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April 28

Some bird

What's this bird? Yellow beak and yellowish (yellow-greenish) patches on the upper side, spotted in Warsaw, Poland. Thanks. Brandmeister^talk 08:57, 28 April 2020 (UTC)[reply]

Common starling. Mikenorton (talk) 09:02, 28 April 2020 (UTC)[reply]

Duplicate Wikidata items?

I came across two Wikidata items, Q23015579 and Q132641 which are both called "Glyptodontidae" where one has the description "subfamily of mammals (fossil)" but the other has the description "family of mammals." However, otherwise they appear to be about the same thing. Are these items duplicates? DraconicDark (talk) 15:42, 28 April 2020 (UTC)[reply]

There's a fairly consistent latin naming system for various Taxonomic ranks, with consistent use of suffixes for specific ranks. In this case, the -idae suffix is for a zoological family, and the -inae suffix is for zoological subfamily. Other than that, I don't know how to solve any knock-on problems from the double naming, but the -idae suffix indicates that that specific word is for a family and not a subfamily. --Jayron 32 15:48, 28 April 2020 (UTC)[reply]

@Jayron32: Upon further inspection, it turns out Q132641 is called "Glyptodontinae" which would fit the naming convention you stated. However, there still appears to be significant overlap between the subjects, and apparently the glyptodonts' classification isn't as clear as I had originally thought. This makes the question of whether to merge the items more complicated. DraconicDark (talk) 16:10, 28 April 2020 (UTC)[reply]

Taxonomic classifications change alarmingly frequently, and the subfamily Glyptodontinae is currently classified under the family Chlamyphoridae, but that doesn't mean there didn't used to be a family named "Glyptodontidae". Whether that family was subsumed into Chlamyphoridae and reclassified as a subfamily, or whether Chlamyphoridae was formerly called Glyptodontidae and just renamed, I couldn't say, both are possible and there are numerous examples of each happening, as well as other permutations (Glyptodontidae containing several subfamilies, including Glyptodontinae as well as several others, and then it was just deleted and its subfamilies added to Chlamyphoridae for example). You'd need an actual taxonomist to be able to make a decision here; I'm just a well-read chemist myself and don't have a lot of training here beyond what I find in Wikipedia articles. --Jayron 32 16:15, 28 April 2020 (UTC)[reply]

Explained at Glyptodon. Jayron32's suggestion of a former family subsumed into another is correct.--Khajidha (talk) 17:37, 28 April 2020 (UTC)[reply]

@Khajidha: So should the Wikidata items be merged? DraconicDark (talk) 18:25, 28 April 2020 (UTC)[reply]

Not being clear as to what Wikidata is or what it does or how it's used, I do not have any opinion on this. --Khajidha (talk) 21:08, 28 April 2020 (UTC)[reply]

Higher/lower voltages with respect to fire safety

In the United States, the standard voltage used is 110V. In many other countries, the voltage is much higher. Is there any consensus in the engineering community as to which one is safer? I would think higher voltages are safer since circuits with a higher voltage do not require as high of a current as lower voltage circuits, but obviously electrical engineering is a bit more complex than that, so I might be wrong. --Puzzledvegetable^{Is it teatime already?} 17:42, 28 April 2020 (UTC)[reply]

The standard in the US (and Canada) is actually 120 V: see Mains electricity by country. However, that's just the nominal voltage; in practice it will vary from one place or time to another, and most of the time a variation of that amount is insignificant. --76.71.6.31 (talk) 20:01, 28 April 2020 (UTC)[reply]

Aren't both of you wrong? The standard in the U.S. is 117 rms. 67.175.224.138 (talk) 22:11, 28 April 2020 (UTC).[reply]

I cited a reference, but it's only Wikipedia. What's yours? --76.71.6.31 (talk) 22:21, 28 April 2020 (UTC)[reply]

Either one can kill you. What kills you is the electric current going through your body; see our article on Electrocution. The higher the current, the sooner it is fatal. The strength of the current is determined by Ohm's Law and is a function of voltage and resistance. Given the resistance of your body, if it is not isolated from ground, the current is directly proportional to the voltage. So at 220V the current will be twice that at 110V, and will be deadly sooner. --Lambiam 18:48, 28 April 2020 (UTC)[reply]

Actually at 220 V you need twice less current to draw the same power, so the fuses typically have less amps. 93.136.9.236 (talk) 23:33, 28 April 2020 (UTC)[reply]

What you need is less, but that doesn't have to do with how the current manifests. Current is determined by Ohm's Law, and your resistance does not differ because you are standing in Europe and not the US. If your body resistance isn't changing, then the current resulting from 220 V traveling through you will be double. --OuroborosCobra (talk) 23:40, 28 April 2020 (UTC)[reply]

Do fuses have different resistances in Europe? 67.175.224.138 (talk) 23:44, 28 April 2020 (UTC).[reply]

For fire safety, the issue is the heat generated; see Joule heating. The heat is proportional to the product of current and voltage. So a voltage that is twice as high but a current that is only half as high produce the same amount of heat in the wiring. The amount of useful work is also proportional to that product; see AC power. ~~So in the end the fire hazard, given the same amount of work, should be equal.~~ --Lambiam 19:02, 28 April 2020 (UTC)[reply]

The rate of heat generation in a resistance is indeed given by the product of the current through it and the voltage between its ends. But the voltage drop across a poor connection in domestic wiring is not equal to the supply voltage, since the connection is in series with some domestic appliance such as an oven or kettle.

Unless the connection is very bad indeed, its resistance will be far lower than that of the appliance. In that case the current in the circuit is approximately the power rating of the appliance divided by the supply voltage (so double the current for half the supply voltage). The voltage dropped by the poor connection is the current times its resistance (so doubled by doubling the current). That means for an appliance of given power, and a connection of given resistance, the heating at the connection is quadrupled by halving the supply voltage, consequently increasing the fire risk.

This is an over-estimate, as we have neglected the fact that the poor connection will reduce the current slightly (though it is easy to correct for that). catslash (talk) 00:50, 29 April 2020 (UTC)[reply]

My current [pun not intended] calculations suggest even a factor of 16, assuming that the resistance of the poor connection is small compared to that of the appliances hooked up. The reason is that (to achieve the same amount of useful work or heating) the appliance resistance increases proportionally to the square of the voltage. The relative contribution of the poor connection to the voltage drop over the serial circuit of poor connection plus appliances then decreases proportionally to the inverse of that square, giving another factor of four. --Lambiam 05:26, 29 April 2020 (UTC)[reply]

I'm not following you. If the source voltage doubles and the load impedance is quadrupled to maintain the same power, then the ratio of load_impedance : I^2R_loss_of_poor_connection does indeed increase by 16X, but isn't that a meaningless ratio to even consider? -- ToE 13:13, 29 April 2020 (UTC)[reply]

I have now redone the algebraic calculations using a computer algebra system. I must have made some error in the calculations done by hand. To a first approximation, the heat generated in a bad (high-resistance) wire is indeed inversely proportional to the square of the source voltage, when in a serial circuit with an appliance of a given power consumption. --Lambiam 12:17, 30 April 2020 (UTC)[reply]

@Lambiam: You have written "... produce the same amount of heat in the wiring." When you use the word "wiring" I will assume you mean the two conductors that join the power source to the appliance; in which case the amount of heat produced in the wiring is not the same regardless of the voltage at the source. As you know, the heat produced in a conductor is the square of the current multiplied by the resistance of the conductor. If an appliance draws a given power P, and we consider first one supply voltage, and then consider doubling that supply voltage but still supplying power P to the appliance; doubling the voltage will cause the current in the wiring to halve. The heat produced with half the current will be only one-quarter of the heat produced in the first case, simply because joule heating is directly proportional to the square of the current providing the resistance is unchanged. So the risk of fire caused by joule heating in the wiring decreases dramatically as supply voltage is increased to an appliance drawing a given power P. Dolphin (t) 15:20, 29 April 2020 (UTC)[reply]

By "wiring" I was thinking of the wires in the walls, which in older buildings are often not in good shape, and by "voltage that is twice as high" I meant the voltage drop across the wiring. However, the latter is not simply proportional to the voltage supplied by the supplier. --Lambiam 12:17, 30 April 2020 (UTC)[reply]

Here in Australia, with 240 volts, power points/wall sockets almost always have switches, meaning that devices can remain plugged in but not be electrically alive. My limited observation is that switches don't generally exist on US wall sockets with 110 volts. While not a direct consequence of the different voltages, perhaps the switches are there because of the higher voltage involved, and definitely make things safer. HiLo48 (talk) 02:39, 29 April 2020 (UTC)[reply]

A frequent feature in American homes is that some of the wall outlets will be controlled by a switch, though not necessarily a direct part of the outlet. The purpose of that setup is to allow floor lamps and such to be turned off by the switch. Outlets that are near water sources, such as sinks, may have a different type of switch, kind of a "breaker" which trips if there's a short. ←Baseball Bugs ^{What's up, Doc?} carrots→ 03:07, 29 April 2020 (UTC)[reply]

I always called them GFIs but wiki calls them RCDs. hmm. Pelirojopajaro (talk) 13:54, 29 April 2020 (UTC)[reply]

Just to make sure that I'm understanding this, the answer to my question is that the heat in a circuit is proportional to the product of the current and the voltage, and not just the current, as I assumed in my question. Therefore, higher voltages are not necessarily less likely to cause a fire. Correct? --Puzzledvegetable^{Is it teatime already?} 14:12, 30 April 2020 (UTC)[reply]

That is correct. Electric power is voltage times current, and heat generated from electricity is directly proportional to power. This type of heating, called Joule heating, is often written as "I²/R", but that's just a rewriting of the variables in terms of current and resistance, voltage is just I/R, so voltage is just as important here. It's important to note that V and I are not independent variables, they are linked via Ohm's law, and changes to one (given a constant resistance) will cause changes in the other. Since Electrical resistance is temperature dependent, the voltage-current relationship is significantly non-linear over a wide enough temperature range. When it is said that "it's the current that kills you, not the voltage", that's also true, but misleading. Current is dependent on the number of electrons whereas voltage is not. A few electrons at an extremely high voltage will produce only a small current. However, if we're dealing with the same system (say a specific wire) then increasing the voltage through that wire will have a direct increase in the current. Ignoring the temperature dependence I mentioned earlier, if you double the voltage across a wire, you'll also double the current through that same wire, meaning that circuit is now twice as dangerous. On the flipside, by changing the material, you can change the current delivered without affecting the voltage. You can get astronomical voltages across good insulators, and get minimal to no current from them (that's what makes them an insulator). The point with this tangent is to note that the voltage and current need to be dealt with separately, but not entirely so, because each affects the other, but not in ways where we can say they are identical. There's a reason why electrical engineers need a good amount of calculus to do their jobs, the real math here gets messy. --Jayron 32 14:28, 30 April 2020 (UTC)[reply]

Being the question 'Higher/lower voltages with respect to fire safety', the answer is 'yes, higher voltage increases considerably fire safety'. That is by the way why Europe moved from 110 V to 240 V at the beginning of the 20th century: this way they had not to replace miles and miles of wire no more able to manage the rapidly growing loads. The point is that doubling the voltage you can double the impedance of your appliance to do the same work, thus halving the necessary current. And as Dolphin pointed above, by halving the current you divide by four the resistive heating of the conductors.

So the reasoning that by doubling the voltage you also double the current if the circuit doesn't change is true but here irrelevant as the circuit must and will change: a toaster is part of the circuit, and a toaster drawing 800 W by 110 V must more than double its impedance to draw 800 W by 240 V, thus more than halving the necessary current.

Generally speaking, by doubling the voltage trasmission losses are reduced by a factor of four, that is why very long overhead power lines are operated with tensions up to 1 MV. 2003:F5:6F0A:C00:5CC0:1EE6:8417:F1F (talk) 19:38, 30 April 2020 (UTC) Marco PB[reply]

Yes, but I understood from Lambiam that the heat is based on the product of the voltage and the current, so the lower current should be irrelevant since it requires a higher voltage, and the heat remains the same. Is this incorrect? --Puzzledvegetable^{Is it teatime already?} 20:31, 30 April 2020 (UTC)[reply]

This is exactly the point, because the heat does not remain the same: heat doesn't depend from the current but instead from the current squared, see Joule_heating and voltage doesn't appear at all in the equation. You halve the current, you divide the heat by four. Suppose the current is 2, then heat is 2² = 4. Half the current = 1 and 1² is 1.

Lambiam is mistaken: heat is not based on the product of voltage and current, but on the product of current squared and resistance (current itself depends on voltage divided by resistance.) 2003:F5:6F0A:C00:5CC0:1EE6:8417:F1F (talk) 21:54, 30 April 2020 (UTC) Marco PB[reply]

They are the same. By Ohm's Law, I = V/R, so V = IR. Then we find for the product of voltage and current: VI = IV = I×IR = I²R, the product of current squared and resistance. No mistake there. My initial mistake was very different. It was not to realize that the relative voltage drop across the wiring is affected by the voltage-dependent resistance of the appliances (hooked together with the wiring resistance in a serial circuit). --Lambiam 07:32, 1 May 2020 (UTC)[reply]

Okay, if you looked at Lambiam's table of 110 vs. 220 V. We saw the R got quadrupled. So should we look at P as in temperature of wire in terms of VI or I²R? The problem with looking in terms of VI is you don't see the change in R when V is doubled and I is halved. This makes I²R more convenient to use? 67.175.224.138 (talk) 14:18, 3 May 2020 (UTC).[reply]

Okay going back to what Jayron said as voltage = I/R, when UK goes from 120 V to 210 V or so, part of the confusion is, is both I and R both changing? Voltage can go up if R is getting smaller with I being the same. Voltage can also go up if I is getting bigger with R being the same. And so and so forth. 67.175.224.138 (talk) 05:27, 1 May 2020 (UTC).[reply]

@Jayron32: and the IP address immediately above have written that voltage is I/R and power is I²/R. Both are incorrect. Ohm’s law tells us that voltage drop across a conductor is IR; and heat is generated by a conductor at the rate of I²R. Dolphin (t) 06:03, 1 May 2020 (UTC)[reply]

Okay, then as IR goes up, the safety is therefore the same? Stronger amps, but stronger resistance, if going up equally. 67.175.224.138 (talk) 06:39, 1 May 2020 (UTC).[reply]

An example with numbers may clear up some things. The assumption is that we have a series circuit consisting of an ideal voltage source and two resistance components. One component is an appliance or combination of appliances, for example an electric heater and a washing machine, consuming together nominally 3kW. The other component is formed by the wiring, rated at 0.02Ω.

For 110V

source voltage: 110V; nominal appliance wattage: 3000W; wiring resistance: 0.02Ω

appliance resistance: 4.03Ω; total resistance: 4.05Ω; current: 27.1A

appliance voltage drop: 109.5V; wiring voltage drop: 0.5V

actual appliance power: 2970W; wiring power: 14.7W

For 220V

source voltage: 220V; nominal appliance wattage: 3000W; wiring resistance: 0.02Ω

appliance resistance: 16.13Ω; total resistance: 16.15Ω; current: 13.6A

appliance voltage drop: 219.7V; wiring voltage drop: 0.3V

actual appliance power: 2990W; wiring power: 3.7W

For a voltage source that is twice as high, the heat generated in the wiring drops to 25%. If the wiring resistance goes up, that heating also goes up to a maximal value of 750W, after which it decreases. The maximal value is reached sooner for the lower source voltage; when the (faulty) wiring resistance reaches 8.1Ω, the still rising 220V wiring power exceeds the already dropping 110V wiring power. --Lambiam 10:51, 1 May 2020 (UTC)[reply]

So notice that R is quadrupled when we went from 110 V to 220 V. Needless to say, I think it was pointless to say heat/temperature/fire safety is determined by the product of I * V. Just as it could be said I²R. In the latter, we saw the R quadrupled. And since R = V/I, a stronger R is a bigger V and smaller I. So I think to answer the original poster's question, we shoulda said this heat/fire safety in terms of R (or V/I), and not as VI or I²R. @Puzzledvegetable:. 67.175.224.138 (talk) 14:38, 3 May 2020 (UTC).[reply]

Weird, it seems like the discussion ended already. I'm curious to know why voltage and amps can't be independent of each other. Can you not increase the voltage of something with the amps staying the same? But as a separate, and and somewhat political question, if 220 V is safer than 120 V from a fire-safety point of view, then why hasn't U.S. attempted to change into that? 67.175.224.138 (talk) 05:29, 2 May 2020 (UTC).[reply]

For any electrical load that is resistive in nature, Ohm's law is applicable and says that the current is directly proportional to the voltage. Fire-safety is only one aspect of an electrical supply system; another is safety with respect to electrocution. The thing that protects us all from electrocution is the insulation surrounding all live conductors; the higher the voltage, the greater the amount of insulation that is required, but there is always a risk that the insulation surrounding an appliance will be damaged, allowing a person to come in contact with a live conductor. Keeping the voltage low is an alternative to using increasing amounts of insulation so it can be argued that people who live in a country where the domestic supply voltage is 110V are safer with respect to electrocution than those who live in a country where 220V is supplied. Dolphin (t) 12:30, 2 May 2020 (UTC)[reply]

The lower domestic supply voltage in the US, and consequent higher current, necessitates thicker wires. Thick wire use a lot of copper, which is expensive. This motivated the use of aluminum wiring in domestic premises in the US in the 1960s and 70s. Achieving and maintaining a good electrical connection to aluminum can be problematic, and poor connections a fire hazard (especially those carrying high currents). In this way the lower voltage did, somewhat indirectly, lead to a fire risk. catslash (talk) 16:24, 2 May 2020 (UTC)[reply]

"The higher the voltage, the greater the amount of insulation is required." So that's like saying the lower the amps/current, the more insulation needed. Insulation is synonymous with resistance right? Weird, I think of current as the amount of electricity, and voltage as the force that pushes the electricity. So UK is like the stronger force that pushes the same amount of current, but will actually be less current, which I visualize as faster electricity, whereas US has weaker force for more electricity. So has anyone said whether resistors/fuses are different in UK and US? 67.175.224.138 (talk) 22:27, 2 May 2020 (UTC).[reply]

The insulation being discussed is what surrounds the wire (rubber/plastic sleeve, etc.), not a poorly-conductive segment of the conductors themselves. DMacks (talk) 06:59, 3 May 2020 (UTC)[reply]

With regards to fire safety, people earlier were saying that fire safety is proportional to IV. But as V goes up, so does R. So that make UK safer for fire safety, but US safer for electrocution? 67.175.224.138 (talk) 22:38, 2 May 2020 (UTC).[reply]

Okay, it looks like with all the said, it can be summarized into a simple paragraph: "Higher voltage is safer for fire hazards, but lower voltage is safer for electrocution." Higher voltage being with lower amps and lower voltage being higher amps. And although this answers with power (IV) it doesn't answer with the 3rd variable R. But among R, there's 2 factors to consider: the wire resistance R, and the insulation around the wire. Does anyone want to take a dab at the economic feasibility, between US and UK, of the wire and insulation around the wire? 67.175.224.138 (talk) 13:26, 3 May 2020 (UTC).[reply]

I am also curious to know why electrocution is a bigger issue, when the amps goes down. Weird. 67.175.224.138 (talk) 06:16, 4 May 2020 (UTC).[reply]

Let's say we are working on an appliance that operates at 240 V and draws a current of 10 A. Let's also imagine the insulation is broken and we accidentally make contact with a live terminal. Suddenly we are exposed to 240 V, but don't imagine the current that surges through our bodies is only 10 A. The current would be so much higher, and we all hope it is high enough to trip a circuit breaker, or some similar safety device, to protect us from injury. Dolphin (t) 12:13, 4 May 2020 (UTC)[reply]

Well, in my city, the local electric trains, run on a 3rd rail, that is 600 V. Would that make the amps super low? Yet this electrocution should be worse than household electricity as no humans to survive electrocution. Are train rails non-ohmic? 67.175.224.138 (talk) 13:19, 4 May 2020 (UTC).[reply]

See electrical injury § pathophysiology catslash (talk) 14:47, 4 May 2020 (UTC)[reply]

I see a 4th variable is Hertz for AC electricity. Does Hertz change as we go from 120 V to 220 V? 67.175.224.138 (talk) 15:31, 4 May 2020 (UTC).[reply]

What exactly does Hertz measure in the context of electricity? What part of the electricity oscillates? Seeing as you mentioned it in the context of AC electricity, I assume it's the frequency of changes in direction of the current, if that even makes sense. Is that correct? And more broadly, what does it mean for the current to change direction, and why does it do that? What's the benefit of that over DC? --Puzzledvegetable^{Is it teatime already?} 12:45, 5 May 2020 (UTC)[reply]

It's an independent variable. Different countries have different standards. According to mains electricity, there are all four combinations or 110ish-volt or 240ish-volt at 50 Hz or 60 Hz. And some special equipment has its own independent of apart from standard line supplies (Amtrak's 25 Hz traction power system for example). DMacks (talk) 16:11, 4 May 2020 (UTC)[reply]