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230v L-N, and 70v L-E

Looking at my lighting earlier and see that my multimeter is measuring 230v from live to neutral at the ceiling light, but only70 ish volts from live to earth. Am I right in thinking the earth may be disconnected somewhere and the 70ish volts im reading is due to capacitance ? I cannot get my head round why its like this   The live is definately at 230v so if the earth was disconnected I should not be getting any reading atall. Where is this 70v reading from ???? and between what ?? I need to check the continuity of the earth tomorrow but it cannot be continuous or i would be getting 230v, even a loose earth connection would still read full voltage would it not ??
  • Easy answer first.

    Others will follow with more info I`m sure so this is just the first thing to consider.


    Yes you can pick up stray capacitance (and inductance and indeed leaky resistance)  on most modern multimeters because their imput impedance is usually around 10 Megohms which makes then accurate because they do not load the circuit. Any such stray pick up will substantially stay there and not collapse.

    If you were to put a load across the terminals (a resistor or a lamp) you would see if this voltage collapses. Yes it does sound like you`ve probably got an earth connection adrift.
  • Try measuring the voltage between L and N when the light is switched off. I'll bet it's not zero.
  • Voltage across L-E when the light is off  reads zero volts; my first thoughts was  this 70v was inductance. Its just I cannot figure why if one lead is on the L at 230v, and the other lead is on E its still reading 70v across them. Am I right in supposing that the induced voltage on the earth, is actually around 160v ? My reading of 70v is due to the  potential between them (230 - 160) =70v ?
  • This has become a classic 'gotcha' with digital meters. The capacitance of modern cables is perhaps 50-100pF between cores per metre. so the earth if it is not properly tied down then it floats in the middle of an L-N voltage divider of two capacitances . For  few metre of cable, and a meter of a few million ohms input impedance you do not measure the half mains voltage but you do see a significant fraction of it. When you put the meter across the other part of the accidental divider you see a similar partial voltage. do not expect the two parts to add up to the total voltage, as you will always read lower on the side loaded by the meter.

    Mike.
  • Try measuring the voltage between L and N when the light is switched off. I'll bet it's not zero.

    :

    Voltage across L-E when the light is off reads zero volts

    Presumably with "lamps in" or equivalent - i.e. the SL is shunted to N by the load. With no load on the SL you're likely to see a similar apparent voltage on the SL when the switch is off.


        - Andy.
  • AJJewsbury:
    Try measuring the voltage between L and N when the light is switched off. I'll bet it's not zero.

    :

    Voltage across L-E when the light is off reads zero volts

    Presumably with "lamps in" or equivalent - i.e. the SL is shunted to N by the load. With no load on the SL you're likely to see a similar apparent voltage on the SL when the switch is off.


    Ah yes, I first discovered this on a 2 A socket, but I have also found it Between the pins of an ordinary lamp holder. Clearly in both circumstances, the load was removed. IIRC, there was a current of about 1 mA.


    That was between L and N, not L and E and in both cases on 2-way lighting.


  • Indeed - see also all the reports of low energy lamps flickering or flashing when switched off.

       - Andy.
  • AJJewsbury:

    Indeed - see also all the reports of low energy lamps flickering or flashing when switched off.


    When my eyes are dark adapted I can see the CFL flashing above my bed. No two-way lighting there. It is very feint and I doubt that it would be seen otherwise.


  • It makes one wonder just how much energy is seeping away even when things are switched 'off'.
  • But current flowing in capacitance is out of phase with the driving voltage, so does not represent any dissipation - you get back the electrons you put in to charge up the cables later in the cycle, (unused!).

    This capacitive leakage  is sometimes called the 'displacement' current, as it is the energy of setting up the electric field in the insulator, and then relaxing it again, like stretching and relaxing an ideal  spring not actually doing any work in it - unlike a resistive leakage which would show at DC. (if you like images then a resistor  is more like a frictional brake or well damped shock absorber, while a capacitance is more like an elastic, and to complete the picture an inductor is more like an inertial mass on low friction sliders, slow to get going and slow to stop but not much actual loss)

    At least in principle you could tune out the capacitive  reactance with an inductor of suitable value, and the have a circulating current oscillating back and forth between them but no net current driving current at all to speak of.

    Of course this  is not very practical (except perhaps on very long transmission lines of known geometry, where compensating reactors are indeed sometimes used,)

    LV the network is slightly inductive anyway, so if anything stray capacitance helps a bit (indeed we add caps on motors and traditionally ballasted light fittings to try and keep amps and volts more or less together).

    Of course the capacitance of cables is slightly lossy so  the phase shift is not a perfect right angle, and there is some, but very little, actual dissipation.

    regards Mike