Bench Power Supply Overload and Plug Fuse Selection
During a school open-day build, a student assembles a 12 V, 1.8 A lamp test rig. The 12 V DC adaptor is marked 12 V, 1.5 A. The mains lead is 2 m of 0.5 mm² flexible cable fitted with a 13 A plug fuse. The student measures current by placing a multimeter, set to the 10 A range, in series with the lamp. With the lamp switched off the meter reads 0.02 A; with it on it reads 1.82 A. After ten minutes the adaptor case is hot, one flexible lead feels warm near the plug, and the lamp flickers when the lead is moved. The teacher is absent and the student asks whether the rig can be left running during the exhibition. The adaptor has no visible fuse or current-limiting device.
1. Identify four unsafe or incorrect practices in the setup and state the correction for each. Prioritise the immediate action before the exhibition.
Read the worked answer
Immediate action: switch off, unplug and tag the rig out of service; do not leave it unattended.
Adaptor overload - 12 V × 1.5 A = 18 W rated, but 12 V × 1.82 A = 21.8 W, about 121% of rating. The case is hot because the adaptor is being run beyond its continuous rating. Replace it with a 12 V supply rated at least 2 A (preferably 3 A), or use a lower-power lamp.
Wrong plug fuse - a 13 A fuse is far too large for a 1.5 A adaptor and 0.5 mm² flexible lead. The fuse protects the flexible cable, not the adaptor; 0.5 mm² is typically rated around 3 A. Fit a 3 A fuse (or 2 A if available) and check the plug termination.
Warm lead and flicker - a healthy 0.5 mm² lead carrying 1.8 A should not feel warm. Warmth plus flicker indicates a high-resistance joint, loose plug screw or broken strands. Replace the lead and re-terminate; do not tape it.
Measurement practice - the 10 A terminal on many multimeters is unfused or has limited protection. Use a clamp meter or the correct current range, and check the meter's CAT rating. A mistake on the 10 A range can cause an arc.
Supervision - a student should not run a mains-powered rig without a teacher or technician. PAT-test or inspect the lead before reuse.
2. Calculate the adaptor's rated output power and the lamp's power at 12 V. Explain why the 13 A fuse does not protect the adaptor or the 0.5 mm² lead.
Read the worked answer
Adaptor rated output: P = V I = 12 × 1.5 = 18 W. Lamp power at 12 V: P = 12 × 1.82 = 21.8 W.
A 13 A fuse is designed to carry 13 A continuously and to rupture only at many times that current. On the primary side the adaptor current is roughly I_primary ≈ 21.8/(240 × 0.8) ≈ 0.11 A, so the fuse is about 100 times the normal current. Even a short-circuit fault may not clear quickly if the adaptor's own impedance limits the current. The fuse therefore does not protect the adaptor against overload, and it does not protect the 0.5 mm² lead, whose capacity is around 3 A. The adaptor may have an internal thermal fuse, but that is a last resort and cannot be assumed. Correct protection is a plug fuse appropriate to the lead (3 A), plus a supply rated for the load. The 13 A fuse only protects against a very high-current fault in the mains lead.
3. The meter reads 0.02 A with the lamp off. Is that a fault? Explain the warm lead and flicker using P = I²R and state what measurement you would take next.
Read the worked answer
A 0.02 A reading with the lamp off is probably not a fault. Many adaptors draw a small no-load current for an indicator LED or control circuit: 0.02 A × 12 V = 0.24 W. It only becomes suspicious if the adaptor is cool and the reading is unstable.
The warm lead is the serious symptom. Power dissipated in a resistance is P = I²R. At 1.82 A, even 0.5 Ω of extra resistance at a poor plug joint dissipates 1.82² × 0.5 ≈ 1.66 W in a small volume, enough to feel warm and to degrade insulation. The flicker shows the resistance is intermittent, which can cause arcing and local heating.
Next measurements: isolate the lead, inspect the plug and cable, measure continuity/end-to-end resistance of each conductor, and measure voltage drop across the lead under load. If the drop exceeds about 0.5 V at 1.8 A, the lead resistance is above about 0.28 Ω and the lead should be replaced. Do not simply replace the fuse with a larger one.