Zs and parallel paths...

[junksamiad]

Hi all,

easy question I’m sure, so hope someone can clear up for me.

If there are parallel paths on a circuit upon which Zs is being measured, will the parallel path cause the Zs to go up or down, and could someone explain why this would be so at all?

I just need a clear understanding of this for my revision.

Thanks again as always!

 

[davesparks]

Parallel paths means you will have multiple routes to earth connected in parallel.

If you think about what type of measurement you are actually doing during a Zs test then using your knowledge of basic electrical science, whether the Zs goes up or down.

 

[TJ Anderson]

The parallel paths (if there is any) will decrease the loop impedance.
A good easy example of where this happens is if you installed a new radial circuit to a boiler.
Your R1+ R2 tests would give you a value and you would add this to Ze to give you an expected value for Zs.
Later when verifying Zs, you would find it was lower than your expected value, quite significantly sometimes.
This is because the boiler has a gas pipe connected to it that is also bonded to the main earthing terminal. It may also have all copper water pipework which is yet another parallel path.
Now when you take Zs, your earth fault loop impedance path back to the MET is not just the cpc but also the boiler pipework via a main protective bonding conductor back to the MET, hence, the lower value.

 

[Sintra]

Your Zs will go down if you have parallel earth paths. For this example let’s say we have a Zs of 0.2 ohms at your consumer unit and a circuit with equal sized conductors for line and cpc. You measure your R1+R2 and get 0.8 ohms so we know then that the line conductor is 0.4 ohms and the cpc is 0.4 ohms. Calculating your Zs for the circuit will give you a total of 1 ohm. Now we add a parallel earth path of 0.4 ohms so now our R2 consists of two resistances of 0.4 ohms. Using the formula 1/Rt=1/R1+1/R2 we can see that our total R2 resistance will be 1/0.4+1/0.4 = 0.2 ohms. Our R1 is still 0.4 ohms so add this to our R2 which is now 0.2 ohms gives us a R1+R2 of 0.6 ohms. Now add this to our Zs at the consumer unit 0.2 ohms and it then gives us a Zs for the circuit of 0.8 ohms.

 

[buzzlightyear]

Simple example - c.p.c. to an immersion heater with copper pipework. As the exposed-conductive-part of the immersion heater is also in contact with the copper cylinder and hence the copper pipework the normal path along the c.p.c. is paralleled by an second path though the cylinder, along the pipework and back to the MET via main bonding.my version .

 

[junksamiad]

Your Zs will go down if you have parallel earth paths. For this example let’s say we have a Zs of 0.2 ohms at your consumer unit and a circuit with equal sized conductors for line and cpc. You measure your R1+R2 and get 0.8 ohms so we know then that the line conductor is 0.4 ohms and the cpc is 0.4 ohms. Calculating your Zs for the circuit will give you a total of 1 ohm. Now we add a parallel earth path of 0.4 ohms so now our R2 consists of two resistances of 0.4 ohms. Using the formula 1/Rt=1/R1+1/R2 we can see that our total R2 resistance will be 1/0.4+1/0.4 = 0.2 ohms. Our R1 is still 0.4 ohms so add this to our R2 which is now 0.2 ohms gives us a R1+R2 of 0.6 ohms. Now add this to our Zs at the consumer unit 0.2 ohms and it then gives us a Zs for the circuit of 0.8 ohms.

Ah, thanks Sintra. So it’s because the resistance is in parallel and therefore we then divide 1 by each resistance and add them together (and divide 1 by the total aswell). Got that now yes. Thanks so much, that’s so much clearer to me.

Why is it said that if the Zs taken during a periodic inspection is higher than that recorded on initial verification then the circuit is deteriorating? If Zs is a measure of impedance on a circuit then how can a higher value equate to deterioration?

 

[telectrix]

Ah, thanks Sintra. So it’s because the resistance is in parallel and therefore we then divide 1 by each resistance and add them together (and divide 1 by the total aswell). Got that now yes. Thanks so much, that’s so much clearer to me.

Why is it said that if the Zs taken during a periodic inspection is higher than that recorded on initial verification then the circuit is deteriorating? If Zs is a measure of impedance on a circuit then how can a higher value equate to deterioration?

if the Zs increases over time, it couldindicate a connection becoming high resistance, due to corrosion or poorly tightened.

 

[Taylortwocities]

Why is it said that if the Zs taken during a periodic inspection is higher than that recorded on initial verification then the circuit is deteriorating? If Zs is a measure of impedance on a circuit then how can a higher value equate to deterioration?

If the Zs rises then it may be because the original Ze has increased (you may been to talk to the DNO about that) or if one or more if the parallel paths has increased. A common example is where one of the original parallel paths is the equipotential bond to the incoming metal water pipe. Let’s say that value was 3ohms.
The water board comes along and replaces the water pipes with plastic. The equipotential bond is basically no longer effective. So you lose one of the parallel paths and the Zs measurement increases.

 

[sinewove]

Ah, thanks Sintra. So it’s because the resistance is in parallel and therefore we then divide 1 by each resistance and add them together (and divide 1 by the total aswell). Got that now yes. Thanks so much, that’s so much clearer to me.

Why is it said that if the Zs taken during a periodic inspection is higher than that recorded on initial verification then the circuit is deteriorating? If Zs is a measure of impedance on a circuit then how can a higher value equate to deterioration?

Could be corrosion of a terminal on live or cpc or a poor connection.
S

 

[Baddegg]

Corrosion or poorly tightened terminals.....another reason these days is old central heating system removed and replaced with all plastic pipes...

 

[Mike Johnson]

Hi all,

easy question I’m sure, so hope someone can clear up for me.

If there are parallel paths on a circuit upon which Zs is being measured, will the parallel path cause the Zs to go up or down, and could someone explain why this would be so at all?

I just need a clear understanding of this for my revision.

Thanks again as always!

Think about the word Impedance and a footpath, in a stable state the path is constant, if you add anything to that path (say increase its width) which allow anything to travel that route easier, then the impedance has been reduced, decrease the width of the path (rusty gate) the impedance has increased.