--- title: "Node equations and Ybus, three-source network" subject: "Power System Analysis II" chapter: 1 key: "semester_5/power_system_analysis_ii:ybus-node-equations-three-sources" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/ybus-node-equations-three-sources site: BCT Engineering --- # Node equations and Ybus, three-source network Power System Analysis II · Chapter 1: Interconnected Power System - Asked in: 2074 Asoj, 8 marks; 2070 Chaitra, 5 marks (asked 2 times) - Video: [BUS ADMITTANCE MATRIX - PART - 01 - FORMATION OF Y-BUS - ADVANTAGES](https://www.youtube.com/watch?v=rWjoK28YMkk) — zenmurali Sachithanandam - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/ybus-node-equations-three-sources - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-1 ## Question For the network given below, obtain node equations and then form admittance matrix (YBus). All impedances and voltages are marked in pu. [Figure: three buses; line 1-2 = 0.02 + j0.04, line 1-3 = 0.01 + j0.03, line 2-3 = 0.0125 + j0.025; source E1 = 1∠0° with X = j0.12 connected to bus 1 through transformer T1 of j0.08; source E2 = 1∠0° with X = j0.13 connected to bus 2 through transformer T2 of j0.12; source E3 = 1∠−36.87° with X = j0.1 connected to bus 3 through transformer T3 of j0.1] ## Short answer Convert each source with its series reactance to a Norton equivalent (current injection and shunt admittance). Compute line admittances. Ybus (pu) = [[20–j55, –10+j20, –10+j30], [–10+j20, 26–j56, –16+j32], [–10+j30, –16+j32, 26–j67]]. ## Answer Convert each voltage source with its series impedance into an equivalent current source (Norton form) at its bus, then write KCL at each bus. #### Step 1: Source branches Each source reactance is in series with its transformer, so the total series reactance to each bus is: | Bus | Source + transformer | Shunt admittance $y_{i0}$ | |---|---|---| | 1 | $j0.12 + j0.08 = j0.20$ | $1/j0.2 = -j5$ | | 2 | $j0.13 + j0.12 = j0.25$ | $1/j0.25 = -j4$ | | 3 | $j0.10 + j0.10 = j0.20$ | $1/j0.2 = -j5$ | Equivalent injected currents $I_i = E_i\,y_{i0}$: $$ \begin{aligned} I_1 &= \frac{1\angle 0^\circ}{j0.2} = -j5\ \text{pu} \\ I_2 &= \frac{1\angle 0^\circ}{j0.25} = -j4\ \text{pu} \\ I_3 &= \frac{1\angle -36.87^\circ}{j0.2} = 5\angle -126.87^\circ = -3 - j4\ \text{pu} \end{aligned} $$ #### Step 2: Line admittances $$ \begin{aligned} y_{12} &= \frac{1}{0.02 + j0.04} = 10 - j20\ \text{pu} \\ y_{13} &= \frac{1}{0.01 + j0.03} = 10 - j30\ \text{pu} \\ y_{23} &= \frac{1}{0.0125 + j0.025} = 16 - j32\ \text{pu} \end{aligned} $$ #### Step 3: Node equations (KCL) $$ \begin{aligned} I_1 &= y_{10}V_1 + y_{12}(V_1 - V_2) + y_{13}(V_1 - V_3) \\ I_2 &= y_{20}V_2 + y_{12}(V_2 - V_1) + y_{23}(V_2 - V_3) \\ I_3 &= y_{30}V_3 + y_{13}(V_3 - V_1) + y_{23}(V_3 - V_2) \end{aligned} $$ Collecting terms: $$ \begin{aligned} -j5 &= (y_{10} + y_{12} + y_{13})V_1 - y_{12}V_2 - y_{13}V_3 \\ -j4 &= -y_{12}V_1 + (y_{20} + y_{12} + y_{23})V_2 - y_{23}V_3 \\ -3 - j4 &= -y_{13}V_1 - y_{23}V_2 + (y_{30} + y_{13} + y_{23})V_3 \end{aligned} $$ #### Step 4: Ybus elements $$ \begin{aligned} Y_{11} &= -j5 + (10 - j20) + (10 - j30) = 20 - j55 \\ Y_{22} &= -j4 + (10 - j20) + (16 - j32) = 26 - j56 \\ Y_{33} &= -j5 + (10 - j30) + (16 - j32) = 26 - j67 \\ Y_{12} &= Y_{21} = -10 + j20 \\ Y_{13} &= Y_{31} = -10 + j30 \\ Y_{23} &= Y_{32} = -16 + j32 \end{aligned} $$ #### Result $$ \begin{bmatrix} -j5 \\ -j4 \\ -3 - j4 \end{bmatrix} = \begin{bmatrix} 20 - j55 & -10 + j20 & -10 + j30 \\ -10 + j20 & 26 - j56 & -16 + j32 \\ -10 + j30 & -16 + j32 & 26 - j67 \end{bmatrix} \begin{bmatrix} V_1 \\ V_2 \\ V_3 \end{bmatrix} $$ **Answer:** $Y_{bus}$ is the 3×3 matrix above (pu); it is symmetric, and its diagonal terms include the source-branch admittances $-j5$, $-j4$, $-j5$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is each voltage source with series impedance handled? A. Ignored B. Converted to a shunt capacitor C. Converted to a Norton equivalent current source E/Z at its bus D. Replaced by an open circuit Answer: C. Ii = Ei * yi0. (Key point: Source conversion) 2. For bus 1, source j0.12 and transformer j0.08 give shunt admittance y10: A. -j10 pu B. j5 pu C. -j0.2 pu D. -j5 pu Answer: D. 1/j0.2 = -j5. (Key point: Source branch) 3. The injected current at bus 3 (E3 = 1 angle -36.87 deg, total j0.2) is: A. -3 - j4 pu B. -j5 pu C. 3 + j4 pu D. -4 - j3 pu Answer: A. I3 = 1 angle -36.87 / j0.2 = 5 angle -126.87 = -3 - j4. (Key point: Current I3) 4. Y11 for the network is: A. -10 + j20 pu B. 20 - j55 pu C. 26 - j56 pu D. 30 - j55 pu Answer: B. Y11 = -j5 + (10 - j20) + (10 - j30) = 20 - j55. (Key point: Y11) ## Related in Power System Analysis II - [Node equations and Ybus, six-bus network](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aybus-six-bus-node-equations) — Power System Analysis II, chapter 1 - [Ybus for network with three sources](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aybus-three-sources-2080) — Power System Analysis II, chapter 1 - [Ybus with generator and transformer impedances](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aybus-generators-transformers-table) — Power System Analysis II, chapter 1 - [G-S two iterations, lines 0.02+j0.11](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Ags-two-iterations-2080) — Power System Analysis II, chapter 2 - [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 modified by mid-line tap changer](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aybus-tap-changer-midpoint) — Power System Analysis II, chapter 1