Why has DIYNOT got a electrics forum?

[Softus]

Total ballony...every single part of it...it shows that you understand VERY little about electricity.

This is (a) unhelpful and (b) wrong, viz:

115V will not give rise to twice the current in a circuit that 230V will. It's just that you need twice the current to achieve the same power output.

This is unequivocally true - quite the opposite of ballony. Are you actually trying to destroy all support for your arguments B_S?

However a more accurate quantity for the ampere can be defined as exactly 6.241 509 629 152 65×10E18 elementary charges per second.

How, in the name of all that is holy, is this relevant to what's being discussed?

Still waiting for your reply to the FULL LOAD CURRENT question - you've gone uncharacteristically silent on that one!

 

[jbonding]

this is one of the reasons i dont even venture to electrics, i went there with a question a long time back and got the responce the original poster is talking of "what the hell are you touching it for,leave it alone,Your stupid, pay me all your money attitude". i agree electric can kill and if you drink enough paint that will also kill you and if you fall down a hole or off a roof that might kill you. You could even get run over at the builders merchants, so theres plenty of ways of dying (wether you knew the time place and whatever lol).should this thread still be in general chat?when does the thread get moved to electrics?

It's general electrics, so belongs in both.

i had a feeling someone would point this out, then thought naaa they wont

 

[Igorian]

Oilman, if you think Power is not constant, then simply consider you PC power supply..

It is rated at say 300W for argument sake...but that 300W is the same at 230V and at the combined total of all the voltages that it produces to run your machine...

Example:

+12V = 14A = 168W
-12V = 2A = 24W
+5V = 10A = 50W
-5V = 2A = 10W
+3V = 15A = 45W
-3V = 1A = 3W

Total = 300W maximum power across the voltages..but at 230V it is still 300W but only 1.3A

If you doubt this, then I suggest you look it up.

This is even more boll*cks than I first thought. There is a confusion here between supply and consumption. I.E., what the PSU CAN supply and what the load will actually use. As you haven't mentioned any load, it doesn't make any sense. If we are talking supply, then yes, the maximum that can be supplied will be 300W at, and across, the voltages you state. If you try and draw more power than can be supplied by a particular rail, it will overload (except a SM PSU is reasonably well protected so more likely that the load won't get the power it requires and hence won't function correctly). However, a load may draw less current and therefore use less power. I'm not sure where the 230V comes in. The PSU could never supply 230V while maintaining the same design, so the argument becomes irrelevant. If you change the design so that it can provide 230V, then it's likely the maximum output would increase, otherwise why bother. I wouldn't want to attach it to my PC tho .

Take a more simple example based on power consumed. 230V and say 460 Ohm load (to keep the maths simple). This will use 230^2/460 = 115W. Now increase the voltage by X 2. 460^2/460 = 460W. Quite a dramatic increase, would you not agree and hardly constant.

So , which is it you refer to supply or demand?

 

[Softus]

i had a feeling someone would point this out, then thought naaa they wont

Thanks for sharing. What was your point again?

 

[jbonding]

i had a feeling someone would point this out, then thought naaa they wont

Thanks for sharing. What was your point again?

read my previous post and you might get the point.

 

[Softus]

read my previous post and you might get the point.

It's either that you agree with Bazdaa, or that you think the topic is in the wrong forum, or both. Is that it? Is that all you have to say?

 

[jbonding]

read my previous post and you might get the point.

It's either that you agree with Bazdaa, or that you think the topic is in the wrong forum, or both. Is that it? Is that all you have to say?

i do agree with bazdaa, i think the topic was in the right forum but after page 1 or 2 you might as well of tuned into GENERAL electrics. i said all i had to say in my first post but had to add it because you couldnt understand my first post

 

[securespark]

Tell that to the families of electrocution victims.

Remember the high profile story of an MP's daughter (?) who was killed because of a poor install?

She was electrocuted when simultaneously touching a live shelf and a dishwasher (I believe).

The shelf was live because a screw had pierced the (meandering) cable feeding a socket below.

Now, the install may well have been safe otherwise, but because of that wandering cable, someone died.

My point is you don't have to make glaringly dangerous eff-ups (like using undersized cables, or chopping out the cpc) to make a potentially dangerous install.

Melodramatic Jim is not.

You don't frighten me either with posts like that. It is no different for the unfortunate relatives of someone who was rundown by a car driven by an incompetant. Melodrama, it is.

Educate people, give them some knowledge and confidence. Don't just try to frighten them off. Getting a Part P qualification is not a guarantee of a safe installation, it just means someone else doesn't check the work.

Oilman, ffs, I'm not trying to frighten you or anyone else ftm. And Jim, I'm sure would agree wholeheartedly when I say that we are trying to educate people in the dangers of electricity. It has to be appreciated that there is more to wiring than just running cables and connecting up each end. This list is by no means exhaustive...... Is the cable sufficient to carry the design current? Has the designer applied the relevant correction factors to the cable? Is the EFL low enough to operate the protective device within the disconnection time? Are the IR readings high enough? Are any RCD's operating within the allowed time and.............etc etc.....

 

[securespark]

however the smple fact is the heart will still be subjected to three different voltages at alternating frequencies...Yes the phase to phase voltage will be 415V, but the frequencies of the supply are still there, and they will effect the heart...unless you change the laws of physics that is incontravertible....now I am no doctor so what the exact effect on the heart would be I have no idea, but a single phase supply has a single frequency of 50Hz, and the heart cannot beat at this frequency....

I am not a Dr, either, but have a good grasp of A "O" Human Biology, and if the human heart is subjected to an electric shock at 50Hz, the heart will attempt to keep up with that rhythm, and fail, literally.
If arrhythmia continues for more than a few seconds, Ventricular Fibrillation or Vfib occurs, which is irregular contractions (or better, a lack of co-ordination of the cardiac muscle of the ventricles, rather than acting in unison), and unless medical intervention comes within a few minutes, the heart will fail, and stop beating altogether. This is closely followed by a loss of conciousness, and if untreated, death. A defib. machine works by "shocking" the cardiac muscle back into its own rhythm.

 

[Big_Spark]

Total ballony...every single part of it...it shows that you understand VERY little about electricity.

This is (a) unhelpful and (b) wrong, viz:

Personal Opinion, your entitled to it, but I stand by my comment.

115V will not give rise to twice the current in a circuit that 230V will. It's just that you need twice the current to achieve the same power output.

This is unequivocally true - quite the opposite of ballony. Are you actually trying to destroy all support for your arguments B_S?

Softus, Why does the increase in current occur? The Power in an electrical circuit is constant, this cannot change unless the parameters of the circuit are changed.

Look:

2000 @ 230V = 8.69A
2000 @ 115V = 17.39A

Current cannot be changed unless either the resistance, capacitance, inductance or VOLTAGE in a circuit changes. Lowering the voltage causes the current to rise, converseley increasing the voltage will cause the currrent to fall. Read Oilmans statement again, it is electrically incorrect. The Potential Difference (Voltage) of a circuit is critical to the amount of energy (Q) required to do the same amount of work in the same amount of time. So if you want to achieve the same with less PD, then you cause an increase in the current, they are intrinsically linked together.

However a more accurate quantity for the ampere can be defined as exactly 6.241 509 629 152 65×10E18 elementary charges per second.

How, in the name of all that is holy, is this relevant to what's being discussed?

Just finishng off the comment I made above this, and the way this thread has gone I did not want to leave open the possibility of someone posting the mopre accurate figure to try to take this thread in yet another direction, but your corrent, it is not that relevent to the topic of discussion.

Still waiting for your reply to the FULL LOAD CURRENT question - you've gone uncharacteristically silent on that one!

No I have not, I have answered that question several times in various posts. The FULL LOAD CURRENT of a circuit is the maximum current that the circuit requires to operate or is capable of producing. In the examples I gave where a person becomes part of the circuit, the FULL LOAD CURRENT is the maximum current that is capable of flowing with respect to the combined resistances in that circuit, ie, the normal circuit resistance AND the human body resistance as the human body is a limiting factor in this by virtue of the high resistance limiting the maximum current that would flow in the undamaged or unbridged circuit.

So, if the circuit has a 30 Ohm resistance on it, when the circuit is undamaged, the FULL LOAD CURRENT is 230V/30 = 7.66A (1761.8W)

However in the examples earlier, the Human now forms a bridge across the Neutral and completes and becomes part of the circuit. Now the TOTAL resistance in that circuit is 30 + 1500 (Average Human dry body resistance) = 1530 Ohms, so 230/1530 = 0.15A (34.5W). This is the FULL LOAD current of that circuit whilst the human forms the bridge, it is no different to putting a normal resistor of 1500 Ohms into the circuit.

Do you understand now Softus?

 

[Big_Spark]

Oilman, if you think Power is not constant, then simply consider you PC power supply..

It is rated at say 300W for argument sake...but that 300W is the same at 230V and at the combined total of all the voltages that it produces to run your machine...

Example:

+12V = 14A = 168W
-12V = 2A = 24W
+5V = 10A = 50W
-5V = 2A = 10W
+3V = 15A = 45W
-3V = 1A = 3W

Total = 300W maximum power across the voltages..but at 230V it is still 300W but only 1.3A

If you doubt this, then I suggest you look it up.

This is even more boll*cks than I first thought. There is a confusion here between supply and consumption. I.E., what the PSU CAN supply and what the load will actually use. As you haven't mentioned any load, it doesn't make any sense. If we are talking supply, then yes, the maximum that can be supplied will be 300W at, and across, the voltages you state. If you try and draw more power than can be supplied by a particular rail, it will overload (except a SM PSU is reasonably well protected so more likely that the load won't get the power it requires and hence won't function correctly). However, a load may draw less current and therefore use less power. I'm not sure where the 230V comes in. The PSU could never supply 230V while maintaining the same design, so the argument becomes irrelevant. If you change the design so that it can provide 230V, then it's likely the maximum output would increase, otherwise why bother. I wouldn't want to attach it to my PC tho .

Take a more simple example based on power consumed. 230V and say 460 Ohm load (to keep the maths simple). This will use 230^2/460 = 115W. Now increase the voltage by X 2. 460^2/460 = 460W. Quite a dramatic increase, would you not agree and hardly constant.

So , which is it you refer to supply or demand?

Igorian....I posted my example as that..an example to show that power in an electrical circuit is constant if no other factors influence the circuit.

the 230V in my example is because the mains supply is rated at 230V, so it is the current drawn from the mains if the PSU was operating at maximum load of 300W. The rest of what you bang on about is irrelevent to this discussion.

In your example you have changed the parameters, consequently the outcome of that circuit has changed. Equally if you changed the input voltage of the PSU in my example, all other parameters would change. I was trying to show Oilman that it is not power that kills as this is a product of the parameters of an electrical circuit, and whilst power remains constant in any electrical circuit unless other factors change, altering the voltage, as you have pointed out, will cause a change (on reflection perhaps the PSU was a bad choice of example after being awake for 17 hours!! and not giving a more indepth explanation).

P= VI, it is a product of these two parameters, Oilmans assertion was incorrect, although I can understand why he may think that way. Change either one, as I have said and did say, and you change Power, that is correct...BUT Power in constant in electrical circuit for a given voltage.

However my example is valid as an example of how power remains constant.

 

[oilman]

.......... Lowering the voltage causes the current to rise, converseley increasing the voltage will cause the currrent to fall. Read Oilmans statement again, it is electrically incorrect. The Potential Difference (Voltage) of a circuit is critical to the amount of energy (Q) required to do the same amount of work in the same amount of time. So if you want to achieve the same with less PD, then you cause an increase in the current, they are intrinsically linked together.

I suggest you use a dictionary to find out about the word "cause".

Incidentally you will not have realised that you have just agreed with my statement but the English language seems to allow you to become confused.

As for FULL LOAD CURRENT, stop using three words where one (CURRENT) will do.

You should buy a program called Stylewriter, it will help you no end.

There are other holes in your last post but it is getting so tiresome trying to correct everything you write, we'll just have to trust that by now people will see the flaws written by someone who would outlaw DIY electrical work, while apparently being unable to publish facts.

 

[Softus]

i said all i had to say in my first post but had to add it because you couldnt understand my first post

You said that you had a feeling, and then a thought. It wasn't that I didn't understand this, it was that I was flabbergasted that your contribution comprised nothing else

 

[ban-all-sheds]

Actually, on the secondary, the current is quite low, so it hurts, but you generally live (and hopefully don't do it again )

It isn't if you want the same power out of it. W=EI, so if you half the volts you need to double the current.

I think Igorian is talking about the primary/secondary of a car ignition system. There the issue isn't really output power, it's a voltage high enough to generate a nice fat spark.

EDIT - ignore this - I posted it without refreshing, so didn't see that it had already been covered. Sorry.

 

[kendor]

not followed this post completely but i'll stick my tuppence worth in by saying that the skin conductance quoted will vary greatly from individual to individual dependant on the point of contact (can be anything from 100,000- 4,000,000 ohms -studies carried out in the 1950's by several researchers ie Dr Yoshio Nakatani) to whether the skin is wet say due to sweat or whether the person has dry skin or not, also the resistance could be variable due to insulation forming from the skin burning.

The rule of 50v @ 50ma being the point of possible fatality that was taught to us sparks at college was a general rule of thumb and not to be taken literally but in saying that protective devices would be designed with this in mind.

Going back to the arguments earlier the sparks involved should be ashamed of themselves quoting resistance when talking of AC circuits, now I want you to repeat after me I M P E D A N C E spells impedance