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I have a wire in normal condition runs 4A. The device I am using is an auto-transformer, and its data sheet recommends to use fuse of 10A. So I went to buy the 10A fuse, and mount it onto a fuse holder. However, my friend recommends to insert a 150k ohm resistor connected in parallel with the 10A fuse.

I don't know what he is talking about, is he doing this correctly? Why do we need a resistor? In normal condition, can we still use the fuse without the resistor? Because he doesn't seem to know what he is talking about. Please refer to the figure below.

The fuse holder can be found here

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  • \$\begingroup\$ It should be noted that even if the fuse blows there will be voltage and current through the attached circuit, so depending on what it does this may or may not be a good idea. \$\endgroup\$
    – PlasmaHH
    Commented Feb 26, 2015 at 9:09

4 Answers 4

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This fuse holder has a small neon bulb in series with the resistor. The neon bulb acts as an indicator that the fuse has blown.

When the fuse is intact, the voltage across the fuse is small. So is the voltage across the neon light with resistor, because it's in parallel with the fuse. When the fuse blows it becomes open circuit, and the supply voltage appears across the neon bulb with resistor. The neon bulb lights up. The purpose of the resistor is to limit the current through the neon bulb.

Wired across the fuse terminals and contained in the plastic cover is a resistor and neon indicator that would presumably light up if the fuse was ruptured. [from Amazon product customer review]

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Your fuse holder already has a resistor and neon bulb in parallel with the fuse. As the fuse is basically a near perfect short in parallel with the resistor and bulb, the resistor and bulb will have little to no current flow. It looks like it is a Brown Black Yellow resistor with Gold or Silver tolerance, which is 100K 5~10%. When the fuse blows, the resistor allows a little current through the neon bulb, lighting it to let you know the fuse blew.

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The FUSE is a short circuit PAST the resistor and bulb for the supply voltage. If/When the fuse blows the voltage will take the parallel path and light the Neon Bulb (probably an NE2) indicating that the fuse is blown. Same circuit is good for DC using an LED and a current limiting resistor.

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    \$\begingroup\$ Welcome to EE.SE. This question has already been answered and the original poster accepted it. \$\endgroup\$
    – winny
    Commented Jan 20, 2017 at 20:29
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There is an additional potential reason for adding in a resistor in parallel - avoiding in-rush current - consider if one one side of the fuse is a powerful battery or voltage source, and on the other has a relatively large discharged capacitor - if you connect it directly, then the current flowing into the capacitor to charge it will be quite large - whereas if you have a resistor, the capacitor can slowly charge through the resistor, and then when you place the fuse in, there isn't any voltage across it that could cause it to instantly blow.


I'm adding this comment to the answer about 5 years on BECAUSE there are potential safety issues which readers should be aware of - RM

This could work BUT would be a very bad idea 1n many cases.
A blown fuse is USUALLY a very good indication that the circuit is unpowered.
A removed fuse even more so.
This action makes neither of the above true in some cases.
The circuit is "live" with the fuse blown or even removed.

In most cases if there was even minimal load then a resistor in parallel with the fuse of high enough value to be safe would not allow the capacitor to charge.
However, with eg a photo flash this MAY work depending on other circuitry - with possibly lethal results. The capacitor could be at high voltage with the fuse blown or removed.

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    \$\begingroup\$ I've never seen a system that relies on inserting the fuse to make power. Normally you'd insert the fuse with no power applied and then switch on. \$\endgroup\$
    – Chris H
    Commented May 16, 2016 at 16:00
  • \$\begingroup\$ @ChrisH then a large value resistor across the switch could be used - or some sort of "precharge" circuit to prevent in-rush currents when applying power \$\endgroup\$ Commented May 16, 2016 at 17:16
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    \$\begingroup\$ Yes. This could be done. In parallel with the fuse as an inrush limiter would go against good design principles. \$\endgroup\$
    – Chris H
    Commented May 16, 2016 at 19:16
  • \$\begingroup\$ This could work BUT would be a very bad idea. A blown fuse is USUALLY a very good indication that the circuit is unpowered. A removed fuse even more so. In most cases if there was even minimal load then a resistor in parallel with the fuse of high enough value to be safe would not allow the capacitor to charge. With eg a photo flash this MAY work depending on other circuitry - with possibly lethal results. \$\endgroup\$
    – Russell McMahon
    Commented Jul 21, 2021 at 23:35

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