Timeline for What exactly is a short circuit?
Current License: CC BY-SA 3.0
34 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 4, 2017 at 13:09 | comment | added | Majenko | @Pacerier Yes. Just ask Samsung. | |
| Oct 4, 2017 at 13:09 | comment | added | Pacerier | @Majenko, Re "explode" Are you serious? Practically speaking does it happen in real world or would the safety net be activated? | |
| Jan 6, 2017 at 22:02 | comment | added | LorenzoDonati4Ukraine-OnStrike | @Majenko Of course that's the problem of ideal models! The are all nice and dandy and easy to understand... until you realize that if they existed they would blow up the universe! :-D | |
| Jan 6, 2017 at 21:54 | comment | added | Majenko | @LorenzoDonati However, in reality, they do have a resistance, and thus there is a potential difference across them. Approximation for circuit design is one thing. Describing how current flows is another entirely. Besides - if you place an ideal conductor across an ideal voltage source the voltage across it will be 0 regardless of the voltage of the source, so I=V/R = 0/0 = 0. So no current, no voltage, and probably no universe either because you just broke it ;) | |
| Jan 6, 2017 at 21:51 | comment | added | LorenzoDonati4Ukraine-OnStrike | @Majenko Of course ideal things don't exist physically. They are useful nonetheless. It's all about the models we use and how well we want to model the physical phenomena. Ideal conductors do exist as models, since all those lines we draw in a schematic connecting components are ideal conductors, because usually (at least at 1st order approximation) we don't care about the resistances of the PCB tracks or connection wires. | |
| Jan 6, 2017 at 21:45 | comment | added | Majenko | @LorenzoDonati Except of course the fact that they don't exist. There is no "ideal" anything. Even superconductors aren't ideal conductors (though they are getting close). | |
| Jan 6, 2017 at 21:44 | comment | added | LorenzoDonati4Ukraine-OnStrike | @Majenko Just a nitpick: "Current flows between differences in potential." Well, excluding the corner case of ideal conductors, i.e. "real" short-circuits, where the resistance is actually zero. V = R * I, if R is zero any amount of current can flow and still we have no voltage drop. In other words, ideal conductors don't need a voltage across them to support a current flow. | |
| S Jan 6, 2017 at 18:57 | history | suggested | MrAP | CC BY-SA 3.0 | improved formatting |
| Jan 6, 2017 at 18:56 | review | Suggested edits | |||
| S Jan 6, 2017 at 18:57 | |||||
| S Jan 6, 2017 at 18:56 | history | suggested | MrAP | CC BY-SA 3.0 | improved formatting |
| Jan 6, 2017 at 18:45 | review | Suggested edits | |||
| S Jan 6, 2017 at 18:56 | |||||
| S Dec 19, 2016 at 19:10 | history | suggested | Lightness Races in Orbit | CC BY-SA 3.0 | Fixed grammar |
| Dec 19, 2016 at 18:55 | review | Suggested edits | |||
| S Dec 19, 2016 at 19:10 | |||||
| Dec 19, 2016 at 18:14 | comment | added | Crowley | @BlueRaja-DannyPflughoeft: What happens if we set both resistance and potential difference to 0? First task: Imagine a piece of cooper wire is ideal superconductor. Second task: Let it lie on the table. Voila, no current flows. Nasty explanation may be: You are dividing hard zero (real voltage) by soft zero (idealised resistance). | |
| Dec 19, 2016 at 17:15 | comment | added | Majenko | I am wondering if anyone actually read the last paragraph of the answer. By the comments I am guessing not. | |
| Dec 19, 2016 at 17:14 | comment | added | user56384 | @Lundin if a conductor could not handle the current, it should be labelled a 'fuse' instead. Similarly, the battery that catches fire should be labelled "incendiary device". Thus, all electronics is illegal. | |
| Dec 19, 2016 at 11:24 | review | Suggested edits | |||
| Dec 19, 2016 at 11:39 | |||||
| Dec 19, 2016 at 8:55 | comment | added | Lundin | Battery on fire assumes that all conductors can handle 5000A. The PCB/the wires might burn before before the battery itself does. | |
| Dec 19, 2016 at 8:51 | comment | added | user20088 | /agree with user23013 5000A would appear only when the Rint of the battery is 0, which would assume the voltage source is ideal - but an ideal voltage source simply wouldn't get on fire; I'd say that mentioning Rint is worthwhile here, since it, in some situations, actually prevent any serious damage from even a complete shorting Rload. | |
| Dec 19, 2016 at 8:51 | comment | added | rackandboneman | ...could lead to being misunderstood as an experiment instruction: "Add LED and 470 Ohm resistor, done. Add 1mOhm Resistor, done. Set battery on fire, done. Add battery (on fire), done." | |
| Dec 19, 2016 at 7:34 | comment | added | user23013 | I think you should just add the internal resistance of the battery to the circuit. Keeping 6.38mA on the intended path doesn't seem realistic. | |
| Dec 19, 2016 at 2:48 | comment | added | BlueRaja - Danny Pflughoeft | @MrAP: "What if we set the resistance and potential difference both to 0?" - That is similar to the question "What happens when two point-particles collide?" (the PE equation says both get infinite velocity). The real answer to both questions is "our equations don't work, but that's fine because that situation is impossible in the real world" | |
| Dec 19, 2016 at 0:20 | comment | added | Matthew Gordon | Deserves a +1 for the 'On Fire' probe alone | |
| Dec 19, 2016 at 0:13 | comment | added | Majenko | Schematics added. | |
| Dec 19, 2016 at 0:12 | history | edited | Majenko | CC BY-SA 3.0 | added 786 characters in body |
| Dec 18, 2016 at 22:51 | comment | added | user59864 | @MrAP: Any elecrtical wire is generally considered a short ciruit; draw a straight line and there you have your schematics. | |
| Dec 18, 2016 at 22:15 | comment | added | MrAP | Could you please add a good diagram for short circuits? | |
| Dec 18, 2016 at 21:28 | comment | added | Tom Carpenter | @MrAP 0/0 is an undefined value. In fact in this case it can be any value. Looking at Ohms law the other way, for a '0' ohm resistor, there will be 0V across it regardless of the current: \$V=IR=I\times0=0\$ holds true for any value of I. | |
| Dec 18, 2016 at 21:23 | comment | added | MrAP | What if we set the resistance and potential difference both to 0?Will the charge not flow(I=0/0)? | |
| Dec 18, 2016 at 21:20 | comment | added | Majenko | If there is no difference in potential then no current can flow. Current flows between differences in potential. | |
| Dec 18, 2016 at 21:17 | vote | accept | MrAP | ||
| Dec 18, 2016 at 21:16 | comment | added | MrAP | If the potentials across the path are the same, the resistance remaining low, then also will it be known as a short circuit and will there be current across in that case? | |
| Dec 18, 2016 at 21:15 | vote | accept | MrAP | ||
| Dec 18, 2016 at 21:16 | |||||
| Dec 18, 2016 at 21:02 | history | answered | Majenko | CC BY-SA 3.0 |