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The thermal and magnetic protection for AC circuit breakers take into account both the timing and the magnitude of the AC current passing through.

Does a standard circuit breaker trip on the current related to the Apparent (kVA) Power, or the current related to the True (kW) Power?

I would understand they act on the current related to the True Power (kW), and the amount of Reactive Power (kVAr) should in some way (?) affect the tripping curve (such as the B, C or D in the miniature circuit breakers) or is there no effect and just the current related to the Apparent Power is which is effectively sensed by the internal coil?

Edit 1:

So Far:

  • The Circuit Breaker should act on RMS Current \$I_{RMS}\$, which is the current acting on the Thermal (Power Dissipation Time Average, over a Bimetalic element) and Magnetic (Magnetic Flow Time Average, magnetizing a ferromagnetic element) actions in a Circuit Breaker (at least, those depicted, whose operation is based on these actions).
  • The Power Factor of the load have no impact on the tripping action, hence the RMS Current, as in the Apparent Power (kVA), is the key factor, regardless the load is Resistive (kW) or Capacitive|Inductive (kVAr).

Edit 2:

Some confusions I have read in answers and comments.

  • Note that a trip in a CB do not imply a fault. They are absolutely different concepts. A trip can be caused by an steadily increase of the load connected to the CB, without involving any electrical fault at all. (The operation is affected and that would constitute a fault, but a design, operational, or management fault, not strictly an electrical one, at least compared with classical faults, discussion which escapes the scope of this question...).
  • A trip, hence, do not imply a drop in the supply voltage. That is a mistake, from the above confusion. We straightforwardly can think on counterexamples of this (which one?).
  • Apparent Power, Reactive Power, True Power, Impedance are definitions, and the question uses them. At the very end, Breakers and Distribution Panels are used to control Power.

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    \$\begingroup\$ The A in kVA is RMS current. That's what trips the breaker. Nobody wants to say that the breaker trips because of the "current related to total power", because it's not the right way to think about what's happening. There's no reason to bring power into it, and "total power" is not a real thing. It's just a lie told by some power meters that measure RMS voltage and current independently. \$\endgroup\$ – Matt Timmermans Feb 20 at 17:01
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    \$\begingroup\$ You are limiting your question to household type of circuit brakers like in the picture. There are bigger circuit breakers, typically for the higher voltages up to 700 kV and even higher, that have separate trigger circuits that can trip on much different conditions than just overcurrent, such as phase and voltage.. The sky is the limit \$\endgroup\$ – Roland Feb 20 at 18:44
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    \$\begingroup\$ FYI, kVA is generally referred to as 'apparent' power, not 'total' power. I would usually expect 'total' power to mean 'real' or 'true' power. \$\endgroup\$ – SomeoneSomewhereSupportsMonica Feb 21 at 4:41
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    \$\begingroup\$ It's always about current, not about power. I understand what's causing the confusion, but this is not how we talk about breakers. \$\endgroup\$ – Mast Feb 21 at 9:43
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    \$\begingroup\$ @Brethlosze var is used for reactive power (that's what the r stands for). Apparent power is V·I, which is what I think you mean when you say "total power". \$\endgroup\$ – Hearth Feb 21 at 18:44
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Thermal and magnetic circuit breakers trip on current, not power.

They know nothing about the voltage in the circuit until the contacts open and start to arc.


enter image description here

Figure 1. A single-phase circuit breaker. Image source Electrical Engineering Portal.

Note that in the single-phase circuit breaker there are only two terminals. When the circuit breaker is closed the voltage difference between the two cable terminals will be very small due to the very low contact resistance. As a result, the circuit-breaker is unaware of the voltage or power being drawn through it.

The three-phase breaker is usually three individual breakers with a mechanical link between them so that they all switch off if there is an over-current on one phase. Usually there will be no cross connection between the phases which would be required if voltage (or power) measurement was being done.


From the OP's self-answer below:

... so, a Circuit Breaker is actually tripping to the current related to the Total Power (kVA).

No, you are assuming that the voltage is constant during the fault. In general the voltage will drop / droop during a fault and that means that it could trip on a wide range of kVA. The circuit breaker trips on current only. You can, if it's useful, calculate the maximum kVA that the breaker can handle at nominal line voltage but that's not what you are asserting.

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  • \$\begingroup\$ So, a CB trip on the current related to the Total Power? \$\endgroup\$ – Brethlosze Feb 19 at 14:06
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    \$\begingroup\$ No. In a fault situation the voltage may collapse. The circuit breaker trips on current. \$\endgroup\$ – Transistor Feb 19 at 15:35
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    \$\begingroup\$ For example, a 20A circuit breaker is still 20A whether operated at 24VAC in a control system, or 240VAC mains, despite one being 480VA and the other 4800VA. \$\endgroup\$ – SomeoneSomewhereSupportsMonica Feb 21 at 4:40
  • \$\begingroup\$ @Transistor Fault may or may not imply a voltage drop. This will depend on the line model. In here were are just talking about the trip of a device with two actions (Thermal and Magnetic). But no, in the operation range of standard CBs, we wont expect a Voltage Drop. Whatever, question do not assume this. \$\endgroup\$ – Brethlosze Feb 21 at 16:49
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    \$\begingroup\$ @Brethlosze Faults almost always result in a voltage drop. If you're cynical enough, you might even say that preventing that voltage drop is the main purpose of circuit breakers, and saving lives is a secondary concern--because that voltage droop is a big problem for power companies. I would expect a voltage drop in almost any fault condition (not necessarily in an overload condition, but definitely in a fault). \$\endgroup\$ – Hearth Feb 21 at 18:42
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Circuit breakers don't trip on power at all. They trip on CURRENT. There is no such thing as "imaginary current" and so the circuit breaker will sense and trip on an over-limit current regardless of whether it's leading or lagging the voltage.

Keep in mind what "imaginary" power really is. There is no such thing, it's merely a conceptualization that follows the convenient complex math that models what is really going on. Imaginary power is really just the voltage leading the current or the voltage lagging the current in a reactive circuit vs. the voltage and current tracking each other in a purely resistive circuit.

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  • \$\begingroup\$ So the CB should break in the current related to the Total Power (?) \$\endgroup\$ – Brethlosze Feb 19 at 13:56
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    \$\begingroup\$ As in Total Power has nothing to do with it! \$\endgroup\$ – StainlessSteelRat Feb 19 at 14:35
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    \$\begingroup\$ @Brethlosze There's no such thing as a "current related to the Total Power". Apparent power is a function of the voltage and current. Real power is a function of the voltage, current and phase shift. The same current is "related to" both of them, and that current is the only thing the breaker sees. \$\endgroup\$ – TooTea Feb 21 at 9:37
  • \$\begingroup\$ @TooTea The current is \$I=P/V\$, where \$P\$ is the Total Power, \$V\$ is the Measured RMS voltage, and \$I\$ is the Measured RMS Current, which is actually the current the CB senses for their Thermal and Magnetic actions. Hence, in numbers, a CB will trip on the current related to the Total Power. \$\endgroup\$ – Brethlosze Feb 21 at 15:53
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    \$\begingroup\$ @Brethlosze Note that \$P = IV\$ which means for the same power, current could be very high if voltage is very low, or vice-versa. One important reason for circuit breakers to exist is that heat generated by current in wiring could cause a fire. Power carried by a wire does not cause as much heating if the current is very low and the voltage is very high, but power composed of high current and low voltage can cause significant heating. A CB carrying 110 amps and 1 volt (110 watts) will trip, the same CB carrying 1 amp and 110 volts (also 110 watts - the same total power) will not trip. \$\endgroup\$ – Todd Wilcox Feb 22 at 12:10
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The thermal breaker trip is based on RMS current. The bimetal is heated by the current, so current waveforms that are peaky (high crest factor) will cause more heating. There is no knowledge of power- neither from a voltage or from a power factor point of view- the same RMS current into a purely reactive load (zero power) will cause the breaker to trip the same as if the RMS current is into a resistive load. The breaker will trip at the same RMS current whether it's connected to a 60VDC supply or a 600VAC supply.

Magnetic tripping is typically intended to quickly open in case of a short and works on current alone.

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  • \$\begingroup\$ Actually, AC and DC current do not trip a breaker in a same way,. These are different equipment. Besides, a AC current with a DC voltage would be deduced from an AC Impedance (?). So, we can not ride of frequency in here, and keep just RMS. This is wrong. \$\endgroup\$ – Brethlosze Feb 21 at 15:46
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    \$\begingroup\$ @Brethlosze The AC breakers are generally not nameplate rated to break DC and should not be used as such, obviously, but that doesn't change the fact that they will respond similarly to the same RMS current. \$\endgroup\$ – Spehro Pefhany Feb 21 at 18:18
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Circuit breakers don't care a whit about power, real or reactive or otherwise. The only thing they care about is the current. If you were to take a breaker rated for 10 amps at 250 volts and run 20 amps through it at 5 volts, it would trip exactly the same as if you ran 20 amps through it at 230 volts, or 20 amps at 120 volts, or 20 amps at 0 volts!

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  • \$\begingroup\$ This do not answer my question. For a given condition, fault or not, it then do trips for the current related to the Total Power (kVA)? \$\endgroup\$ – Brethlosze Feb 21 at 15:20
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    \$\begingroup\$ No, it does not trip total power or real power or reactive power or anything--it trips purely and only on current. \$\endgroup\$ – Hearth Feb 21 at 18:20
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    \$\begingroup\$ 20 amps at 0 volts is actually convenient if you're trying to demonstrate how a circuit breaker functions \$\endgroup\$ – user253751 Feb 22 at 16:57
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It is correct to say Circuit Breakers "don't know anything about power", and "the only thing they care about is the current", however, in a real-world application, the supply voltage is constant, so it is also correct to say circuit breakers open on kVA, not kW, because kVA is always current-multiplied-by-voltage (constant), whereas kW is not. Note: 'imaginary' current does actually exist, however it is the fraction of total current which is flowing in and out of an energy storage element, ie, an inductor (magnetic storage), or capacitor.

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    \$\begingroup\$ The supply voltage is not constant, though. That's kind of the definition of a fault condition. If you try to pull a kiloampere through house wiring, you're not going to get the nominal grid voltage, and you'll probably get a fire to boot if the circuit breaker isn't still able to trip at the drastically reduced voltage! \$\endgroup\$ – Hearth Feb 20 at 5:33
  • \$\begingroup\$ @Hearth - light bulb moment! People are discussing the two different modes of operation of circuit breakers, and coming to differing conclusions, it seems: If an overload condition occurs resulting in perhaps three times rated current, a circuit breaker willl trip after a certain known time (thermal operation - internal element reaches a certain temperature), with voltage remaining more or less constant. I think this is what the OP is asking about, from the phrasing, and so for design purposes in this condition, it is accurate to assume constant voltage. \$\endgroup\$ – michaeldthorpe Feb 20 at 6:15
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    \$\begingroup\$ Be careful with the wording. kVA is "measured voltage" times "measured current". kW is based on the instantaneous product of voltage and current. The "imaginary" current is not imaginary in any real sense of the word, it is simply current that is out of phase with respect to the voltage. \$\endgroup\$ – Elliot Alderson Feb 20 at 16:30
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    \$\begingroup\$ kVA=kW+kVAR Are we on the same page, @ElliotAlderson? \$\endgroup\$ – michaeldthorpe Feb 21 at 11:03
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    \$\begingroup\$ No, \$\text{VA} = \sqrt{\text{W}^2 + \text{VAR}^2}\$ You measure the voltage. Then you measure the current. Multiply and you get VA. To get watts you must multiply the instantaneous voltage times the instantaneous current. \$\endgroup\$ – Elliot Alderson Feb 21 at 12:45
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Circuit Breaker is actually tripping to the current related to the Total Power (kVA).

What do you mean by "current related to the Total Power (kVA)"? Do you mean the part of the current which is in phase with the voltage? If that where true a circuit breaker would never trip with a pure imaginary load. And that's obviously not what happens in reality. So, as has already been stated several times there is no relation to power. It's only the current which triggers tripping of a circuit breaker.

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  • \$\begingroup\$ Welcome to SE/EE! Please take the tour and read some of the other help pages to learn how to use this site. It is not a forum, so you don't post an answer if you have a question to any other post. Post it as a comment to that post. \$\endgroup\$ – the busybee Feb 20 at 17:41
  • \$\begingroup\$ Unfortunately you need a reputation of 50 before you can post comments. A couple of good questions or answers will get you there. \$\endgroup\$ – Transistor Feb 20 at 21:13
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Most answers and comments focus on the ability of a circuit breaker to turn off on a short circuit fault condition, as indicated by markings like: 50kA, but this is only half the story.

The other important function of the type of circuit breaker as shown by the poster is to protect wires and motors from overheating, as indicated by the I-T curves and by characterizations as "slow acting" or "fast acting". For this task the circuit breaker tripping circuit utilizes a bimetal that heats up from high currents, where "high" means a value close to the maximum operating current. Heat generated in the bimetal is proportional to I-squared, so lower currents have almost no effect on the bimetal. The bimetal performs an itegration of current over time to arrive at an estimate of the energy. When exceeding some threshold, the curved bimetal trips the circuit breaker.

To answer the exact question: the bimetal measures actual currents without regard to voltage, let alone phase angle, so the voltage is the normal line voltage and you may say that the circuit breaker trips on apparent power, commonly called kVA, Not real power as understood as the opposite of apparent power, commonly called Watt, W, kW.

For example, to protect the circuit of a 1 kW motor, you must also know its power factor, e.g. 85%. For simplicity, assume single phase, 1 kV. The normal operating current will be 1 A divided by 85% hence 1.2 A. Multiply by the voltage to arrive at real versus apparent power. You need to look at the I-T curves for 1.2 A. You don't want to trip if the motor humms away nice and smoothly for, say, one hour, so you will find out that your circuit breaker must be a bit bigger than 1.2 A. On the other hand, suppose your motor blocks or stalls, then the current will go up quite a bit, and the circuit breaker better trips after not too long, or your motor or wires may emit some smoke, or worse.

When tripping on overload, the voltage is just the normal line voltage when your lights don't flicker. In this case you may measure (three-phase) power.

When tripping on a short circuit, the voltage goes to zero, the lights will go out, and it makes no sense to measure power in the affected phase, as others correctly commented.

To make things a bit more complicated is that Fast and Slow is normally not selected to wait for a long time to switch off on overload, but to not trip on the inrush current of a motor when it is starting up.

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  • \$\begingroup\$ This is the FIRST answer who understand the quid of this question. \$\endgroup\$ – Brethlosze Feb 26 at 2:52
  • \$\begingroup\$ @Brethlosze is there still a part of your question unanswered? \$\endgroup\$ – Roland Mar 1 at 13:38
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Circuit Breakers trips on Current, not Power. Though we can have an Imaginary Current with respect some reference phasor (i.e. a voltage), the Circuit Breaker is not measuring the relative phase between current and voltage.

Because the Circuit Breaker measures the excitation over a set of coils, both over the Thermal and the Magnetic action, none of them takes into account the phase between voltage and current, so, a Circuit Breaker is actually tripping to the current related to the Total Power (kVA).

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  • \$\begingroup\$ that would make some presumptions about how the V is related to the current, wouldn't it? (ps I didn't downrate your answer) \$\endgroup\$ – Pete W Feb 19 at 14:25
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    \$\begingroup\$ @Brethlosze: Circuit breakers don't know anything about power. They are much like fuses. A 1 ampere fuse will blow at 1 ampere - whether the circuit is powered from 1V, 12V, or 50V. It blows depending on the current only. \$\endgroup\$ – JRE Feb 19 at 15:08
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    \$\begingroup\$ V being proportional to I is only true in some circuits. And certainly not in a fault condition, where even if V is proportional to I, the constant of proportionality will change! \$\endgroup\$ – Hearth Feb 19 at 16:46
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    \$\begingroup\$ "... so, a Circuit Breaker is actually tripping to the current related to the Total Power (kVA)." No, you are assuming that the voltage is constant during the fault. In general the voltage will drop / droop during a fault and that means that it could trip on a wide range of kVA. The circuit breaker trips on current only. You can, if it's useful, calculate the maximum kVA that the breaker can handle at nominal line voltage but that's not what you are asserting. \$\endgroup\$ – Transistor Feb 19 at 17:19
  • \$\begingroup\$ @Hearth Impedance $V/I=Z$ and Total Power $P=VI$ holds for ALL circuits, no matter if you are in fault or not, no matter if it is constant in time or not. \$\endgroup\$ – Brethlosze Feb 21 at 15:07

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