electric shock voltage levels

[radweld]

The only way half the voltage will flow is on the negative or positive cycle of the AC sine wave, 115v.

I would have thought the full 230v flows (earth fault loop path) and as long as the circuit Zs if below BS 7671 tabulated values for that circuit, the protective devices should open and clear the fault.

Voltage never, ever, flows anywhere.

 

[Spark123]

If you are talking about the Sine wave then it goes from 340v to -340v, not sure where you're getting the 115v from??

If we talk about a RMS voltage of 230v across a potential divider with equal resistances as PaulC says, will end up with half the RMS voltage at the centre point. It is never quite that simple though

 

[Spark123]

You don't need to worry about the 50v rule where disconnection occurs quickly, within 0.4s for the given scenario.

Death from electrocution is un-likely but possible when 50v is applied to the body for less than 0.4 second.

Death or serious injury subsequent to the shock is possible. A fall in which the head hits a hard object or scalding from a toppled kettle of boiling water need to be considered as possible accidents consequential to a mild non fatal electric shock.

The regs allowed this situation for years, it was up until 2008 that sockets on a TN system (not for outdoor use etc etc etc) didn't have to be protected by an RCD 30mA.
The two main concepts of EEBADS were reduce the touch voltage to 50v or disconnect the supply quickly enough to remove danger, as ADS is still.

 

[Astra99]

Sorry to nit-pick, but if the RMS voltage is 230, the peak is plus or minus 325V. 340V would give an RMS of 240.

For those who have long forgotten, RMS multiplied by the square root of 2 (1.414) gives you the peak which, for a sinewave is expressed as plus or minus.

RMS multiplied by 2 (or peak multiplied by root 2) gives the peak-to-peak voltage

Anyone else suddenly remember A-level physics lessons?

 

[radweld]

Sorry to nit-pick, but

RMS multiplied by 2 (or peak multiplied by root 2) gives the peak-to-peak voltage

RMS multiplied by 2 root(2) (or peak multiplied by 2) gives the peak-to-peak voltage. Or at least it did in my A-level physics lessons.

 

[Astra99]

Thanks Radweld! Memory playing tricks again! It has been over 40 years.

 

[echoes]

While we're in the classroom, may I raise my hand and remember that the r.m.s. value can only be expressed as 1/sqrt(2) x Vpk if the waveform is sinusoidal; something that is often forgetten.

 

[Stoday]

To be pedantic as is the wont of several posters on this forum, the potential divider effect does not come into effect immediately. The full voltage (up to 325V) applies for a few microseconds whilst the effect of the short flows through the copper wires at around 150m/μS.

If the distance between the distribution transformer and the short is long enough, i.e. around 1500km, the short will appear to be an open circuit. Happily, only National Grid has to worry about such phenomena.

 

[JohnW2]

To be pedantic as is the wont of several posters on this forum....If the distance between the distribution transformer and the short is long enough, i.e. around 1500km, the short will appear to be an open circuit. Happily, only National Grid has to worry about such phenomena.

If pedantic is the name of the game, isn't 1500km more than twice the length of the UK, hence probably not even of much concern to the UK national grid?

Kind Regards, John.

 

[vibetime]

just wanted to say thanks for the answers to my post especially paul c for the detailed clear answer its all become clear !!!!

 

[Paul_C]

For those who have long forgotten, RMS multiplied by the square root of 2 (1.414) gives you the peak which, for a sinewave is expressed as plus or minus.

Any for anyone else who has forgotten, or never learned, the main reason for the widespread use of RMS values is that they result in the same amount of power as the equivalent level of DC. For example, 100V DC across 20 ohms results in a current of 5A, and a power dissipation of 500W. An AC supply of 100V RMS connected across the same 20-ohm resistance results in a current flow of 5A RMS, and also gives the same power dissipation of 500W.

The RMS (Root Mean Square) is not - as sometimes seems to be believed - the mathematical average of one half cycle of a sinewave. The RMS value is 0.707 (the reciprocal of the square-root of 2) times the peak value, but the average value is approx. 0.637 times the peak value. Of course, the mathematical average value for a complete cycle of pure sinewave is zero, unless there is a DC bias applied.

 

[Stoday]

Intuitively I would expect the average value (of a half cycle) to be equivalent to the d.c. value. I know it is not, but I don't know why.

I've always assumed that it's one of those conundrums where intuition does not give the righr answer. Like, for example drilling. Why should there be a significant difference between turning the drill with the work fixed be different from turning the work and keeping the drill fixed. I do know the answer to that.

 

[JohnW2]

Intuitively I would expect the average value (of a half cycle) to be equivalent to the d.c. value. I know it is not, but I don't know why.

Probably the simplest reason to understand "why" is to realise that power is proportional to voltage squared, not voltage - with a DC voltage of E across a resistance of R, the power dissipated in the resistance is E²R. Consider the following simple 'stepped' AC waveform, with a voltage E for half of each half cycle and 0.5E for the other half:

Untitled

• JohnW2
• 9 Mar 2011
• 1

Considering just the first half-cycle, the average voltage is obviously just the average of E and 0.5E, namely 0.75E.
During the first half of this half cycle, power is E²R. During the second half, it is (0.5E)²R, namely 0.25E²R. The average power over that half cycle is therefore the average of E²R and 0.25E²R, namely 0.625E²R (and not 0.75E²R).

Does that help at all? The same genertal principle obvioulsy applies to a sine wave, but needs more complex maths to illustrate.

Kind Regards, John.
edit: typos corrected (a couple of 'R's omited from expressions), with aologies

 

[JohnW2]

Probably the simplest reason to understand "why" is to realise that power is proportional to voltage squared, not voltage .... etc. etc.

Following my illustration/explanation, I perhaps should have added a brief explanation as to how this leads to the concept of the 'RMS' (Root Mean Square) value of voltage (or current).

As I explained, since power is proportional to voltage squared (i.e. E²), what we are interested in (to determine power) is effectively the average of E² (not the average of E). We therefore work out the average value of E² (the 'mean square' value) and then take the square root to get it back into units of volts (rather than volts squared) - hence 'root mean square'.

Kind Regards, John.