Help needed with understanding IR

[keitkeiter]

the problem question I have states:

"the following line to earth values were recorded for six circuits in an installation. What would be the expected overall value when measured at the origin and what action should be taken, if any?"

I understand that an IR reading higher 1 ohm is a pass (though you'd want 20+ and it to be in the 100s for a new circuit). But what I am stumped on is how I work out the origin?

do I add the resistance values together and divide it by the 500v from the IR meter?

Any help would be greatly appriated

 

[Richard Burns]

Each circuit is in parallel with the others.
Therefore you calculate resistances in parallel.
1/R = 1/R1 + 1/R2 + 1/R3 etc.
Where R is resistance at origin and R1, R2,R3, etc. are the individual circuits.

 

[Pete999]

the problem question I have states:

"the following line to earth values were recorded for six circuits in an installation. What would be the expected overall value when measured at the origin and what action should be taken, if any?"

I understand that an IR reading higher 1 ohm is a pass (though you'd want 20+ and it to be in the 100s for a new circuit). But what I am stumped on is how I work out the origin?

do I add the resistance values together and divide it by the 500v from the IR meter?

Any help would be greatly appriated

 

[keitkeiter]

Amazing, thank you both very much. Great illustration of the calculator there. This formula will now be engraved into my mind. Cheers guys

 

[telectrix]

brill pete. that method works. on my calc. i use x to -1 as it don't have 1/x

 

[Pete999]

brill pete. that method works. on my calc. i use x to -1 as it don't have 1/x

Glad you both found it useful.

 

[The Ghost]

Rt = 1/R1 + 1/R2 + 1/R3 etc

even?

 

[The Ghost]

Or...
product over sum method
10x20x30x40/10+20+30+40 =240000/100=2400
then (reciprocal) 1/2400 = 4.166666667 =4.8 ohms reduced to the most significant number. My preferred method btw.

 

[Richard Burns]

even?

No I think not.
Also 1/2400 = 0.000416, product over sum is for two resistances only.
You would need to do each pair (10 and 20, 30 and 40) and then do each result again to get the answer.

 

[ELECNEWT]

I always thought that the Product Over Sum method only worked for resistances taken two at a time.

For resistances 10 20 30 and 40 ohms in parallel.

For the first pair 10 x 20 / (10+20) =6.666 ohms

For the second pair 6.666 x 30 / (6.666+30) = 5.454 ohms

For the third and final pair 5.454 x 40 / (5.454 +40) = 4.8 ohms

Total resistance is 4.8 ohms

 

[Risteard]

I understand that an IR reading higher 1 ohm is a pass … But what I am stumped on is how I work out the origin?

Firstly I presume you meant one Megohm (one million Ohms) as opposed to one Ohm.

Secondly the minimum values in the Regulations are for the entire installation (i.e. after you have done your calculation if you don't measure the whole installation in one go). (For a larger installation it can be subdivided.)

 

[Ian1981]

Scientific calculator or phone is your friend
As Pete has posted in his video

 

[johnduffell]

As above you can't accept a lower value than 1mohm on any part of the installation whether combined or not. If you measure 20 circuits and get 0.1mohm combined that could be a fault on one circuit.
However if you break it down and they are all 1.5mohm then that's classed as ok.

 

[telectrix]

brill pete. that method works. on my calc. i use x to -1 as it don't have 1/x

wtf is elecnewt giving me a dumb. i tried that method pete linked to and it gave me 4.8 ohms in about 30 seconds, as opposed to doing all that lowest common denominator what i vaguely remember from secondary school in the early 60's.

 

[telectrix]

I always thought that the Product Over Sum method only worked for resistances taken two at a time.

For resistances 10 20 30 and 40 ohms in parallel.

For the first pair 10 x 20 / (10+20) =6.666 ohms

For the second pair 6.666 x 30 / (6.666+30) = 5.454 ohms

For the third and final pair 5.454 x 40 / (5.454 +40) = 4.8 ohms

Total resistance is 4.8 ohms

correct, but too long winded. would need a whole beermat to work it out that way.

 

[GBDamo]

I am showing my ignorance but where would I apply this figure.

I understand what it is and how it is calculated but where, say on an EICR, would you ever record it.

Also on a typical EIC it is not unusual to have readings of >200MOhms (instrument limitation). The true figure could be much higher. How would you calculate this?

Is it just a feel for the numbers? This looks close best calculate it?

Say we have IR results of 2MO + 3MO + 4MO The value for the installation would be 0.93MOhms. Whilst each individual figure is a pass, although dodgy, the overall IR for the installation is a fail.

Where is this recorded and by what criterion does it fail against?

Thanks OP for starting this, of all the tests its IR I find most tricky.

 

[ELECNEWT]

Sorry Tel.
I did not mean to give you a dumb. I was actually trying to give Pete a like. But I was doing it on my cheap smartphone with big fingers. I agree that the product over sum method is very cumbersome and prone to errors.

 

[johnduffell]

Say we have IR results of 2MO + 3MO + 4MO The value for the installation would be 0.93MOhms. Whilst each individual figure is a pass, although dodgy, the overall IR for the installation is a fail.

there's no such thing as an overall IR failure, if you do the IR overall and it's >1mohm then that guarantees the whole installation is alright (subject to 2 way switching etc) as you can't have any circuit <1mohm if the global is >1mohm. That is a big time saver given that many installs would be OK.
However the converse is not true, so if the global IR is 0.93 you have to break it down in order to know if there's a fault. Same as any fault finding.

 

[Risteard]

by what criterion does it fail against?

643.3.2

 

[Risteard]

there's no such thing as an overall IR failure, if you do the IR overall and it's >1mohm then that guarantees the whole installation is alright (subject to 2 way switching etc) as you can't have any circuit <1mohm if the global is >1mohm. That is a big time saver given that many installs would be OK.
However the converse is not true, so if the global IR is 0.93 you have to break it down in order to know if there's a fault. Same as any fault finding.

Not quite correct.

The insulation resistance for the entire installation in parallel must meet the minimum requirements of BS7671. For a larger installation this is subdivided into individual distribution boards with all final circuits connected.

See 643.3.2