Easy Volt Drop Calculations Using Test Readings

by Steve on January 26, 2011

Electrical author and lecturer Chris Kitcher is a master of explaining the most complicated regulations and formulas and simplifying them for us mere mortals.  Volt drop calculations are no exception.  Every time I read one of his articles I learn something new.

As we know it is now a requirement of BS7671 that volt drop is verified for each circuit (Reg 612.14).  But normally checking or calculating voltdrop in cables involves knowing the length of run, conductor sizes or applying loads to circuits and taking readings.

Here Chris explains how to calculate volt drop for an existing circuit using R1+R2 or R1+RN without leaving the comfort of your van.

Calculating Volt Drop using test readings by Chris Kitcher

Mearuring circuit volt drop under load

Voltage drop can be verified by measuring the voltage at the origin of the supply and then again at the furthest point on the circuit.

Of course this requires the circuit to be loaded up to its maximum operating current for this method to provide accurate results.

Measurement is a simple process when it is a fixed load such as an immersion heater or a known fixed load, but It can be more difficult if we are dealing with socket outlets, cookers or lighting circuits.

Calculation is easier

Calculation is by far a much simpler process which has an added advantage of being able to be carried out while still sitting in your van!

Using the test readings

To carry out a test we must have available the R1 + R2 value for each circuit.  This is really all we need to carry out a very accurate voltage drop calculation.

As a check we can simply multiply the R1 + R2 value by either the load current, if it is known, or by the rating of the circuit protective device (U=IxR), this must then be multiplied by a factor of 1.20.  This is the factor from table 9C in the on site guide to correct for the rise in temperature when the circuit is under load.

Allow for temperature rise under load

The resistance of the copper used in BS (BASEC) cables will rise by 2% for each 5°C rise in conductor temperature.  If we are using 70°C thermoplastic cable the temperature of the conductor is likely to rise from room temperature of around 20°C, to its maximum operating temperature of 70°C.

This of course gives us a temperature difference of 50°C which will result in a rise in conductor resistance of 20%.  As we know If we multiply anything by 1.20 it will increase its value by 20%, that’s why we must use the multiplier.

Using R1+R2

Often the R2 (CPC) will be a conductor with a smaller CSA than the Live conductors, this of course will result in the calculation showing a higher voltage drop than there would be in reality (because the current flows in R1 & RN.  We hope!).

Providing the R1+R2 calculation gives a result of less than the permitted value of 3% (6.9v) or 5% (11.5v) depending on the type of circuit, then all is good.   Remember this is only a check.  But where the R2 calculation for a smaller CPC gives a higher than permitted volt drop it will be worth trying again using RN instead.

As an example lets take a circuit which is wired using a 2.5/1.5 twin and earth cable.  It has an R1 + R2 value of 0.6Ω and the circuit is protected by a 20 amp device.

Using the R1 + R2 value of 0.6Ω we can now calculate the voltage drop for the circuit. 0.6 x 20 x 1.20 = 14.4 volts.  Clearly this is far too high, however if we use R1 + RN we may end up with an acceptable result.

Of course many test certificates will not show the value for RN but don’t despair.

How to calculate RN

The problem is we do not have a reading for RN but a simple calculation will give you all of the information you need.

CSA line                            x (R1=R2) =R2
CSA line + CSA cpc

To put figures to this:

2.5          = 0.625 x 0.6 = 0.375
2.5+1.5

0.375Ω is the resistance of R2, therefore if we subtract this value from the

R1 + R2 value we will have the value of R1.

0.6Ω – 0.375Ω = 0.225Ω

The resistance of the 2.5 line conductor is 0.225Ω

Therefore the 2.5 Neutral conductor will be the same.

If we now double this value we will have R1 + RN.

0.225 + 0.225 = 0.45Ω

Now we can carry out the voltage drop calculation using 0.45Ω as the resistance value.

0.45 x 20 x 1.20 = 10.8 volts and this shows the volt drop to be ok.

Conclusion

As you can see, this is a simple calculation which once it has been performed a couple of times will become second nature.  It is also much easier than trying to measure the volt drop by applying loads to circuits.

And as previously suggested, the added bonus is that in can be carried out in the comfort of your van or anywhere else for that matter.

Many thanks to Chris Kitcher for this article – Steve.

Reference  to BS7671  App 4.6  and On-Site Guide Appendix 6  to find the Volotage Drop Calculation.

Chris’s latest book A Practical Guide to the 17th Edition of the Wiring Regulations and reviews.

An interview with Chris Kitcher



ElectriciansBlog.co.uk

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{ 3 comments… read them below or add one }

albert keates October 29, 2011 at 12:51

Could not agree more; even rocket science and brain surgery rely on the breadth of experience and knowledge of their practicioners, and, inevitably, circumstances vary.
Be thankful that, unlike the gas industry, “not prescribed” is not necessarily proscribed.

Chris Kitcher September 18, 2011 at 17:32

My veiw is that common sense has to be used as all installations are different. You have to consider that whenever we calculate volt drop we are always assuming that the conductors are at their maximum temperature, of course this will not always been the case.
For new installations the volt drop must be calculated before installing the cables and for that reason should always be within the specified limits. However if the volt drop fails using the calculation during a periodic inspection I would always consider the following.
Has the rating of the protective device been used in the calc ? if so what is the likelyhood of the maximum current flowing in the circuit? As an example if a 3kW immersion heater was to be connected to a circuit protected by a 20A device, and the 20 was used in the calculation then the volt drop would be far greater than if the 13A load was used. As it is a fixed load it will not ever draw more than 13A
If my calculation was above to the acceptable limit and I had a cause for concern I would carry out a direct measurement with the circuit under load wherever possible.
Any volt drop above the permitted limit must be recorded on the PIR, what is recorded is possibly a cause for debate, depending on the age of the installation. Remember pre 17th the permitted volt drop was 9.2 volts for lighting now it is 6.9 volts, for an older installation maybe a 4 would be suitable, if it was safe when installed and all other parts of the circuit comply, is it unsafe now? My view is that all jobs should be considered individually as in reality our industry is not really as precise as we would perhaps like it to be, partly due to the changes over the years. Changing the words in a document would not in my view make an installation which has been safe for years unsafe.

Steve September 18, 2011 at 16:16

Question from David Elgood
What do we do about the volt drop on an installation if it’s not exceptable?

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