Eight situations where the readings do not sit comfortably. Each one gives you a set of results and a judgement to make. Decide before you open the answer, because the reasoning is the part worth having and it lands better once you have committed to something.
These are pitched at someone who has been doing this for years and has not had cause to think about the underlying theory in a while. Nothing here is a trick question. Every one of them is a situation that turns up, and most of them are places where a circuit can be signed off and still not do what the certificate says it will.
1. A ring final circuit with one socket reading high
You are on initial verification of a domestic ring final circuit. End to end, r1 is 0.62 ohms, rn is 0.62 ohms and r2 is 1.04 ohms. With the ends cross connected line to neutral, every socket reads a uniform 0.31 ohms. Cross connected line to earth, five sockets read 0.41 ohms and one double socket in the hallway reads 0.95 ohms.
What is causing the 0.95 ohm reading at the hallway socket?
Show the answer and the reasoning
Answer: The hallway double socket is wired as an unfused spur off the ring
On a healthy ring, every point on the loop reads (r1 + r2) / 4, which here is (0.62 + 1.04) / 4 = 0.415 ohms. The five sockets at 0.41 ohms confirm the ring itself is sound.
A single outlet reading appreciably higher than that value is not on the loop. The extra resistance is the there and back length of the spur conductor, added in series to the ring value at the point the spur was taken off.
The 1.04 ohm r2 against a 0.62 ohm r1 is the expected ratio for 2.5 mm2 line conductors with a 1.5 mm2 protective conductor, so the cable itself is what you think it is.
Why the others are wrong
- A bridged interconnection pulls readings below (r1 + r2) / 4, not above it.
- A bad probe contact would not repeat at one specific outlet and nowhere else.
- Reversed polarity does not change loop resistance at all.
Where this is written down
Continuity of ring final circuit conductors is a required test under BS 7671 Regulation 643.2. The (r1 + r2) / 4 relationship and the spur diagnosis are set out in IET Guidance Note 3.
Settled rule
Go deeper: Ring final circuit testing, step by step, for the three step continuity procedure and what each cross connection is actually proving.
2. Measured Zs is inside the table value, and the circuit still fails
A commercial ring circuit on a 32 A Type B MCB to BS EN 60898 is tested at an ambient 20 degrees C. Ze is 0.22 ohms, R1 + R2 is 0.95 ohms, so measured Zs is 1.17 ohms. The tabulated maximum Zs in BS 7671 Table 41.3 is 1.37 ohms.
Is the circuit compliant for energising as it stands?
Show the answer and the reasoning
Answer: No, it exceeds the temperature corrected limit of about 1.10 ohms
The values in Table 41.3 are worked out for conductors sitting at their maximum operating temperature, 70 degrees C for a general purpose thermoplastic cable. You are testing a cold circuit.
Copper gains roughly 20 per cent in resistance across that temperature rise, so BS 7671 Appendix 3 applies a 0.8 factor to compare a cold measurement against a hot table value. That gives 0.8 x 1.37 = 1.096 ohms.
At 1.17 ohms the circuit is above that. Under a genuine fault, with the conductors already warm from load, Zs would climb past 1.37 ohms and the 0.4 second disconnection time would not be met.
This is the single most common way a circuit passes on the certificate and would not have disconnected in service.
Why the others are wrong
- Comparing a cold reading straight against a hot table value is the mistake the whole scenario is built on.
- An upstream RCD is not a fix for an earth fault loop impedance that is too high for the overcurrent device.
- The disconnection time for a final circuit up to 63 A on a TN system is 0.4 seconds, not 0.1.
Where this is written down
The 0.8 factor for comparing a measurement taken at ambient temperature against a tabulated maximum is given in BS 7671 Appendix 3. Table 41.3 itself is derived with Cmin taken as 0.95.
Settled rule
Go deeper: The 0.8 rule and temperature correction, for where the factor comes from and how to apply it to a measured reading.
3. Insulation resistance on a circuit full of electronics
Periodic inspection of an office lighting circuit. The circuit contains DALI dimmers and emergency battery packs, and the client will not have the luminaires dismantled. A 500 V DC test between line and earth gives 0.08 megohms.
What is the correct next step?
Show the answer and the reasoning
Answer: Link line and neutral together and test between the live conductors and earth at 250 V DC
A 500 V test across equipment that cannot be disconnected is measuring the equipment, not the wiring. The 0.08 megohm reading is very probably the dimmer input filters and the battery packs, and you risk damaging them.
Where connected equipment is likely to influence the result or be damaged, BS 7671 has you link the live conductors together and test between them and the protective conductor at 250 V DC.
Linking line and neutral removes any voltage across the equipment while still proving the insulation to earth, which is the part that matters for safety.
The minimum acceptable value is 1.0 megohm. Anything below that is investigated, not simply recorded.
Why the others are wrong
- 1000 V DC will destroy the semiconductors you were trying to protect.
- Coding a defect before you have applied the correct test procedure is premature.
- A clamp meter reading does not satisfy the insulation resistance requirement and does not involve safe isolation.
Where this is written down
BS 7671 Regulation 643.3.3 covers testing where connected equipment is likely to influence the result or be damaged, including the 250 V DC test with live conductors linked. Test voltages and minimum values are in Table 64.
Settled rule
Go deeper: Insulation resistance testing, 250 V and 500 V, for test voltages, minimum values, and how to handle circuits with equipment left connected.
4. Safe isolation, and the step people skip
A 400 V three phase board is being isolated to change a feeder breaker. The isolator is locked off. The approved two pole voltage indicator is checked on a proving unit, then applied across L1-L2, L2-L3, L3-L1, and each phase to neutral and to earth. Everything reads dead.
What is the final step before you touch a conductor?
Show the answer and the reasoning
Answer: Prove the voltage indicator again on the proving unit
The sequence is prove, test, prove. You check the indicator on a known source, you test the conductors, and then you check the indicator again.
The reason is simple and unglamorous. If the indicator failed at some point during the dead testing, every reading you just took would also have shown dead, and you would have no way of knowing.
Re-proving afterwards is what turns a set of dead readings into evidence. Without it you have an instrument of unknown condition and a board you are about to put your hands into.
Why the others are wrong
- A photograph records what you did. It does not tell you the instrument still works.
- An electrode resistance test has nothing to do with isolation.
- Energising a load to check is not a test method, and on a partly isolated board it is dangerous.
Where this is written down
HSE Guidance Note GS38 sets out the requirements for test equipment and the practice of proving the indicator before and after use. The duty to work on equipment made dead sits in the Electricity at Work Regulations 1989, Regulation 14.
Settled rule
Go deeper: Safe isolation in the real world, for the full sequence, GS38 probe requirements, and installations with more than one source.
5. Which prospective fault current goes on the certificate
At the intake of a three phase 400 V TN-C-S commercial installation, the instrument gives a prospective short circuit current between lines of 4.8 kA, and a prospective earth fault current of 2.9 kA.
What value is recorded as the prospective fault current?
Show the answer and the reasoning
Answer: 4.8 kA
The figure recorded is the highest prospective fault current that could flow at that point, because it is the figure the protective devices have to be able to interrupt.
Here that is the 4.8 kA short circuit value. Recording the smaller earth fault figure would understate what a device at the origin has to break.
The two values are alternatives, not contributions. They describe two different faults, and only one of them happens at a time.
Why the others are wrong
- Recording only the earth fault current hides the larger duty.
- Adding the two together has no physical meaning.
- Doubling the line to line value is not a recognised method for anything.
Where this is written down
BS 7671 Regulation 643.7.3.201 requires the prospective short circuit current and the prospective earth fault current to be measured, calculated or determined by another method at the origin and at other relevant points.
Settled rule
Go deeper: Prospective fault current, measured and calculated, for how PSCC and PEFC are taken on single and three phase supplies, and the multipliers people misuse.
6. Is the five times test still required
A 30 mA Type A RCBO on a domestic socket circuit is tested with a calibrated multifunction tester. At the rated residual operating current it trips in 28 ms at 0 degrees and 32 ms at 180 degrees. The apprentice sets the instrument up for a 150 mA test.
Is the five times test needed for the certificate?
Show the answer and the reasoning
Answer: No, that test was removed for general RCDs to BS EN 61008 and 61009
Amendment 2 to BS 7671 simplified this. A general, non delayed RCD is verified with an alternating current test at the rated residual operating current, and it must operate within 300 ms.
The five times test is no longer required for certification of these devices. Your 28 ms and 32 ms readings already demonstrate compliance comfortably.
The integral test button still has to be operated, but it proves the mechanism, not the tripping current. It is not a substitute for an instrument test.
S type, time delayed devices are different again: they are expected to operate between 130 ms and 500 ms.
Why the others are wrong
- The 40 ms limit belonged to the five times test that was removed.
- The change was not restricted to one device type, and Type AC has separate problems of its own.
- A button press does not verify the residual current at which the device operates.
Where this is written down
The requirement to test at a current equal to or higher than five times the rated residual operating current was removed by BS 7671:2018+A2:2022. The current requirements appear in the notes to Regulations 643.7.1 and 643.8.
Settled rule
Go deeper: RCD selection and why Type AC is finished, for Type A, F and B selection, DC saturation, and what changed for testing.
7. Supplementary bonding resistance in a special location
A leisure centre sauna room has simultaneously accessible extraneous conductive parts and exposed conductive parts, and supplementary equipotential bonding has been installed between them. The circuit is protected by a 32 A Type C MCB.
What is the maximum permissible resistance between the bonded parts?
Show the answer and the reasoning
Answer: 0.156 ohms
The requirement is that the resistance between the parts satisfies R is less than or equal to 50 / Ia, where Ia is the current that operates the protective device in the required time.
For a Type C device, Ia is taken as ten times the rated current, so 10 x 32 = 320 A.
That gives 50 / 320 = 0.156 ohms. The logic is that even with the fault present, the voltage between the two parts a person can touch at the same time stays below 50 V until the device clears.
Why the others are wrong
- 0.05 ohms is the rule of thumb figure people quote for main bonding continuity, not a limit derived from the device.
- 1.66 ohms comes from using a 30 mA RCD figure in the formula.
- 1666 ohms is 50 divided by 0.03, which is the same error taken one step further.
Where this is written down
BS 7671 Regulation 415.2.2 gives the condition R is less than or equal to 50 / Ia for AC systems, where Ia is the operating current of the protective device.
Settled rule
Go deeper: Maximum Zs calculator, for device operating currents and disconnection thresholds for each device type and rating.
8. A main bonding conductor to the gas service
A 10 mm2 copper main bonding conductor runs 18 metres from the main earthing terminal to the incoming metallic gas service. Tested with the wandering lead method, the reading including the test lead is 0.14 ohms. The lead on its own reads 0.09 ohms.
Is the reading acceptable?
Show the answer and the reasoning
Answer: Yes, the conductor itself is 0.05 ohms once the lead is subtracted
Subtract the lead: 0.14 minus 0.09 gives 0.05 ohms for the conductor.
Check it against what the cable should be. Copper of 10 mm2 is about 1.83 milliohms per metre, so 18 metres is roughly 0.033 ohms. The measurement is the right order of magnitude, with the balance in the two terminations.
That comparison is the useful part. A single number tells you very little until you know what the cable ought to read, and doing the arithmetic is what turns a reading into a verification.
The 0.05 ohm figure commonly quoted for bonding continuity is a guidance benchmark, not a limit in BS 7671, so treat a reading close to it as a prompt to check the terminations rather than an automatic pass or fail.
Why the others are wrong
- 0.01 ohms is not achievable over 18 metres of 10 mm2 and is not a requirement.
- There is no 10 metre limit on a bonding conductor that is correctly sized.
- The connection is made on the consumer side of the meter, within 600 mm of it, so before the meter is the wrong side.
Where this is written down
The 0.05 ohm expectation for a bonding conductor is guidance from IET Guidance Note 3, not a limit stated in BS 7671. Conductor resistance per metre is tabulated in the IET On-Site Guide.
Engineering judgement
Go deeper: Earthing and bonding in panels and installations, for conductor sizing, connection points, and what a bonding measurement should look like.