📘 Derivations of Unsymmetrical Fault Currents
We derive LG, LL and LLG fault currents using symmetrical components. Assume prefault voltage V = 1 pu.
🔹 1️⃣ Single Line-to-Ground (LG) Fault Derivation
Assume phase 'a' is grounded: Va = 0 Ib = Ic = 0 Using symmetrical components: Ia = I1 + I2 + I0 For LG fault: I1 = I2 = I0 Why? Because boundary condition forces equal sequence currents. Now: Ia = 3I1 Sequence network connection: Z1, Z2, Z0 connected in series. Voltage equation: V = I1 Z1 + I2 Z2 + I0 Z0 Since I1 = I2 = I0: V = I1 (Z1 + Z2 + Z0) Therefore: I1 = V / (Z1 + Z2 + Z0) Total fault current: Ia = 3I1I_fault = 3V / (Z1 + Z2 + Z0)
🔹 2️⃣ Line-to-Line (LL) Fault Derivation
Assume phase b and c shorted: Ib = -Ic Ia = 0 Symmetrical component condition: I0 = 0 I1 = -I2 Zero sequence current absent. Sequence network: Z1 and Z2 in series. Voltage equation: V = I1 Z1 + I2 Z2 Since I2 = -I1: V = I1 (Z1 + Z2) Thus: I1 = V / (Z1 + Z2) Phase fault current: I_fault = √3 I1I_fault = √3 V / (Z1 + Z2)
🔹 3️⃣ Double Line-to-Ground (LLG) Fault Derivation
Assume b and c shorted to ground. Boundary conditions: Va ≠ 0 Vb = 0 Vc = 0 Sequence currents satisfy: I1 + I2 + I0 = 0 Sequence network: Z2 and Z0 in parallel Then in series with Z1. Equivalent impedance: Z_eq = Z1 + (Z2 Z0)/(Z2 + Z0) Thus: I1 = V / Z_eq Total fault current: Ia = 3I1I_fault = 3V / (Z1 + (Z2 Z0)/(Z2 + Z0))
📊 Key Comparison from Derivation
| Fault Type | Sequence Network Connection | Formula |
| LG | Z1 + Z2 + Z0 (Series) | 3V/(Z1+Z2+Z0) |
| LL | Z1 + Z2 | √3V/(Z1+Z2) |
| LLG | Z1 + (Z2 || Z0) | 3V/(Z1+(Z2Z0)/(Z2+Z0)) |
🎯 Important Conceptual Insights
- LG fault → all three sequence currents equal
- LL fault → zero sequence absent
- LLG fault → mixed parallel connection
- Grounding strongly affects Z0
Fault Derivation = Boundary Conditions + Sequence Network Logic
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