Cable Choice for a Remote 12 V Pump
A farmer wants to run a 12 V, 120 W submersible pump from a battery shed to a water tank 35 m away. The cable route needs two copper conductors, so the total conductor length is 70 m. The farmer has a coil of 2.5 mm² copper cable and proposes to use it because it is already available. The resistivity of copper is 1.7 × 10⁻⁸ Ω m. The pump is rated at 12 V and 120 W, and the farmer expects it to start reliably. A supplier recommends a maximum voltage drop of 5% at the pump terminals. The pump motor has a starting current about three times its running current, and the battery voltage falls to 11.5 V under starting load. The farmer wants to know whether the existing cable is adequate and, if not, what size to buy.
1. Calculate the running current, the cable resistance, the voltage drop and the power wasted in the 2.5 mm² cable. Decide whether the cable meets the 5% rule.
Read the worked answer
Running current:
[ I=\frac{P}{V}=\frac{120}{12}=10\ \text{A} ]
Cable resistance for total length 70 m:
[ R=\frac{\rho L}{A}=\frac{(1.7\times10^{-8})(70)}{2.5\times10^{-6}} ]
[ R=0.476\ \Omega ]
Voltage drop:
[ V_{\text{drop}}=IR=(10)(0.476)=4.76\ \text{V} ]
Pump terminal voltage:
[ 12-4.76=7.24\ \text{V} ]
Power wasted in the cable:
[ P_{\text{cable}}=I^2R=(10)^2(0.476)=47.6\ \text{W} ]
The allowed drop is 5% of 12 V, or 0.60 V. The cable gives 4.76 V, far above this, so it is not adequate. The pump would receive only about 7.2 V while running, and much less during starting.
2. Select a suitable standard cable size from 4, 6, 10, 16 and 25 mm², and explain why the starting current makes the choice more critical than the running calculation alone.
Read the worked answer
For a 0.60 V drop at 10 A, the maximum allowed resistance is
[ R_{\max}=\frac{0.60}{10}=0.060\ \Omega ]
The required area is
[ A=\frac{\rho L}{R_{\max}}=\frac{(1.7\times10^{-8})(70)}{0.060} ]
[ A=1.98\times10^{-5}\ \text{m}^2=19.8\ \text{mm}^2 ]
The next standard size above this is 25 mm². A 16 mm² cable would give (R=0.074\ \Omega), a drop of 0.74 V, which fails the 5% rule at running current.
Starting current is about 30 A. With 25 mm² cable, (R\approx0.048\ \Omega), so the starting drop is about (30\times0.048=1.4\ \text{V}). From an 11.5 V battery the pump would see about 10.1 V at start, which is marginal but much better than with 2.5 mm² cable, where the starting drop would be about 14 V and the pump would almost certainly fail to start. The cable must therefore be chosen for starting voltage, not only running voltage.