Compressor and light bar on a long tub run: sizing the cable before blaming the relay
A 12 V ute has a 240 W compressor (nameplate 22 A) and a 60 W LED bar added in the tub. The installer ran one 4 mm² positive cable 6.2 m from the battery to a 30 A relay at the rear, used the chassis rail as the return, and fitted a 25 A blade fuse at the battery end. On a 38 °C day the compressor ran for six minutes, then slowed. The fuse held, but the relay terminal block was discoloured and the cable insulation under a chassis clip felt soft. With the compressor running the meter showed 12.5 V at the battery posts, 10.9 V across the compressor terminals and 0.15 V between the battery negative post and the compressor body. Copper resistivity is 0.0175 Ω·mm²/m at 20 °C, rising about 0.4 % per °C; conductor temperature is estimated at 75 °C.
1. Calculate the expected cable voltage drop at 22 A and decide whether the measured 1.6 V is explained by cable size, by a bad connection, or both. Then state the smallest standard cable size that would hold total drop at or below 3 % of 12.5 V.
Read the worked answer
Cable resistance
- One way 6.2 m, so the loop is 12.4 m.
- R at 20 °C = 0.0175 × 12.4 ÷ 4 = 0.0543 Ω
- Temperature factor at 75 °C = 1 + 0.004 × (75 − 20) = 1.22
- R hot = 0.0543 × 1.22 = 0.066 Ω
- Drop at 22 A = 0.066 × 22 = 1.46 V
Add the measured 0.15 V earth return: 1.61 V, which is exactly the 12.5 − 10.9 = 1.6 V on the meter. So the drop is explained by the cable and the chassis return; you should not go hunting for a mystery bad joint in the positive run. The discoloured relay block is a separate fault (contact heating).
Sizing to 3 %
3 % of 12.5 V = 0.375 V. If the earth is redone as a dedicated cable with a realistic 0.05 V drop, the positive run may lose 0.325 V:
- R max hot = 0.325 ÷ 22 = 0.0148 Ω
- R max at 20 °C = 0.0148 ÷ 1.22 = 0.0121 Ω
- A = 0.0175 × 12.4 ÷ 0.0121 = 17.9 mm²
That means 20 mm² as the next standard size, or 16 mm² if you shorten the run and keep the return in good condition. A cheaper engineering answer is to move the relay forward and keep only the compressor tails long. Always design from the nameplate current, not the marketing wattage.
2. The relay terminal block is discoloured and the fuse is 25 A. Explain why the fuse did not protect the cable and relay, and specify the protection and relay rating you would fit.
Read the worked answer
Why the fuse held
A blade fuse protects against short circuit and gross overload; it does not police a cable that is only just big enough. Two things let heat build without blowing it:
- 4 mm² clipped to a chassis rail in a 38 °C engine bay is derated well below its free-air rating. Its real continuous capacity here is roughly 20–25 A, so 22 A sits at or above the limit - enough to soften insulation over six minutes, not enough to open a 25 A fuse, which will carry its rating indefinitely and needs well over 100 % for a long time.
- The compressor is inductive. Inrush is several times 22 A for a fraction of a second, which erodes fuse elements and relay contacts over months but is too brief to blow a correctly rated fuse.
What to fit
- Cable upsized to 16 mm² (or 10 mm² with a short run) and a fuse at the battery end sized to protect that cable: 30 A for 16 mm², 20 A if 4 mm² is retained for a short run.
- A 30 A relay is marginal for a 22 A motor with inrush. Fit a 40–60 A automotive relay or a proper contactor, with a suppression diode across the coil, and crimp rather than solder the heavy terminals.
- Replace the discoloured block and any spade that has lost tension. Discolouration means contact resistance, which means local heat, which means more resistance - it will not recover on its own.