Voltage drop increase the current?

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marcosgue

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Hello all, this's my scenario,
Temporary panel 240V, single phase feeding lighting in construction commercial building, the lighting load is supply through 20A, single pole circuit breaker, the breaker tripping time to time.
I've checked and I got measurement 21-22amp in the circuit, seem to me evident overload. I've already fixed adding another home run and divide the total load in two circuits. When I told my supervisor about this situation, he says the breaker was tripped due to voltage drop because when the voltage decrease, the current increase and trip the breaker.
I think he's wrong and also I don't believe what he says is the best example to show the relationship between voltage and current.
Because we learn everyday in this forum I'd like hear the opinion of the experts.
Thanks in advance
 
Well I'm not an expert, but I do know that a motor will try to draw more current to make up for a voltage drop (within reason).

On the other hand a resistive load is constant and if the voltage drops the current drops.

So, the answer is, it depends on the loads.
 
A lighting load which uses an active driver circuit (LED, fluorescent, etc.) can be designed to draw a constant power to the lamp regardless of small fulctuations in input voltage. So a 5% drop in voltage at the fixture could cause a 5% increase in the current.
On the other hand a 5% voltage drop will not cause a 10% increase in current!

Incandescent bulbs will deliver less power, but the current will not decrease in exact proportion to the voltage change since the resistance of the filament drops as it cools.

Regardless of the exact cause of the current, even a continuous load of only 20A is not going to be allowed on a 20A breaker.
 
As oldsparky said it depends on the load type. Did you calculate what the amount of load that was actually connected to the 20 amp circuit?
 
yes, they're LED luminaries and 100VA each, and the 20A breaker was carrying 24 of them, because they're continuous load shouldn't be more than 80% the rating of the circuit.
with this load in the circuit is possible "voltage drop" increase the current to trip the breaker or this's overload condition?
 
211112-1955 EST

marcosgue:

Considering nothing else you are overloaded at steady state conditions.

But more important in your design you must consider inrush current of the lights. This may be rather large.

.
 
Well gar like you say this's an evident overload condition that cause trip the breaker,
with this type of load and this specific situation, is possible the current due voltage drop trip the 20A breaker?
 
marcosgue said:
Well gar like you say this's an evident overload condition that cause trip the breaker,
with this type of load and this specific situation, is possible the current due voltage drop trip the 20A breaker?
Click to expand...
The connected load is 20 amps, your current readings exceed that by an amp or two. If it's not a VD issue raising the current then what could it be?
 
could be overload condition since the connected load shouldn't exceed 16A or increase the breaker size up to 25A
 
I say forget VD. Your overloaded, period.
If the lights are incandescent bulbs, like the old 130V construction bulbs, the current on them goes down with the voltage.
We see this VD as bulbs that are dimmer at the end of a 100 bulb string.
 
In this case, voltage drop was the straw that broke the camel's back.

As others have noted, the design was overloaded from the start. You should have no more than 16A if continuous load on a 20A circuit, but as designed this circuit was loaded right up to the full 20A at nominal ratings.

On top of this the LED drivers were probably acting as constant power loads.
For such loads as input voltage goes down the current goes up. Thus your nominal 20A load became an actual 22A load.

I bet you find that once the circuit is divided up that the total current goes down. The reduced current in each home run will mean less voltage drop.

Jon
 
Hv&Lv said:
Too many lights on the circuit.
Click to expand...
The number of lights on this circuit is 24 which have a total current of 20 amps. His readings are 1 to 2 amps above that. There is another factor that is pushing the load above the 20 amps.
 
infinity said:
The number of lights on this circuit is 24 which have a total current of 20 amps. His readings are 1 to 2 amps above that. There is another factor that is pushing the load above the 20 amps.
Click to expand...
But that's assuming the 100VA rating is exact and correct. I suspect a +/-10% tolerance is not unusual in such luminaires, especially since most of them these days are manufactured in China.
 
marcosgue said:
yes, they're LED luminaries and 100VA each,
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LED's & Inductive luminaries can't use VA for load calcs NEC 220.18(B).

UL always required rated Volts & Amp markings, which BTW include efficiency losses not reported by nameplate VA, or Watts after power factor; not the same as ballast factor.

See UL WHITE BOOK - (FKSZ)
DRIVERS FOR LIGHT-EMITTING-DIODE ARRAYS, MODULES AND CONTROLLERS

The proliferation of LED products sold with a (CE) mark, missing nameplate amperage, or UL / NRTL listing, does not mean you are excused or allowed to use VA markings for load calcs.

Rather, once installed in jurisdictions adopting NFPA-70, these LED products are subject to construction defect law, you are subject to joint & several liability, and the building becomes an un-insurable crime scene, regardless of how many years your client paid insurance premiums.
 
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Jon456 said:
But that's assuming the 100VA rating is exact and correct. I suspect a +/-10% tolerance is not unusual in such luminaires, especially since most of them these days are manufactured in China.
Click to expand...
You might be right given that the numbers are so close to the 20 amp limit of the circuit. In your opinion could VD contribute to increased circuit current?
 
Well Infinity, first in this type of continuos load installation is not right connect 20A load in 20A breaker. secondly, the inrush current I don't think cause tripping the breaker because the load is permanently connected and this current rises at the time to turn the circuit on.
to best of my knowledge to maintain the constant power, if the voltage decrease the current have to increase P=ExI, in the case of VD the fluctuation of current depend of variation of the load, let's say if the current goes down then the voltage should be down also V=IxR. Can you correct me?
one add information this temp circuit use NM #10awg
 
infinity said:
The number of lights on this circuit is 24 which have a total current of 20 amps. His readings are 1 to 2 amps above that. There is another factor that is pushing the load above the 20 amps.
Click to expand...
LEDs don’t work like that
 
marcosgue said:
Well Infinity, first in this type of continuos load installation is not right connect 20A load in 20A breaker. secondly, the inrush current I don't think cause tripping the breaker because the load is permanently connected and this current rises at the time to turn the circuit on.
to best of my knowledge to maintain the constant power, if the voltage decrease the current have to increase P=ExI, in the case of VD the fluctuation of current depend of variation of the load, let's say if the current goes down then the voltage should be down also V=IxR. Can you correct me?
one add information this temp circuit use NM #10awg
Click to expand...
That formula is true for constant power devices like motors, but to be true pf needs to be added to that DC formula to work somewhat with AC circuits.
For resistive loads as mentioned earlier that formula is useless.
LEDs aren’t AC. They power a DC driver.
Look at the pf on the driver.


Add that to your formula

Edit to add..
Look at the voltage rating on the led.
I bet it says something like “90-120” or even “90-277” volts.
Your voltage drops below 90 leds will generally start blinking or shut off.
Your applying a formula trying to make it work with constants when your forgetting about the variables.
 
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