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nestordt · Feb 22, 2009, 09:52 AM

Hello all:

I want to get started installing aprox 6 electric outlets in my basement but still have a doubt about it.
After making the study of which would be the best route/circuit to use, I've found I can only use one, from two going to the basement: circuit having the garage door (and light), garage lights: flourescents and light bulb, garage outlet (rarely used), 2 outlets (bsmnt washroom being one).
The question is: would be too much for a circuit to hold all the described plus the 6 new outlets planned to install? Would power in new outlets be affected every time garage door is working? My garage door is 1/2HP.

The other circuit coming down to the basement seems to have a big load already (5 outlets in living and dinning, 1st floor wshroom light and outlet, 2 lights in rec room, 1 light in bsmt washroom, 1 outlet in utility room), so it is discarded.

Thanks,

nestordt · Feb 22, 2009, 09:54 AM

In addition, I can say I am aware of the load every circuit must hold (120x20x80%, 120x15x80%), and number of devices recommended, but never heard about how a garage opener can affect performance of other devices according to how many the circuit holds. Thanks!

donf · Feb 22, 2009, 09:55 AM

What is the amperage on the existing circuit (Breaker size)? What is the amperage draw of the garage door opener?

donf · Feb 22, 2009, 09:59 AM

Nestordt,

You are not aware!

Power to "Continuous loads" < defined as an active load for 3 or more continuous hours> is derated by 80%. For example a water heater and lights.

As well as "Plugged loads"

nestordt · Feb 24, 2009, 05:03 PM

Thanks Don. The breaker of the circuit holding garage door, garage lights, garage outlet, and basement washroom, is 15 amp. Such circuit has 5 devices (2 lights in garage).
I am familiar with alternate and continuous currents concepts but after being said by an electrician that W=VxAmp can be used as an aproximation, I sent my question out here to confirm it.
My garage opener is 1/2 HP, which after found formulas and equivalences would give me a 3.1 AMP (in electric motors 1 HP = 746 watts, then 1/2 HP = 373 watts. If watt= volt x amp --> 1/2 HP would give me 3.1 amp) PLEASE correct me if this formulas/equivalences are false.

Thanks again for the feedback.

Missouri Bound · Feb 24, 2009, 06:13 PM

It's unlikely you will have a problem... however... expect the lights to blink a bit when the opener operates... and if you happen to have a computer on a circuit with a motor you [I]will[I] have some issues down the line.

stanfortyman · Feb 24, 2009, 07:40 PM

Originally Posted by donf

Power to "Continuous loads" < defined as an active load for 3 or more continous hours> is derated by 80%. For example a water heater and lights.

As well as "Plugged loads"

Just a quick correction.

A continuous load is one that is expected to be on for three hours or more.
Most residential lighting does not fit this description. In fact, not much in a home does.

Certain things though, like fixed electric heat or a water heater as Don mentioned, must also be considered "continuous loads".

I'm not sure what "plugged loads" are, but something being plugged in has no bearing on it being a continuous load.

Stratmando · Jun 26, 2009, 03:14 PM

If you are installing outlets, I am guessing walls and ceiling may be exposed, I would try to get its own or a conduit to one of the downstairs boxes for future,
Don't forget TV, Tel, Computer, Speakers, alarm, intercomm or other before walls go up, doen't get cheaper or easier than now, 3 way switching for lights?