--- title: "Two-conductor open circuit fault" subject: "Power System Analysis II" chapter: 5 key: "semester_5/power_system_analysis_ii:open-conductor-fault-series" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/open-conductor-fault-series site: BCT Engineering --- # Two-conductor open circuit fault Power System Analysis II · Chapter 5: Unsymmetrical Faults on Power Systems - Asked in: 2083 Baishakh (new course), 5 marks - Video: [Open Conductor Fault: Two Conductors Open (Series Fault)](https://www.youtube.com/watch?v=2G-Q7P6afas) — E Gyan - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/open-conductor-fault-series - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-5 ## Question Show that for the two-conductor open circuit fault in a 3-phase system, the sequence networks are to be connected in series to simulate the fault. ## Short answer For a two-open-conductor fault (phases b and c open, phase a healthy), boundary conditions Ib=Ic=0 and Va=0 give equal sequence currents I0=I1=I2 and sequence voltages summing to zero. This forces the positive, negative and zero sequence networks between the break points to connect in series. The healthy-phase fault current is Ia=3E/(Z1+Z2+Z0). ## Answer In a **two-conductor open fault**, two phases (say b and c) are broken between points $p$ and $q$ while phase a stays healthy. Writing the fault conditions in symmetrical components shows that the three sequence networks, seen between $p$ and $q$, must be joined **in series**. #### Fault conditions ```text p q a --------------------------- (healthy, Ia flows) b -----x x-------- (open, Ib = 0) c -----x x-------- (open, Ic = 0) ``` Let $V_{a}, V_{b}, V_{c}$ be the series voltage drops across the break (from $p$ to $q$) in each phase. - Phases b and c are open: $I_b = 0,\ I_c = 0$ - Phase a is healthy (no break): $V_a = 0$ (drop across the break in phase a is zero) #### Sequence currents $$ \begin{bmatrix} I_{a0} \\ I_{a1} \\ I_{a2} \end{bmatrix} = \frac{1}{3}\begin{bmatrix} 1 & 1 & 1 \\ 1 & a & a^2 \\ 1 & a^2 & a \end{bmatrix}\begin{bmatrix} I_a \\ 0 \\ 0 \end{bmatrix} $$ so $$ I_{a0} = I_{a1} = I_{a2} = \frac{I_a}{3} $$ The **same current flows** through all three sequence networks. #### Sequence voltages Since $V_a = V_{a0} + V_{a1} + V_{a2} = 0$: $$ V_{a0} + V_{a1} + V_{a2} = 0 $$ #### Sequence network equations Seen from the break terminals $p$-$q$, each sequence network has a Thevenin form. Only the positive sequence network contains a source (the pre-fault open-circuit voltage $V_{pq}^{0}$ or the equivalent driving voltage $E$): $$ \begin{aligned} V_{a1} &= E - Z_1 I_{a1} \\ V_{a2} &= -Z_2 I_{a2} \\ V_{a0} &= -Z_0 I_{a0} \end{aligned} $$ Adding and using $V_{a0}+V_{a1}+V_{a2}=0$ and $I_{a0}=I_{a1}=I_{a2}$: $$ I_{a1} = \frac{E}{Z_1 + Z_2 + Z_0} $$ #### Interpretation Equal currents in all three networks and voltages that add to zero are exactly the conditions of three elements in **series** in a closed loop: ```text p1 +--[ +ve seq N/W, Z1, E ]--+ q1 | | p2 +--[ -ve seq N/W, Z2 ]-----+ q2 (series loop: | | q1->p2, q2->p0, p0 +--[ zero seq N/W, Z0 ]----+ q0 q0->p1) ``` Hence the positive, negative and zero sequence networks are connected **in series** between the fault points to simulate a two-conductor open fault, and the healthy-phase current is $I_a = 3I_{a1} = \dfrac{3E}{Z_1+Z_2+Z_0}$. This is the series-fault counterpart of the single line-to-ground shunt fault. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For phases b and c open and phase a healthy, which conditions hold? A. Ib = Ic = 0 and the series voltage drop in phase a, Va = 0 B. Ia = 0 and Vb = Vc C. Ia = Ib = Ic D. Vb = Vc = 0 and Ia = 0 Answer: A. Open phases carry no current, and the healthy phase has no voltage drop across the break. (Key point: Fault conditions of two-conductor open) 2. Why do the three sequence currents become equal? A. Because Z0 = Z1 = Z2 B. Because the voltages are equal C. Because all three networks have an EMF D. Because only Ia is nonzero, so each Ia_k = Ia/3 Answer: D. With Ib = Ic = 0 the transformation gives Ia0 = Ia1 = Ia2 = Ia/3. (Key point: Equal sequence currents) 3. Which voltage condition, together with equal currents, requires a series connection? A. Va1 = Va2 = Va0 B. Va0 + Va1 + Va2 = 0 C. Va1 = -Va2 and Va0 = 0 D. Va0 = Va1 + Va2 Answer: B. The healthy phase has zero drop, so the sequence drops sum to zero, as around a series loop. (Key point: Voltage condition) 4. What is the current in the healthy phase a? A. Ia = E/(Z1 + Z2 + Z0) B. Ia = 3E/(Z1 + Z2 + Z0) C. Ia = E/(3(Z1 + Z2 + Z0)) D. Ia = 0 Answer: B. Ia1 = E/(Z1 + Z2 + Z0) and Ia = 3Ia1. (Key point: Healthy-phase current) ## Related in Power System Analysis II - [SLG fault: sequence networks in series](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aslg-sequence-networks-series) — Power System Analysis II, chapter 5, asked 5× - [Sequence networks decoupled for static circuit](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Astatic-circuit-sequence-decoupled) — Power System Analysis II, chapter 4 - [Sequence networks with parallel transformer banks](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Asequence-networks-parallel-transformers-2079) — Power System Analysis II, chapter 4 - [DLG fault sequence network connection](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Adlg-sequence-network-connection) — Power System Analysis II, chapter 5, asked 3× - [Sequence networks of Y load, neutral current](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Astar-load-sequence-networks-neutral) — Power System Analysis II, chapter 4, asked 2× - [L-G fault in phase B](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Alg-fault-phase-b) — Power System Analysis II, chapter 5 ## Related in other subjects - [Equivalent ABCD of two networks in series](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aseries-abcd-equivalent) — Power System Analysis I, chapter 5 - [Series connection of two-port networks](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Atwo-port-series-connection) — Filter Design, chapter 4 - [Murray loop test for open and short circuit faults](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Amurray-loop-open-short-circuit) — Power System Analysis I, chapter 2, asked 5× - [Fault level: two 40 MVA generators, 2 Ω feeder](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-two-40mva-2-ohm-feeder) — Power Plant Equipment, chapter 4, asked 5×