You can no longer post new replies to this discussion. If you have a question you can start a new discussion
To prevent overvoltage in the neutral conductor, what must be provided in the circuit for protection against overvoltage?
Former Community Member
To prevent overvoltage in the neutral conductor In the case of a connection between the phase conductor and the neutral conductor, what must be provided in the circuit for protection against overvoltage?
OK, if you have a L-N fault, the N counductor at the point of the fault will go up to something like half the normal L-N voltage and hopefully the resulting fault current will cause short-circuit protection will operate. You wouldn't really want to prevent N being pulled up to a higher voltage, as that would prevent the protective device operating as intended.
OK, if you have a L-N fault, the N counductor at the point of the fault will go up to something like half the normal L-N voltage and hopefully the resulting fault current will cause short-circuit protection will operate. You wouldn't really want to prevent N being pulled up to a higher voltage, as that would prevent the protective device operating as intended.
Or is this a question about SPDs?
- Andy.
So many thanks. I mean when there is L-N fault that you explained completely. But why you said "will go up to something like half the normal L-N voltage"? How did you calculate this?
Assume that the L and N have equal cross-sectional area. Assume that the L and N are approximately equal lengths. Let the resistance of the L up to the fault be (some unknown) value R. From the earlier assumptions, assume that the resistance of N is also R.
By shorting L to N, we're connecting a resistance of 2R across the supply. A current of I = V/2R will flow, where V is the supply voltage. The voltage across the L (from the supply to the short) will be IR = (V/2R)R = V/2. The voltage across the N will also be V/2. So the dead short point will be at V/2 above the supply ground, and V/2 below the supply voltage.
Assume that the L and N have equal cross-sectional area. Assume that the L and N are approximately equal lengths. Let the resistance of the L up to the fault be (some unknown) value R. From the earlier assumptions, assume that the resistance of N is also R.
By shorting L to N, we're connecting a resistance of 2R across the supply. A current of I = V/2R will flow, where V is the supply voltage. The voltage across the L (from the supply to the short) will be IR = (V/2R)R = V/2. The voltage across the N will also be V/2. So the dead short point will be at V/2 above the supply ground, and V/2 below the supply voltage.
