ELECNEWT
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- Reaction score
- 72
I can understand that when supplementary equipotential bonding is applied, you can establish a bonding connection from very near to origin of the fault and so the voltage at the extraneous part (say a radiator) will be very close to the voltage at the fault.
Taking the example of a circuit with a 20 amp breaker the current to operate the breaker within 5 seconds is 60 amps. Textbooks state that if the resistance between exposed or extraneous conductive parts is 0.83 ohms or less then supplementary bonding is not required.
I understand that 50volts divided by 60 amps gives the resistance value of 0.83 ohms.
I have tried drawing a realistic circuit diagram to prove that 0.83 ohms is safe, but I just tie myself in knots.
Would any member be willing to find the time to draw such a diagram (similar to Richard’s hint hint) showing the position of the 0.83 ohms and an alternative diagram using the value of say 5 ohms?
Taking the example of a circuit with a 20 amp breaker the current to operate the breaker within 5 seconds is 60 amps. Textbooks state that if the resistance between exposed or extraneous conductive parts is 0.83 ohms or less then supplementary bonding is not required.
I understand that 50volts divided by 60 amps gives the resistance value of 0.83 ohms.
I have tried drawing a realistic circuit diagram to prove that 0.83 ohms is safe, but I just tie myself in knots.
Would any member be willing to find the time to draw such a diagram (similar to Richard’s hint hint) showing the position of the 0.83 ohms and an alternative diagram using the value of say 5 ohms?