Would voltage drop affect a breaker from opening during a short circuit or fault condition?
At short circuit, the voltage at the breaker is near 0, thats the reason of a "short" circuit.
So voltage drop doesnt seem to affect, since voltage is "0".
In fault condition you have to be more specific, there are several types of faults conditions. phase to phase, phase to ground, two phase to ground, open phase, and so on.
Would voltage drop affect a breaker from opening during a short circuit or fault condition?
The breaker in question is a 1P- 277V- 20AMP GE TEY,thanks....
we're talking about a relatively small amount of voltage drop, ( 3% to 10%, and thats a high value as it is)
Your talking about the normal running load voltage drop.
Under short circuit conditions the voltage drop will increase dramatically as the load increases dramatically. But ultimately it is about impedance.
Resistance/impedance is current limiting, so yes if it doesn't allow enough current to flow in a fault it can keep a breaker from opening, but with out the specifics of the circuit, this can not be chiseled in stone.
Circuit wiring consists of:
Est length-240'-275'
#12 THHN CU-in the form of 12/2-MC cable.
277v load-Est to be 4-6amps of flourescent lighting.
The condition from the information I received (I was not directly involved) is a- phase to ground condition.
091002-0617 EST
macmikeman:
The resistance of #14 cooper at 68 deg F. is 2.525 ohms. Double this and the loop resistance is 5.05 ohms. Assume 120 V to the 2000 ft loop and the current is 23.8 A. Now answer your question.
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Resistance/impedance is current limiting, so yes if it doesn't allow enough current to flow in a fault it can keep a breaker from opening, but with out the specifics of the circuit, this can not be chiseled in stone.
Circuit wiring consists of:
Est length-240'-275'
#12 THHN CU-in the form of 12/2-MC cable.
277v load-Est to be 4-6amps of flourescent lighting.
At 275' your circuit would produce a bolted fault current of 2505.6 amps, way more than enough to open that breaker in less time it would take to damage the conductors.
At 240' it would produce a 2871 amp bolted fault.
This is an interesting thing to think about.
In many breakers it does take some small level of voltage to power the instantaneous trip coil. So it seems like- in theory -that a short circuit with low enough impedance between it and the breaker could limit the available voltage to such a low value that there is not enough voltage present to trip the instantaneous coil?
Of course the thermal overload, which does not require nearly as much voltage (more current dependent), would eventually trip the breaker provided the short circuit current was above its trip levels.
The voltage is only at zero at the point of the fault, not at the breaker, the resistance of the conductors, between the breaker and the point of fault will only lower the voltage to the point of the voltage drop of the applied fault current, it will be the voltage drop of the conductors ahead of this breaker, that will determine the voltage at the breaker.
The voltage is only at zero at the point of the fault, not at the breaker, the resistance of the conductors, between the breaker and the point of fault will only lower the voltage to the point of the voltage drop of the applied fault current, it will be the voltage drop of the conductors ahead of this breaker, that will determine the voltage at the breaker.
Thanks Gar, that does the trick. Like I said, I never attempted to try it out myself to see, I considered the very idea of it pretty stupid actually. When you think about it, 23.8 amps on a 15 would of course trip the breaker, but such a load is not all that uncommon after hacks get their hands on a lighting circuit......