12 V LED Work Lamp on a Long Two-Core Cable
A maintenance team installs a 24 W, 12 V LED work lamp at the end of a 40 m run of two-core flexible cable. The cable datasheet gives 0.0075 Ω per metre per conductor. The 12 V sealed lead-acid battery reads 12.0 V at its terminals while the lamp is on, and a digital multimeter at the lamp terminals reads 10.8 V. The lamp is noticeably dimmer than when tested with a 1 m lead. A spare 2 A blade fuse is available; the existing fuse holder is rated 10 A. The battery is switched by a relay at the battery end. The lamp data states 12 V nominal and 24 W, with no minimum operating voltage. The team must decide whether the lamp is faulty, whether the cable is acceptable, and what fuse rating to fit. They also need to explain the measurement to an apprentice without simply blaming the lamp.
1. Calculate the steady lamp current, the cable loop resistance, and the expected lamp-terminal voltage. Is the lamp faulty? Show the steps.
Read the worked answer
First find the lamp resistance from its rating: R_L = V²/P = 12²/24 = 6 Ω. Each conductor has R = 0.0075 × 40 = 0.30 Ω, so the go-and-return cable resistance is 0.60 Ω. With the lamp connected the total loop is 6.60 Ω. Current I = 12.0/6.60 = 1.82 A. Cable drop = 1.82 × 0.60 = 1.09 V, so expected lamp voltage = 12.0 − 1.09 = 10.91 V. The measured 10.8 V agrees within about 0.1 V, which is reasonable for meter and contact resistance. Therefore the lamp is not faulty; it is operating at reduced voltage because of the long cable. A common mistake is to use 24/12 = 2 A and then expect 2 × 0.6 = 1.2 V drop; the current is lower because the lamp is not receiving 12 V. The dimness is explained by cable drop, not a defective LED.
2. Recommend a fuse rating. Explain why the 10 A holder/fuse is not a safe choice and whether the 2 A spare is adequate.
Read the worked answer
The steady current is 1.82 A, but an LED driver can have a short inrush pulse. A 2 A fuse is only about 10% above the steady current and may nuisance-blow, especially in a warm enclosure; it also gives little margin for manufacturing tolerance. The 10 A fuse is definitely wrong: it would not protect the lamp flex or the LED driver, whose internal wiring may be rated around 2–3 A. The correct approach is to select the smallest standard fuse above the lamp manufacturer’s marked maximum and above the cable current rating. The 0.0075 Ω/m conductor is roughly 2.5 mm², so a 3 A fuse is a sensible choice if the manufacturer permits it; the 2 A spare can be used only as a temporary test. Check that the holder, flex, and connector are rated for the chosen fuse. Never fit a 10 A fuse simply because the holder accepts it.
3. The team needs a practical fix that keeps the 12 V lamp. Compare replacing the cable with 4 mm² or 6 mm² conductors and give one non-cable alternative.
Read the worked answer
Using ρ = 0.0172 Ω mm²/m at 20 °C: for 4 mm², R per conductor = 0.0172 × 40 / 4 = 0.172 Ω, loop = 0.344 Ω; total = 6.344 Ω; I = 12/6.344 = 1.89 A; drop = 0.65 V; lamp voltage is about 11.35 V. For 6 mm², R per conductor = 0.0172 × 40 / 6 = 0.115 Ω, loop = 0.229 Ω; total = 6.229 Ω; I = 1.93 A; drop = 0.44 V; lamp voltage is about 11.56 V. The 6 mm² version is better but harder to terminate; 4 mm² is usually enough. A non-cable alternative is to move the battery and relay within a few metres of the lamp, use a 230 V LED lamp of similar output, or use a 24 V battery and lamp. A boost converter at the lamp can hold 12 V but wastes energy and does not fix the cable loss. Also check the relay contacts, since a poor contact adds to the drop.