Bay 4 sub-main: undersized cable or a hidden series resistance?
During a night shift at a coaching depot, the 230 V single-phase sub-main feeding pit lighting and socket outlets in Bay 4 dims visibly whenever a 2 kW portable heater is switched on. The sub-main is 40 m of 4 mm2 copper twin-and-earth drawn in conduit from the depot main board; take copper resistivity as 0.0172 ohm·mm2/m. With the heater on, a true-RMS clamp meter reads 8.6 A and the socket voltage falls from 231 V off-load to 217 V, while the main board busbar holds steady at 232 V. Insulation resistance of the sub-main, tested at 500 V DC with all loads disconnected, is 180 Mohm. The socket earth loop impedance is 1.9 ohm and the 30 mA RCD on that way has never tripped. The bay is due to be released for planned maintenance in 6 hours. The shift supervisor wants a decision now: keep the circuit energised until then, or isolate it?
1. Calculate the voltage drop that the sub-main conductors should produce at 8.6 A, compare it with the measured 14 V, and state what the difference tells you about the circuit.
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Loop resistance, not one-way resistance. Current goes out on the line and returns on the neutral, so use 2L: R = rho × 2L / A = 0.0172 × (2 × 40) / 4 = 0.344 ohm.
Expected drop = 0.344 × 8.6 = 2.96 V, about 1.3% of 231 V - comfortably inside the 4% normally allowed for a sub-main.
Measured drop = 231 − 217 = 14 V, which is 6.1%.
Apparent loop resistance = 14 / 8.6 = 1.63 ohm, i.e. about 1.29 ohm more than the cable can account for.
Interpretation. Roughly 79% of the loop resistance is not in the copper. It sits in a loose terminal screw, a corroded joint in a conduit box, a poorly made crimp or a partially broken conductor. Insulation resistance of 180 Mohm rules out an insulation or leakage fault, so this is a series defect. The energy dissipated in that hidden resistance is I²R = 8.6² × 1.29 ≈ 95 W concentrated in a joint the size of a thumbnail. That is a fire risk, not a voltage-quality nuisance. The cable is not undersized; a fault exists in the loop.
2. The supervisor proposes replacing the 4 mm2 run with 10 mm2 cable during the shutdown. Evaluate this proposal technically and commercially.
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What 10 mm2 would achieve. New conductor resistance = 0.0172 × 80 / 10 = 0.138 ohm, so the drop due to the cable falls to 0.138 × 8.6 = 1.18 V - a real improvement, but it applies only to the part of the 1.63 ohm that actually lives in the cable.
Why it does not solve the reported symptom. If the 1.29 ohm of excess resistance is in a terminal at the board, a joint box, or the socket itself, it survives the cable change untouched. The drop would still be about 1.18 + (1.29 × 8.6) ≈ 12.3 V, still over 5%, and the hot joint would still be dissipating about 95 W. Replacing cable while leaving the defect is the classic 'treat the symptom' error.
The one case where replacement works. If the resistance turns out to be a damaged section of the old 4 mm2 conductor itself (crushed under a conduit clip, water ingress, strand breakage), then renewing the run removes the defect - but you still have to locate it first, because the same defect may be at the terminations you intend to reuse.
Commercial point. The 4 mm2 already complies (1.3% drop). Spending on 10 mm2 buys nothing if the joint is bad, and the same money spent on an electrician for one shift finding and remaking the joint fixes the actual problem.
3. State the tests you would carry out in the 6-hour window and, with reasons, whether the circuit may stay energised until then.
Read the worked answer
Decision: isolate now. A ~95 W hot spot inside a conduit box or terminal, in a depot with rolling stock and combustible material nearby, is an ignition source. Neither the healthy 180 Mohm insulation reading nor the silent RCD is reassurance: a series resistance produces no residual current, so the RCD was never going to see it. Keep the bay on temporary lighting from another way if needed.
Tests, in order:
- Isolate, lock off, prove dead.
- Low-resistance ohmmeter end-to-end on line and neutral, and on the CPC; a reading well above 0.344 ohm for the phase-neutral loop localises the section.
- Under load (with a temporary supply), measure millivolts across each terminal and joint. A healthy terminal shows a few millivolts; several hundred millivolts identifies the culprit directly.
- Thermographic scan of the board, conduit boxes and socket before shutdown if the camera is available.
- Visual inspection for discoloured or embrittled insulation, arcing marks, green corrosion.
- After remaking the joint: repeat the drop test (target under 3% at full load), repeat the earth loop impedance, and trip-test the 30 mA RCD.