Pump Circuit: Calculated vs. Measured Voltage Drop
A 1.5 kW, 230 V single-phase centrifugal pump serves a hillside reservoir. Its controller sits at the end of a 75 m cable run from the service panel, fed with 3.5 mm² THHN copper in PVC conduit. Nameplate current is 7.7 A. With the pump running, the technician measures 232 V line-to-neutral at the panel and 208 V at the controller terminals. The pump runs hot, starts sluggishly, and the owner has already replaced two start capacitors in eight months. The motor itself was megger-tested and is sound. The conduit contains only the two circuit conductors and the equipment grounding conductor, and no other loads are connected to the circuit.
1. Is the 24 V drop explained by the conductor length and size? Show the calculation.
Read the worked answer
No - the conductor accounts for only about a quarter of it.
Take 3.5 mm² copper at roughly 5.3 Ω/km at operating temperature:
- V_drop = 2 × I × L × R (factor 2 for the outgoing and return path)
- V_drop = 2 × 7.7 A × 0.075 km × 5.3 Ω/km ≈ 6.1 V (about 2.6 % of 230 V)
So the cable run should cost you ~6 V, and the measured drop is 24 V. Roughly 18 V of the drop is unaccounted for by conductor resistance, which means it is being dropped somewhere that is not a 75 m length of 3.5 mm² copper.
2. What does the discrepancy indicate physically, and what is the immediate risk?
Read the worked answer
Convert the drop into loop resistance and compare:
- Measured: 24 V ÷ 7.7 A ≈ 3.1 Ω of loop resistance
- Expected from the cable: 2 × 0.075 × 5.3 ≈ 0.8 Ω
- Excess: about 2.3 Ω concentrated at one or more points
That much resistance in a joint dissipates P = I²R ≈ (7.7)² × 2.3 ≈ 135 W in a spot the size of a thumb. A loose terminal screw, a corroded splice, an oxidized aluminium-to-copper connection, or a back-stabbed device will do this. The joint heats, carbonizes, resistance climbs further, and the failure accelerates - a classic fire initiator inside a controller or junction box. Note also that a low-voltage megger or continuity test will usually pass, because the defect only shows up under load.
The sluggish starting is a separate symptom of the same resistance: at roughly 6× FLA inrush (≈46 A) the same 2.3 Ω would collapse the terminal voltage far below the ~195 V the cable alone would give, which is why capacitors keep failing.
3. What sequence of checks and what repair would you specify?
Read the worked answer
- Load the circuit and measure successively - panel terminals, any junction boxes, the controller line terminals, the disconnect. The point where the voltage takes a big step is the defect.
- Thermograph or feel every termination immediately after a loaded run; the hot joint identifies itself.
- Re-terminate with correct torque, replace back-stabbed or wire-nut splices with listed connectors, and separate dissimilar metals with listed connectors.
- Re-measure after repair and confirm the loaded drop is back near the calculated ~6 V.
- Revisit the design. At 7.7 A running the 3.5 mm² is adequate, but on starting the drop is marginal at 75 m. If the pump is critical, upsize the run (5.5 mm² roughly halves the drop) rather than accept repeated capacitor failures. Do not simply increase the breaker - that addresses none of the above.