The circuit will both be at higher voltage (well the parts of it) and will self-adjust. The reasoning behind this is pretty simple: First, the phases are out of phase, or to explain it simply, their voltage peaks happen at different times, so while one is at its maximum voltage, the two other phases will be at lower voltage than the one at maximum is, so basically, the load will be at the voltage between phases. Since we don't have the neutral conductor, the current which was going through it will have to go through one of the phases and that's how the circuit will "self-adjust". You can see that from the simulation here.
When the switch is closed, all 3 loads are at the same voltage of 5 V compared to the ground. Once the switch is open, the voltages at the loads will change. So the lightest load will drop the least voltage and the greatest load will drop the most voltage.
Tl; Dr
The order of breaker trips is usually laid out in a selectivity calculation. Usually the closer to the source (generator in this case), the later the breaker will trip. The order of trip is not determined by the breaker's capacity, but by it's time current curve. This is roughly speaking a logarithmic multiplication of the overload current and the time it lasts.
However, in a ring-net system like mentioned in the question, it is almost impossible to be selective enough using "time current" to cover all configurations and possible faults. In such cases there are systems which determine the "location" (Zone selectivity) of the short circuit by measuring the direction of the current, as mentioned in Transistor's answer or in a more digital way in modern breakers. There are systems where breaker communicate this data to make a final determination which ones should open. Or other methods. This page from ABB has some nice graphic examples about advanced selectivity techniques.
Order of protection
More explained below, but looking to the diagram the following order of protections should kick in, depending on the seriousness of the fault:
- Earth fault monitoring warning / alarm
- Earth fault monitoring trip (optional)
- Differential current protection trip
- Overload protection as per time current curve and possibly zone selectivity
- Short circuit protection according zone selectivity
More on time current curves
From this source, with more detailed explanation.
Time-current curves are used to show how fast a breaker will trip at
any magnitude of current. The following illustration shows how a
time-current curve works. The figures along the bottom (horizontal
axis) represent current in amperes. The figures along the left side
(vertical axis) represent time in seconds.
To determine how long a breaker will take to trip at a given current,
find the level of current on the bottom of the graph. Draw a vertical
line to the point where it intersects the curve. Then draw a
horizontal line to the left side of the graph and find the time to
trip. For example, in this illustration a circuit breaker will trip
when current remains at 6 amps for 0.6 seconds.
It can be seen that the higher the current, the shorter the time the
circuit breaker will remain closed. It can be seen from the
time-current curve on the following page that actual time-current
curves are drawn on log-log paper, and the horizontal line is in
multiples of the breaker’s continuous current rating. From the
information box in the upper right hand corner, note that the
time-current curve illustrated on the following page defines the
operation of a CFD6 circuit breaker.
For this example a 200 ampere trip unit is selected.
Overload vs Short circuit
An overload constitutes an excess of AC current during an amount of time. Those are thermal protected according the time current curves noted above. Short circuits however are more violent and are magnetic protected. The above mentioned sources (and google) also give you more info about this subject, which is an important background.
Diagram in question
The above part hopefully answers the text part of the question, which specifically mentions over-current trips. However, the diagram attached to the question actually has an earth fault. This means 1 or more phases is connected to earth and doesn't necessarily creates an overload or short circuit current.
Measurement of earth faults
The insulation value of the installation's insulation level is usually monitored and alarmed by an earth fault protection system. Depending on the installation requirements an earth fault can lead to a trip, or not. I found a data sheet for such a device on the market, which gives some more background on this subject.
Single phase earth fault
This should not give any overload or short circuit currents. However, it may affect the insulation of the installation cables. When 1 phase is connected to earth, it's phase-earth potential becomes 0V. The other 2 phases will have an potential voltage to earth multiplied by √3. If the installation is not capable of handling this increase of voltage, the fault should be terminated by opening breaker.
From this article:
These insulation levels are discussed as follows:
100% level:
Cables in this category may be applied where the system is provided
with relay protection which normally clears ground faults within 1
minute This category is usually referred to as the grounded systems.
133% level:
Cables in this category may be applied where the system is provided
with relay protection which normally clears ground faults within 1
hour This category is usually referred to as the low resistance
grounded, or ungrounded systems.
173% level:
Cables in this category may be applied where the time needed to
de-energize the ground fault is indefinite This level is recommended
for ungrounded and for resonant grounded systems.
If the installation uses cable in the first, 100% category, the earth fault measurement should cause a breaker trip. In the other 2 cases it will just raise an alarm to the plant's operators and they should intervene within the time limitations noted above. Usually earth faults are logged to not exceed the cable's service life.
Differential protection
If there are any currents to earth or mismatch in phase currents, which doesn't necessarily case an overload or short circuit, there is a differential protection which should trip breakers. The question already includes a differential measurement around the fault, which means that this should act and not any over-current relay, only if there is in fact a current to earth. A nice read on this subject can be found here.
More reading
It can be a nice start to read about the various levels of protection in this wikipedia article and use google on the terms used.
Best Answer
Unfortunately not.
An Residual Current Breaker trips only if the sum of the current on all four wires is nonzero. This residual current means there's a leak to earth, possibly through a human, and the breaker trips.
In your case, the loss of neutral meant the supplied voltages became unequal, some too high, but this won't make the total current nonzero.
There may be other devices which can protect you, but the basic RCB won't.