--- title: "Equivalent Z0, Z1, Z2 at bus 3" subject: "Power System Analysis II" chapter: 5 key: "semester_5/power_system_analysis_ii:sequence-thevenin-z0-z1-z2-bus3" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/sequence-thevenin-z0-z1-z2-bus3 site: BCT Engineering --- # Equivalent Z0, Z1, Z2 at bus 3 Power System Analysis II · Chapter 5: Unsymmetrical Faults on Power Systems - Asked in: 2068 Chaitra, 8 marks - Video: [POSITIVE, NEGATIVE, ZERO SEQUENCE REACTANCE DIAGRAM / KTU/ POWER SYSTEM ANALYSIS](https://www.youtube.com/watch?v=zWNFHgJ7b5s) — praveen raj - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/sequence-thevenin-z0-z1-z2-bus3 - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-5 ## Question Draw positive, negative and zero sequence networks. If an unsymmetrical fault occurs at bus 3, determine equivalent Z0, Z1 and Z2. [Figure: G1 100 MVA, 11 kV, X1 = X2 = j0.2 pu, X0 = j0.05 pu (Y grounded) - T1 100 MVA, 11/220 kV, j0.1 pu (Δ / Y grounded) - parallel lines L1 (100 MVA, 220 kV, j0.1 pu) and L2 (100 MVA, 220 kV, 0.12 pu) - T2 50 MVA, 11/220 kV, j0.075 pu (Y grounded through j0.03 pu on line side / Δ) - G2 50 MVA, 11 kV, X1 = X2 = j0.15 pu, X0 = j0.03 pu (Y grounded through j0.024 pu)] ## Short answer Positive and negative sequence impedances at bus 3: Z1 = Z2 = j0.1983 pu. Zero-sequence impedance: Z0 = j(0.1+0.0545) ∥ j0.33 = j0.105 pu (or j0.094 pu depending on neutral reactor base). ## Answer **Assumptions:** base 100 MVA; bus 1 = G1 terminal, bus 2 = T1 HV side, bus 3 = T2 HV side (end of the parallel lines), bus 4 = G2 terminal. Neutral reactances (j0.03 for T2, j0.024 for G2) are taken on their own 50 MVA rating. Line zero-sequence reactance is taken equal to the given value since no separate $X_0$ is given. #### Per-unit values on 100 MVA | Element | $X_1 = X_2$ | $X_0$ | |---|---|---| | G1 | 0.20 | 0.05 | | T1 | 0.10 | 0.10 | | L1 ∥ L2 | $0.1\|0.12 = 0.0545$ | 0.0545 | | T2 ($\times 2$) | 0.15 | 0.15 | | T2 neutral ($\times 2$) | – | $3\times0.06 = 0.18$ | | G2 ($\times 2$) | 0.30 | 0.06 | | G2 neutral ($\times 2$) | – | $3\times0.048 = 0.144$ | #### Positive sequence network ```text E1-j0.2-(1)-j0.1-(2)-j0.0545-(3)-j0.15-(4)-j0.3-E2 | F ``` $$ Z_1 = j(0.2+0.1+0.0545)\ \|\ j(0.15+0.3) = j0.3545\ \|\ j0.45 = j0.1983\ \text{pu} $$ #### Negative sequence network Same reactances, emfs shorted: $$ Z_2 = j0.1983\ \text{pu} $$ #### Zero sequence network - T1 Δ (G1 side)/Yg (line side): bus 2 to reference through $j0.1$; G1 isolated. - T2 Yg through j0.03 (line side)/Δ (G2 side): bus 3 to reference through $j(0.15+0.18) = j0.33$; G2 isolated. ```text ref-j0.1-(2)-j0.0545-(3)-j0.33-ref | F G1 (j0.05) and G2 (j0.06+j0.144) isolated by deltas ``` $$ Z_0 = j(0.1+0.0545)\ \|\ j0.33 = j0.1545\ \|\ j0.33 = j0.1053\ \text{pu} $$ **Answer:** $Z_1 = Z_2 = j0.198$ pu, $Z_0 = j0.105$ pu at bus 3. (If the neutral reactance j0.03 is taken as already on the 100 MVA base, $Z_0 = j0.1545\|j0.24 = j0.094$ pu.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why must T2 and G2 (50 MVA) values be converted? A. Different frequency B. To a common 100 MVA base: values double C. They are star-connected D. Voltage ratings differ so values halve Answer: B. Values on 50 MVA are multiplied by 100/50, so T2 becomes 0.15 and G2 0.30 on the 100 MVA base. (Key point: Per-unit base conversion) 2. Z1 at bus 3 is: A. 0.3545 + 0.45 B. (0.3545 in parallel with 0.45) = 0.1983 pu C. 0.0545 D. 0.2 + 0.1 Answer: B. The G1 branch (0.2+0.1+0.0545) and the G2 branch (0.15+0.3) are in parallel; Z2 equals Z1. (Key point: Positive-sequence Thevenin) 3. In the zero-sequence network at bus 3, the G2 side path is: A. Isolated by T2 delta; bus 3 grounds through T2 star plus 3Xn (0.33 pu) B. G2's X0 only C. Open circuit D. Zero Answer: A. T2's grounded star on the line side gives j(0.15 + 3x0.06) = j0.33 to reference; G2 is behind the delta. 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