Calculating R1+R2 on a ring...

[junksamiad]

Hi guys,

I just need help with something really basic that I’m struggling with in terms of calculating resistance (as opposed to measuring).

Just a hypothetical question here, not a real situation, I just need to understand the theory.

If I had a socket, just 1, on a radial, running from DB at a length of 10m, then the resistance of the line conductor (2.5mm let’s say) would be 7.41 x 10 / 1000 = 0.74 ohms. I presume that’s correct.

However, if I then decided to make this socket a ring, and I ran a 2.5mm T&E back to the consumer unit, my resistance would have doubled. So it would then be 7.41 x 20 / 1000 = 1.48 ohms.

Let’s say hypothetically (for the sake of the math), my CPC was the same conductor size, then my R1+R2 would be 2.96 ohms?

If calculating resistance on a ring- as opposed to measuring resistance on a ring- do I still have to divide by 4?

Any help much appreciated as always.

 

[telectrix]

yous way out. firstkly (7.41 x 10)/1000 = 0.074. secondly, if you wer to make it a ring, then it's 2 cables in parallel, so the resistance would be halved.

if you measure the 2 2.5's before ringing, then you have 0.148, when ringed you divide by 4 because you are measuring at the mid point which means divide by 2. then as there are 2 cables parallel, divide by 2 again. i.e.divide by 4.try watching john ward on youtube. and remember the difference between r1 and R1. the use if uppper case or lower case letters is critical.

 

[telectrix]

remember that r1, r2, rN are end-end values and only apply to ring circuits. R1, R2, Rn are the valuse used to calculate Zs.

 

[Grant1987]

Hi guys,

I just need help with something really basic that I’m struggling with in terms of calculating resistance (as opposed to measuring).

Just a hypothetical question here, not a real situation, I just need to understand the theory.

If I had a socket, just 1, on a radial, running from DB at a length of 10m, then the resistance of the line conductor (2.5mm let’s say) would be 7.41 x 10 / 1000 = 0.74 ohms. I presume that’s correct.

However, if I then decided to make this socket a ring, and I ran a 2.5mm T&E back to the consumer unit, my resistance would have doubled. So it would then be 7.41 x 20 / 1000 = 1.48 ohms.

Let’s say hypothetically (for the sake of the math), my CPC was the same conductor size, then my R1+R2 would be 2.96 ohms?

If calculating resistance on a ring- as opposed to measuring resistance on a ring- do I still have to divide by 4?

Any help much appreciated as always.

Hia mate, the best way to obtain test results is to actually carry out the necessary tests it’s all too easy to calculate them you will fall into the bad habit of calculating results and from my experience if you calculate results and then actually carry out the tests you’re readings will differ unless you use torques screwdriver and every connection is tightened to same strength? Just test everything and don’t calculate you know it’s all as you see it not what you imagine it

 

[junksamiad]

Thanks Grant1987 and Telectrix.

I’m happy with the idea that for ring’s you divide by 4 when measuring, it’s just I was struggling to understand why this is done, hence why I was focusing on the theoretical side of calculating.

Your advice about measuring at the midpoint has clarified for me somewhat Telectrix. So it’s essentially a case is it that if the current is being drawn by a socket two ways, the resistance decreases? Then that decrease compounds again because the two ways (I.e., two cables) essentially increase the CSA of the cable?

I have seen a few explanations in books but am still struggling to follow their explanations. I will take your advice and watch John Ward and I’m sure it will come together further in my head.

 

[telectrix]

Thanks Grant1987 and Telectrix.

I’m happy with the idea that for ring’s you divide by 4 when measuring, it’s just I was struggling to understand why this is done, hence why I was focusing on the theoretical side of calculating.

Your advice about measuring at the midpoint has clarified for me somewhat Telectrix. So it’s essentially a case is it that if the current is being drawn by a socket two ways, the resistance decreases? Then that decrease compounds again because the two ways (I.e., two cables) essentially increase the CSA of the cable?

I have seen a few explanations in books but am still struggling to follow their explanations. I will take your advice and watch John Ward and I’m sure it will come together further in my head.

if you consider the mid point of a ring, the current is shared equaly by each leg. so that equates to a divide by 2. the r1 reading is both legs in series do that equates to another dvide by 2. when you measure R1+R2 using the figure 8 connections, then all points will give the same reading , but in practice, the load sharing is only equal at the mid point. part way round the ring, the sorter leg will carry a bigger share of the load due to it's lower impedance.

 

[Intoelectrics]

Hia mate, the best way to obtain test results is to actually carry out the necessary tests it’s all too easy to calculate them you will fall into the bad habit of calculating results and from my experience if you calculate results and then actually carry out the tests you’re readings will differ unless you use torques screwdriver and every connection is tightened to same strength? Just test everything and don’t calculate you know it’s all as you see it not what you imagine it

I have disagreed with you because I believe its good practice to test & calculate. It gives you a better understanding of what test results to expect and also a mental image of how a circuit might function. Yes you will have discrepancies between testing and calculating, due to connections, testing equipment etc... but if you calculate as well and compare the results the dependencies should not be that far off?

 

[Deleted account]

No

Hia mate, the best way to obtain test results is to actually carry out the necessary tests it’s all too easy to calculate them you will fall into the bad habit of calculating results and from my experience if you calculate results and then actually carry out the tests you’re readings will differ unless you use torques screwdriver and every connection is tightened to same strength? Just test everything and don’t calculate you know it’s all as you see it not what you imagine it

working out what the resistance of a circuit prior to installation is essential so you can verify that the circuit in question is fit for purpose.
Also it is in my opinion good to do a couple of quick calcs even in your head to ascertain what results you should be getting and if they are way out the it’s something that must be noted or corrected.

Edit @Intoelectrics you got there first.

 

[junksamiad]

Brilliant, thanks Telectrix, that’s about cleared things up for me where I can follow the logic. I see now yes. And this is why you don’t divide by anything further on the test on a radial right, because the load is not shared but instead stops at the last point?

I would also always calculate too beforehand so I agree. I mean, why would you not if it can feasibly verify your tests?

 

[telectrix]

the purpose of calculating resistance of a circuit is at the design stage. by using the milli ohms per metre and the length of the circuit, you need to ensure that the resistance is low enough to comply with ADS - the OCPD tripping within the specified parameters. e.g. you design a circuit using 2.5mm cable and the calculated resistance is too high to get the required value of Zs, then you redeign using a larger csa cable. then you go on to ensure that the cable selected is suitable for the design load current, then calculate the volt drop. when all calculations are within specified limits, then you're good.

 

[Fitzy]

Search for Sparky Ninja on you tube, he has done an excellent video on testing a rfc at the different points of the ring and the values based on the different lengths of each leg at each point.

 

[Grant1987]

I have disagreed with you because I believe its good practice to test & calculate. It gives you a better understanding of what test results to expect and also a mental image of how a circuit might function. Yes you will have discrepancies between testing and calculating, due to connections, testing equipment etc... but if you calculate as well and compare the results the dependencies should not be that far off?

Ok I understand fully the calculating side of it my point was to ensure you test it fully as you should and to not just test a little bit and then calculate the rest assuming the results, so you’re saying you test the whole installation and then compare you’re results to the calculated results? So what I said is correct you should always test fully and not just calculate readings.