Blown Fuse While Measuring Adapter Current
While testing a 5 V, 2 A USB power adapter, a student wants to measure the output current with a digital multimeter. The load is a 10 Ω, 5 W resistor, so the expected current is about 0.5 A. The meter is set to the 200 mA DC current range and the probes are placed in series with the load. The display shows 0.00 and then the meter goes dead; the 200 mA fuse has blown. The student says the fuse was unnecessary and wants to replace it with a piece of wire, then use the 10 A range. The meter's 10 A range has a shunt of about 0.01 Ω; the 200 mA range has a shunt of about 1 Ω. The adapter is rated 5 V ±5%.
1. Explain why the fuse blew. Include the expected current, the meter range, and what the 0.00 reading means.
Read the worked answer
Expected current = V / R = 5 V / 10 Ω = 0.50 A. The 200 mA range can carry at most 0.20 A, so 0.50 A is about 2.5 times the range. The fuse opened to protect the meter's shunt and circuit.
The meter should have been set to the 10 A range before the circuit was broken. The 0.00 reading is not a zero-current measurement; it is the open circuit caused by the blown fuse. The student also needs to check that the leads are in the correct current terminals, because the 10 A terminal is often separate.
2. Explain the correct method for measuring current and why bypassing the fuse is dangerous.
Read the worked answer
Current is measured in series: break the circuit, connect the meter's current terminal and COM in series with the load, and select a range above the expected current. Start with the highest current range if the current is unknown, then switch down.
The fuse is a safety device. Bypassing it with wire removes protection from the shunt, leads, and adapter. If the load or adapter faults, the meter or leads can overheat, burn, or explode. Replace the fuse with the correct type and rating. Never use a wire link or a higher-rated fuse just to make the meter work.
3. If the student uses the 10 A range correctly, what reading is expected and what is the burden voltage? Why might the adapter output change?
Read the worked answer
On the 10 A range, burden voltage = I × shunt = 0.50 A × 0.01 Ω = 0.005 V, or 5 mV. This is negligible, so the reading should be about 0.50 A.
The adapter output may sag slightly under load because of its regulation and cable resistance. A 5 V ±5% rating allows 4.75–5.25 V. If the current reads very differently, check the lead resistance, poor contacts, and whether the adapter is genuinely rated for 2 A. Do not use the 200 mA range for this load even though its resolution is better; the current exceeds its limit.