--- title: "Ybus with generator and transformer impedances" subject: "Power System Analysis II" chapter: 1 key: "semester_5/power_system_analysis_ii:ybus-generators-transformers-table" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/ybus-generators-transformers-table site: BCT Engineering --- # Ybus with generator and transformer impedances Power System Analysis II · Chapter 1: Interconnected Power System - Asked in: 2076 Chaitra, 4 marks - Video: [Formation of bus admittance matrix using inspection method with transformer 3 #powersystemanalysis](https://www.youtube.com/watch?v=Jxg-7k2s9Vw) — EEE, SJBIT - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/ybus-generators-transformers-table - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-1 ## Question Form the Ybus matrix for the following network. All impedance values are in per unit system. [Figure: G1 through T1 to bus 1; G2 through T2 to bus 2; line L1 between buses 1 and 2, L2 between buses 1 and 3, L3 between buses 2 and 3] | Component | G1 | G2 | T1 | T2 | L1 | L2 | L3 | |---|---|---|---|---|---|---|---| | Z, pu | j0.1 | j0.12 | j0.1 | j0.12 | 0.03+j0.08 | 0.02+j0.05 | 0.025+j0.06 | ## Short answer Treat generator plus transformer reactances as shunt admittances to ground. Compute line admittances from impedances. Diagonal Y_ii sum of connected admittances; off-diagonal Y_ik = –y_ik. Ybus = [[11.0061–j33.2003, –4.1096+j10.9589, –6.8966+j17.2414]; [–4.1096+j10.9589, 10.0267–j29.3268, –5.9172+j14.2012]; [–6.8966+j17.2414, –5.9172+j14.2012, 12.8137–j31.4426]] pu. ## Answer Each generator reactance is in series with its transformer, so it forms a shunt branch from its bus to the reference (the generator emf is a current injection). Bus 3 has no generator. #### Admittances | Branch | Impedance (pu) | Admittance (pu) | |---|---|---| | G1 + T1 at bus 1 | $j0.1 + j0.1 = j0.2$ | $-j5$ | | G2 + T2 at bus 2 | $j0.12 + j0.12 = j0.24$ | $-j4.1667$ | | L1 (1-2) | $0.03 + j0.08$ | $4.1096 - j10.9589$ | | L2 (1-3) | $0.02 + j0.05$ | $6.8966 - j17.2414$ | | L3 (2-3) | $0.025 + j0.06$ | $5.9172 - j14.2012$ | Example: $y_{12} = \dfrac{0.03 - j0.08}{0.03^2 + 0.08^2} = \dfrac{0.03 - j0.08}{0.0073} = 4.1096 - j10.9589$. #### Elements $$ \begin{aligned} Y_{11} &= -j5 + y_{12} + y_{13} = 11.0061 - j33.2003 \\ Y_{22} &= -j4.1667 + y_{12} + y_{23} = 10.0267 - j29.3268 \\ Y_{33} &= y_{13} + y_{23} = 12.8137 - j31.4426 \end{aligned} $$ **Answer:** $$ Y_{bus} = \begin{bmatrix} 11.0061 - j33.2003 & -4.1096 + j10.9589 & -6.8966 + j17.2414 \\ -4.1096 + j10.9589 & 10.0267 - j29.3268 & -5.9172 + j14.2012 \\ -6.8966 + j17.2414 & -5.9172 + j14.2012 & 12.8137 - j31.4426 \end{bmatrix}\ \text{pu} $$ ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Generator G1 (j0.1) and transformer T1 (j0.1) at bus 1 give a shunt admittance of: A. -j10 B. -j5 C. -j0.2 D. j5 Answer: B. Series impedance is j0.2, so the admittance is 1/j0.2 = -j5. (Key point: Series source branch) 2. y12 = 1/(0.03 + j0.08) is: A. 4.1096 - j10.9589 B. 0.03 - j0.08 C. 11.0061 - j33.2 D. -4.1096 + j10.9589 Answer: A. Divide by 0.03^2 + 0.08^2 = 0.0073. (Key point: y12 of L1) 3. Why does Y33 contain no shunt term? A. Bus 3 has no generator or shunt element B. Bus 3 is the slack bus C. Shunts are always zero D. Y33 is always zero Answer: A. Y33 = y13 + y23 only because bus 3 has no source branch. (Key point: Bus 3 diagonal) 4. Y23 equals: A. 5.9172 - j14.2012 B. -5.9172 + j14.2012 C. -j14.2012 D. 12.8137 - j31.4426 Answer: B. Mutual admittance is the negative of the line 2-3 admittance. 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