---
title: "Reactance diagram: G, T1, j18 Ω line, T2, M"
subject: "Power System Analysis I"
chapter: 3
key: "semester_4/power_system_analysis_i:reactance-diagram-g-t1-t2-m"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/reactance-diagram-g-t1-t2-m
site: BCT Engineering
---

# Reactance diagram: G, T1, j18 Ω line, T2, M

Power System Analysis I · Chapter 3: Computational Technique

- Asked in: 2068 Bhadra, 5 marks
- Video: [REACTANCE DIAGRAM](https://www.youtube.com/watch?v=Cp6K6JkhM3g) — Wisdom is Power
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/reactance-diagram-g-t1-t2-m
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3

## Question

Draw a reactance diagram of the electric power system given below and express all reactances in per unit system by choosing appropriate base values. G1: 14 kV, 30 MVA, 10%; M: 11 kV, 25 MVA, 13%; T1: 66/11 kV, 25 MVA, 12%; T1: 13.8/66 kV, 25 MVA, 12% (as printed; one of these is T2). [Figure: G → T1 → line j18 Ohms/phase → T2 → motor M.]

## Short answer

Using 25 MVA and 13.8/66/11 kV bases: generator reactance 0.0858 pu, both transformers 0.12 pu, line j18 ohm gives 0.1033 pu, and motor 0.13 pu. Draw a single loop from generator through T1, line, T2 to motor with these reactances to neutral.

## Answer

The two transformers are read as **T1: 13.8/66 kV** (at the generator) and **T2: 66/11 kV** (at the motor), both 25 MVA, 12 %.

**Choice of base:** 25 MVA (transformer rating), 66 kV on the line. Then the base is 13.8 kV in the generator zone and 11 kV in the motor zone.

$$
Z_{b,line} = \frac{66^2}{25} = 174.24\ \Omega
$$

#### Per-unit reactances
$$
\begin{aligned}
X_G &= 0.10\times\frac{25}{30}\times\left(\frac{14}{13.8}\right)^2 = 0.0858\ \text{pu}\\
X_{T1} &= 0.12\ \text{pu (rating = base)}\\
X_{line} &= \frac{18}{174.24} = 0.1033\ \text{pu}\\
X_{T2} &= 0.12\ \text{pu}\\
X_M &= 0.13\times\frac{25}{25}\times\left(\frac{11}{11}\right)^2 = 0.13\ \text{pu}
\end{aligned}
$$

| Element | Base kV | X (pu, 25 MVA) |
|---|---|---|
| G (30 MVA, 14 kV) | 13.8 | 0.0858 |
| T1 | 13.8/66 | 0.1200 |
| Line j18 Ω | 66 | 0.1033 |
| T2 | 66/11 | 0.1200 |
| Motor M | 11 | 0.1300 |

#### Reactance diagram
```text
      j0.12    j0.1033    j0.12
 +----/\/\/----/\/\/------/\/\/----+
 |     T1       line       T2      |
j0.0858                          j0.13
 |                                 |
(Eg)                              (Em)
 |                                 |
 +---------------------------------+
            reference bus
```

**Answer:** On 25 MVA (13.8 / 66 / 11 kV): $X_G = 0.0858$, $X_{T1} = 0.12$, $X_{line} = 0.1033$, $X_{T2} = 0.12$, $X_M = 0.13$ pu.

## Check yourself

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

1. Which bases are used?
   A. 30 MVA with 14 kV everywhere
   B. 25 MVA with 66 kV on the line, 13.8 kV generator side and 11 kV motor side
   C. 25 MVA with 11 kV everywhere
   D. 66 MVA with 25 kV
   Answer: B. Transformer rating gives MVA base; its ratios give zones. (Key point: Base selection)

2. Why is X_G = 0.0858 pu?
   A. 0.10 x 25/30 x (14/13.8)^2
   B. 0.10 x 30/25
   C. 0.10 unchanged
   D. 0.10 x (13.8/14)^2 x 30/25
   Answer: A. Convert MVA and voltage. (Key point: Generator conversion)

3. What is the line pu reactance?
   A. 18/66
   B. 18/25
   C. 18 x 174.24
   D. 18/174.24 = 0.1033 pu
   Answer: D. Zb = 66^2/25 = 174.24. (Key point: Line reactance)

## Related in Power System Analysis I

- [Reactance diagram: 30 MVA, 6.6 kV base, three lines](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-30mva-g1-motor) — Power System Analysis I, chapter 3
- [Per unit circuit of three-generator system, 30 MVA base](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Apu-circuit-30mva-g1-g2-g3) — Power System Analysis I, chapter 3
- [Reactance diagram: three generators, 30 MVA 6.6 kV base](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-30mva-three-generators) — Power System Analysis I, chapter 3
- [Reactance diagram: 50 MVA, 13.8 kV base, three generators](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-50mva-13-8kv) — Power System Analysis I, chapter 3
- [Per unit values: 100 MVA generator, two motors](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aper-unit-100mva-6kv-two-motors) — Power System Analysis I, chapter 3
- [Reactance diagram: two generators, two motors, j50 Ω line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-two-gens-two-motors) — Power System Analysis I, chapter 3

## Related in other subjects

- [30 MVA salient pole motor maximum power](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Asalient-motor-30mva-max-power) — Electric Machines II, chapter 2, asked 3×
- [Generator-reactor system fault MVA](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Agenerator-reactor-fault-mva) — Power System Analysis II, chapter 3
- [Fault at receiving end of 132 kV lines](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afault-132kv-receiving-end-2071) — Power System Analysis II, chapter 3
- [Short circuit MVA of feeder circuit breaker](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afeeder-cb-short-circuit-mva) — Power System Analysis II, chapter 3, asked 2×