--- title: "L-G fault of unloaded generator: equal currents" subject: "Power System Analysis II" chapter: 5 key: "semester_5/power_system_analysis_ii:lg-fault-unloaded-generator-derivation" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/lg-fault-unloaded-generator-derivation site: BCT Engineering --- # L-G fault of unloaded generator: equal currents Power System Analysis II · Chapter 5: Unsymmetrical Faults on Power Systems - Asked in: 2080 Bhadra, 8 marks - Video: [Single Line to Ground Fault on an Unloaded Generator/LG Fault](https://www.youtube.com/watch?v=WtJvn4aLcTE) — Dr.Nagaraja Kumari - 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-unloaded-generator-derivation - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-5 ## Question Draw positive sequence, negative sequence and zero sequence network of synchronous generator. Show that all three sequence currents of faulty phase are equal in case of single line to ground fault of unloaded synchronous generator. Also find the expression of fault current. ## Short answer Boundary conditions I_b=I_c=0 and V_a=Z_f I_a give I_a0=I_a1=I_a2=I_a/3, so all three sequence networks are in series. The fault current for a bolted single line-to-ground fault is I_f = 3E_a/(Z1+Z2+Z0); with fault impedance Z_f it is I_f = 3E_a/(Z1+Z2+Z0+3Z_f). ## Answer For an L-G fault on phase a of an unloaded generator, $I_b = I_c = 0$, which gives $I_{a0} = I_{a1} = I_{a2} = I_a/3$; the three sequence networks are in series and $I_f = 3E_a/(Z_1+Z_2+Z_0)$ (bolted) or $3E_a/(Z_1+Z_2+Z_0+3Z_f)$ with fault impedance. **Generator:** star connected, balanced EMFs $E_a$, $a^2E_a$, $aE_a$; neutral grounded through $Z_n$. #### Sequence networks of an unloaded generator The generator EMFs are balanced (positive sequence only), so with $Z_0 = Z_{g0} + 3Z_n$: $$ V_{a0} = -Z_0I_{a0}, \qquad V_{a1} = E_a - Z_1I_{a1}, \qquad V_{a2} = -Z_2I_{a2} $$ ```text Positive Negative Zero F1 o F2 o F0 o |Ia1 ^ |Ia2 ^ |Ia0 ^ Z1 Z2 Z0 | | | (Ea) | | | | | N1 o N2 o N0 o ``` - **Positive sequence:** contains the EMF $E_a$ and $Z_1$ (sub-transient reactance $X_d''$ for fault studies). No current flows in $Z_n$. - **Negative sequence:** no EMF, impedance $Z_2$, connected to the neutral directly. - **Zero sequence:** no EMF, impedance $Z_0 = Z_{g0} + 3Z_n$; the current through the neutral is $3I_{a0}$. #### 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 are equal $$ \begin{bmatrix} I_{a0}\\I_{a1}\\I_{a2} \end{bmatrix} = \frac13\begin{bmatrix} 1&1&1\\1&a&a^2\\1&a^2&a \end{bmatrix}\begin{bmatrix} I_a\\0\\0 \end{bmatrix} \quad\Rightarrow\quad I_{a0} = I_{a1} = I_{a2} = \frac{I_a}{3} $$ 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 ($Z_f = 0$): $I_f = \dfrac{3E_a}{Z_1+Z_2+Z_0}$. #### 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the sequence networks of an unloaded generator, which network contains the EMF? A. Negative sequence only B. Zero sequence only C. All three D. Positive sequence only Answer: D. Generator EMFs are balanced, so only the positive sequence network has a source. (Key point: Sequence networks of generator) 2. A generator neutral is grounded through Zn. What zero sequence impedance appears in the zero sequence network? A. Z0 = Zg0 + 3Zn B. Z0 = Zg0 + Zn C. Z0 = Zg0 D. Z0 = Zg0 + Zn/3 Answer: A. The neutral carries 3Ia0, so its drop 3Zn Ia0 appears as 3Zn in the zero sequence network. (Key point: Zero sequence impedance with neutral impedance) 3. Why does Ib = Ic = 0 give Ia0 = Ia1 = Ia2? A. Because Z0 = Z1 = Z2 B. Because each sequence current is (1/3)(Ia + 0 + 0) = Ia/3 C. Because the neutral is isolated D. Because the fault impedance is zero Answer: B. With only Ia nonzero, every row of the transformation gives Ia/3, so the same current flows in all three sequence networks. (Key point: Equal sequence currents) 4. What is the L-G fault current with fault impedance Zf at the terminals of an unloaded generator? A. Ea/(Z1 + Z2 + Z0 + Zf) B. 3Ea/(Z1 + Z2 + Z0) C. 3Ea/(Z1 + Z2 + Z0 + 3Zf) D. Ea/(Z1 + Z2 + Z0 + 3Zf) Answer: C. Ia1 = Ea/(Z1 + Z2 + Z0 + 3Zf) and If = Ia = 3Ia1. 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