---
title: "G-S voltage and power flow, 2-bus"
subject: "Power System Analysis II"
chapter: 2
key: "semester_5/power_system_analysis_ii:gs-two-bus-power-flow-new"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/gs-two-bus-power-flow-new
site: BCT Engineering
---

# G-S voltage and power flow, 2-bus

Power System Analysis II · Chapter 2: Load Flow Analysis

- Asked in: 2082 Bhadra (new course), 4+3 marks
- Video: [Gauss Seidal Method Solved Problem -2 |  GS Method | Power flow Analysis | Power System Analysis](https://www.youtube.com/watch?v=MQBtptz5VaY) — Simple n Easy Learning
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/gs-two-bus-power-flow-new
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-2

## Question

Using G-S load flow analysis, compute the voltage at bus 2 of figure shown below if V1 = 1∠0° pu. Find also power flow between bus 1 and 2. (all values are in pu). [Figure: bus 1 with source V1 and load Sd1; line Z = j0.75 between buses 1 and 2; bus 2 with load Sd2 = 0.5 + j1.0 and injection S2 = j1.0 (shunt capacitor)]

## Short answer

Net bus 2 injection is –0.5 + j0 pu. Gauss-Seidel converges to V2 = 0.9115∠–24.29° pu. Power flow from bus 1 to 2: S12 = 0.500 + j0.226 pu.

## Answer

Bus 1 is the slack bus ($V_1 = 1\angle 0^\circ$). At bus 2 the net injected power is the capacitor injection minus the load:

$$
S_2 = P_2 + jQ_2 = j1.0 - (0.5 + j1.0) = -0.5 + j0\ \text{pu}
$$

So bus 2 is a PQ bus with $P_2 = -0.5$, $Q_2 = 0$. The load $S_{d1}$ at bus 1 does not affect $V_2$.

#### Y-bus
$y_{12} = 1/j0.75 = -j1.3333$ pu:

$$
Y_{bus} = \begin{bmatrix} -j1.3333 & j1.3333 \\ j1.3333 & -j1.3333 \end{bmatrix}
$$

#### Gauss-Seidel equation for bus 2
$$
\begin{aligned}
V_2^{(k+1)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{(k)*}} - Y_{21}V_1\right] \\
&= \frac{1}{-j1.3333}\left[\frac{-0.5}{V_2^{(k)*}} - j1.3333\right] \\
&= 1 - \frac{j0.375}{V_2^{(k)*}}
\end{aligned}
$$

**Iteration 1** (flat start $V_2^{(0)} = 1\angle 0^\circ$):

$$
V_2^{(1)} = 1 - \frac{j0.375}{1} = 1.0000 - j0.3750 = 1.0680\angle -20.556^\circ
$$

**Iteration 2:**

$$
V_2^{(2)} = 1 - \frac{j0.375}{1.0000 + j0.3750} = 0.8767 - j0.3288 = 0.9363\angle -20.556^\circ
$$

Continuing until the change is below $10^{-4}$ pu:

| Iteration $k$ | $V_2^{(k)}$ (rectangular) | $V_2^{(k)}$ (polar) |
|---|---|---|
| 1 | 1.0000 - j0.3750 | $1.0680\angle -20.556^\circ$ |
| 2 | 0.8767 - j0.3288 | $0.9363\angle -20.556^\circ$ |
| 3 | 0.8594 - j0.3750 | $0.9376\angle -23.575^\circ$ |
| 4 | 0.8400 - j0.3666 | $0.9165\angle -23.575^\circ$ |
| 5 | 0.8364 - j0.3750 | $0.9166\angle -24.150^\circ$ |
| 6 | 0.8326 - j0.3733 | $0.9125\angle -24.150^\circ$ |
| ... | ... | ... |
| 10 | 0.8308 - j0.3749 | $0.9115\angle -24.289^\circ$ |
| 11 | 0.8308 - j0.3750 | $0.9115\angle -24.294^\circ$ |

$$
V_2 = 0.9115\angle -24.29^\circ\ \text{pu}
$$

**Check (exact solution):** with $Q_2 = 0$ and a lossless line, $|V_2| = \cos\delta_2$ and $P_2 = 1.3333|V_2|\sin\delta_2 = -0.5$, so $\sin 2\delta_2 = -0.75$, giving $\delta_2 = -24.30^\circ$ and $|V_2| = 0.9114$ pu, which matches.

#### Power flow between buses 1 and 2
$$
\begin{aligned}
I_{12} &= \frac{V_1 - V_2}{j0.75} = 0.5000 - j0.2256 = 0.5486\angle -24.29^\circ\ \text{pu} \\
S_{12} &= V_1I_{12}^{*} = 0.5000 + j0.2256\ \text{pu} \\
S_{21} &= V_2(-I_{12})^{*} = -0.5000\ \text{pu}
\end{aligned}
$$

Line loss $= S_{12} + S_{21} = j0.2257$ pu (reactive only, as the line has no resistance).

**Answer:** $V_2 \approx 0.911\angle -24.29^\circ$ pu; power sent from bus 1 to bus 2 is $S_{12} = 0.500 + j0.226$ pu, and bus 2 receives $0.5$ pu real power with zero reactive power.

## Check yourself

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

1. What is the net injected power at bus 2?
   A. S2 = j1.0 - (0.5 + j1.0) = -0.5 pu
   B. S2 = 0.5 + j1.0 + j1.0
   C. S2 = 0.5 + j2.0
   D. S2 = -0.5 - j2.0
   Answer: A. The capacitor injects +j1.0 and the load draws 0.5 + j1.0, leaving -0.5 + j0 so bus 2 is PQ. (Key point: Net bus 2 injection)

2. What does the update formula for V2 reduce to?
   A. V2 = 1 - j0.375 / V2*
   B. V2 = 1 + j0.375 V2
   C. V2 = 0.5 / V2*
   D. V2 = 1 - 0.375 V2*
   Answer: A. (1/-j1.3333)[-0.5/V2* - j1.3333] = 1 - j0.375/V2*, since 0.5/1.3333 = 0.375. (Key point: G-S equation)

3. What are the converged voltage and the nature of the line loss?
   A. V2 = 0.9115 at -24.29 deg; loss is reactive only (j0.2257)
   B. V2 = 1.0 at 0 deg; no loss
   C. V2 = 1.068 at -20.6 deg; real loss
   D. V2 = 0.9115 at +24.29 deg; real loss
   Answer: A. The line has only reactance so the loss is purely reactive, and the converged V2 matches the exact solution 0.9114 at -24.30 deg. (Key point: Result and loss)

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