---
title: "Impedance and p.u. reactance diagram, 5000 VA base"
subject: "Power System Analysis I"
chapter: 3
key: "semester_4/power_system_analysis_i:impedance-diagram-5000va-250v"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/impedance-diagram-5000va-250v
site: BCT Engineering
---

# Impedance and p.u. reactance diagram, 5000 VA base

Power System Analysis I · Chapter 3: Computational Technique

- Asked in: 2082 Kartik (new course), 2+6 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/impedance-diagram-5000va-250v
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3

## Question

Figure below shows a single line diagram of a power system. Draw its impedance diagram without showing their values. Also, redraw the circuit with per unit values in the reactance diagram. Take 5,000 VA base and common system base voltage of 250 V. [Figure: generators G1 (2000 VA, 250 V, Z = j0.2 p.u.) and G2 (2000 VA, 250 V, Z = j0.3 p.u.) in parallel on a common bus, feeding transformer T1 (4000 VA, 250/800 V, Z = j0.2 p.u.), then a line Z = 40 + j150 Ω, then transformer T2 (8000 VA, 1000/500 V, Z = j0.06 p.u.) supplying a load.]

## Short answer

With 5 kVA base and base voltages 250 V (gen), 800 V (line), 400 V (load), the per-unit reactances are: G1 j0.5, G2 j0.75, T1 j0.25, line 0.3125+j1.1719, T2 j0.0586. Magnetizing branches are neglected.

## Answer

#### (a) Impedance diagram (no values)
Each generator is shown as an EMF behind its internal impedance. Each transformer is shown as series leakage impedance with a shunt magnetising branch. The line is a series $R + jX$ (short line), and the load is an impedance.

```text
  G1        T1                  T2
 (~)-Zg1-+-Zt1-+--R+jX line--+-Zt2-+-- Load
         |     |             |     |   Z_L
 (~)-Zg2-+    Ym             Ym     |
  G2     |     |             |     |
 ref ----+-----+-------------+-----+---
```

#### (b) Selecting the bases
- $S_b = 5000$ VA for the whole system.
- Generator section: $V_b = 250$ V.
- Line section: through T1 (250/800 V), $V_b = 250\times800/250 = 800$ V.
- Load section: through T2 (rated 1000 V on the line side), $V_b = 800\times500/1000 = 400$ V.

#### Per-unit values
Use $Z_{new} = Z_{old}\,\dfrac{S_{b,new}}{S_{old}}\left(\dfrac{V_{old}}{V_{b,new}}\right)^2$.

$$
\begin{aligned}
G_1&: j0.2\times\frac{5000}{2000}\times1 = j0.5\\
G_2&: j0.3\times\frac{5000}{2000} = j0.75\\
T_1&: j0.2\times\frac{5000}{4000}\times\left(\frac{250}{250}\right)^2 = j0.25
\end{aligned}
$$

Line: the base impedance is

$$
Z_b = \frac{800^2}{5000} = 128\ \Omega
$$

$$
Z_{line} = \frac{40 + j150}{128} = 0.3125 + j1.1719\ \text{p.u.}
$$

T2 (rated 1000 V, while the base on its side is 800 V):

$$
j0.06\times\frac{5000}{8000}\times\left(\frac{1000}{800}\right)^2 = j0.0586
$$

#### Reactance (per-unit) diagram
```text
  j0.5         j0.25   0.3125+j1.172   j0.0586
 (G1)--+------/\/\---------/\/\--------/\/\----Load
       |                                       |
 (G2)--+ j0.75                                 |
       |                                       |
 ------+------------ reference ----------------+
```

| Element | p.u. value |
|---|---|
| G1 | j0.50 |
| G2 | j0.75 |
| T1 | j0.25 |
| Line | 0.3125 + j1.1719 |
| T2 | j0.0586 |

The magnetising branches are neglected in the reactance diagram. The load impedance is not given, so it is shown as a block. Its p.u. value would be $Z_L/Z_b$ with $Z_b = 400^2/5000 = 32\ \Omega$.

## Check yourself

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

1. The base voltage in the line section, through T1 (250/800 V), is
   A. 250 V
   B. 800 V
   C. 400 V
   D. 1000 V
   Answer: B. V_b = 250 x 800/250 = 800 V. (Key point: Line section base voltage)

2. The load section base voltage via T2 (1000/500 V) is
   A. 800 V
   B. 500 V
   C. 400 V
   D. 250 V
   Answer: C. V_b = 800 x 500/1000 = 400 V. (Key point: Load section base voltage)

3. G1 (2000 VA, j0.2) on 5000 VA base is
   A. j0.08
   B. j0.5
   C. j0.2
   D. j1.25
   Answer: B. 0.2 x 5000/2000 = 0.5. (Key point: G1 p.u.)

4. The line base impedance is 800^2/5000 =
   A. 128 ohm
   B. 32 ohm
   C. 12.8 ohm
   D. 1280 ohm
   Answer: A. Z_b = 128 ohm, giving (40+j150)/128 = 0.3125 + j1.1719. (Key point: Line p.u.)

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