Geyser Circuit Losing 19 V: Cable or Joint?
A single-storey house in Bhubaneswar is supplied at 230 V, single phase. A 2 kW, 230 V storage geyser is fed from the distribution board through a 32 m run of 1.0 mm² copper twin-core cable, part of which is buried in a wall conduit shared with two lighting circuits. With all loads switched off, a digital multimeter reads 228 V at the geyser terminal box. With the geyser switched on, the same meter reads 209 V and a clamp meter on the live conductor reads 9.6 A. The owner reports that the lights dim briefly whenever the geyser starts, and that the neutral terminal inside the geyser connector block feels warm and its screw turns half a turn before it tightens. Take copper resistivity as 0.0172 Ω·mm²/m; the manufacturer's current rating for 1.0 mm² copper in conduit is 11 A, and the customary design limit for voltage drop on a final circuit is 3%.
1. Calculate the voltage drop contributed by the cable at 9.6 A, deduce the resistance of the loose neutral connection, and state whether the cable alone explains the measured drop.
Read the worked answer
Total measured drop = 228 − 209 = 19 V.
Cable: R per conductor = ρL/A = 0.0172 × 32 / 1.0 = 0.550 Ω. Current goes out and returns, so loop resistance = 2 × 0.550 = 1.101 Ω. Drop = 9.6 × 1.101 = 5.3 V, about 2.3% of 228 V.
So the cable explains only 5.3 V of 19 V. The remaining 13.7 V sits across the loose neutral terminal: R = V/I = 13.7 / 9.6 = 1.43 Ω.
Check by heat: P = I²R = 9.6² × 1.43 ≈ 132 W dissipated in one terminal - far beyond what a sound joint can dissipate, which is exactly why it feels warm. This is a fire risk, not an inconvenience.
Conclusion: two faults act together - a long 1.0 mm² run and, more seriously, a high-resistance joint. Replacing the cable alone would still leave a hot terminal and a 13.7 V drop.
2. Recommend the cable size, protective devices and circuit arrangement, supporting the choice with the 3% drop criterion.
Read the worked answer
3% of 228 V = 6.8 V, i.e. maximum loop resistance 6.8/9.6 = 0.71 Ω, i.e. 0.355 Ω per conductor.
- 1.0 mm²: 0.550 Ω per conductor - fails.
- 1.5 mm²: 0.0172 × 32/1.5 = 0.367 Ω - just outside the target.
- 2.5 mm²: 0.220 Ω, giving 9.6 × 0.440 = 4.2 V = 1.85% - comfortably inside 3%, with margin for a larger geyser later.
Recommendation: a dedicated geyser circuit in 2.5 mm² copper (rating roughly 17–18 A in conduit), protected by a 16 A MCB and a 30 mA RCD, with a 20 A DP switch beside the geyser and a proper terminal block. Do not share the conduit with lighting circuits. Cut back any discoloured conductor end, re-terminate with a ferrule and torque the screw properly instead of simply tightening the old joint, and verify earth continuity from the geyser body to the earth bar.
3. State two measurements you would take after the repair, the values you would accept, and why each matters.
Read the worked answer
1) Loaded voltage at the geyser terminal box. With 228 V at the DB and 9.6 A flowing, expect 228 − 4.2 ≈ 224 V. Accept anything above about 218 V (3%). A reading back near 209 V proves the joint or cable is still faulty, regardless of how tight it feels.
2) Insulation resistance at 500 V DC with the geyser isolated and all switches on. Expect ≥ 1 MΩ, typically above 50 MΩ for new PVC cable; below 1 MΩ means moisture or mechanical damage.
Supporting checks: run the geyser loaded for ten minutes and then feel/scan the terminals for warming (a hot joint after repair means the work is not finished), and record all readings on the invoice. Written readings protect the customer and give the technician evidence if the same complaint returns.