# Some questions from the electrically ignorant **URL:** \ **Category:** Factual Questions\ **Created:** [February 27, 2009, 4:51pm UTC](https://boards.straightdope.com/t/some-questions-from-the-electrically-ignorant/487679 "2009-02-27T16:51:09Z")\ **Posts on this page:** 19\ **Page:** 2
**Author:** ![HongKongFooey](https://avatars.discourse-cdn.com/v4/letter/h/c4cdca/32.png) [@HongKongFooey](https://boards.straightdope.com/u/HongKongFooey)\ **Post date:** [March 1, 2009, 12:21am UTC](https://boards.straightdope.com/t/some-questions-from-the-electrically-ignorant/487679/21 "2009-03-01T00:21:16Z")
> [@Q.E.D](#): > > Not when you’re standing right next to a phase-to-phase fault at 480 V and your hand is holding the meter probe which dislodged the improperly fastened wire which caused it. Did you actually experience this? ---
**Author:** ![Q.E.D](https://avatars.discourse-cdn.com/v4/letter/q/51bf81/32.png) [@Q.E.D](https://boards.straightdope.com/u/Q.E.D)\ **Post date:** [March 1, 2009, 12:31am UTC](https://boards.straightdope.com/t/some-questions-from-the-electrically-ignorant/487679/22 "2009-03-01T00:31:11Z")
> [@HongKongFooey](#): > > Did you actually experience this? Yep. The equipment I was working on fared far worse than I did, but nothing some new control wires are a fresh coat of paint couldn’t fix. I got away with little more than singed finger hair and a temporary ringing in the ears. After I bitched about it, they stopped using fuse holders with terminals meant for 10 AWG min. on 18 AWG wires. :rolleyes: ---
**Author:** ![HongKongFooey](https://avatars.discourse-cdn.com/v4/letter/h/c4cdca/32.png) [@HongKongFooey](https://boards.straightdope.com/u/HongKongFooey)\ **Post date:** [March 1, 2009, 12:40am UTC](https://boards.straightdope.com/t/some-questions-from-the-electrically-ignorant/487679/23 "2009-03-01T00:40:53Z")
That’s interesting, I’ve had close calls in the past but so far nothing that bad. I had to fight for years to get proper arc-flash protection; they were OK with uniforms but didn’t want to get gloves or face shields. I finally got the gloves and fusible leads but no face protection yet. ---
**Author:** ![Q.E.D](https://avatars.discourse-cdn.com/v4/letter/q/51bf81/32.png) [@Q.E.D](https://boards.straightdope.com/u/Q.E.D)\ **Post date:** [March 1, 2009, 1:36am UTC](https://boards.straightdope.com/t/some-questions-from-the-electrically-ignorant/487679/24 "2009-03-01T01:36:30Z")
If you really want proper PPE, I’d say you should buy it yourself. I know, your company should provide it, but unless you want to be a whistleblower and get OSHA (or whatever the Canadian equivalent is) you can’t really rely on that. ---
**Author:** ![HongKongFooey](https://avatars.discourse-cdn.com/v4/letter/h/c4cdca/32.png) [@HongKongFooey](https://boards.straightdope.com/u/HongKongFooey)\ **Post date:** [March 1, 2009, 2:08am UTC](https://boards.straightdope.com/t/some-questions-from-the-electrically-ignorant/487679/25 "2009-03-01T02:08:20Z")
That’s good advice. Our safety is very good overall, it’s just hard sometimes to get managers to understand the dangers of medium and high voltage. I have friends working in various mines, you wouldn’t believe the conditions their electrical guys have to deal with. ---
**Author:** ![Lumpy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lumpy/32/446_2.png) [@Lumpy](https://boards.straightdope.com/u/Lumpy)\ **Post date:** [March 1, 2009, 4:29pm UTC](https://boards.straightdope.com/t/some-questions-from-the-electrically-ignorant/487679/26 "2009-03-01T16:29:04Z")
So I gather that ohmage (if that’s a word) isn’t just the passive loss of an imperfect conductor, but is directly related to the working load on an electric circuit? ---
**Author:** ![HongKongFooey](https://avatars.discourse-cdn.com/v4/letter/h/c4cdca/32.png) [@HongKongFooey](https://boards.straightdope.com/u/HongKongFooey)\ **Post date:** [March 1, 2009, 4:40pm UTC](https://boards.straightdope.com/t/some-questions-from-the-electrically-ignorant/487679/27 "2009-03-01T16:40:46Z")
> [@Lumpy](#): > > So I gather that ohmage (if that’s a word) isn’t just the passive loss of an imperfect conductor, but is directly related to the working load on an electric circuit? Resistance is the usual term used for opposition to current in a DC circuit, impedance for A.C. It gets confusing because some devices such as semi-conductors have ‘resistances’ that vary, ignore them for now. In laymen’s terms you can apply a fixed resistance value to most items. The resistance in a wire, for instance, might have 3 ohms over a given length. If you heat the wire up the resistance will go up, if you make the wire longer or thinner the resistance goes up. It’s just a measure of how much it resists current flow with a given voltage applied across it. ---
**Author:** ![HongKongFooey](https://avatars.discourse-cdn.com/v4/letter/h/c4cdca/32.png) [@HongKongFooey](https://boards.straightdope.com/u/HongKongFooey)\ **Post date:** [March 1, 2009, 5:10pm UTC](https://boards.straightdope.com/t/some-questions-from-the-electrically-ignorant/487679/28 "2009-03-01T17:10:27Z")
> [@Lumpy](#): > > My electric range top works because the coils resist the flow of electricity and turn it into heat. OK, so why does an insulator block electricity altogether, rather than turn it into heat? Is an insulator not the same thing as a very high resistor? The heating element in an appliance resists some of the current flow but lets most of it through; a resistor in a circuit is there to _control_ the flow of current. The insulator around the conductor is chosen to resist most of the electric potential across it. If you put a large enough voltage across it it will break down. Everything conducts at some point, air conducts at something like 3kV/inch or so. The difference between a conductor and an insulator has lot to do with how you intend to use it to make electricity useful. ---
**Author:** ![Crafter\_Man](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/crafter_man/32/458_2.png) [@Crafter\_Man](https://boards.straightdope.com/u/Crafter_Man)\ **Post date:** [March 1, 2009, 8:12pm UTC](https://boards.straightdope.com/t/some-questions-from-the-electrically-ignorant/487679/29 "2009-03-01T20:12:52Z")
> [@HongKongFooey](#): > > Resistance is the usual term used for opposition to current in a DC circuit. FWIW, I have never really liked this definition. Let’s say I have a constant current source adjusted for 100 mA, and it’s being used to supply current to a load. If I stick a resistor in series with the constant current source, is the resistor opposing the current? No - the current remains the same. When someone asks me what a resistor is, I tell them it is simply a component that obeys V = IR, where R is a constant. It need not obey this equation exactly; as long as the current and voltage have an approximate linear relationship, then we can call it a resistor. ---
**Author:** ![HongKongFooey](https://avatars.discourse-cdn.com/v4/letter/h/c4cdca/32.png) [@HongKongFooey](https://boards.straightdope.com/u/HongKongFooey)\ **Post date:** [March 1, 2009, 8:18pm UTC](https://boards.straightdope.com/t/some-questions-from-the-electrically-ignorant/487679/30 "2009-03-01T20:18:57Z")
> [@Crafter\_Man](#): > > FWIW, I have never really liked this definition. Let’s say I have a constant current source adjusted for 100 mA, and it’s being used to supply current to a load. If I stick a resistor in series with the constant current source, is the resistor opposing the current? No - the current remains the same. So what did your constant-current source do to compensate for the added resistance and maintain 100mA in the loop? ---
**Author:** ![Crafter\_Man](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/crafter_man/32/458_2.png) [@Crafter\_Man](https://boards.straightdope.com/u/Crafter_Man)\ **Post date:** [March 1, 2009, 8:24pm UTC](https://boards.straightdope.com/t/some-questions-from-the-electrically-ignorant/487679/31 "2009-03-01T20:24:02Z")
> [@HongKongFooey](#): > > The heating element in an appliance resists some of the current flow but lets most of it through The heating element isn’t really “resisting” the current. All you can really say is that the heating element has resistance (unit: ohms), and that V = IR. It’s a bit more complicated than this, of course, as a more complete model would have to include temperature coefficients, source resistance, etc. But if we ignore that stuff for the moment, all we have to say is that V = IR and P = IV. > [@HongKongFooey](#): > > The insulator around the conductor is chosen to resist most of the electric potential across it. The insulator doesn’t “resist” the voltage - the voltage is there. The insulator simply has very high resistance (unit: ohms), and the system is designed so that very little current will flow through it when the insulator is subjected to the system voltage (peak voltage on one side of the insulator vs. ground on the other side of the insulator, peak voltage on one side of the insulator vs. a different peak voltage on the other side of the insulator, etc.). ---
**Author:** ![Crafter\_Man](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/crafter_man/32/458_2.png) [@Crafter\_Man](https://boards.straightdope.com/u/Crafter_Man)\ **Post date:** [March 1, 2009, 8:27pm UTC](https://boards.straightdope.com/t/some-questions-from-the-electrically-ignorant/487679/32 "2009-03-01T20:27:57Z")
> [@HongKongFooey](#): > > So what did your constant-current source do to compensate for the added resistance and maintain 100mA in the loop? It increased its output voltage. ---
**Author:** ![HongKongFooey](https://avatars.discourse-cdn.com/v4/letter/h/c4cdca/32.png) [@HongKongFooey](https://boards.straightdope.com/u/HongKongFooey)\ **Post date:** [March 1, 2009, 8:31pm UTC](https://boards.straightdope.com/t/some-questions-from-the-electrically-ignorant/487679/33 "2009-03-01T20:31:49Z")
> [@Crafter\_Man](#): > > It increased its output voltage. If it hadn’t increased it’s voltage the current through the resistor would have dropped because the added resistance limited the current flow at the original applied voltage. ---
**Author:** ![HongKongFooey](https://avatars.discourse-cdn.com/v4/letter/h/c4cdca/32.png) [@HongKongFooey](https://boards.straightdope.com/u/HongKongFooey)\ **Post date:** [March 1, 2009, 8:42pm UTC](https://boards.straightdope.com/t/some-questions-from-the-electrically-ignorant/487679/34 "2009-03-01T20:42:45Z")
> [@Crafter\_Man](#): > > The insulator doesn’t “resist” the voltage - the voltage is there. This is ridiculous, if the insulator around a wire didn’t ‘resist’ the voltage applied across it it would break down and conduct. This is not a good habit for insulators. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [March 1, 2009, 9:23pm UTC](https://boards.straightdope.com/t/some-questions-from-the-electrically-ignorant/487679/35 "2009-03-01T21:23:59Z")
> [@](#): > > This is ridiculous, if the insulator around a wire didn’t ‘resist’ the voltage applied across it it would break down and conduct. This is not a good habit for insulators. The insulator did not resist the voltage. It did withstand the voltage. ---
**Author:** ![HongKongFooey](https://avatars.discourse-cdn.com/v4/letter/h/c4cdca/32.png) [@HongKongFooey](https://boards.straightdope.com/u/HongKongFooey)\ **Post date:** [March 1, 2009, 9:32pm UTC](https://boards.straightdope.com/t/some-questions-from-the-electrically-ignorant/487679/36 "2009-03-01T21:32:37Z")
> [@Chronos](#): > > The insulator did not resist the voltage. It did withstand the voltage. Youse guys is confusin’ me. It appears to me that it withstood the voltage across it _because_ it offered enough resistance to the applied voltage to prevent it from establishing any meaningful flow of current through it. The two ways of saying it seem the same to me. What’s better, conventional or electron current flow? 😛 ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [March 1, 2009, 9:37pm UTC](https://boards.straightdope.com/t/some-questions-from-the-electrically-ignorant/487679/37 "2009-03-01T21:37:09Z")
> [@Crafter\_Man](#): > > > [@HongKongFooey](#): > > > > Resistance is the usual term used for opposition to current in a DC circuit. > > FWIW, I have never really liked this definition. Let’s say I have a constant current source adjusted for 100 mA, and it’s being used to supply current to a load. If I stick a resistor in series with the constant current source, is the resistor opposing the current? No - the current remains the same. Your objection doesn’t make sense to me. The constant current source is external to the resistor, but resistance is a property of the resistor. You shouldn’t have to rely on a particular source to descibe that property. In your example, you’ve still got to somehow create enough voltage across the resistor to cause the current flow. For example, if you have a shorted resistor with the 100 mA flowing through the short, and you open the short, you won’t immediately have 100 mA flowing through the resistor. Current from the current source will have to flow to build up charge to cause a voltage across the resistor. > [@Chronos](#): > > The insulator did not resist the voltage. It did withstand the voltage. Yeah, a resistor “resists” current. It seems more accurate to me to say that a conductor resists voltage. ---
**Author:** ![Napier](https://avatars.discourse-cdn.com/v4/letter/n/ce73a5/32.png) [@Napier](https://boards.straightdope.com/u/Napier)\ **Post date:** [March 1, 2009, 9:53pm UTC](https://boards.straightdope.com/t/some-questions-from-the-electrically-ignorant/487679/38 "2009-03-01T21:53:14Z")
> [@Lumpy](#): > > So I gather that ohmage (if that’s a word) isn’t just the passive loss of an imperfect conductor, but is directly related to the working load on an electric circuit? I don’t remember hearing “ohmage”, but the phenomenon you are talking about is the passive loss of an imperfect conductor, or it is a working load, or those are the same thing, or something else altogether. It depends on what you wanted the circuit to do in the first place. Systems that deliver electrical power for some specific purpose are always somewhat lossy, including resistive losses (and also some time dependent losses for AC systems but never mind that now). This wastes energy, and could start a fire if enough power is getting turned into heat in your wiring. This is a pretty passive loss. But, if you do this in the filament of an incandescent lamp, which is a short wire connected right across the mains, you WANT it hot, and so this passive loss IS the working load, and much of it turns into light. Anything that runs on DC and takes a fairly constant current when the voltage is constant could be said to have a resistance; that is, while it runs it will look like a resistor of that value from the point of view of the circuit. In this case you might refer to it as a load of so many ohms. In some situations like electric railways that use the motors as generators to slow the train going down a hill, you would put huge banks of power resistors on top of the car where they can be cooled by air flow, and you would pick resistances that were something like the motor resistance. You would likely think of this as converting power from one form to another, and whether a loss or not depends on how much you wanted the new form of power. ---
**Author:** ![Crafter\_Man](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/crafter_man/32/458_2.png) [@Crafter\_Man](https://boards.straightdope.com/u/Crafter_Man)\ **Post date:** [March 1, 2009, 10:10pm UTC](https://boards.straightdope.com/t/some-questions-from-the-electrically-ignorant/487679/39 "2009-03-01T22:10:09Z")
**ZenBeam** : I guess I have a problem with saying a resistor is always “opposing” or “resisting” some quantity. I just don’t look at them that way. When I see a resistor in a circuit, the only thing I think of is V = IR and P = IV. To me, a resistor is just a mathematical object that serves a function. [Previous page](https://boards.straightdope.com/t/some-questions-from-the-electrically-ignorant/487679.md?page=1)