ft from the house there is an exterior power outlet on a pole (no light on the pole, just the outlet). The line from the circuit breaker is completely dedicated to this one outlet the whole way -- currently no forks or splices. It is a 12 gauge UF-B wire run connected to a 20 amp gfci breaker. It is not in conduit and looks about 2 to 2.5 feet deep.

I am wondering if I can continue the line from the power outlet 50ft from the house on more 12 gauge UF-B for another 150ft to the shed -- which will have 2 or 3 power outlets and two LED lights for the inside. Or I am wondering if it would be required for me to have a dedicated line and a subpanel in the shed -- and maybe heavier wire.

Thanks!

Answer 1

Generically I'd knee-jerk to conduit and 30A 240V service with a sub-panel, but if your needs are going to be adequately met by 20A 120V power and the loading is sufficiently low that the voltage drop from 200 ft of 12Ga wire won't be a huge issue, yes, you can do that.

Given the relative expense of digging a ditch once, and never having to dig it again if the small cost of conduit is added, I think direct burial is a terrible idea, so I'd use conduit and THWN at least for the extra 150 ft, and I'd contemplate using heavier wire than 12Ga for that 150ft to reduce voltage drop from the long run. At which point I'd be looking hard at just digging another 50 ft and doing the whole thing to a higher standard, but you may not think the way I do.

Harper will stop in to say the 3% voltage drop rule isn't a rule (and it's not, merely a guideline), but if you get close to 20A at 200 feet on 12Ga copper you'll be down 12 volts or more - over 10% of 120V service, and generally not a good plan. But if the actual shed use will be down below 10A the 5% or so loss is unlikely to be a problem, until someone loads it up fully despite your planned use.

Answer 2

In the USA you get to play fast and loose with this. If you simply extend with 12 AWG, you'll have 0.6% voltage drop per amp that you draw... so a 1/4 amp battery charger won't give any drop at all, but a 14 amp lawnmower will give you about 8% drop - not good. For permanently installed wiring, I like to keep voltage drop under about 5-6%. So you're pretty much maxing out at 10 amps in my book. On the other hand, this is a far superior situation than that other thing you would do, running two 100' 16 AWG extension cords; that gives you voltage drop in the twenties of percent for the same lawnmower.

If you bumped to 10 AWG copper wire, at 16A max you'd be at 4.20% for this 150' section of the run, added to the 2.31% drop you already have for the 50' section, that'd be 6.51% @ 16A and proportionately less at fewer amps, so about 5% at 12A. I could probably live with that.

Since it's a GFCI protected single circuit, you can trench it at 12" of cover.

If you're in Canada, this is a mess because you're not allowed to have more than 3% voltage drop at 80% of breaker trip (16A). The 50' run already has 2.31%, and you'd pretty much need industrial service feeder to get 150' at 0.69% voltage drop @ 16A. So in Canada, you'd want to homerun from the house using 4 AWG aluminum to get under 3% drop. On the upside, Tim Hortons.

Answer 3

The code doesn't prohibit what you propose, it does (oddly) advise that minimum required size of wire may result in unacceptable voltage loss due to the resistance of the wire.

The voltage loss is dependent on the actual load, there are various "voltage loss calculators online that you can use to compare the effects of load and length. If you only really power lights and battery chargers it could work, but I would in no way recommend or be involved in such an installation.

Voltage loss is most often mitigated by using larger wire (that has less resistance), at very least start by using #10 or maybe even #8 for the rest of the circuit.