Timeline for What causes a fuse to blow, the current or the power?
Current License: CC BY-SA 4.0
22 events
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| Sep 30, 2020 at 21:56 | comment | added | J... | @luiscolorado Yes, but you never use a fuse as the only element in the circuit. The fuse is never limiting current - it's designed to introduce as small a v-drop as needed to perform its function, so the voltage drop across the fuse will always be the same for a given current, regardless of the operating voltage of that circuit - because the full operating voltage is never applied to the fuse directly (unless there's a fault, in which case it will blow). Any fuse that you directly apply rated voltage across will blow, because they're basically negligible resistors over their rated range. | |
| Sep 30, 2020 at 19:44 | comment | added | luiscolorado | @J... You are correct, it is silly to apply conditions out of the specs of the components. I'm just responding to the comments of Brian Dummond in the context of the main question. Brian claims incorrectly that the fuse blows "at 6V, 12V or 240V, the fuse still blows at 20A." As discussed, you can't change the voltage without affecting the current, so a voltage high enough would make the fuse melt or arc. | |
| Sep 30, 2020 at 19:32 | comment | added | J... | @luiscolorado It's like saying the weight of an object doesn't affect the accuracy of a scale - if you put 2kg on a scale the indicator will raise by 2kg, whether going from 0kg to 2kg, or 100kg to 102kg. The mass doesn't matter. Until you put 1000kg on a scale rated for 20kg and the whole thing is crushed. Obviously that's not a reasonable evaluation of the linearity of the scale. Same with putting 1kV across a 10V fuse. That's just silly - it doesn't have any impact on the I^2t discussion. | |
| Sep 30, 2020 at 19:26 | comment | added | J... | @luiscolorado In a fuse it is, because the voltage across a fuse will only ever depend on the current through it, and a fuse blows on current. Obviously no component specifications apply when taken outside of the design range - for a fuse used within its design voltage limitations the current and time are the only parameters that matter. | |
| Sep 30, 2020 at 19:18 | comment | added | luiscolorado | @J... As you can see in the article, there are separate I^2t limits for arcing and melting. A higher voltage doesn't necessarily mean arcing. We can't simply say "voltage is irrelevant for fuse behavior", but we can say "voltage is irrelevant for fuse behavior only under certain circumstances." About your statement "energy is energy - the voltage doesn't matter", I'm not sure what you mean. Voltage, current and resistance are related. Power can be calculated as I^2R, but I is V/R. If you vary voltage, the current changes. The current is not independent of the voltage. | |
| Sep 30, 2020 at 18:49 | comment | added | J... | @luiscolorado That note is strictly for arcing considerations. If the voltage is too high, the blown fuse could still sustain an arc if it is not designed to expect a higher voltage on the terminals. This has nothing to do with the blow time since a fuse carrying an arc will already have blown. Energy is energy - the voltage doesn't matter. | |
| Sep 30, 2020 at 18:46 | comment | added | luiscolorado | @J..., As you can see at that site, I^2t applies when the fuse is operating under the nominally rated voltage: "System voltage exceeding the fuse’s rated voltage may result in fuse damage." The voltage is relevant, but it's excluded from most discussions because it is assumed that voltage will not exceed the established limits. | |
| Sep 30, 2020 at 18:40 | comment | added | luiscolorado | @slebetman It depends on the resistance of the main load. Let's say that a circuit is designed to be fed with 6 V, the fuse's resistance is 1 ohm, and the current is 1A. That means that the resistance of the load would be 5 ohms. The fuse is dissipating 1W in normal conditions. If you put the same circuit on 240V, the current would be I=V/(R1+R2)=240/(1+5)=40A. The fuse is using now I^2*R1=40^2*1=1600 W. More than most microwave ovens. The fuse will probably melt. | |
| Sep 23, 2020 at 23:51 | comment | added | slebetman | @luiscolorado Say you apply 240V to a 6V circuit. And say the resistance of the fuse is something like 1 ohm. And say the circuit is currently consuming 1A. The voltage accross the fuse is V = IR, which is 1A * 1 ohm which is 1V for a 240V circuit or a 6V circuit | |
| Sep 23, 2020 at 18:24 | comment | added | Kevin Keane | @R..GitHubSTOPHELPINGICE That question really is moot, isn't it? If a fuse blows essentially immediately, then asking whether voltage affects the time it takes becomes meaningless. Of course, in reality there is no such thing as "immediate" and, yes, the laws don't change, so it is still true, even if we are now only talking about milliseconds or nanoseconds or less. | |
| Sep 23, 2020 at 8:33 | comment | added | alex.forencich | Well, if you want to get pedantic, ultimately it's heat that causes it to blow. The temperature is a function of several things, including not only the current, but also how long it is applied and the thermal characteristics of the fuse. The fuse does not heat up instantly; it takes some time to warm up, therefore it is possible to significantly exceed the current limit without blowing the fuse if the pulse is short enough. This is also related to the energy dissipated into the fuse - volts (across fuse) times amps times time, also written as I^2 T after factoring in the fuse resistance. | |
| Sep 23, 2020 at 3:20 | comment | added | R.. GitHub STOP HELPING ICE | Is this true in the case where the load's resistance is very close to zero, i.e. where the fuse is blowing because of a short? While not the only mode of failure, it's a very common one. | |
| Sep 22, 2020 at 23:35 | comment | added | J... | @luiscolorado It's not relevant. Fuses rate by I^2t - what matters is the current through them and the time that the current is applied. | |
| Sep 22, 2020 at 22:23 | comment | added | luiscolorado | @MikeBrockington is correct. The voltage is relevant. I think that most people ignore the voltage because they assume that it remains fairly constant, which is true in the vast majority of the cases. It is incorrect to say that the fuse will blow all the same at 6V, 12V, or 240V. If you apply 240V to a circuit designed for 6V, the fuse will probably blow because the current in the circuit would be multiplied by 40 (240V/6V), thus the fuse would dissipate 40^2=1600 times more power. | |
| Sep 22, 2020 at 15:44 | comment | added | MikeB | @pipe Simple application of Kirchoff really - the fuse has a very small resistance, but could not possibly function if it was a superconductor. It therefore always has a some voltage across it, proportional to BOTH the input voltage and the resistance of the rest of the circuit. | |
| Sep 22, 2020 at 9:10 | comment | added | user16324 | @Kevin That's trivially true, arising because the fuse resistance R is temperature depnendent, but still doesn't imply a dependence on the supply voltage. | |
| Sep 22, 2020 at 4:06 | comment | added | Kevin Keane | @pipe I think Mike Brockington is right, strictly speaking. The fuse's resistance, together with the load, forms a voltage divider. If the fuse's resistance increases due to heat, and the load's resistance remains unchanged, the voltage across the fuse will also increase slightly, heating the fuse further and thus accelerating the blowing process. In practice, this effect is hopefully minimal. Especially because a fuse is supposed to heat up before blowing. | |
| Sep 21, 2020 at 17:08 | comment | added | user16324 | @pipe it doesn't of course. The voltage drop across the fuse depends on the current through it, not the supply voltage. | |
| Sep 21, 2020 at 16:48 | comment | added | pipe | @MikeBrockington How would the fuse know what the supply voltage is? No matter if the supply is 10 or 1000 volts, the voltage drop will be the same. | |
| Sep 21, 2020 at 16:15 | comment | added | MikeB | The voltage does indeed have a slight effect - as the fuse warms up, the resistance will increase, and at that point the voltage drop will also rise in proportion to the supply voltage. | |
| Sep 21, 2020 at 11:48 | vote | accept | Jackson Harvey | ||
| Sep 21, 2020 at 10:03 | history | answered | user16324 | CC BY-SA 4.0 |