--- title: "Sequence networks of Y load, neutral current" subject: "Power System Analysis II" chapter: 4 key: "semester_5/power_system_analysis_ii:star-load-sequence-networks-neutral" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/star-load-sequence-networks-neutral site: BCT Engineering --- # Sequence networks of Y load, neutral current Power System Analysis II · Chapter 4: Unbalanced System Analysis - Asked in: 2073 Shrawan, 8 marks; 2071 Shrawan, 10 marks (asked 2 times) - Video: [Sequence Network Of Impedance Load | Sequence Impedance | Part 2 II PSA](https://www.youtube.com/watch?v=Dd5duuZ8UFA) — University Concept - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/star-load-sequence-networks-neutral - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-4 ## Question Deduce the sequence networks for balanced Y connected load with neutral grounded through impedance Zn and show also that current through neutral impedance is three times the zero sequence current. ## Short answer For a balanced Y load with neutral impedance Zn, the positive and negative sequence networks each consist of the phase impedance Zs. The zero sequence network includes Zs plus 3Zn in series because neutral current equals 3Ia0. The neutral current In = Ia+Ib+Ic = 3Ia0, as positive and negative components cancel. ## Answer For a balanced star load with neutral grounded through $Z_n$, the sequence networks are three separate circuits with impedances $Z_1 = Z_2 = Z_s$ and $Z_0 = Z_s + 3Z_n$, and the neutral carries $I_n = 3I_{a0}$. #### Neutral current is three times zero sequence current By the symmetrical component definitions: $$ \begin{aligned} I_a &= I_{a0} + I_{a1} + I_{a2} \\ I_b &= I_{a0} + a^2 I_{a1} + a I_{a2} \\ I_c &= I_{a0} + a I_{a1} + a^2 I_{a2} \end{aligned} $$ Adding and using $1 + a + a^2 = 0$: $$ \begin{aligned} I_n &= I_a + I_b + I_c \\ &= 3I_{a0} + (1+a^2+a)I_{a1} + (1+a+a^2)I_{a2} \\ &= 3I_{a0} \end{aligned} $$ So only zero sequence currents flow through the neutral, and the neutral current is $3I_{a0}$. Positive and negative sequence currents form balanced sets that cancel at the star point. #### Circuit and phase equations ```text Ia --> Zs a o-----/\/\---+ Ib --> Zs | b o-----/\/\---+ n Ic --> Zs | c o-----/\/\---+ | In = Ia+Ib+Ic Zn | ground ``` Voltage of each phase to ground = drop in $Z_s$ + drop in $Z_n$: $$ \begin{aligned} V_a &= Z_s I_a + Z_n (I_a + I_b + I_c) \\ V_b &= Z_s I_b + Z_n (I_a + I_b + I_c) \\ V_c &= Z_s I_c + Z_n (I_a + I_b + I_c) \end{aligned} $$ In matrix form $V_p = Z_p I_p$ with $$ Z_p = \begin{bmatrix} Z_s+Z_n & Z_n & Z_n \\ Z_n & Z_s+Z_n & Z_n \\ Z_n & Z_n & Z_s+Z_n \end{bmatrix} $$ #### Transformation to sequence quantities Put $V_p = A V_s$ and $I_p = A I_s$: $$ A V_s = Z_p A I_s \;\Rightarrow\; V_s = A^{-1} Z_p A\, I_s = Z_s^{seq} I_s $$ with $A = \begin{bmatrix} 1&1&1\\1&a^2&a\\1&a&a^2 \end{bmatrix}$, $A^{-1} = \frac13\begin{bmatrix} 1&1&1\\1&a&a^2\\1&a^2&a \end{bmatrix}$. Write $Z_p = Z_s U + Z_n J$, where $U$ is the unit matrix and $J$ is the matrix of all ones. Then $A^{-1}(Z_sU)A = Z_sU$. For $J$: every row of $J A$ is $[3\ \ 0\ \ 0]$ (since column sums of $A$ are $3, 1+a^2+a=0, 1+a+a^2=0$), and $A^{-1}$ times that gives $$ A^{-1} J A = \begin{bmatrix} 3&0&0\\0&0&0\\0&0&0 \end{bmatrix} $$ Hence $$ Z_{seq} = A^{-1} Z_p A = \begin{bmatrix} Z_s+3Z_n & 0 & 0 \\ 0 & Z_s & 0 \\ 0 & 0 & Z_s \end{bmatrix} $$ #### Sequence networks ```text Positive Negative Zero a1 --Ia1--> a2 --Ia2--> a0 --Ia0--> | | | Zs Zs Zs | | | n1 (ref) n2 (ref) 3Zn | n0 (ref) ``` - $V_{a1} = Z_s I_{a1}$, $V_{a2} = Z_s I_{a2}$, $V_{a0} = (Z_s + 3Z_n) I_{a0}$. - The impedance $Z_n$ appears as $3Z_n$ in the zero sequence network only because the voltage drop across it is $Z_n \times 3I_{a0}$, while the network is drawn for one phase carrying $I_{a0}$. - Solidly grounded neutral: $Z_n = 0$; ungrounded neutral: zero sequence network open between $Z_s$ and the reference. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a balanced star load with grounded neutral, the neutral current In equals: A. 3 Ia0 B. Ia0 C. Ia1 + Ia2 D. 3 Ia1 Answer: A. In = Ia+Ib+Ic; positive and negative terms cancel because 1+a+a^2=0, leaving 3Ia0. (Key point: Neutral current) 2. Why do positive and negative sequence currents not flow through Zn? A. They are blocked by Zs B. They form balanced sets that sum to zero at the star point C. They have a different frequency D. They flow only in the delta Answer: B. Balanced sets sum to zero at the star point, so only in-phase zero sequence currents add in the neutral. (Key point: Why only zero sequence) 3. The sequence impedances of the star load with neutral impedance Zn are: A. Z1 = Z2 = Zs + 3Zn, Z0 = Zs B. Z0 = Z1 = Z2 = Zs + Zn C. Z1 = Z2 = Zs, Z0 = Zs + 3Zn D. Z0 = Zs + Zn, Z1 = Z2 = Zs Answer: C. Only the zero sequence network includes the neutral impedance, as 3Zn. (Key point: Sequence impedances) 4. Zn appears as 3Zn in the zero sequence network because: A. Zn is connected in three parallel paths B. Zn is tripled by the transformation matrix A^-1 only C. The load has three phases so Zs is tripled too D. The drop across Zn is Zn x 3Ia0 while the network carries only Ia0 per phase Answer: D. Drop = Zn(3Ia0) = (3Zn)Ia0, so 3Zn is the equivalent in a per-phase network. (Key point: Why 3Zn) ## Related in Power System Analysis II - [Sequence impedances of star load with Zn](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Astar-load-sequence-zn) — Power System Analysis II, chapter 4, asked 2× - [Sequence impedance of load with mutual coupling](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Amutually-coupled-load-sequence-impedance) — Power System Analysis II, chapter 4 - [Zero sequence Thevenin impedance at bus 2](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Azero-sequence-thevenin-bus2) — 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 - [Sequence network of loaded generator with Zn](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aloaded-generator-sequence-network-zn) — Power System Analysis II, chapter 4 - [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 ## Related in other subjects - [Neutral current and power, 380 V unbalanced load](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aunbalanced-star-neutral-current-380v) — Basic Electrical Engineering, chapter 8 - [Star system with three loads, neutral current](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Astar-three-loads-400v-neutral-current) — Basic Electrical Engineering, chapter 8 - [Balanced circulating current: 10 kV alternator](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Acirculating-current-10kv-10-ohm) — Switchgear and Protection, chapter 7 - [Current through inductor, equivalent impedance](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Acurrent-through-inductor-equivalent-impedance) — Basic Electrical Engineering, chapter 2