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Most small switch mode power supplies have an input resistor (typical 10 ohm) in front of the rectifier.

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Mostly this resistor is covered by a shrink tube. The top wire do not have an insulation, so it can't be a safety precaution. Why is this resistor covered by a shrink tube?

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Perhaps for some (small) heat proofing as it sits directly against the capacitor next to it? – David Jan 9 at 14:28
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Or maybe the opposite? The black surface of the heatshrink may radiate the heat better than the light colour coating of the resistor. – transistor Jan 9 at 14:33
    
MAY be thermistors made to change resistance under load current - sleeving would increase their sensitivity to heating. – Russell McMahon Jan 9 at 14:40
    
@David: It still generates the same amount of heat. – Michael Jan 9 at 17:37

It's because the resistor is intended to be the "weak link" in case something in the power supply fails shorted -- in other words, a cheap fuse. It also serves to limit the inrush current. The resilient heat-shrink tubing is there to contain any fragments resulting from its (possibly violent) failure, and to prevent the free ends from moving around and creating additional shorts.

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The sleeve will also add a small amount of distance between the bent wire and the opposite metal end cap of the resistor even after inserting, soldering and handling, the metal end caps are only insulated by the 'paint' coating and are a weak link in the withstand voltage of the device if it sees mains related voltages across the resistor in some situations.

I also support the added dissipation and shrapnel protection scenarios.

EDIT:

Inrush currents (when charging the HV DC link) on a series input resistor may result in significant voltage across the resistor.

Lighting spikes and load dump voltage transients may also exceed the optimistic design limits and the extra little air gap may just prevent a dozen failed units after every thunderstorm.

Having the protection there after the resistor fails (especially if it is a fusible protection components) is doubly important as the insulation breakdown would now form a short circuit across the already toasted resistor and deliver the supply voltage across possibly other failed (short circuit) components with undesirable results.

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Note that there won't be any significant voltage across the resistor before it fails. – Dave Tweed Jan 9 at 15:02

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