--- title: "Bus classification and 4-bus Ybus" subject: "Power System Analysis II" chapter: 1 key: "semester_5/power_system_analysis_ii:bus-classification-ybus-four-bus" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/bus-classification-ybus-four-bus site: BCT Engineering --- # Bus classification and 4-bus Ybus Power System Analysis II · Chapter 1: Interconnected Power System - Asked in: 2076 Asoj, 8 marks - Video: [What is power/load flow analysis?? Why Swing, PV, and PQ buses?? Why numerical methods are used??](https://www.youtube.com/watch?v=JKXIGilK23Y) — A Serious Circuit - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/bus-classification-ybus-four-bus - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-1 ## Question Explain bus classification in power flow analysis with their known and unknown quantities and hence formulate the bus admittance matrix for four bus system. ## Short answer Buses are slack (known |V|, δ; unknown P, Q), PV (known P, |V|; unknown Q, δ), or PQ (known P, Q; unknown |V|, δ). For a four-bus system, Ybus diagonal elements are sum of admittances at each bus, off-diagonals are negative mutual admittances, and zero if no direct connection. ## Answer In load-flow analysis each bus has four quantities: $P$, $Q$, $|V|$ and $\delta$. Two are specified and two are found by the solution. Buses are classified according to which two are known. #### Bus classification | Bus type | Known | Unknown | Example | |---|---|---|---| | **Slack / swing / reference bus** | $\lvert V\rvert$, $\delta$ (usually $\delta = 0$) | $P$, $Q$ | Largest generating station | | **PV / generator / voltage-controlled bus** | $P$, $\lvert V\rvert$ (with $Q_{min}$, $Q_{max}$) | $Q$, $\delta$ | Generator buses, buses with SVC | | **PQ / load bus** | $P$, $Q$ | $\lvert V\rvert$, $\delta$ | Load buses (about 80% of buses) | - **Slack bus:** only one in the system. Its P and Q make up the difference between scheduled generation and load + losses (losses are unknown before the solution). Its voltage angle serves as reference. - **PV bus:** real power output is set by the turbine and voltage magnitude by the AVR. If the computed Q exceeds its limits, Q is fixed at the limit and the bus becomes a PQ bus. - **PQ bus:** net injection $P = P_G - P_D$, $Q = Q_G - Q_D$ is known; voltage is found. #### Ybus for a four-bus system Take a four-bus network with lines 1-2, 1-3, 2-3, 2-4 and 3-4. Each line has series admittance $y_{ik}$ and shunt $y'_{ik}/2$ at each end; total shunt at bus $i$ is $y_{i0}$. Applying KCL: $$ \begin{aligned} I_1 &= y_{10}V_1 + y_{12}(V_1 - V_2) + y_{13}(V_1 - V_3) \\ I_2 &= y_{20}V_2 + y_{12}(V_2 - V_1) + y_{23}(V_2 - V_3) + y_{24}(V_2 - V_4) \\ I_3 &= y_{30}V_3 + y_{13}(V_3 - V_1) + y_{23}(V_3 - V_2) + y_{34}(V_3 - V_4) \\ I_4 &= y_{40}V_4 + y_{24}(V_4 - V_2) + y_{34}(V_4 - V_3) \end{aligned} $$ $$ \mathbf{Y}_{bus} = \begin{bmatrix} y_{10} + y_{12} + y_{13} & -y_{12} & -y_{13} & 0 \\ -y_{12} & y_{20} + y_{12} + y_{23} + y_{24} & -y_{23} & -y_{24} \\ -y_{13} & -y_{23} & y_{30} + y_{13} + y_{23} + y_{34} & -y_{34} \\ 0 & -y_{24} & -y_{34} & y_{40} + y_{24} + y_{34} \end{bmatrix} $$ General rule: $Y_{ii} = y_{i0} + \sum_{k\ne i} y_{ik}$ and $Y_{ik} = -y_{ik}$ ($= 0$ if no line, e.g. $Y_{14}$). These elements are then used in the power-flow equations $$ P_i - jQ_i = V_i^*\sum_{k=1}^{4} Y_{ik}V_k $$ which are solved for the unknowns listed in the table above. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which quantities are known at the slack bus? A. P and Q B. P and |V| C. |V| and delta D. Q and delta Answer: C. The slack bus has specified voltage magnitude and angle; P and Q are found, absorbing losses. (Key point: Slack bus) 2. At a PV bus the unknowns are: A. P and |V| B. Q and delta C. |V| and delta D. P and Q Answer: B. P and |V| are specified; Q and angle are computed. (Key point: PV bus) 3. What happens at a PV bus if the computed Q exceeds its limit? A. It becomes the slack bus B. Q is fixed at the limit and the bus is treated as PQ C. Its P is set to zero D. The load flow stops Answer: B. With Q held at the limit, |V| can no longer be held, so the bus acts as PQ. (Key point: PV to PQ switch) 4. Why must a slack bus exist? A. Losses are unknown before the solution, so one bus supplies the balance of P and Q B. To keep Ybus symmetric C. To make all voltages equal 1 pu D. Because PQ buses have no angle Answer: A. Total losses are unknown in advance, so the slack bus P and Q make up the difference and its angle is the reference. (Key point: Why slack is needed) ## Related in Power System Analysis II - [Bus admittance matrix and its elements](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aybus-definition-elements) — Power System Analysis II, chapter 1 - [Ybus for 3-bus system (2078)](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aybus-three-bus-2078) — Power System Analysis II, chapter 1 - [Ybus of five-bus system](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aybus-five-bus) — Power System Analysis II, chapter 1 - [Bus quantities and decoupled load flow steps](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Abus-types-decoupled-steps) — Power System Analysis II, chapter 2 - [G-S load flow with given Ybus, losses](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Ags-ybus-j12-two-iterations) — Power System Analysis II, chapter 2 - [Ybus, Zbus and 4-bus Ybus formulation](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aybus-zbus-four-bus) — Power System Analysis II, chapter 1 ## Related in other subjects - [Bus-bar reactor scheme with four generators](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Abus-bar-reactor-four-generators) — Power Plant Equipment, chapter 4, asked 2× - [Fault level: four 25 MVA generators, bus reactor](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-four-25mva-bus-reactor) — Power Plant Equipment, chapter 4