--- title: "L-G fault through Zf: currents and voltages" subject: "Power System Analysis II" chapter: 5 key: "semester_5/power_system_analysis_ii:lg-fault-with-zf-derivation" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/lg-fault-with-zf-derivation site: BCT Engineering --- # L-G fault through Zf: currents and voltages Power System Analysis II · Chapter 5: Unsymmetrical Faults on Power Systems - Asked in: 2070 Chaitra, 10 marks - Video: [LG Fault On Unloaded Synchronous Generator With Fault Impedance Zf | Unsymmetrical Faults | PSA](https://www.youtube.com/watch?v=TAGwkFv1DwQ) — Learn2Earn 4Life - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/lg-fault-with-zf-derivation - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-5 ## Question What is the difference between symmetrical components of positive, negative and zero phase sequence? A 3-phase synchronous generator with its neutral solidly grounded and operating at no load develops an L-G fault in one of the phase have fault impedance Zf. Derive expressions for the fault currents and the line to ground voltage at the location of the fault at all the phases. ## Short answer Positive sequence has a-b-c rotation and generator source; negative has a-c-b; zero has no rotation, all phasors in phase. For an L-G fault through Zf on phase a with solidly grounded neutral, the three sequence networks connect in series with 3Zf. Sequence currents: I0=I1=I2=Ea/(Z1+Z2+Z0+3Zf). Fault current If=3Ea/(Z1+Z2+Z0+3Zf). Faulted phase voltage Va=ZfIf; healthy phase voltages Vb=I1[(a^2-a)Z2+(a^2-1)Z0+3a^2Zf], Vc=I1[(a-a^2)Z2+(a-1)Z0+3aZf]. ## Answer #### Difference between positive, negative and zero sequence components | Feature | Positive sequence | Negative sequence | Zero sequence | |---|---|---|---| | Magnitudes | Equal | Equal | Equal | | Phase displacement | 120° | 120° | 0° (all in phase) | | Phase order | a-b-c (same as system) | a-c-b (reverse) | No rotation | | Relation to phase a | $V_{b1} = a^2V_{a1}$, $V_{c1} = aV_{a1}$ | $V_{b2} = aV_{a2}$, $V_{c2} = a^2V_{a2}$ | $V_{b0} = V_{c0} = V_{a0}$ | | Source in generator | Yes (EMF) | No | No | | Return path needed | No | No | Neutral/ground ($I_n = 3I_0$) | | Effect | Useful power, torque | Reverse torque, rotor heating | Ground currents, interference | #### L-G fault through $Z_f$ on phase a (solidly grounded, no load) Generator equations: $V_{a1} = E_a - Z_1I_{a1}$, $V_{a2} = -Z_2I_{a2}$, $V_{a0} = -Z_0I_{a0}$ (solid grounding: $Z_0 = Z_{g0}$). #### Boundary conditions (fault on phase a through $Z_f$) ```text a o----------+ b o--- (Ib=0) c o--- (Ic=0) | Ia = If Zf | ground ``` $$ I_b = 0, \qquad I_c = 0, \qquad V_a = Z_fI_a $$ #### Sequence currents $$ \begin{bmatrix} I_{a0}\\I_{a1}\\I_{a2} \end{bmatrix} = \frac13\begin{bmatrix} 1&1&1\\1&a&a^2\\1&a^2&a \end{bmatrix}\begin{bmatrix} I_a\\0\\0 \end{bmatrix} \quad\Rightarrow\quad I_{a0} = I_{a1} = I_{a2} = \frac{I_a}{3} $$ So the **same current flows in all three sequence networks**, which means they are in series. #### Voltage condition $$ V_a = V_{a0} + V_{a1} + V_{a2} = Z_fI_a = 3Z_fI_{a1} $$ Substituting the network equations: $$ -Z_0I_{a1} + E_a - Z_1I_{a1} - Z_2I_{a1} = 3Z_fI_{a1} $$ $$ I_{a1} = I_{a2} = I_{a0} = \frac{E_a}{Z_1 + Z_2 + Z_0 + 3Z_f} $$ #### Fault current $$ I_f = I_a = 3I_{a1} = \frac{3E_a}{Z_1 + Z_2 + Z_0 + 3Z_f} $$ For a bolted fault ($Z_f = 0$): $I_f = \dfrac{3E_a}{Z_1+Z_2+Z_0}$. The currents in the healthy phases are $I_b = I_c = 0$. #### Interconnection of sequence networks ```text F1 --(Ea)--[Z1]-- N1 == F2 --[Z2]-- N2 == F0 --[Z0]-- N0 | | +------------------------[ 3Zf ]------------------------+ common current Ia1 = Ia2 = Ia0 ``` The three networks are connected in **series** with $3Z_f$: $N_1$ is joined to $F_2$ (the "==" links), $N_2$ is joined to $F_0$, and $N_0$ returns to $F_1$ through $3Z_f$. The common current $I_{a1}$ flows through all of them. Both conditions ($I_{a0} = I_{a1} = I_{a2}$ and $V_{a0}+V_{a1}+V_{a2} = 3Z_fI_{a1}$) are satisfied only by this series connection. #### Line-to-ground voltages at the fault Let $I_1 = I_{a1} = I_{a2} = I_{a0} = \dfrac{E_a}{Z_1+Z_2+Z_0+3Z_f}$. **Phase a:** $$ V_a = Z_fI_a = 3Z_fI_1 = \frac{3Z_fE_a}{Z_1 + Z_2 + Z_0 + 3Z_f} $$ **Phase b:** $$ \begin{aligned} V_b &= V_{a0} + a^2V_{a1} + aV_{a2} \\ &= -Z_0I_1 + a^2(E_a - Z_1I_1) - aZ_2I_1 \end{aligned} $$ Replacing $a^2E_a = a^2I_1(Z_1 + Z_2 + Z_0 + 3Z_f)$: $$ V_b = I_1\left[(a^2 - a)Z_2 + (a^2 - 1)Z_0 + 3a^2Z_f\right] $$ **Phase c:** similarly, with $aE_a = aI_1(Z_1+Z_2+Z_0+3Z_f)$: $$ V_c = I_1\left[(a - a^2)Z_2 + (a - 1)Z_0 + 3aZ_f\right] $$ For a bolted fault ($Z_f = 0$): $V_a = 0$ and $$ V_b = \frac{E_a\left[(a^2 - a)Z_2 + (a^2 - 1)Z_0\right]}{Z_1 + Z_2 + Z_0}, \quad V_c = \frac{E_a\left[(a - a^2)Z_2 + (a - 1)Z_0\right]}{Z_1 + Z_2 + Z_0} $$ ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How do zero sequence components of the three phases differ from positive and negative ones? A. They are 120° apart B. They have the reverse phase order a-c-b C. They are equal in magnitude and all in phase (0° displacement) D. They are 180° apart Answer: C. Zero sequence components Va0 = Vb0 = Vc0 are identical in magnitude and phase, so they need a neutral or ground return. (Key point: Zero sequence components) 2. What is the main effect of negative sequence currents in a machine? A. Useful torque B. Reverse-rotating field causing rotor heating and braking torque C. Ground currents and telephone interference D. Voltage regulation Answer: B. Negative sequence has reverse phase order, giving a reverse-rotating field, reverse torque and rotor heating. (Key point: Negative sequence effect) 3. For a solidly grounded no-load generator with L-G fault on a through Zf, what is If? A. Ea/(Z1 + Z2 + Z0 + Zf) B. 3Ea/(Z1 + Z2 + Z0 + 3Zf) C. 3Ea/(Z1 + Z2 + Z0) D. Ea/(Z1 + Z2 + Z0 + 3Zf) Answer: B. Ia1 = Ea/(Z1 + Z2 + Z0 + 3Zf) and If = Ia = 3Ia1. (Key point: Phase-a fault current) 4. What is Va at the fault with fault impedance Zf? A. Va = 0 always B. Va = Zf Ia = 3Zf Ea/(Z1 + Z2 + Z0 + 3Zf) C. Va = Ea D. Va = Zf Ia/3 Answer: B. The boundary condition is Va = Zf Ia, and Ia = 3Ia1, so Va = 3Zf Ia1; it is zero only for a bolted fault. (Key point: Faulted phase voltage) ## Related in Power System Analysis II - [Fault currents of isolated-neutral 380 V generator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Agenerator-faults-380v-isolated-neutral) — Power System Analysis II, chapter 5 - [Sequence components of load phase voltages](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aload-phase-voltages-components) — Power System Analysis II, chapter 4 - [L-G fault at bus 1 through Zf](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Alg-fault-bus1-parallel-lines) — Power System Analysis II, chapter 5 - [SLG fault at bus 3, currents and voltages](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aslg-fault-bus3-three-generators) — Power System Analysis II, chapter 5 - [Line currents with isolated and grounded neutral](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Astar-impedance-unbalanced-voltages-neutral) — Power System Analysis II, chapter 4 - [L-G fault and 3-phase fault, 30 MVA generator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Alg-fault-30mva-voltages-3phase) — Power System Analysis II, chapter 5 ## Related in other subjects - [Advantages of solidly grounded generators](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Asolidly-grounded-advantages) — Power Plant Design, chapter 4 - [Delta load (15+j20) phase and line currents](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Adelta-load-15j20-220v-phasors) — Basic Electrical Engineering, chapter 8