Calculators
Calculate cable impedance and loop impedance
Loop impedance decides whether enough current flows under fault conditions for the protective device to disconnect. Two things are regularly overlooked, namely the warmer conductor under fault and a protective conductor smaller than the line conductor.
Result
- Loop impedance, line to protective conductorThe governing figure for the disconnection condition.
- —Ω
- Resistance of one conductor (one way)
- —Ω
- Reactance of one conductor (one way)
- —Ω
- Impedance for a three-phase short circuit
- —Ω
- Share of reactance in the loopBelow 10 % it may be neglected, above that not.
- —%
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A guide, not a design to standard. We give no warranty for the correctness of the results. The applicable standards and case-by-case verification govern.
The grey values are an example. Simply enter your own.
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Calculating is the easy part.
A formula gives you a number. Designing a plant also demands installation method, grouping, discrimination, the standards in force and a look at the installed base. That is what we take on.
Method and questions
Cable impedance
Resistance and reactance of a cable, plus the loop through the protective conductor.
Discuss a projectWorked example
The default case the calculator starts with, worked through once. Change the values above and it recalculates immediately.
Inputs
- One-way cable length
- 60 m
- Line conductor cross-section
- 16 mm²
- Protective conductor cross-section
- 16 mm²
- Conductor material
- Copper (κ = 56 m/Ω·mm²)
- Conductor temperature
- 80 °C
- Inductive reactance per kilometre
- 0.08 Ω/km
Result
- Loop impedance, line to protective conductor
- 0.166 Ω
- Resistance of one conductor (one way)
- 0.083 Ω
- Reactance of one conductor (one way)
- 0.005 Ω
- Impedance for a three-phase short circuit
- 0.083 Ω
- Share of reactance in the loop
- 5.79 %
Formula and standards
Formula
- R = L / (κ · A) · [1 + α · (ϑ − 20 °C)]
- X = X′ · L
- Loop Z = √[(R_L + R_PE)² + (2 · X)²]
- Three-phase Z = √(R_L² + X²)
Assumptions and standards
- The disconnection condition is calculated with a heated conductor, because the loop impedance is then highest and the fault current smallest.
- The inductive reactance per unit length comes from the cable data sheet and is entered here.
- The upstream impedance of network and transformer is not included. It adds to this and dominates on short runs.
Why does the protective conductor count?
Because the fault current returns through it. If the protective conductor is half the size of the line conductor, its resistance is twice as high. The loop impedance rises by half compared with equal cross-sections, and the fault current falls accordingly.
When may I ignore the reactance?
As long as its share of the loop stays below roughly ten percent, which is the case at small cross-sections. From about 50 mm² it becomes noticeable, and at 240 mm² it co-determines the impedance.
Why 80 °C and not 20 °C?
Because the disconnection condition has to hold in the worst case. A warm conductor has more resistance, hence less fault current. Calculating at 20 °C gives too large a fault current and therefore too optimistic an answer.
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