Short Circuit Current Calculation For Cable AS/NZS 3008

Welcome to our load fault current calculator! This page about the calculation of the fault current through the cable to a load given:
Voltage (V) [Single Phase or 3-Phase needs to be specified]
Source fault current (kA)
Cable size (mm^2) [Based on AS/NZ 3008]
Cable Length (Km) 
Cable impedance (Ω) [Use design codes to find R_cable and X_cable]
Steps:
Search AS/NZS 3008 and find resistance/km value at 75 degrees for the specified cable area (mm^2) in Table 35 of AS/NZS 3008 Electrical installations - Selection of cables 
Repeat for reactance/km on Table 30 for multi-core PVC circular conductors 
Multiply impedance/Km by length (remember values are in ohms/kilometre!) 
Input respective values and produce load fault value
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CalculationInputs﻿﻿V_1P
:240volts​
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﻿﻿I_S_fault 
:10kA​
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﻿﻿R_cable
:16.5​
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﻿﻿X_cable
:0.111ohms​
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﻿﻿Length
:0.3km​
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Output﻿﻿I_1P_load fault  
:24.1833466ampere​
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﻿﻿I_3P_load fault
:48.3666933ampere​
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﻿﻿Z_1P_source
:0.023960739ohms​
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﻿﻿Z_cable
:4.95011201ohms​
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ExplanationShort circuit fault current is the amount of current that flows through a circuit when a short circuit occurs. It is an important factor to consider in electrical engineering, as it determines the size of protective devices such as circuit breakers and fuses.
When a short circuit occurs, the resistance in the circuit drops to nearly zero, causing the current to increase significantly. The short circuit fault current is the maximum current that will flow through the circuit at the point of the fault. Calculating the short circuit fault current is important to ensure that the protective devices are properly sized to handle the high current.
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Single Phase fault current:
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I1ϕ−load−fault=V1ϕ−sourceZsource+2⋅ZcableI_{ 1\phi  - load - fault}=  \frac{ V_{1 \phi - source} }{ Z_{source} + 2   \cdot Z_{cable} }  I1ϕ−load−fault​=Zsource​+2⋅Zcable​V1ϕ−source​​
Relation between 3-Phase and Single-Phase voltage:
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V1ϕ=V3ϕ−sourceIsourcefaultV_{1 \phi} = \frac {V_{3 \phi  -source} } {I_{source_fault} }V1ϕ​=Isourcef​ault​V3ϕ−source​​
3 Phase fault current: 
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I3ϕ−load−fault=Vsource3 (Zsource+Zcable)I_{ 3\phi  - load - fault}=  \frac{ V_{source} }{\sqrt{3} \:(  Z_{source} +   Z_{cable} )}  I3ϕ−load−fault​=3​(Zsource​+Zcable​)Vsource​​
Source Impedance: 
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Z1ϕ=V1ϕ−sourceIsourcefaultZ_{1 \phi } =  \frac{ V_{1\phi-source} }{ I_{source fault} } Z1ϕ​=Isourcefault​V1ϕ−source​​
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Z3ϕ=V3ϕ−source3⋅Isource faultZ_{3\phi} = \frac {V_{3\phi-source}}{ \sqrt 3 \cdot I_{source \: fault}  }Z3ϕ​=3​⋅Isourcefault​V3ϕ−source​​
Cable Impedance:
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Zcable=Rcable2+Xcable2Z_{cable} =   \sqrt{ R^{2} _{cable} +  X^{2} _{cable} }Zcable​=Rcable2​+Xcable2​​
Where:
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I1ϕ−load−fault=Single phase fault current (A) I3ϕ−load−fault=Three phase fault current (A) V1ϕ=Single phase voltage (V) V3ϕ−source=Three phase source voltage (V) I source−fault=Three phase fault current (A) Z1ϕ=Single phase source impedance (Ω) Z3ϕ=Three phase source impedance (Ω) Zcable=Cable impedance (Ω)Rcable=Cable resistance (Ω)Xcable=Cable reactance (Ω)I_{ 1\phi  - load - fault}  = Single \: phase \: fault \: current \: (A) \\\ \\ I_{ 3\phi  - load - fault}  = Three \: phase \: fault \: current \: (A) \\\ \\ V_{1\phi} = Single \: phase \: voltage \: (V) \\\ \\V_{3 \phi  -source} = Three \: phase \: source \: voltage \: (V) \\\ \\ I_{    \: source - fault} =  Three \: phase \: fault \: current \: (A)\\\ \\ Z_{1\phi } = Single \: phase \: source \: impedance \: (Ω) \\\ \\ Z_{3 \phi} = Three \: phase \: source \: impedance \: (Ω)\\\ \\  Z_{cable} = Cable \: impedance  \:(Ω) \\ R_{cable} = Cable \: resistance \: (Ω) \\X_{cable} = Cable \: reactance \: (Ω) I1ϕ−load−fault​=Singlephasefaultcurrent(A) I3ϕ−load−fault​=Threephasefaultcurrent(A) V1ϕ​=Singlephasevoltage(V) V3ϕ−source​=Threephasesourcevoltage(V) Isource−fault​=Threephasefaultcurrent(A) Z1ϕ​=Singlephasesourceimpedance(Ω) Z3ϕ​=Threephasesourceimpedance(Ω) Zcable​=Cableimpedance(Ω)Rcable​=Cableresistance(Ω)Xcable​=Cablereactance(Ω)
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