--- title: "Q at PV bus and initial Jacobian" subject: "Power System Analysis II" chapter: 2 key: "semester_5/power_system_analysis_ii:nr-q2-jacobian-p4" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/nr-q2-jacobian-p4 site: BCT Engineering --- # Q at PV bus and initial Jacobian Power System Analysis II · Chapter 2: Load Flow Analysis - Asked in: 2070 Asar, 4+8 marks - Video: [Newton Raphson Power Flow Example Part 2](https://www.youtube.com/watch?v=ZngdGkck1-8) — GeneralPAC: Power System Tutorials - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/nr-q2-jacobian-p4 - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-2 ## Question Figure below shows the one line diagram of a power system with generators at buses 1 and 2. The line admittances are marked in pu in the diagram. Bus number, active power generation (PG), reactive power generation (QG), active power load (PL), reactive power load (QL), Bus voltage and bus type are tabulated below. All values are in pu. i) Determine reactive power at bus-2 (use initial guess for unknown voltages) ii) Determine Jacobian matrix (J⁰) for the first iteration of N-R load flow method. [Figure: y12 = −j40, y13 = −j20, y23 = −j20; bus 2 with P2 = 4.0 p.u and |V2| = 1.05; load 5 + j4 p.u at bus 3] | Bus No. | PG | QG | PL | QL | Bus voltage | Bus Type | |---|---|---|---|---|---|---| | 1 | ? | ? | 0 | 0 | 1.0∠0° | Slack | | 2 | 4.0 | ? | 0 | 0 | 1.05 | PV | | 3 | 0 | 0 | 5.0 | 4.0 | ? | PQ | ## Short answer With flat guesses V2=1.05∠0°, V3=1∠0°, Q2=QG2=3.15 pu. The first-iteration Jacobian for unknowns δ2, δ3, |V3| is J0 = [[63, −21, 0], [−21, 41, 0], [0, 0, 39]]. ## Answer Bus 1 is the slack bus ($1\angle 0^\circ$), bus 2 a PV bus ($P_2 = 4.0$ pu, $|V_2| = 1.05$ pu) and bus 3 a PQ bus ($P_3 = -5$, $Q_3 = -4$ pu). Initial guesses: $V_2^{0} = 1.05\angle 0^\circ$, $V_3^{0} = 1\angle 0^\circ$. #### Y-bus With $y_{12} = -j40$, $y_{13} = -j20$, $y_{23} = -j20$: $$ Y_{bus} = \begin{bmatrix} -j60.0000 & j40.0000 & j20.0000 \\ j40.0000 & -j60.0000 & j20.0000 \\ j20.0000 & j20.0000 & -j40.0000 \end{bmatrix} \Rightarrow |Y_{11}| = |Y_{22}| = 60,\ |Y_{33}| = 40\ (\theta = -90^\circ),\ |Y_{12}| = 40,\ |Y_{13}| = |Y_{23}| = 20\ (\theta = 90^\circ) $$ #### i) Reactive power at bus 2 $$ \begin{aligned} Q_2 &= -\sum_{j=1}^{3}|V_2||V_j||Y_{2j}|\sin(\theta_{2j} - \delta_2 + \delta_j) \\ &= -\left[(1.05)(1)(40)\sin 90^\circ + (1.05)^2(60)\sin(-90^\circ) + (1.05)(1)(20)\sin 90^\circ\right] \\ &= -\left[42 - 66.15 + 21\right] = 3.1500\ \text{pu} \end{aligned} $$ Since $Q_{L2} = 0$, $Q_{G2} = 3.1500$ pu. #### ii) Jacobian for the first iteration Unknowns $\delta_2$, $\delta_3$, $|V_3|$; equations $\Delta P_2$, $\Delta P_3$, $\Delta Q_3$. With $Y_{ij} = |Y_{ij}|\angle\theta_{ij}$, $V_i = |V_i|\angle\delta_i$ and $\phi_{ij} = \theta_{ij} - \delta_i + \delta_j$: $$ \begin{aligned} P_i &= \sum_{j=1}^{n} |V_i||V_j||Y_{ij}|\cos\phi_{ij} \\ Q_i &= -\sum_{j=1}^{n} |V_i||V_j||Y_{ij}|\sin\phi_{ij} \end{aligned} $$ Jacobian elements (polar form, $j \ne i$): $$ \begin{aligned} \frac{\partial P_i}{\partial \delta_i} &= \sum_{j\ne i} |V_i||V_j||Y_{ij}|\sin\phi_{ij} \\ \frac{\partial P_i}{\partial \delta_j} &= -|V_i||V_j||Y_{ij}|\sin\phi_{ij} \\ \frac{\partial P_i}{\partial |V_i|} &= 2|V_i||Y_{ii}|\cos\theta_{ii} + \sum_{j\ne i} |V_j||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial P_i}{\partial |V_j|} &= |V_i||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial Q_i}{\partial \delta_i} &= \sum_{j\ne i} |V_i||V_j||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial Q_i}{\partial \delta_j} &= -|V_i||V_j||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial Q_i}{\partial |V_i|} &= -2|V_i||Y_{ii}|\sin\theta_{ii} - \sum_{j\ne i} |V_j||Y_{ij}|\sin\phi_{ij} \\ \frac{\partial Q_i}{\partial |V_j|} &= -|V_i||Y_{ij}|\sin\phi_{ij} \end{aligned} $$ $$ \begin{aligned} \frac{\partial P_2}{\partial \delta_2} &= (1.0500)(1.0000)(40.0000)\sin(90.00^\circ) + (1.0500)(1.0000)(20.0000)\sin(90.00^\circ) = 63.0000 \\ \frac{\partial P_2}{\partial \delta_3} &= -(1.0500)(1.0000)(20.0000)\sin(90.00^\circ) = -21.0000 \\ \frac{\partial P_2}{\partial |V_3|} &= (1.0500)(20.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial P_3}{\partial \delta_2} &= -(1.0000)(1.0500)(20.0000)\sin(90.00^\circ) = -21.0000 \\ \frac{\partial P_3}{\partial \delta_3} &= (1.0000)(1.0000)(20.0000)\sin(90.00^\circ) + (1.0000)(1.0500)(20.0000)\sin(90.00^\circ) = 41.0000 \\ \frac{\partial P_3}{\partial |V_3|} &= 2(1.0000)(40.0000)\cos(-90.00^\circ) + (1.0000)(20.0000)\cos(90.00^\circ) + (1.0500)(20.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial Q_3}{\partial \delta_2} &= -(1.0000)(1.0500)(20.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial Q_3}{\partial \delta_3} &= (1.0000)(1.0000)(20.0000)\cos(90.00^\circ) + (1.0000)(1.0500)(20.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial Q_3}{\partial |V_3|} &= -2(1.0000)(40.0000)\sin(-90.00^\circ) - (1.0000)(20.0000)\sin(90.00^\circ) - (1.0500)(20.0000)\sin(90.00^\circ) = 39.0000 \end{aligned} $$ $$ [J^{0}] = \begin{bmatrix} \frac{\partial P_2}{\partial \delta_2} & \frac{\partial P_2}{\partial \delta_3} & \frac{\partial P_2}{\partial |V_3|} \\ \frac{\partial P_3}{\partial \delta_2} & \frac{\partial P_3}{\partial \delta_3} & \frac{\partial P_3}{\partial |V_3|} \\ \frac{\partial Q_3}{\partial \delta_2} & \frac{\partial Q_3}{\partial \delta_3} & \frac{\partial Q_3}{\partial |V_3|} \end{bmatrix} = \begin{bmatrix} 63 & -21 & 0 \\ -21 & 41 & 0 \\ 0 & 0 & 39 \end{bmatrix} $$ The off-diagonal blocks are zero at the flat start because the network has no resistance and all angles are zero. For use in the first iteration the mismatch vector is $[\Delta P_2,\ \Delta P_3,\ \Delta Q_3]^T = [4.00,\ -5.00,\ -3.00]^T$ (since $P_2^0 = P_3^0 = 0$ and $Q_3^0 = -1.0000$). **Answer:** $Q_2 = 3.1500$ pu; $J^{0} = \begin{bmatrix} 63 & -21 & 0 \\ -21 & 41 & 0 \\ 0 & 0 & 39 \end{bmatrix}$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For this system (slack bus 1, PV bus 2, PQ bus 3), which variables form the unknown vector in N-R? A. delta2, delta3, |V3| B. delta2, delta3, |V2|, |V3| C. |V2|, |V3|, Q3 D. delta1, delta2, delta3 Answer: A. PV bus 2 has fixed |V2| so only its angle is unknown; PQ bus 3 has both angle and magnitude unknown, giving a 3x3 Jacobian. (Key point: Unknowns of N-R) 2. What is Q2 computed at the flat start with |V2| = 1.05, |V1| = |V3| = 1? A. 3.15 pu B. -3.15 pu C. 4.0 pu D. 0 Answer: A. Q2 = -[42 - 66.15 + 21] = 3.15 pu, and since Q_L2 = 0, Q_G2 = 3.15 pu. (Key point: Q at PV bus) 3. Why do dP/d|V| and dQ/d(delta) entries come out zero at the first iteration? A. The network is lossless (angle of Y = +/-90 deg) and all angles are zero, so the cosine terms vanish B. Because V is 1 pu C. Because bus 3 is a load bus D. Because the slack bus has no unknowns Answer: A. With purely reactive Y and zero angles, phi = +/-90 deg so the cos terms are zero, decoupling P-delta and Q-V. (Key point: Zero off-diagonal blocks) 4. Which Jacobian matrix results for this problem? A. [63 -21 0; -21 41 0; 0 0 39] B. [60 -20 0; -20 40 0; 0 0 40] C. [63 -21 0; -21 41 0; 0 0 41] D. [41 -21 0; -21 63 0; 0 0 39] Answer: A. dP2/ddelta2 = 1.05*(40+20) = 63, dP3/ddelta3 = 20 + 21 = 41 and dQ3/d|V3| = 2*40 - 20 - 21 = 39. 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