Sizing a new work-light circuit on a 24 V tipper
A 24 V tipper truck is in for a work-light installation. The customer wants two 24 V LED work lamps drawing 15 A in total, mounted on the rear chassis cross-member. The cable route from the fuse panel in the cab to the lamps is 6 m one-way, run inside split conduit along the chassis rail where surface temperature reaches 70 °C in summer. On the shelf you have 2.5 mm² and 6 mm² thin-wall cable, 20 A and 25 A blade fuses, and the body-builder guide, which specifies a maximum 3 % volt drop on lighting circuits. Copper resistivity is 0.0175 Ω·mm²/m at 20 °C and the temperature coefficient is 0.00393 per °C. The fuse panel has two spare 30 A ignition-fed outlets.
1. Calculate the volt drop for both cable sizes at the working temperature and decide which cable to fit. Show your reasoning, including anything that makes the real-world result worse than the simple calculation.
Read the worked answer
Loop length = 12 m (6 m out and 6 m back).
2.5 mm²
- R at 20 °C = 0.0175 × 12 ÷ 2.5 = 0.0840 Ω
- Correction to 70 °C = 1 + (0.00393 × 50) = 1.197 → R = 0.1005 Ω
- Drop = 15 A × 0.1005 Ω = 1.51 V = 6.3 % - well over the 3 % limit.
- It also fails thermally: 2.5 mm² thin-wall in conduit at 70 °C ambient derates to roughly 12–15 A, at or below the load.
6 mm²
- R at 20 °C = 0.0175 × 12 ÷ 6 = 0.0350 Ω → at 70 °C = 0.0419 Ω
- Drop = 15 A × 0.0419 Ω = 0.63 V = 2.6 % - inside the 3 % limit, with ample current capacity after derating.
Fit 6 mm².
Why the real result is worse: these lamps are constant-power LED drivers, so as terminal voltage falls they draw more current to hold output, which increases the drop again. Add connector, fuse and terminal resistance (typically 0.1–0.3 V) and the 2.5 mm² option would leave the lamps under about 92 % of supply and the cable running hot inside the conduit.
2. Specify the fuse rating and its location, and answer the customer who says 'just put a 25 A fuse in so it stops blowing'.
Read the worked answer
The fuse protects the cable, not the lamp.
Coordination rule: load current ≤ fuse rating ≤ derated cable capacity.
- Load = 15 A continuous → a 20 A fuse gives margin for inrush without nuisance blowing.
- 6 mm² derated for 70 °C ambient and bundling is comfortably above 20 A.
- A 25 A fuse would not protect the 6 mm² as intended if a partial short or a damaged section reduces its capacity, and it buys nothing: a correctly sized circuit does not blow fuses. If fuses are blowing, the fault is a short, a damaged cable or a lamp drawing more than specified - find it, don't uprate the protection.
Location: at the source end - in the spare outlet of the fuse panel, so the unprotected length of cable is the shortest practical (a few hundred mm at most, protected by the panel's own feed). Never fuse at the load end: the whole run then has no short-circuit protection.
Also fuse the relay coil circuit separately (typically 5 A) and take the work-light feed through a relay rather than a 15 A switch, so the switch only carries coil current.
3. The customer now wants a third identical lamp added to the same run, taking the load to 22.5 A. Does the existing 6 mm² and 20 A fuse still work? Give your decision with numbers.
Read the worked answer
Cable: fails the volt-drop rule.
- Drop = 22.5 A × 0.0419 Ω = 0.94 V = 3.9 % of 24 V - outside the 3 % limit, and the added connector losses push it further.
- Options: run a second 6 mm² pair for the third lamp (keeps each run under 15 A), or go to 10 mm² (R at 20 °C = 0.0210 Ω; at 70 °C = 0.0251 Ω; drop = 22.5 × 0.0251 = 0.57 V = 2.4 %).
Fuse: must change. 22.5 A continuous against a 20 A fuse is a guaranteed nuisance blow. With 10 mm², a 25 A fuse is now correctly coordinated (load 22.5 A ≤ 25 A ≤ derated cable capacity). With two separate 6 mm² runs, each keeps its own 20 A fuse.
Verification after the change: measure volt drop at each lamp connector with all lamps lit and the engine off, then again at 28 V with the engine running; check terminal and fuse temperatures by hand/IR after 5 minutes; confirm each lamp still sees at least the manufacturer's minimum operating voltage.