I have a fuse holder that is rated for 32V DC, but I need to hook it up to some capacitors that are rated at 5 F and 50 volts. Will the fuse just blow right away even if the trip current isn't achieved, or will the fuse holder still function normally? If it will work normally, would Ohm's law apply and I just use a fuse rated lower than what I need?
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3\$\begingroup\$ Depends. Are you charging the caps up to its rated specs? \$\endgroup\$– tangrsApr 19, 2017 at 4:08
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\$\begingroup\$ What type of fuse holder it is? Please add datasheet or type identification. \$\endgroup\$– ChupacabrasApr 19, 2017 at 5:33
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5\$\begingroup\$ Are you actually going to be putting 50V on the caps, or is that just their rating? What actual potential difference will you be applying to the circuit? \$\endgroup\$– TripeHoundApr 19, 2017 at 7:54
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3 Answers
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Below 300v or so, at normal air pressure, the voltage is too low to initiate a spark across any practical gap. This points to the 32v rating of the fuseholder not being governed by breakdown issues.
The fuse itself is a different matter, as this starts to open with a current flowing, the perfect condition for maintaining current flow with an arc. Are the fuses themselves rated at 32v, or some other voltage?
As 32v is in the range of voltages that various regulation bodies deem as 'safe to touch', I would at least consider the possibility that this fuseholder does not pass touch tests for high voltage, so is relegated to being used at touch safe voltages.
If this is a one-off for home use (so you don't need to get it insurance rated), and your 50v supply is isolated (so any accidental contact problems with the fuse holder are mitigated, however note that 50v is rated as touch-safe by some jusidictions), and the actual fuse in the holder has a voltage rating of 50v or above, then it should be OK to use. Please add a picture of the holder, or a link to its specifications, to be sure.
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\$\begingroup\$ It is definitely more difficult to extinguish a 50V 10A arc than a 5V 10A arc. \$\endgroup\$ Apr 19, 2017 at 14:41
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The voltage rating of a physical component is specified such that if the maximum voltage is reached the component will not suffer catastrophic failure. If the fuseholder rated for 32V is brought up to 50V then the fuse may not matter since the current could either arc or creep at that voltage.
answered Apr 19, 2017 at 4:15
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\$\begingroup\$ Some components rated for use in a directly mains connected application will be tested for far HIGHER voltages than what is on the label, actually... compare the Class I to Class IV specifications commonly found in test equipment... \$\endgroup\$ Apr 19, 2017 at 9:41
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A fuse that has not yet blown will have close to 0 V across it; the voltage rating is irrelevant at this point. The problem comes after the fuse starts to blow due to overcurrent. If the available voltage is greater than the fuse's voltage rating, it may fail to open, open more slowly than expected, or catch on fire.
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\$\begingroup\$ I think you meant "fail to closed". \$\endgroup\$ Apr 19, 2017 at 7:45
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3\$\begingroup\$ No, I think he means "fail to open" as in "fail to become open circuit" rather than "fail into a closed circuit" \$\endgroup\$– FinbarrApr 19, 2017 at 8:30
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1\$\begingroup\$ Oh I get it now. English is difficult :) \$\endgroup\$ Apr 19, 2017 at 11:42