---
title: "N-R Jacobian and first iteration (new course)"
subject: "Power System Analysis II"
chapter: 2
key: "semester_5/power_system_analysis_ii:nr-jacobian-ybus-j12-new"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/nr-jacobian-ybus-j12-new
site: BCT Engineering
---

# N-R Jacobian and first iteration (new course)

Power System Analysis II · Chapter 2: Load Flow Analysis

- Asked in: 2083 Baishakh (new course), 4+3 marks
- Video: [Newton Raphson Power Flow Example Part 1](https://www.youtube.com/watch?v=-XeG8WAjS7E) — 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-jacobian-ybus-j12-new
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-2

## Question

The single line diagram of power system network is shown in the figure below. The Y bus of the system and bus data are shown in the table. (Use N-R method). (i) Form the Jacobian Matrix. (ii) Find the unknown values after the first iteration. Ybus = [−j12 j6 j6; j6 −j12 j6; j6 j6 −j12]. [Figure: line 1-2 is 100 km, line 1-3 is 200 km, line 2-3 is 400 km]

Table: Bus data

| Bus | Bus type | Injected P | Injected Q | Bus voltage magnitude | Voltage phase angle |
|---|---|---|---|---|---|
| 1 | Slack | - | - | 1 | 0 |
| 2 | PV | 0.6 | - | 1 | - |
| 3 | PQ | −0.5 | −0.4 | - | - |

## Short answer

Jacobian at flat start: J^0 = [[12, -6, 0], [-6, 12, 0], [0, 0, 12]] pu. After first iteration: Δδ2 = 0.0389 rad (2.23°), Δδ3 = -0.0222 rad (-1.27°), Δ|V3| = -0.0333 pu; V2 = 1∠2.23°, V3 = 0.9667∠-1.27°.

## Answer

Bus 1 is the slack bus ($1\angle 0^\circ$), bus 2 a PV bus ($P_2 = 0.6$ pu, $|V_2| = 1$ pu) and bus 3 a PQ bus ($P_3 = -0.5$, $Q_3 = -0.4$ pu). Flat start: $V_2^0 = V_3^0 = 1\angle 0^\circ$. The line lengths are not needed since $Y_{bus}$ is given: $|Y_{ii}| = 12\angle -90^\circ$, $|Y_{ij}| = 6\angle 90^\circ$.

#### (i) Jacobian matrix
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, $j \ne i$):

$$
\begin{aligned}
\frac{\partial P_i}{\partial \delta_i} &= \sum_{j\ne i} |V_i||V_j||Y_{ij}|\sin\phi_{ij} \\
\frac{\partial P_i}{\partial \delta_j} &= -|V_i||V_j||Y_{ij}|\sin\phi_{ij} \\
\frac{\partial P_i}{\partial |V_i|} &= 2|V_i||Y_{ii}|\cos\theta_{ii} + \sum_{j\ne i} |V_j||Y_{ij}|\cos\phi_{ij} \\
\frac{\partial P_i}{\partial |V_j|} &= |V_i||Y_{ij}|\cos\phi_{ij} \\
\frac{\partial Q_i}{\partial \delta_i} &= \sum_{j\ne i} |V_i||V_j||Y_{ij}|\cos\phi_{ij} \\
\frac{\partial Q_i}{\partial \delta_j} &= -|V_i||V_j||Y_{ij}|\cos\phi_{ij} \\
\frac{\partial Q_i}{\partial |V_i|} &= -2|V_i||Y_{ii}|\sin\theta_{ii} - \sum_{j\ne i} |V_j||Y_{ij}|\sin\phi_{ij} \\
\frac{\partial Q_i}{\partial |V_j|} &= -|V_i||Y_{ij}|\sin\phi_{ij}
\end{aligned}
$$

At the flat start every $\phi_{ij} = \pm 90^\circ$, so all $\cos$ terms vanish:

$$
\begin{aligned}
\frac{\partial P_2}{\partial \delta_2} &= (1.0000)(1.0000)(6.0000)\sin(90.00^\circ) + (1.0000)(1.0000)(6.0000)\sin(90.00^\circ) = 12.0000 \\
\frac{\partial P_2}{\partial \delta_3} &= -(1.0000)(1.0000)(6.0000)\sin(90.00^\circ) = -6.0000 \\
\frac{\partial P_2}{\partial |V_3|} &= (1.0000)(6.0000)\cos(90.00^\circ) = 0.0000 \\
\frac{\partial P_3}{\partial \delta_2} &= -(1.0000)(1.0000)(6.0000)\sin(90.00^\circ) = -6.0000 \\
\frac{\partial P_3}{\partial \delta_3} &= (1.0000)(1.0000)(6.0000)\sin(90.00^\circ) + (1.0000)(1.0000)(6.0000)\sin(90.00^\circ) = 12.0000 \\
\frac{\partial P_3}{\partial |V_3|} &= 2(1.0000)(12.0000)\cos(-90.00^\circ) + (1.0000)(6.0000)\cos(90.00^\circ) + (1.0000)(6.0000)\cos(90.00^\circ) = 0.0000 \\
\frac{\partial Q_3}{\partial \delta_2} &= -(1.0000)(1.0000)(6.0000)\cos(90.00^\circ) = 0.0000 \\
\frac{\partial Q_3}{\partial \delta_3} &= (1.0000)(1.0000)(6.0000)\cos(90.00^\circ) + (1.0000)(1.0000)(6.0000)\cos(90.00^\circ) = 0.0000 \\
\frac{\partial Q_3}{\partial |V_3|} &= -2(1.0000)(12.0000)\sin(-90.00^\circ) - (1.0000)(6.0000)\sin(90.00^\circ) - (1.0000)(6.0000)\sin(90.00^\circ) = 12.0000
\end{aligned}
$$

$$
[J^{0}] = \begin{bmatrix} 12 & -6 & 0 \\ -6 & 12 & 0 \\ 0 & 0 & 12 \end{bmatrix}
$$

#### (ii) Unknowns after the first iteration
Calculated powers at the flat start: $P_2^0 = P_3^0 = 0$ and $Q_3^0 = -[6 - 12 + 6] = 0$. Mismatches:

$$
\begin{aligned}
\Delta P_2 &= 0.6 - 0 = 0.6 \\
\Delta P_3 &= -0.5 - 0 = -0.5 \\
\Delta Q_3 &= -0.4 - 0 = -0.4
\end{aligned}
$$

$$
\begin{bmatrix} 0.6 \\ -0.5 \\ -0.4 \end{bmatrix} = \begin{bmatrix} 12 & -6 & 0 \\ -6 & 12 & 0 \\ 0 & 0 & 12 \end{bmatrix}\begin{bmatrix} \Delta\delta_2 \\ \Delta\delta_3 \\ \Delta|V_3| \end{bmatrix}
$$

From the first two rows: $12\Delta\delta_2 - 6\Delta\delta_3 = 0.6$ and $-6\Delta\delta_2 + 12\Delta\delta_3 = -0.5$, giving

$$
\begin{aligned}
\Delta\delta_2 &= \frac{0.6(12) + 6(-0.5)}{144 - 36} = \frac{4.2}{108} = 0.0389\ \text{rad} = 2.228^\circ \\
\Delta\delta_3 &= \frac{12(-0.5) + 6(0.6)}{108} = \frac{-2.4}{108} = -0.0222\ \text{rad} = -1.273^\circ \\
\Delta|V_3| &= -0.4/12 = -0.0333\ \text{pu}
\end{aligned}
$$

So $V_2^{1} = 1.0000\angle 2.228^\circ$ pu and $V_3^{1} = 0.9667\angle -1.273^\circ$ pu.

With these voltages ($P_i - jQ_i = V_i^{*}\sum_k Y_{ik}V_k$):

$$
\begin{aligned}
Q_2 &= 0.2154\ \text{pu} \\
P_1 &= -0.1044\ \text{pu}, \quad Q_1 = 0.2060\ \text{pu}
\end{aligned}
$$

**Answer:** $J^0$ as above; after one iteration $\delta_2 = 2.228^\circ$, $|V_3| = 0.9667$ pu, $\delta_3 = -1.273^\circ$, $Q_2 = 0.2154$ pu, $S_1 = -0.1044 + j0.2060$ pu.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is the Jacobian 3x3 for this system?
   A. Unknowns are delta2, delta3 and |V3|; PV bus 2 has no Q equation or |V| unknown
   B. Unknowns are all three angles
   C. Every bus contributes P and Q equations
   D. The slack bus adds one row
   Answer: A. Equations are P2, P3 and Q3 for unknowns delta2, delta3 and |V3|. (Key point: Jacobian size)

2. What is dQ3/d|V3| at the flat start?
   A. 12
   B. 6
   C. 24
   D. 0
   Answer: A. -2(1)(12)sin(-90) - 6 - 6 = 24 - 12 = 12. (Key point: Flat-start Jacobian)

3. Given 12 d2 - 6 d3 = 0.6 and -6 d2 + 12 d3 = -0.5, what is delta2 after the first iteration?
   A. About 0.0389 rad (2.23 deg)
   B. About -0.0222 rad
   C. About 0.05 rad
   D. About 0.0333 rad
   Answer: A. Solving gives d2 = 4.2/108 = 0.0389 rad, and d3 = -2.4/108 = -0.0222 rad. (Key point: Solving the correction)

4. Why is the Q3 mismatch equal to -0.4 at the flat start?
   A. Calculated Q3 = 0 so DeltaQ3 = -0.4 - 0
   B. Calculated Q3 = -0.4
   C. Specified Q3 is zero
   D. Bus 3 is a PV bus
   Answer: A. Calculated Q3 = -[6 - 12 + 6] = 0, so the mismatch equals the specified Q3 of -0.4 pu. (Key point: Mismatch at flat start)

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