Voltage Drop on a 12 V LED Strip Run
A shop installs 5 m of 12 V LED strip rated 4.8 W/m in a customer's false ceiling. The 12 V, 5 A adapter sits at one end, and the strip is fed only at its far end through 14 m of 0.75 mm² copper cable (resistivity about 0.0175 Ω·mm²/m). At switch-on the near end is bright, but the far end measures 10.2 V and looks dim; the adapter terminals measure 12.1 V. The customer insists the adapter is too small and wants a 10 A unit. The strip draws 2 A, and the connectors are warm but not hot. The technician has 14 m of 2.5 mm² cable and one adapter lead, and can make one ceiling visit. The shop gives a 30-day workmanship warranty. The customer also asks whether a 24 V strip would be better.
1. Calculate the strip current, the cable voltage drop and the expected voltage at the strip. Decide whether fitting a 10 A adapter solves the problem.
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Total load: P = 5 m × 4.8 W/m = 24 W. Current: I = P/V = 24/12 = 2 A. The existing 5 A adapter is already rated above 2 A, so a 10 A adapter cannot raise the voltage at the strip; it only changes the adapter's current capability.
Cable loop length = 2 × 14 m = 28 m. Resistance: R = ρl/A = 0.0175 × 28 / 0.75 = 0.653 Ω. Voltage drop = I × R = 2 × 0.653 = 1.31 V. Expected strip voltage = 12.1 − 1.31 = 10.79 V, close to the measured 10.2 V once connector and joint resistance are included.
The dim far end is a voltage-drop problem, not an adapter-current problem. A larger adapter would still leave about 1.3 V across the cable and could increase fault current. The correct fix is to reduce loop resistance or split the current.
2. Recommend the most practical fix using the available materials, with calculations. State one trade-off and one reason a 24 V strip is not a same-visit solution.
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Replace the 14 m run with 2.5 mm² cable. New resistance: R = 0.0175 × 28 / 2.5 = 0.196 Ω. Drop = 2 × 0.196 = 0.392 V. Strip voltage ≈ 12.1 − 0.39 = 11.7 V, about 0.9 V better than the present 10.8 V.
Feeding both ends with two 0.75 mm² runs is an alternative: each conductor would carry about 1 A, giving a drop of about 1 × 0.653 = 0.65 V and a strip voltage near 11.45 V. That requires a second cable and joint; with one adapter lead and one visit, the 2.5 mm² replacement is more certain.
Trade-off: thicker cable costs more and is stiffer to route in the ceiling. A 24 V strip would need a new strip and a 24 V supply or converter, so it is not a same-visit repair. Check that connectors are rated for 2 A and re-check polarity.
3. What measurement would confirm the cable is the main culprit, and what checks should be made before re-energizing?
Read the worked answer
Measure the adapter output and the strip input simultaneously under load with a DMM. A difference of about 1.3 V confirms cable and connector loss. Measure each joint drop; a warm joint with more than about 0.2 V drop should be remade.
Before energizing: isolate the mains supply, verify adapter polarity against the strip markings, inspect the cable for damage, confirm joints are mechanically secure and insulated, check that the adapter is ventilated, and confirm no short circuit from the supply side. Record before-and-after voltages on the job sheet because the workmanship warranty may be claimed.