Voltage Drop on a Long Pump Run
A 12.6 V battery in a truck bed feeds a 12 V sprayer pump rated at 8 A. The pump sits 18 ft from the battery, and both the supply and the return conductor are 16 AWG copper, so the current travels 36 ft of wire. The resistance of 16 AWG copper is about 4.02 ohms per 1,000 ft; 14 AWG is 2.53, 12 AWG is 1.59, and 10 AWG is 1.00 ohm per 1,000 ft. Assume the pump continues to draw about 8 A. The installer wants the pump to see no more than a 3% drop from battery voltage, because the pump's own manual warns that low supply voltage shortens motor life.
1. Compute the conductor resistance, the voltage reaching the pump, and the percent drop.
Read the worked answer
Resistance: 4.02 Ω/1,000 ft = 0.00402 Ω/ft × 36 ft ≈ 0.145 Ω
Drop: V = IR = 8 A × 0.145 Ω ≈ 1.16 V
Pump voltage: 12.6 − 1.16 ≈ 11.4 V
Percent drop: 1.16 ÷ 12.6 ≈ 9.2%
That is about three times the 3% allowance. The pump is running roughly 0.6 V under its rating, which is exactly the condition the manual warns about.
2. How much power is being lost in the wiring, and how does it compare with the power the pump actually receives?
Read the worked answer
- Wire loss: P = I²R = 8² × 0.145 = 64 × 0.145 ≈ 9.3 W
- Power delivered to the pump: 11.4 V × 8 A ≈ 91.5 W
- Total drawn from the battery: ≈ 100.8 W
About 9% of the battery energy is heating the wire, not spraying. Note that this lost power goes up with the square of current: if the pump were replaced with a 12 A unit on the same wire, the loss would jump to 64 × (12/8)² ≈ 21 W. That is the reason to size the wire for the largest load the circuit will ever carry, not just today's.
3. Determine the smallest gauge from the listed sizes that meets the 3% limit, and state the judgment.
Read the worked answer
Allowable drop = 0.03 × 12.6 = 0.378 V Allowable resistance = 0.378 ÷ 8 A = 0.0473 Ω over 36 ft Per 1,000 ft: 0.0473 ÷ 36 × 1,000 ≈ 1.31 Ω/1,000 ft
Compare the table:
- 12 AWG: 1.59 → drop = 8 × 1.59 × 36/1,000 ≈ 0.46 V (3.6%) ✗
- 10 AWG: 1.00 → drop = 8 × 1.00 × 36/1,000 ≈ 0.29 V (2.3%) ✓
Judgment: 12 AWG is the intuitive answer and it is wrong by a small but real margin; 10 AWG is the smallest listed size that satisfies the spec. Alternatively, move the battery closer - halving the run to 18 ft total gets 16 AWG to about 4.6%, still short, so on this layout the gauge change is the fix.