Symptom in, cause out. Each of these gives you the readings somebody actually took, and the job is to say what they mean before you look at the answer.

Most of them turn on one measurement that does not fit the obvious story. That is the habit worth building: when two readings disagree, the disagreement is the finding, and the next question is which of the two is telling you the truth.

17. Zs is four times what it should be, and Zln is perfect

A 20 A radial socket circuit in a shop on a TN-C-S supply. Ze at the board is 0.14 ohms. At the last socket on the radial, Zs reads 4.62 ohms against a tabulated maximum of 2.19 ohms for a 20 A Type B device. At the same socket, the line to neutral loop impedance reads 0.38 ohms.

Where is the fault, and what is it?

  1. A loose or high resistance connection in the circuit protective conductor
  2. A damaged connection on the incoming supply neutral, the PEN conductor
  3. Thermal damage to the line conductor where it passes through cavity insulation
  4. An oxidised contact in the supply cut-out fuse carrier
Show the answer and the reasoning

Answer: A loose or high resistance connection in the circuit protective conductor

Both readings share the line conductor and the supply. Zs is the line out and the protective conductor back. Zln is the line out and the neutral back.

The line to neutral reading is normal, so the line conductor, the supply line and the neutral return are all sound. Whatever is adding four ohms is in the only part of the path the second measurement does not use, which is the protective conductor.

In practice that is a screw that was never tightened in an intermediate back box, a corroded fly lead, or a conductor left with its sleeve trapped under the terminal.

This is the most useful trick in loop testing and it costs one extra measurement: take Zln at the same point, and it tells you which half of the loop to go and look at.

Why the others are wrong

  • A supply neutral or cut-out fault raises Ze and Zln together, everywhere in the installation.
  • A damaged line conductor would show up in both readings, not one.
  • Nothing in the supply can raise the protective conductor path alone.

Where this is written down

BS 7671 Regulation 643.7.3.201 covers the verification of earth fault loop impedance. The comparison method is set out in IET Guidance Note 3.

Settled rule

18. An RCD that trips at busy times, with a perfect insulation test

A 30 mA Type A RCD on an open plan office sub-board trips two or three times a week, usually mid-morning when the office fills up. Dead insulation resistance, live conductors linked to earth at 500 V, reads over 200 megohms. Live, with a true RMS leakage clamp round line and neutral together on the load side, there is a steady 18.6 mA flowing during the working day.

What is actually happening, and what fixes it?

  1. Accumulated leakage from equipment power supplies is too close to the trip threshold; split the load across several RCBOs
  2. Moisture in the floor grommets is causing an intermittent neutral to earth fault; dry out the screed
  3. Third harmonic current is saturating the RCD core; fit a neutral choke
  4. The RCD mechanism is worn; replace it with the same device
Show the answer and the reasoning

Answer: Accumulated leakage from equipment power supplies is too close to the trip threshold; split the load across several RCBOs

The two measurements are not in conflict. Insulation resistance is a DC test and it says the insulation is intact. The clamp is measuring an AC current that flows through capacitors that are meant to be there.

Every switch mode power supply has filter capacitors from line and neutral to earth, and each workstation contributes a milliamp or two by design. Thirty of them add up to the 18.6 mA you measured.

The guidance figure is that standing leakage downstream of an RCD should not exceed 30 per cent of its rating, which is 9 mA for a 30 mA device. At double that, the RCD is already most of the way to its threshold before anything goes wrong, and a kettle or a monitor powering up finishes the job.

The device is not faulty and there is nothing to repair. Splitting the load so that no single RCD carries more than its share of the leakage is the fix.

Why the others are wrong

  • Over 200 megohms rules out damp insulation.
  • Harmonic current in the neutral passes through the core with the line current and does not create an imbalance unless it is leaving to earth.
  • An identical replacement inherits the same 18.6 mA.

Where this is written down

BS 7671 Regulation 531.3.2 addresses protective conductor current downstream of an RCD, with the working figure that it should not exceed 30 per cent of the rated residual operating current.

Settled rule

19. Balanced voltages, badly unbalanced currents

A 15 kW 400 V delta connected motor on an extraction fan, full load current 28 A, trips its overload after about ten minutes. Line to line voltages at the motor terminal box are balanced: 401, 399 and 400 V. Running currents are 26, 27 and 44 A. With the motor isolated, a micro-ohmmeter across the windings gives 1.18, 1.18 and 0.89 ohms.

What has failed?

  1. Shorted turns within the third phase winding
  2. A broken rotor bar
  3. Single phasing from a blown supply fuse
  4. The fan is mechanically overloaded
Show the answer and the reasoning

Answer: Shorted turns within the third phase winding

The supply is healthy. Balanced line voltages at the terminal box rule out the incoming supply, the fuses and the contactor in one measurement.

That leaves the machine, and the winding resistances say where. One winding is 25 per cent lower than the other two, which means it has fewer effective turns than it started with.

Shorted turns behave like a short circuited secondary on a transformer. Large circulating currents flow in the shorted loop, the flux from that phase is distorted, and the line current feeding it climbs, which is the 44 A.

The overload is doing its job. Fitting a larger one, which is the temptation when a motor keeps tripping, would let the winding cook.

Why the others are wrong

  • A broken rotor bar makes the current swing slowly at slip frequency; it does not change a static winding resistance.
  • Single phasing would give one line no current at all.
  • A mechanical overload raises all three currents together.

Where this is written down

BS EN 60034-1 covers the rating and performance of rotating machines. BS EN 60947-4-1 covers contactors and motor starters, including overload relays.

Settled rule

20. One office cooking its equipment, the office next door browning out

A commercial unit on a three phase four wire TN-C-S supply. In one office, LED drivers and IT equipment keep failing. In the next, computers keep shutting down. At the main board, line to neutral reads 278, 265 and 163 V. Line to line stays balanced at 400, 400 and 401 V. Between the neutral bar and the main earthing terminal, under load, there is 68 V.

What is causing this?

  1. A high resistance or open circuit supply neutral, a floating neutral
  2. Leading power factor from correction capacitors
  3. Insulation breakdown between one phase and the containment
  4. High prospective fault current on the distribution network
Show the answer and the reasoning

Answer: A high resistance or open circuit supply neutral, a floating neutral

The clue is that line to line voltages are untouched while line to neutral voltages are all over the place. Line to line comes from the transformer and cannot move. Line to neutral depends on where the neutral point sits.

With a sound neutral the star point is held at earth potential and every phase gets its share. Lose the neutral, or add resistance to it, and the star point drifts towards whichever phases are most heavily loaded, so lightly loaded phases rise and heavily loaded ones fall.

That is exactly the pattern in front of you, and the 68 V between neutral and earth is the size of the drift, measured directly.

This is a supply side fault and it is dangerous: the neutral bar is at 68 V to earth, and if the connection opens completely the lightly loaded phase heads towards 400 V. It is one to report to the distributor rather than to work around.

Why the others are wrong

  • Power factor correction shifts current relative to voltage; it does not move the star point.
  • A phase to earth fault would collapse that phase and operate a protective device.
  • Prospective fault current is a property of the supply impedance and does not shift a steady state voltage.

Where this is written down

BS 7671 Section 431 covers protection of the neutral conductor and Regulation 537.1.2 covers isolation and switching arrangements including the neutral. A fault on the distributor side of the origin is reported to the distributor.

Settled rule

21. An AFDD that trips whenever the angle grinder runs

A 32 A arc fault detection device on a workshop ring circuit trips the moment a 2 kW corded angle grinder is switched on. It holds all day under a 3 kW resistive heater. Insulation resistance across the circuit is over 500 megohms. A scope on the line shows broadband noise from about 10 to 50 MHz while the grinder runs, and you can see the brushes sparking.

Why is it tripping, and what do you do about it?

  1. The brush arcing looks like a series arc fault to the detector; check the brushes and suppression, or run the tool from a circuit where an AFDD is not required
  2. Motor inrush is operating the magnetic element; fit a Type D device
  3. The grinder has a broken earth conductor causing arcs to earth; rewire the plug
  4. The device is wired the wrong way round; swap supply and load
Show the answer and the reasoning

Answer: The brush arcing looks like a series arc fault to the detector; check the brushes and suppression, or run the tool from a circuit where an AFDD is not required

An AFDD does not measure current or leakage. It listens for the signature of an arc: broadband radio frequency noise together with characteristic flat spots as the current crosses zero.

A universal motor with worn brushes produces exactly that. It is a real arc, in the tool rather than in the wiring, and the detector is not malfunctioning when it responds to it.

So the first move is at the tool: brush condition, commutator condition, and whether the suppression components are still doing anything. Old power tools are a well known source of this.

If the tool is sound and it still trips, the honest answer is that this device and this load do not get along, and the tool goes on a circuit where an AFDD is not required. What you do not do is remove protection from a circuit that needs it.

Why the others are wrong

  • Inrush operates a magnetic trip, which is a different mechanism with a different signature.
  • A double insulated tool has no protective conductor to break.
  • A reversed device would misbehave with every load, not just this one.

Where this is written down

BS EN 62606 sets the general requirements for arc fault detection devices. BS 7671 Regulation 421.1.7 covers where AFDDs are required.

Engineering judgement

22. PLC inputs turning themselves on when the drive accelerates

In a packaging plant, 24 V DC digital inputs on a PLC register false pulses, but only while a 30 kW variable frequency drive is ramping up a conveyor. On the input terminal there are transient spikes up to 16 V peak to peak riding on the DC. The 24 V control multicore is unshielded and runs for 25 metres in the same trunking as the unshielded drive output cables.

What is the mechanism, and what does the standard require?

  1. Capacitive and inductive coupling from the PWM output; separate the control and power cables, or use screened drive cable with a 360 degree gland
  2. A floating 0 V reference; bond 0 V to the building lightning conductor
  3. The input optocouplers are overheating in the trunking; add cooling
  4. Volt drop on the 24 V supply during starting; increase the conductor size
Show the answer and the reasoning

Answer: Capacitive and inductive coupling from the PWM output; separate the control and power cables, or use screened drive cable with a 360 degree gland

A drive output is not a sine wave. It is a train of steep edges, and a fast edge on a conductor is an aerial: it couples into anything running beside it, through the capacitance between the cables and through the changing magnetic field around them.

That gives you a 16 V spike on a 24 V logic input, which is more than enough to cross the threshold and read as a genuine signal. The timing tells you the same story, because it only happens while the drive is switching hardest.

The requirement is separation of circuits in different voltage bands, and the practical fix is either physical distance between the control and the motor cables, or screened motor cable with the screen bonded through 360 degrees at both ends. A screen bonded by a pigtail at one end achieves very little at these frequencies.

Note that both these cables were compliant on their own. What was wrong was that they were in the same trunking.

Why the others are wrong

  • Bonding a 0 V reference to a lightning down conductor invites surge current straight into the PLC and is dangerous.
  • A thermal problem would not track the acceleration ramp.
  • Volt drop causes a sagging supply, not positive spikes.

Where this is written down

BS 7671 Regulation 528.1 covers the proximity of circuits of different voltage bands. BS EN 61800-3 sets the EMC requirements for adjustable speed drive systems.

Settled rule

23. A motor that hums and shakes but will not turn

A 7.5 kW 400 V star connected motor on a blower is stopped for a break. On restart it does not turn, it hums loudly, it vibrates badly, and the breaker trips after about four seconds. Clamped during the attempt: 92 A, 0 A, 92 A.

What has failed?

  1. The middle pole of the contactor is open circuit, so the motor is single phasing at standstill
  2. The blower bearings have seized
  3. Two supply lines have been swapped
  4. The supply voltage is too low and the rotor has stalled
Show the answer and the reasoning

Answer: The middle pole of the contactor is open circuit, so the motor is single phasing at standstill

Zero current on one line, and equal current on the other two, means the motor is connected across a single pair of lines. Current goes out on one and comes back on the other, which is what the readings show.

A three phase motor turns because three windings, fed in sequence, produce a field that rotates. Across a single pair you get a field that pulses in place instead, which can be seen as two equal fields rotating in opposite directions. At standstill they cancel, so the starting torque is zero.

The motor cannot start, so it sits at locked rotor current, which is why it draws 92 A, hums at twice mains frequency, shakes, and trips.

Once running, the same motor would keep going on two lines while quietly overheating, which is why single phasing is more often found as a burnt out winding than as a motor that will not start.

Why the others are wrong

  • A seized bearing gives locked rotor current on all three lines.
  • Swapped lines reverse the direction; the motor still starts.
  • Low voltage gives a weak start with balanced currents.

Where this is written down

BS EN 60947-4-1 covers contactors and motor starters. BS 7671 Section 552 covers rotating machines.

Settled rule

24. An RCD that trips when you switch on a circuit with nothing connected

A dual RCD domestic board. You add a new 6 A lighting circuit to RCD 1. Every switch is off and no fittings or lamps are connected. Switching the MCB on trips RCD 1 immediately. Dead testing gives over 200 megohms between line and neutral, but the neutral to earth insulation resistance reads 0.00 megohms, and a low resistance test between the circuit neutral and earth gives 1.2 ohms. Other circuits on RCD 1 are running fridges and standing loads.

Why did it trip with no load on the new circuit?

  1. Return current from the other circuits on RCD 1 is diverting through the neutral to earth fault, unbalancing the RCD
  2. Closing the MCB put a switching spike onto the earth bar
  3. The new neutral has picked up an induced voltage from a nearby circuit
  4. The 6 A MCB has a faulty thermal element
Show the answer and the reasoning

Answer: Return current from the other circuits on RCD 1 is diverting through the neutral to earth fault, unbalancing the RCD

The fault is a near short between neutral and earth: 1.2 ohms, which is why the insulation reading is zero rather than merely low.

While the MCB is off, that fault is connected to nothing. Closing the MCB connects the faulty neutral to the RCD neutral bar, and the bar is not at zero volts, because the other circuits on that RCD are pushing current along it.

A fraction of a volt across 1.2 ohms is enough to send hundreds of milliamps down the fault to earth. That current left through the RCD in the line conductors of the other circuits and came back outside it, so the device sees an imbalance and trips.

This is why an RCD tripping is not evidence about the circuit you just switched on. The fault is on the new circuit, but the current that trips the device belongs to its neighbours. Isolate the neighbours and the new circuit will sit there quite happily with the same fault on it.

Why the others are wrong

  • Switching an unloaded lighting circuit produces nothing of the sort.
  • An induced voltage on an open conductor cannot supply the current an RCD needs to see.
  • A faulty MCB does not operate the RCD above it.

Where this is written down

BS 7671 Regulation 643.3 covers insulation resistance testing between live conductors and between live conductors and earth. Regulation 531.3.2 covers protective conductor current downstream of an RCD.

Settled rule