Simon Barker:
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.
Simon Barker:
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.
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