electric shock voltage levels
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Paul_C
The voltage which will appear briefly in your kettle example will depend upon the resistance of the conductive paths involved. If the resistance between the live side of the supply and the point of fault contact is equal to the resistance from the metal casing back to the earthed side of the supply, for example, then the voltage which appears on the kettle's casing will be half the supply voltage, or about 115V.
You won't be able to kill your wife that way if you have an rcd fitted. Have you considered the slow poisons?
With a fault of near zero impedance the metalwork will rise to mains voltage but maximum fault current will flow and ensure the fastest disconnection possible (also dependant on protective device characteristics)
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As Paul_C says, the voltage at the fault will rise to approximately half the supply voltage due to the action of the potential divider formed by the earth fault loop impedances.
Prior to the faulty circuit being disconnected, there will be a touch voltage of approximately half the supply voltage between the kettle and a metal tap. The purpose of main and supplementary equipotential bonding conductors is to reduce the magnitude of this voltage during the period of the earth fault. See http://electrical.theiet.org/wiring-matters/22/bonding.cfm?type=pdf
Prior to the faulty circuit being disconnected, there will be a touch voltage of approximately half the supply voltage between the kettle and a metal tap. The purpose of main and supplementary equipotential bonding conductors is to reduce the magnitude of this voltage during the period of the earth fault. See http://electrical.theiet.org/wiring-matters/22/bonding.cfm?type=pdf
You are presumably assuming that, unlike me, one doesn't have a TT system! I fear that my kettle would end up closer to 230V than 115V for that 30ms whilst one was praying that the RCD did what it said on the tin (with TT, I doubt that the fault current would be high enough for the MCB to trip).
Kind Regards, John
You don't need to worry about the 50v rule where disconnection occurs quickly, within 0.4s for the given scenario.
Death from electrocution is un-likely but possible when 50v is applied to the body for less than 0.4 second.
Death or serious injury subsequent to the shock is possible. A fall in which the head hits a hard object or scalding from a toppled kettle of boiling water need to be considered as possible accidents consequential to a mild non fatal electric shock.
could anyone explain the potential divder earth part, which means half supply voltage is shown at the fault ?
in either case of 230 or 115 volts at the fault is that saying i would recieve a shock at one of the two voltages but if lots of fault currunt was to flow i would be ok as the shock duration is so short ie 0.4 s ?
in either case of 230 or 115 volts at the fault is that saying i would recieve a shock at one of the two voltages but if lots of fault currunt was to flow i would be ok as the shock duration is so short ie 0.4 s ?
The only way half the voltage will flow is on the negative or positive cycle of the AC sine wave, 115v.
I would have thought the full 230v flows (earth fault loop path) and as long as the circuit Zs if below BS 7671 tabulated values for that circuit, the protective devices should open and clear the fault.
P
Paul_C
JohnW2 said:
That's why I said it would be 115V if the respective resistances were equal, and that the actual voltage will depend upon those resistances. As you say, with TT the earth-side resistance is likely to be much greater than the live side. Perhaps this should be counted as another strike against TT, which - in my mind rather sensibly - is not permitted in some other parts of the world.
All of the conductors forming the circuit have a certain amount of resistance, albeit usually very small. In your fault example, the actual current which flows will be determined by the total resistance in the loop. To take an example: If the resistance from the live side of the supply to the point of the fault in the kettle were 0.5 ohm, and the resistance from the case of the kettle, via the earthing system, to the other side of the supply were also 0.5 ohm, you'd have a total loop resistance of 1 ohm (assuming the fault itself to be of zero resistance, and before anyone picks it up, I'm assuming the supply transformer impedance itself is included in the two 0.5-ohm measurements).
From Ohm's Law, 230V / 1 ohm results in a current of 230 amps. You can then use Ohm's Law again to determine the voltage which appears across each individual resistance. 230A x 0.5 ohm is 115V. Hence 115V appears on the kettle case until the circuit is broken.
If the total loop were still 1 ohm, but the live side was 0.25 ohm and the earth side 0.75 ohm, then you'd still have a fault current of 230A, but this time the voltage appearing on the kettle would be higher: 230A x 0.75 ohm = 172.5V.
I'm using Ohm's Law there as the explanation, but you don't actually have to use it directly, you can just use the proportional resistance involved, as for any given current voltage varies inversely with resistance.
So equal resistances results in half supply voltage at the fault point. If the earth-side resistance is two-thirds of the total loop resistance, the voltage at the fault point will be two-thirds of the supply voltage, and so on.
As John highlighted, with a TT system the resistance on the earth side of the circuit formed by the fault is considerably higher than the resistance on the live side, so the voltage appearing on the kettle case will be much nearer to the full supply voltage. With TN-S or TN-C-S, it will be much lower as the two resistance values involved are much closer in value to each other.
P
Paul_C
The source is still supplying the full 230V, but the voltage (with respect to earth) which appears at the fault point depends entirely upon the potential divider effect of the resistances in the complete loop.
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