24 V Pump Feeder Voltage Drop and Cable Sizing
A 24 V battery bank with an internal resistance of 0.05 Ω feeds a 240 W pump controller through a 30 m run of two-core cable. Each conductor has a resistance of 0.012 Ω per metre. The controller draws a steady 10 A while running and trips below 21.6 V at its terminals. On site the battery reads 24.0 V open-circuit, but when the pump starts the controller display shows 16.4 V and it trips. The electrician's first remedy is to run a second identical pair of cores in parallel. The store holds 2.5, 4, 6, 10 and 16 mm² copper cable (resistivity 0.0175 Ω·mm²/m), and the specification allows no more than a 5 % drop at the load terminals with the battery at 24 V.
1. Calculate the actual terminal voltage and the cable loss, and confirm whether the observed 16.4 V is consistent with the data.
Read the worked answer
Cable loop resistance (out and back):
R_cable = 2 × 30 m × 0.012 Ω/m = 0.72 Ω
Total circuit resistance = 0.72 + 0.05 = 0.77 Ω
At 10 A:
V_load = 24 − (10 × 0.77) = 24 − 7.7 = 16.3 V
That is consistent with the 16.4 V display within measurement tolerance. The controller trips because 16.3 V is far below its 21.6 V threshold. Cable loss is 10² × 0.72 = 72 W, roughly 26 % of the power delivered at the terminals - the cable is dissipating more than a quarter of what the pump receives. The battery terminal voltage is only 24 − 0.5 = 23.5 V, so the battery itself is not the culprit; the conductor size is.
2. Evaluate the proposed fix of doubling the cable, then size the conductor that meets the 5 % drop specification.
Read the worked answer
Doubling the run halves the cable resistance to 0.36 Ω, so total resistance becomes 0.41 Ω and V_load = 24 − 4.1 = 19.9 V. Still below 21.6 V, so the pump would continue to trip - the doubling is an improvement but not a solution.
For a 5 % drop the total drop must be ≤ 1.2 V at 10 A:
R_total ≤ 1.2 / 10 = 0.12 Ω R_cable ≤ 0.12 − 0.05 = 0.07 Ω (loop) R per conductor ≤ 0.035 Ω
Using R = ρL/A:
A ≥ (0.0175 × 30) / 0.035 = 15 mm² → select 16 mm²
Check the next size down, 10 mm²: per conductor R = 0.0175 × 30 / 10 = 0.0525 Ω, loop 0.105 Ω, total 0.155 Ω, drop 1.55 V (6.5 %) - fails. Always check the standard size below the calculated area before ordering.
3. A colleague suggests matching the load to the source resistance for maximum power transfer. Assess that idea for this feeder.
Read the worked answer
For maximum power transfer the load must equal the source resistance seen by the load, i.e. R_L = 0.05 + 0.72 = 0.77 Ω. Then:
I = 24 / (2 × 0.77) = 15.6 A P_load = 24² / (4 × 0.77) ≈ 187 W Efficiency = 50 %
This is irrelevant and harmful here. The pump is a constant-power load that needs at least 21.6 V; with a 0.77 Ω load the terminal voltage would be only about 12 V, so it would never run. Maximum power transfer maximises power in the load but deliberately burns half the source power in the source, which is acceptable in a matched RF stage but never in a power distribution feeder. The correct objective is minimum source and cable resistance, i.e. the smallest voltage drop, which is why 16 mm² is chosen although a matched 0.77 Ω load would extract more watts.