---
title: "p.u. impedance diagram: G, T1, line, T2, M"
subject: "Power System Analysis I"
chapter: 3
key: "semester_4/power_system_analysis_i:pu-impedance-diagram-20mva"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/pu-impedance-diagram-20mva
site: BCT Engineering
---

# p.u. impedance diagram: G, T1, line, T2, M

Power System Analysis I · Chapter 3: Computational Technique

- Asked in: 2072 Asoj, 4 marks
- Video: [Problem #2 | Impedance & Reactance diagram | PSA](https://www.youtube.com/watch?v=Mq3pAkxuIjc) — FOR STUDENTS
- 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/pu-impedance-diagram-20mva
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3

## Question

Draw the p.u. impedance diagram for the system shown below. [Figure: G (6.6 kV, 20 MVA, X = 0.15 p.u.) → T1 (11/132 kV, 25 MVA, X = 0.1 p.u.) → line 10 + j60 Ω → T2 (132/11 kV, 20 MVA, X = 0.1) → motor M (11 kV, 15 MVA, X = 0.15 p.u.).]

## Short answer

Use 20 MVA base with base voltages 6.6 kV, 79.2 kV, 6.6 kV. Per-unit values are XG=0.15, XT1=0.2222, Zline=0.0319+j0.1913, XT2=0.2778, XM=0.5556. Draw the series impedance diagram from generator to motor with a common reference bus.

## Answer

Choose the generator rating as base: **20 MVA, 6.6 kV** in the generator circuit. (The 6.6 kV generator feeds the 11 kV winding of T1, so the transformer and motor values need voltage correction.)

#### Base voltages
- Generator zone: $V_{b1} = 6.6$ kV
- Line zone (T1 11/132 kV): $V_{b2} = 6.6\times\frac{132}{11} = 79.2$ kV
- Motor zone (T2 132/11 kV): $V_{b3} = 79.2\times\frac{11}{132} = 6.6$ kV

$$
Z_{b2} = \frac{79.2^2}{20} = 313.63\ \Omega
$$

#### Per-unit values
$$
\begin{aligned}
X_G &= 0.15\ \text{pu}\\
X_{T1} &= 0.1\times\frac{20}{25}\times\left(\frac{11}{6.6}\right)^2 = 0.2222\ \text{pu}\\
Z_{line} &= \frac{10 + j60}{313.63} = 0.0319 + j0.1913\ \text{pu}\\
X_{T2} &= 0.1\times\frac{20}{20}\times\left(\frac{132}{79.2}\right)^2 = 0.2778\ \text{pu}\\
X_M &= 0.15\times\frac{20}{15}\times\left(\frac{11}{6.6}\right)^2 = 0.5556\ \text{pu}
\end{aligned}
$$

#### Impedance diagram
```text
     j0.2222  0.0319+j0.1913  j0.2778
 +---/\/\/-------/\/\/--------/\/\/---+
 |    T1          line          T2    |
j0.15                              j0.5556
 |                                    |
(Eg)                                 (Em)
 |                                    |
 +------------------------------------+
            reference bus
```

**Answer:** On 20 MVA (base 6.6 / 79.2 / 6.6 kV): $X_G = 0.15$, $X_{T1} = 0.2222$, $Z_{line} = 0.0319 + j0.1913$, $X_{T2} = 0.2778$, $X_M = 0.5556$ pu.

## Check yourself

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

1. What base is used and why?
   A. 25 MVA, 11 kV
   B. 20 MVA, 6.6 kV from generator rating
   C. 15 MVA, 11 kV
   D. 100 MVA
   Answer: B. Generator rating as base in generator circuit. (Key point: Base)

2. Why is X_T1 multiplied by (11/6.6)^2?
   A. T1 11 kV winding connects to 6.6 kV base zone
   B. T1 is 25 MVA
   C. Line impedance
   D. Motor
   Answer: A. Rated 11 kV differs from the base 6.6 kV. (Key point: T1 correction)

3. What is Zb in the line zone?
   A. 79.2^2/20 = 313.63 Ω
   B. 132^2/20
   C. 6.6^2/20
   D. 11^2/25
   Answer: A. Vb = 6.6 x 132/11 = 79.2 kV. (Key point: Line zone)

4. X_M value?
   A. 0.15
   B. 0.2
   C. 0.5556
   D. 0.2778
   Answer: C. 0.15 x 20/15 x (11/6.6)^2 = 0.5556. (Key point: Motor reactance)

## Related in Power System Analysis I

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