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Undersea cables use steel for strength and additional conductor to power repeaters. Fiberoptic needs repeaters every couple hundred kilometers, so everything c
by lmilcin 5y ago
Undersea cables use steel for strength and additional conductor to power repeaters.
Fiberoptic needs repeaters every couple hundred kilometers, so everything crossing any ocean is actively powered.
The damage to the cable would come from difference in potential at the ends of the cable. The cable might be at the ocean bottom but it could still get fried if you apply high voltage that could kill repeaters. And without those the cable is useless.
Basically, greatly simplifying, in a solar storm any long conductor might be subjected to high potential difference (potential difference == voltage). It does not matter if the middle is exposed but enough of it needs to be exposed at the ends. So a lot will depend on details of how the cable is actually terminated. And it should be possible to engineer it (ie terminations) so that it is safe from a storm.
- ngrilly 5y agoSuper clear! Thanks!
- marcosdumay 5y agoI don't think the steel support is a real problem. It's conductive, but not that much. One ohm/m is enough to severely limit anything on the kV range over hundreds of km. Steel cable support is often much more resistive than that.
- withinboredom 5y agoThey never said steel was an issue.
- marcosdumay 5y ago> Undersea cables use steel for strength
- withinboredom 5y agoBut they never said it was an issue. I don’t understand how you got that it was.
- lmilcin 5y agoFirst of all, there is also copper conductor in it. But the most important is, I think, you are mistaking voltage and current. Resistance will limit current, but it will do nothing to voltage (except for fast transients depending on capacitive properties of the line -- sorry, there must always be some exceptions to every rule). 1M volts put to a conductor will still be 1M volts regardless of wether the conductor is 1ohm or 1M ohm. Remember, electronics is mostly fried by voltage, not by current. For high current to damage electronics at nominal voltage it must have an internal fault -- short circuit, for example, that causes it to draw more power than necessary. High potential, on the other hand, causes electrons to go all sorts of ways they are not supposed to go, causing damage to all sorts of things that are not supposed to be penetrated by electrons. You can damage sensitive electronics by waving your hand over it. You don't even have to touch it -- just the fact that you are moving some electrons on your hand causes electromagnetic field (moving charge == electromagnetic field) that causes potential to be induced in internal electronics (conductor in changing electromagnetic field == induction). Just read about why exactly electronics is distributed in those funny metalized bags.
- marcosdumay 5y agoElectronics devices are full of protective features against low-current high-voltage events. That's exactly because you can fry some parts just by touching them, and people do not want to entire devices to be that sensitive. As a rule, you need both voltage and current to kill a device. For MOSFETs, that's indeed very little current, but most components are far more forgiving, and the MOSFETs are always protected. Also, resistance causes the voltage to drop. How much it drops depends on the details of the device protection, but you certainly won't get a 1MV drop inside the device if you have many Mohm of resistance outside of it.
- deleted 5y ago[deleted]