Breakers off, but still 6v?
- Thread starter trevmai
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it might be an induced voltage through two phase wires being close together. Unless your instalation is wired with single cores in conduit this is most likely to be in the CU, as T&E is resistant to induction.
If you put any load across it (e.g. a bulb) the voltage will probably drop to zero
If you put any load across it (e.g. a bulb) the voltage will probably drop to zero
gcresser said:
Pish.
It's just a bit of capacative coupling. Nothing to worry about.
You are using a high impedence tester, which will detect these tiny currents. A low impedence tester will not. You will also soo the voltage dissapear if you connect a load across the wires and retest. (such as a lamp, or even a resistor)
<Bah!>
Capacitors are everywhere. You just don't see them.
Flat twin and earth has capacitance between live and earth and also between earth and neutral. A ball park figure is about 100 pF per metre. There is virtually no direct capacitance between live and neutral because the earth wire is in the way. There is leakage from live to earth but none to neutral AS LONG AS THE EARTH WIRE IS ACTUALLY EARTHED. If it's floating, the leakage will go first to the earth wire and then on to neutral. In this case if neutral is being used as switched live you will get a small current at the load, not enough to light a filament bulb but enough to light a neon if the bulb is out.
It IS possible to get capacitive leakage into an isolated live wire from another live wire outside the cable. There is no earth wire in the way this time. The earth wire does reduce the effect by shunting some of this leakage away to earth.
I was about to ask the same thing but JohnD beat me to it. It's impossible. The measurements must have been made under different conditions or else the meter is on the blink.
Flat twin and earth has capacitance between live and earth and also between earth and neutral. A ball park figure is about 100 pF per metre. There is virtually no direct capacitance between live and neutral because the earth wire is in the way. There is leakage from live to earth but none to neutral AS LONG AS THE EARTH WIRE IS ACTUALLY EARTHED. If it's floating, the leakage will go first to the earth wire and then on to neutral. In this case if neutral is being used as switched live you will get a small current at the load, not enough to light a filament bulb but enough to light a neon if the bulb is out.
It IS possible to get capacitive leakage into an isolated live wire from another live wire outside the cable. There is no earth wire in the way this time. The earth wire does reduce the effect by shunting some of this leakage away to earth.
I was about to ask the same thing but JohnD beat me to it. It's impossible. The measurements must have been made under different conditions or else the meter is on the blink.
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JohnD said:
If that's what you meant by 'resistant to induction' then your terminology was incorrect. Any current carrying conductor running parallel to T&E will induce a voltage in the T&E with respect to the current carrying conductor's reference point. However, since the flux linkage increases with parallel distance and decreases with perpendicular distance, if the current carrying conductor is not equidistant from both P and N (i.e. running alongside the T&E) then there will be a phase shift and difference in the linkage. On an unloaded cable this can read as a voltage between P and N.
In response to space cat, inductive coupling is dominant at low frequency, as opposed to capacitive coupling. So while capacitive coupling will have the effect you describe, we're interested mainly in the magnetic fields around the conductors.
aptsys wrote
I was wondering when somebody would mention magnetic coupling. I've carefully avoided the term 'induction' with respect to capacitors because it can only lead to confusion. In my book induction is magnetic. As aptsys says, it can happen when one CURRENT CARRYING conductor runs parallel to another. No current, no induction. So how much voltage might you get?
If we stay with our 100 pF per metre for the moment, the self inductance of the cable will be about 0.25 uH per metre. (That's a very crude estimate.) This gives us 2.5 uH for a 10m length of cable, an impedance of just under one milli-ohm at 50 Hz. If we now shove ten amps into it the voltage drop due to self inductance will be insignificant. The voltage induced into any parallel cable cannot be bigger and in practice it will be much less. (It is true that a current carrying conductor far away from the return path will have higher self inductance than flat twin.)
Now for the capacitive effect. No current is required, just voltage. Two cables running in close proximity over 10m will have a coupling capacitance of about 1nF (another crude estimate), an impedance of some 3 megohms at 50 Hz. The leakage current from live to neutral will therefore be about 80 uA. This tiny current leaking into a wire connected through a load to neutral will put only millivolts across the load - unless that 'load' is a very high impedance multimeter!
I was wondering when somebody would mention magnetic coupling. I've carefully avoided the term 'induction' with respect to capacitors because it can only lead to confusion. In my book induction is magnetic. As aptsys says, it can happen when one CURRENT CARRYING conductor runs parallel to another. No current, no induction. So how much voltage might you get?
If we stay with our 100 pF per metre for the moment, the self inductance of the cable will be about 0.25 uH per metre. (That's a very crude estimate.) This gives us 2.5 uH for a 10m length of cable, an impedance of just under one milli-ohm at 50 Hz. If we now shove ten amps into it the voltage drop due to self inductance will be insignificant. The voltage induced into any parallel cable cannot be bigger and in practice it will be much less. (It is true that a current carrying conductor far away from the return path will have higher self inductance than flat twin.)
Now for the capacitive effect. No current is required, just voltage. Two cables running in close proximity over 10m will have a coupling capacitance of about 1nF (another crude estimate), an impedance of some 3 megohms at 50 Hz. The leakage current from live to neutral will therefore be about 80 uA. This tiny current leaking into a wire connected through a load to neutral will put only millivolts across the load - unless that 'load' is a very high impedance multimeter!
I'm too thick for that 
But remembering we're taking about T&E, and a voltage between P and N conductors, on an isolated cable with an earthed CPC, I still say it doesn't seem to pick it up from adjacent cables.
Is there a reason why it should?
But remembering we're taking about T&E, and a voltage between P and N conductors, on an isolated cable with an earthed CPC, I still say it doesn't seem to pick it up from adjacent cables.
Is there a reason why it should?
Going back to this question:
There are capacitors between the external live wire and all three wires in the T&E. There are also capacitors within the T&E between its own cores. The external capacitors will leak current into all three cores but ---
The earth core is earthed so the tiny current will drop no voltage that you can measure on your meter. The neutral core is still connected back at the fuse box so, once again, the tiny current will drop no voltage. The live wire is isolated. The only place that current can go from this wire is through the internal capacitor to the earth wire or through your meter.
The internal capacitor will undoubtedly be bigger than the external one because the wires are closer together over a greater distance. This means it's impedance is lower than that of the external one. (Think potential divider here.) If the earth and neutral were also isolated you would get a bigger meter reading.
There are capacitors between the external live wire and all three wires in the T&E. There are also capacitors within the T&E between its own cores. The external capacitors will leak current into all three cores but ---
The earth core is earthed so the tiny current will drop no voltage that you can measure on your meter. The neutral core is still connected back at the fuse box so, once again, the tiny current will drop no voltage. The live wire is isolated. The only place that current can go from this wire is through the internal capacitor to the earth wire or through your meter.
The internal capacitor will undoubtedly be bigger than the external one because the wires are closer together over a greater distance. This means it's impedance is lower than that of the external one. (Think potential divider here.) If the earth and neutral were also isolated you would get a bigger meter reading.
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aptsys said:
it's got an "s" on the end in this country.
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