Fuse

Curricula: KS3, GCSE, Sek I, cycle 4

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Aim

See how a fuse protects a circuit from overload and from a short circuit, and why its rating is matched to the load.

The bench

A 6 V source, a 0.5 A fuse and an ammeter lead to two buses. Between the buses are three branches: at the bottom a 6 V / 0.5 W lamp, in the middle a 5.1 Ω load behind an open switch, at the top a push button — it short-circuits the buses for as long as it is held down. The lamp is lit and the ammeter reads about 83 mA.

Procedure

1. Close the load switch. Look at the fuse, the lamp and the ammeter.
2. Open the switch. Replace the fuse link: clear the "Blown" checkbox in the fuse's settings panel.
3. Set the rated current to 2 A and close the load switch again. Write down the ammeter reading.
4. Press and hold the button. Look at the fuse and at the source's power indicator.
5. Release the button, open the load switch, set the rated current back to 0.5 A and turn the source on with its power button.
6. Press the button again. Look at the fuse and at the source's power indicator.

Expected result

The 5.1 Ω load adds about 1 A to the lamp current — slightly more than twice the 0.5 A rating of the fuse link. A real fast-acting fuse link would still last for seconds at such a current: the heat escapes through the end caps, and the closer the current is to the threshold, the longer it lasts. Here it blows after half a second to several seconds, differently for each fuse link. The circuit breaks, the lamp goes out and the ammeter reads 0. A blown fuse link stays open until it is replaced. At a current below about 1.9 times the rating the fuse link does not blow at all.

The 2 A fuse link holds the same current: the ammeter reads about 1.0 A. But in a short circuit the current is about 7 A — only 3.4 times its rating, and such a fuse link needs a tenth of a second. The source's overload protection is quicker, at 20 ms: the power indicator flashes red, the source is switched off and the fuse link is intact. An over-rated fuse link does not protect the circuit.

With the 0.5 A fuse link the short-circuit current, about 6 A, is 12 times its rating: the fuse link blows in a few milliseconds, before the source's overload protection, and the source stays on.

Questions

1. Why is a fuse connected in series with the load rather than in parallel?
2. What happens if you fit a fuse link with a rating much higher than the load current?
3. How is a fuse better than the source's overload protection, which switches off the whole source?
4. Why does the fuse link blow in seconds at double the current, but in milliseconds in a short circuit?

Answers

1. In series the whole load current flows through the fuse, and when the fuse link blows, the circuit is broken. A fuse in parallel would short-circuit the load with its small resistance and blow at once, and after that the load would stay connected to the source without protection.
2. The fuse link lets through currents too large for the load and its wires, so the circuit is left without protection. On this bench the 2 A link holds the 1 A overload and survives the 7 A short circuit, only 3.4 times its rating: the source's overload protection switches off first.
3. Each branch can have its own fuse with a rating matched to its load, so a fault disconnects only that branch and the rest of the circuit keeps working. On this bench the 0.5 A link blows in a few milliseconds and the source stays on; the source's protection trips only above 3 A and switches off everything connected to it.
4. The heat released in the wire is \(P = I^2 R\). Up to about 1.9 times the rating the heat escapes through the end caps before the wire reaches its melting point; at twice the rating the heating is only slightly greater, so the wire creeps up to melting over seconds. In a short circuit, 12 times the rating, the heating is 144 times the rated value, and the wire melts in milliseconds before the heat can escape.