Basic few questions

[basejumper31]

Hi folks,

I'm working my way through the start of a Trevor Linsley book for the 2365 lvl 2 ahead of starting the course in September, and I have some basic questions which I wondered if someone might be able to give me feedback on.

I can't post in the training section yet as I haven't started the course yet.


p.65 of the book,

the following example is given:



"The insulation resistance measured between phase conductors on a 400v supply was found to be 2MΩ. Calculate the leakage current."

(and then it gives the working and the answer, as follows)


"I = V/R



Therefore: I = 400V / 2 x 10[SUP]6 [/SUP]Ω = 200 x 10[SUP]-6 [/SUP]A = 200 µA"



My questions –

1. why does the author use 2 x 10[SUP]6 [/SUP]Ω instead of just saying 2000000Ω ?
   
   
2. Why does it follow that 400/2 x 10[SUP]6[/SUP] equates to 200 x 10[SUP]-6[/SUP]? This may just be because I am no longer familiar with algebraic maths, and have got used to doing everything in decimals, but even so, why choose 200 x 10[SUP]-6[/SUP]? Why not say 2 x 10[SUP]-4[/SUP] instead?
   
   
3. Why do they state the answer in microAmps? If I understand SI notation correctly, 200 microAmps is the same as 2 milliamps, so would it not make more sense to give the answer in terms of milliamps? Or is it because it is a convention to give answers in terms of whole numbers? or is it because milliamps don't exist / are generally not used, in the same way that kilometers are used, but you never hear people talking about megameters, they'd say a thousand kilometers instead?
   
   
4. Should I not worry too much about these kind of details, and just accept it for what it is, and assume I'll pick up these kind of conventions as I go along, or is it good practice to want to pull things apart if I don't understand them?
   
   Thanks in advance :/
   

 

[Richard Burns]

Using scientific notation can make it easier to visualise the numbers in comparison to other numbers.
e.g. writing 20000 and 200000 could be easily confused but 2 x 10[SUP]4 [/SUP]and 2 x 10[SUP]5[/SUP] can be easily distinguished.
Also if you say you have 2 MΩ then it is easy to say this is 2 (x 10[SUP]6[/SUP])Ω simply substituting the M for 10[SUP]6[/SUP] which is readily understood.

In the calculation showing that 400/2 is 200 and that 1/10[SUP]6[/SUP] is 10[SUP]-6[/SUP] demonstrates a simpler means of calculation without resorting to calculators.
If the answer were given as 2 x 10[SUP]-4[/SUP] then the calculation would not be so obvious.

In a similar manner to MΩ above using µ for 10[SUP]-6[/SUP] is standard notation and easy to understand, using 0.2 mA would also be correct but does not derive so easily from the example.
I would say though that usually (for an electrician) milliamps is much more familiar.

I think they are just trying to ensure it is clear how the answers are derived. They could have said: using the figures above and I=V/R the result is 0.2mA but this would not explain why.


Generally they are trying to make things clear and easy to understand and you are trying to make things familiar to you.
However in either case they are correct and being aware of many different methods of expressing the same thing is valuable experience to allow you to readily pick up on whatever is thrown at you later.

 

[basejumper31]

Thank you for taking the time to answer this Richard, appreciated.

 

[telectrix]

richard, how do you get all the mathematical symbols in the text? i can't ise the codes as laptop hasn't got separate numerical.

 

[Brightspark2]

richard, how do you get all the mathematical symbols in the text? i can't ise the codes as laptop hasn't got separate numerical.

Copy and paste them from Google...

Thats the cheats way I do...

 

[Silly Sausage]

ΩΩΩΩΩΩ

Don't know how Richard does it, but I use 'Character Map' from Windows.
You'll find it in System Tools. I have it pinned on the Taskbar.

ΩΩΩΩΩΩ

 

[Richard Burns]

richard, how do you get all the mathematical symbols in the text? i can't ise the codes as laptop hasn't got separate numerical.

The superscript / subscript can be done from the advanced editing, for the symbols (as I have a number pad) I just use the ALT codes.

 

[Richard Burns]

Just tried and you can use the on screen keyboard utility as well ALT234ALT as mouse clicks puts in Ω.

 

[telectrix]

i found this "type mathematical symbols". seems pretty good.

 

[basejumper31]

The above was really useful to me too - got character map pinned to my start bar now also, thanks to those who asked & answered about that

Another basic question - I read in this C&G 2365 lvl 2 textbook that if a person gets a shock of 50mA (but no mention of a voltage level) or over, it is usually fatal. Later it talks about the fact that when doing PAT testing, during the Earth Bonding Test a current of about 25A is sent through the earth pin, via a probe to any exposed metalwork on the item (assuming it is not double insulated).

So if a PAT tester accidentally touched the live part of some PAT test equipment when it was sending that 25A current down the earth pin, would he/she be potentially killed? Or is it that, because of the very un-conductive (insulative I guess) properties of the human body, that very little of the 25A current would actually travel through the body? Basically, am I right in thinking you'd need a whopping great current (or rather current and voltage combination) to actually cause 50mA to travel through the human body, resulting in a potentially fatal shock?

Why doesn't the book specify a certain voltage level when it specifies 50mA as a potentially fatal threshold?

Surely saying 50mA is potentially fatal, without specifying a voltage, is like saying if you get hit by a 20 tonne truck it would kill you, but not specifying the necessary/minimum speed the truck would have to be traveling at in order for it to be dangerous?

 

[basejumper31]

*a bit* like the truck analogy - I realise it's not exactly like that

 

[Richard Burns]

If you consider the human body as a resistor ( generally of approx 1000Ω in dry conditions) then the current that would flow through the body would be dependent on voltage in accordance with I = V/R.
If you have 230 V then approximately 0.23 A or 230 mA would flow through a human body, if the voltage were 55 V then approximately 0.55 A or 55mA would flow, in both cases this is potentially fatal. (This is where RCD protection to 30mA comes in, to protect a person from a fatal current)
However you can see that the current is dependent on voltage but it is the current that can cause the physiological effects such as heart arrhythmia, muscle contractions and death.
Because it is the current that is the key point this is the item referred to.

I should point out that the 1000Ω is only an approximation and if the body is wet then the resistance reduces considerably, which is why there are additional requirements for locations where the body may be wet or immersed in water.

 

[basejumper31]

Good morning (well, noon low lol) I am continuing my way through a Trevor Linsley text book ahead of the start of my college course 2365 lvl 2.

Linsley includes 'check your understanding' sections after each chapter. I'm completing the section after chapter one. Most of the questions he has provided answers for in the rear of the book. But for some he just says "the answers are to be found in the text of the chapter itself". Well I have checked the text of the chapter thoroughly, and I can't find a specific answer to this question, so I thought I'd post it on here and just see if anyone might be able/willing to offer any useful comments about my answer and how adequate it is.

p.59, Question 43:

Use bullet points to list a step-by-step safe electrical isolation procedure for isolating a circuit in a three-phase distribution fuse board.

My answer:

● Test the circuit with a voltage indicator to see that it is live
● Isolate it (i.e. disconnect it from all power sources)
● Test the circuit again to ensure that it is dead (no current)
● Connect the voltage indicator to a proving unit to ensure that it is not faulty.
● Secure the isolation (i.e. Remove all relevant fuses and use padlocks if necessary/appropriate)
● Place warning signs as needed to indicate work is being done
● Commence work

Linsley does state these steps in the chapter, it's just he doesn't say anything about three-phase, or fuse boxes in particular (well, he does in respect of the fact that they can sometimes have asbestos in them, but that's about it).

I'm just wondering if there's anything extra which needs to be done in terms of safe isolation procedure when it comes to three-phase systems as opposed to single phase.

Anyway thanks for reading, I hope everyone's had a good week and looking forward to the weekend!

 

[Richard Burns]

Taking a full approach to safe isolation
Correctly identify the supply you wish to isolate.
Seek permission to isolate the power from the responsible person so that you do not present a danger from unexpected loss of power.
Isolate the supply.
Lock off the supply and retain the key with you.
Post warning notices on the locked off supply.
Prove the voltage indicator is working on a known live source or proving unit.
Test using an approved voltage indicator to GS38 to ensure that all conductors are dead (L1/L2, L1/L3, L2/L3, L1/N, L1/E, L2/N, L2/E, L3/N, L3/E, N/E).
Prove the voltage indicator is working on a known live source or proving unit.
Consider the risks of adjacent live parts.
(issue a permit to work, if required)
Start work on the dead circuit.

 

[Former Sponsor TM]

1) Identify circuit or supply needing isolation;
2) Ask client if OK to isolate;
4) Lock off the isolation with an appropriate lock (keep key on your person or locked away)
5) Put up a warning sign and a barrier if needed warning others that you are working,
5) Check your voltage indicator on a known voltage supply (approved proving unit or incoming supply)
6) Check between: L1 - CPC, L2 - CPC, L3 - CPC, L1 - N, L2 - N, L3 - N, N - CPC, L1 - L2, L1 - L3, L2 - L3
7) Check voltage tester again a known voltage supply
8) Commence work