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Zs, to test or calculate?

Paul Brewerton
A large contractor working on our site have told me yesterday that it is their policy not to live test final circuits where they cannot use a plug? In order to reduce risk, they will now only calculate Zs, on circuits where they would have to open an enclosure, such as FCU's and light fittings.
The control measures we insist are in place, are IP2X equipment, GS38 leads, two man rule with second man having resus training, among others. So I feel the risk has been reduced to as low as reasonably practicable. An d my instinct tells me that a measured Zs reading must be more accurate than a calculated one, since it will include all parallel paths under test.
The contractor is happy to live test distribution circuits, so it seems they want to pick and choose.
They also state that this is how things are now, and have worked at many different sites, Cross Rail, Heathrow, various MOD sites etc, and that they al accept this as common practice.

I like some opinions to find out what's going on out there on other sites. 
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    mapj1:

    A loop impedance test of the kind that uses the  50Hz AC supply and switches in a load and looks at the voltage drop will necessarily include all the same impedances, both real and imaginary, that  will appear in a real fault, as that too uses the AC supply.

     


    But how is that possible? The reactive component of voltage drop in the circuit being measured cannot be determined from an arithmetic subtraction of the voltage drop of the instrument test resistor from the supply voltage. As far as I understand it my MFT places a 59 ohm resistor as a dummy load, tries to establish the volt drop across it over a couple of half cycles and then determines the circuit “impedance” from the voltage difference between that and the supply voltage. Where the reactive component is pronounced surely that is not possible as the VD across the reactance is in quadrature.


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  • 0
    mapj1:

    A loop impedance test of the kind that uses the  50Hz AC supply and switches in a load and looks at the voltage drop will necessarily include all the same impedances, both real and imaginary, that  will appear in a real fault, as that too uses the AC supply.

     


    But how is that possible? The reactive component of voltage drop in the circuit being measured cannot be determined from an arithmetic subtraction of the voltage drop of the instrument test resistor from the supply voltage. As far as I understand it my MFT places a 59 ohm resistor as a dummy load, tries to establish the volt drop across it over a couple of half cycles and then determines the circuit “impedance” from the voltage difference between that and the supply voltage. Where the reactive component is pronounced surely that is not possible as the VD across the reactance is in quadrature.


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