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No RCD tel, but why can't current leak line to neutral through a breakdown in insulation. Not a dead short, but enough to drop voltage across?
Because it would have tripped the MCB. Suppose the circuit has R1+Rn of one ohm (pure guess but close enough for the explanation) that R1=Rn, Zdb is low and that the faulty insulation is concentrated in one area near the end. The difference between voltages L-N and L-E was 245-210 = 35 volts, which would have to be the voltage drop across Rn between the DB and the fault. The current through the fault would be 35/0.5 = 70 amps. If it hadn't tripped the MCB for whatever reason (e.g. the circuit was wrongly connected to a 63A MCB) the heat dissipated would be 70 x 210 = 14.7kW which would rapidly start a fire. You can't dissipate the heat of twelve 2-slice toasters at a fault without it getting, er, toasted. Since a further 35V would also be dropped across R1, the supply would have to be 245+35 = 280V which would be causing other problems elsewhere. FWIW the 'insulation' resistance would be 210/70 = 3 ohms, pretty close to what one would legitimately call a dead short.
All lights that weren't working were disconnected for testing the voltage
This is key. With the lights connected, the voltage would probably have collapsed to a much lower level. The neutral is almost certainly open-circuit at a terminal and what you are reading is more or less a ghost voltage.
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