A 3 kW Heater, 25 Volts Missing and a Warm Switch Plate
A small hotel wants a 3 kW, 230 V storage water heater in a first-floor bathroom. The cable route from the distribution board in the ground-floor lobby is 24 m, and the installer proposes 1.5 mm² copper PVC-insulated single-core wires in a conduit that already carries three other loaded circuits; the ambient temperature inside the riser is about 40 °C. At commissioning the busbar reads 233 V, but a digital multimeter at the heater terminals shows 208 V while the element is on. The nameplate gives 3 kW at 230 V, and a 16 A MCB feeds the circuit. The owner complains that hot water takes far longer than at his other property, which has an identical heater, and the 16 A switch plate feels warm after 20 minutes.
1. Using these readings, calculate the load current and the actual voltage drop, then decide whether the 1.5 mm² cable alone explains the missing 25 V. Show the arithmetic.
Read the worked answer
Load current
I = P / V = 3000 / 230 ≈ 13.0 A at rated voltage. The element's hot resistance is about R = V²/P = 230²/3000 ≈ 17.6 Ω, so at 208 V it would draw roughly 208/17.6 ≈ 11.8 A.
Measured drop
233 − 208 = 25 V, i.e. 25/233 ≈ 10.7 %. The usual design limit from origin to point of use is about 5 % (roughly 11.5 V here).
What 1.5 mm² should have given
R per conductor = ρL/A = 0.0172 × 24 / 1.5 ≈ 0.275 Ω. Two conductors in series ≈ 0.55 Ω. At 13 A: Vd ≈ 13 × 0.55 ≈ 7.2 V (about 3.1 %).
Conclusion
The cable's own resistance accounts for roughly 7 V, not 25 V - the measured drop is over three times the calculated value. That points to additional series resistance at a termination: a slack screw, an oxidised or nicked conductor, a wrong ferrule, a joint between dissimilar metals, or a route longer than stated. The warm switch plate supports this, since heat at a terminal is I²R at a bad joint and not cable heating. Upsizing the cable without finding the joint would hide the symptom and leave a hot spot in the wall.
2. Check the proposed 1.5 mm² cable against the load with the applicable derating, and state the minimum size you would accept and why.
Read the worked answer
Thermal check with derating
Using typical tabulated ratings for copper PVC-insulated conductors in conduit (values vary slightly between tables, so confirm against the one you are working to):
- 1.5 mm² ≈ 17.5 A base.
- Grouping with three other loaded circuits ≈ 0.65.
- Ambient 40 °C ≈ 0.87.
- Iz ≈ 17.5 × 0.65 × 0.87 ≈ 9.9 A.
Two failures follow:
- Iz (9.9 A) is below the load current Ib (13.0 A) - the cable is overloaded.
- The protective device rule Ib ≤ In ≤ Iz fails badly: In = 16 A is far above Iz = 9.9 A, so the MCB will never protect the cable.
Next sizes
- 2.5 mm²: 24 × 0.65 × 0.87 ≈ 13.6 A. It just covers Ib, but In (16 A) still exceeds Iz, so the rule still fails unless the MCB is reduced to 13 A or the grouping improves.
- 4 mm²: 32 × 0.65 × 0.87 ≈ 18.1 A, which satisfies Ib (13 A) ≤ In (16 A) ≤ Iz (18.1 A).
Recommendation: 4 mm² on the existing 16 A MCB, or 2.5 mm² with a 13 A device and the heater moved to a less crowded conduit. Fixed water heaters should also have a 30 mA RCCD and a double-pole switch with indicator.
3. State the order of remedial work and how you would verify the repair numerically.
Read the worked answer
Order of work
- Isolate, lock off, prove dead at the heater and at the board.
- Inspect and remake both terminations - board end and heater end - with the correct lugs, cleaned conductors and a torque driver set to the terminal marking.
- Cut back any heat-discoloured or nicked conductor; a discoloured copper end is a joint that has already been running hot and must not be reused.
- Re-size the run to 4 mm² (or 2.5 mm² plus a 13 A device), keeping the earth conductor continuous and sleeved.
- Re-check polarity and the earth path to the tank and the inlet/outlet pipes.
Verification with numbers
- Conductor resistance: for 4 mm² over 24 m, R = 0.0172 × 24 / 4 ≈ 0.103 Ω per conductor. Measure each core with a low-resistance ohmmeter and compare against this figure - a reading several times higher means a joint is still bad.
- Insulation resistance: ≥ 1 MΩ at 500 V DC for L–E, N–E and L–N.
- Voltage under load: re-energise, let the element draw current, and read at the heater terminals; the drop from the busbar should now be well inside 5 % (about 11 V or less).
- Thermal check: after 30 minutes, no terminal or switch plate should be noticeably warmer than ambient. Any warmth means the joint is still resistive.
- RCCD trip test at IΔn and 5 × IΔn before handover.