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#212881 02/25/14 05:01 PM
Joined: Oct 2002
Posts: 830
Sorry to bother you guys again, but another question come up. Sorry to be so dumb, but there's a lot in the code:( The emergency lights that I may have to use (28 of them) on this job I have, have nameplate ratings of 120 volts, 5.2 watts, Max. amps .052, and VA 6.2 These are recess emergency lights that the architect has drawn in for this project. If I size the circuit according to the "Max amps" it comes up to 1.5 amps. ( 28 lights x .052 amps), and if I size it according to VA's it comes up to 173.6 amps, which is not feasible I know. If using VA's to size them I will have to put more circuits of course. Which is the way the code requires? Hope it makes since. Grandson just got in my lap, so I may have made a mistake:(

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Joined: Jun 2004
Posts: 1,273
Your eyesight is going...

0.52 Amps would be a typical figure -- never 0.052.

"Max amps" means that you're charging up the battery while running the lamp. This is an interesting figure -- but irrelevant for circuit design. Such a surge, by definition, is an intermittent load.

Putting 28 EXITS on one circuit is only possible if the circuit is 277VAC... And that's really pushing it. I saw such a circuit in the Plans and Specifications for an Albertsons in the Bay Area.

It was converted, in the field, to three 120VAC circuits. None of the Albertsons supplied EM fixtures was able to tolerate 277VAC. While the electronic ballast could handle that voltage -- the battery system could not.

Due to mass production economics, it makes no sense to manufacture 277VAC battery packs for the itsy bitsy EM market -- when the 120VAC stuff is still flying off the shelf -- and 120VAC is ALWAYS available whenever 277VAC is the Service voltage.

Beyond that, any D-B scheme is not going to want EM circuits daisy-chained forever. You're going to suffer brutal voltage drops -- unless you're willing to insanely up-size your branch conductors.

EM fixtures are ALWAYS spread out. They are best served by dedicated circuits in most commercial settings. That way smart switching -- and overrides -- can be effected at the distribution panel/ lighting control circuit.

So, without even looking at the print, I'd say that a practical scheme is going to require three to four dedicated circuits. Another option is to have fewer circuits but branch them away/ split them for voltage drop the moment they leave the distribution panel. DON'T attempt to daisy-chain them on and on.

Examples: Perimeter lighting -- exterior facing illumination that wraps the building.

Plan A -- Branch circuit starts at panel and wraps the building from one direction -- clockwise. Voltage drop by the end is brutal. Pipe and wire up-sized to #10.

Plan B -- Branch circuit starts at panel -- then splits its load in half in a j-box / gutter right over the panel. One leg heads off clockwise -- the other heads off counter-clock wise. While both are tapping a single breaker, each #12 conductor sees only its half of the lumiere load.

Two half-wraps complete the perimeter illumination demand. Neither is required to up-size pipe or conductors.


You need to go with Plan B ... with the EMs replacing the wrap logic/ layout expressed above.


BTW, you should face max amps only once: at start. After the building burns down, it will face max amps a second time.

Joined: Jul 2004
Posts: 9,788
Likes: 14
Tesla he said 5.2 watts at 120v


He also said 6.2va max
6.2/120= .0516666 so that is where .052 comes from.

I am betting LEDs.

Doing it all on one circuit sounds NEC compliant but whether it is a good design is another question.

Greg Fretwell
Joined: Oct 2002
Posts: 830
First of all, I apologize. The information that was given to me was: Supply voltage (120 volts), AC input watts (5.8), Max. amps (.052), VA 6.2
Everything was right except the AC input watts.

2nd. I said "emergency lights, not exit lights (they are like recess can lights but has battery back up. Exit lights are another story.

Thanks again..

Joined: Jun 2004
Posts: 1,273
Watts = Amps x Volts (RMS-AC) x power factor

5.8 W = 0.052 A x 110 VAC x pf ?

VA = 6.20

So the power factor is... 93.5%

(The formula wheel assumes that the power factor is One.)

I've never seen an EM fixture with consumption quite that low... Jumped to a confusion.

My bad.

Last edited by Tesla; 02/25/14 11:01 PM.

Joined: Jul 2004
Posts: 9,788
Likes: 14
I was just guessing LED. Don't those things feed power back into the grid by now wink

Seriously I am always amazed by a new, better LED product but not surprised.

Greg Fretwell
Joined: Apr 2002
Posts: 7,333
Likes: 7

Beside the math, don't forget 700.12 (f) (exception 1)
IF this is the same job you referred to in another post.

Joined: Oct 2000
Posts: 2,722
Broom Pusher and
The Apparent Power draw of 6.2 VA per Fixture sounds correct for the Battery Charging Load of a "Normally Dark" Emergency Lighting Fixture.

The Lamp is not Illuminated until the Line Input has an outage, so the only Load connected during normal operation will be the Charging Circuit.

As such, the total connected Load for the (28) Emergency Lighting Fixtures will be:

6.2 VA * 28 = 173.6 VA (174 VA)...
174 VA = 1.45 Amps at 120 VAC

Figure in LCL, and the maximum demand Load will now be:

174 VA * 1.25 = 218 VA...
218 VA = 1.82 Amps at 120 VAC

Unless Specifications or Code Article requires a Dedicated Emergency Lighting Circuit, connect the Emergency Lighting Fixtures to the same Branch Circuit(s) used for the General Lighting in the same area - with the EM Lighting connected at the Line Side of the Switches for the General Lighting
("Ahead Of" the Switches - i.e.: fed directly from the Branch Circuit Breaker at the Panelboard).
Do the same with the Exit Signage as well.

--Scott (EE)

Scott " 35 " Thompson
Just Say NO To Green Eggs And Ham!

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