Caution, long cable!
The critical parameter internal resistance or when Ohm's law does not allow the required current to be generated...
The internal resistance of a circuit is of major importance in the case of a fault, therefore it represents a critical parameter. What good is the best reserve capacity in a power pack if Ohm's law does not permit the required current to be generated? The cable resistance is a major influence which is frequently underestimated.
The following example clarifies the issue:
A device with a current consumption of 5 A is installed 30 m [outward and return cable = 60 m] away from the control cabinet. A 10 A power supply unit, a cable with 1 mm² wire diameter and a 6 A miniature circuit breaker with a C tripping characteristic are used to protect the cable and the connected device.
Calculation of the internal resistances:
- Internal resistance, power supply unit 30 mΩ
- Connection technology (terminals) 20 mΩ
- Miniature circuit breaker 20 mΩ
- Short circuit (in the affected device) 45 mΩ
- Cable 60 m 1mm2
The cable resistance is calculated using the formula: R = 1.07 Ohm
The total resistance of the application results then as: 1.19 Ohm
In case of a short circuit, this results in a short-circuit current of 20.2 A
If we now consider the tripping curve of the selected fuses, we can determine that this runs with the available factor I/IN (20.2/6) = 3.37 in the overcurrent range.
Usually, the miniature circuit breaker would trip in 3 to 10 seconds at this current level.
This does not correspond with the required performance, but rather over-loads or damages the cable used.