Number of circuits

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bigpapa400lb

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How many 120v, 2 wire, 20 ampere branch circuits are required to supply 52 parking lot lights which are rated 500 VA each?

I get 17 or 16.9 answer key says 18 circuits?
 
Here's what I tried. 52 x 500VA = 26kVA. Then take 26kVA / (1500VA/circuit) = 17.33 round up to 18. Not sure why you would use 1500 VA instead of 120x20=2400VA per circuit but it yields the answer they were looking for.

One quick question I have is when calculating number of circuits, do you use 1.25 for continuous loads? Or is that only for sizing the conductors and OCPD?
 
boku0003 said:
Here's what I tried. 52 x 500VA = 26kVA. Then take 26kVA / (1500VA/circuit) = 17.33 round up to 18. Not sure why you would use 1500 VA instead of 120x20=2400VA per circuit but it yields the answer they were looking for.

One quick question I have is when calculating number of circuits, do you use 1.25 for continuous loads? Or is that only for sizing the conductors and OCPD?
Click to expand...

I used 125% and the answer key does refer to 210.19(A)
 
14 amps

14 amps

14A is what my answer would be based on 52 x 500VA x 1.25 = 32.5kVA then divide by 120x20A to get 13.54A, round up to 14 circuits.

I was just throwing 1500VA out there as it is what you use for other circuits (small appliance and bathroom...) and it gets an answer of 18 amps. If I was taking an exam and 14 A was not one of the answer selections, I'd start trying other options.
 
I disagree with the other that 14 circuits are needed. We have to assume that these lights will burn continuously or for 3 hours or more. So at 2400 watts for a 20 amp circuit we must take 80% and we get 1920 watts per 20 amp circuit. Now with 1920 as the max for the circuit we can only get 3 lights (1500 watts) per 20 amp circuit. We have 52 lights so 52/3= 17.3 or 18 circuits. Answer key is correct, IMO

This is not a simple problem as total wattage being used since the circuits must be limited to 1500 watts simply because adding the 4th light would be at 2000 watts which would be over the 80%
 
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:grin:
Hats off to Dennis.
I (and I assume others) got caught up in the "total" calculations as to the number of circuits and didn't think it through as you did. Good job!
 
roger3829 said:
52 fixtues times 500va = 26000va

26000va/120v = 216.67a

20a @80% = 16a/circuit

216.67a/16 = 13.54 circuits or 14
Click to expand...

The above is mathmatical correct.

However Dennis is actually correct based on the physical limitations. We can't split the load from a fixture. So 3 fixtures per circuit would be the max.:)
 
augie47 said:
:grin:
Hats off to Dennis.
I (and I assume others) got caught up in the "total" calculations as to the number of circuits and didn't think it through as you did. Good job!
Click to expand...

Yep your right I took the total VA and not the VA for each circuit.

Good catch Dennis.:)

Chris
 
boku0003 said:
One quick question I have is when calculating number of circuits, do you use 1.25 for continuous loads? Or is that only for sizing the conductors and OCPD?
Click to expand...

You definitely have to consider 1.25 for continuous load which is the same as 80% of the max. load. Thus, as stated before, 2400 watts time 80% is 1920watts. So now we know why we can't load the circuit more than 1920 watts thus making 3 lights the max. we can use on the cir.

Here is where the 125% comes in. We start at the 1920 and multiply by 1.25 and we get 2400 watts. It just depends on your starting point. :) You cannot load the circuit more than 125% or you take 80% of the max load. The answer is the same 1920 watts.

Hope this helps. But it probably just muddied the waters
 
Dennis Alwon said:
I disagree with the other that 14 circuits are needed. We have to assume that these lights will burn continuously or for 3 hours or more. So at 2400 watts for a 20 amp circuit we must take 80% and we get 1920 watts per 20 amp circuit. Now with 1920 as the max for the circuit we can only get 3 lights (1500 watts) per 20 amp circuit. We have 52 lights so 52/3= 17.3 or 18 circuits. Answer key is correct, IMO

This is not a simple problem as total wattage being used since the circuits must be limited to 1500 watts simply because adding the 4th light would be at 2000 watts which would be over the 80%
Click to expand...

This is my calculation:
52x500= 26000VA
26000VAx1.25= 32500VA
32500/1920(2400x80%)= 16.9 = 17 circuits
But the way you calculate make since
 
bigpapa400lb said:
This is my calculation:
52x500= 26000VA
26000VAx1.25= 32500VA
32500/1920(2400x80%)= 16.9 = 17 circuits
But the way you calculate make since
Click to expand...

bigpapa, Since I made a mistake the 1st time, perhaps I should not comment, but, it appears to me that you have taken the "continuous" in twice in your calculations. If you multiply the 26000 x 1.25 that covers it, likewise if you changed your 2400 to 1920 you have covered it.
Once is fine.
The reason the most of us missed it was that we looked at the total load to determine the number of circuits rather than looking at the capability of each individual 20 amp circuit (1920 continuous)
 
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