---
title: "Two sources via 0 - j5 Ω: power flow"
subject: "Power System Analysis I"
chapter: 3
key: "semester_4/power_system_analysis_i:two-sources-0-j5"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/two-sources-0-j5
site: BCT Engineering
---

# Two sources via 0 - j5 Ω: power flow

Power System Analysis I · Chapter 3: Computational Technique

- Asked in: 2070 Magh, 6 marks
- 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/two-sources-0-j5
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3

## Question

Two ideal voltage sources designated as machines 1 and 2 are connected to each other via a line with impedance of 0 - j5 Ω. If E1 = 100∠0° V and E2 = 100∠30° V, determine the magnitude and the direction of power flow.

## Short answer

Current is 10.35∠15° A. Real power 1000 W flows from machine 1 to machine 2. Both machines absorb 267.95 var each; the capacitive line supplies 535.9 var.

## Answer

The line impedance $Z = 0 - j5\ \Omega$ is a pure **capacitive** reactance. Let I flow from machine 1 to machine 2. Then $S_1 = E_1I^*$ is the power supplied by machine 1 and $S_2 = E_2I^*$ is the power received by machine 2.

#### Current
$$
\begin{aligned}
E_1 - E_2 &= 100 - (86.60 + j50) = 13.40 - j50\\
I &= \frac{13.40 - j50}{-j5} = 10 + j2.679 = 10.353\angle15^\circ\ \text{A}
\end{aligned}
$$

#### Powers
$$
\begin{aligned}
S_1 &= E_1I^* = 100(10 - j2.679) = 1000 - j267.95\ \text{VA}\\
S_2 &= E_2I^* = (100\angle30^\circ)(10.353\angle-15^\circ) = 1000 + j267.95\ \text{VA}\\
S_Z &= |I|^2Z = 107.18(-j5) = -j535.90\ \text{VA}
\end{aligned}
$$

Check with $P = \frac{|E_1||E_2|}{X}\sin(\delta_1 - \delta_2)$ using $X = -5$: $P = \frac{100\times100}{-5}\sin(-30^\circ) = +1000$ W, from 1 to 2.

#### Magnitude and direction
| Quantity | Result |
|---|---|
| Real power | 1000 W flows from **machine 1 to machine 2** (1 generates, 2 consumes) |
| Machine 1 reactive | $Q_1 = -267.95$ var, so it **absorbs** 267.95 var |
| Machine 2 reactive | receives +267.95 var, so it **absorbs** 267.95 var |
| Capacitive line | generates 535.90 var (= 267.95 + 267.95) |

Note: with an inductive link, real power flows from the leading machine (2) to the lagging one. A capacitive link reverses this, so power flows from machine 1 to machine 2 even though $E_2$ leads. The line capacitance supplies reactive power equally to both ends, because the voltage magnitudes are equal.

**Answer:** P = 1000 W flows from machine 1 to machine 2. Each machine absorbs 267.95 var, all supplied by the capacitive line (535.9 var).

## Check yourself

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

1. What does Z = 0 - j5 Ω represent?
   A. A pure inductive reactance
   B. A pure capacitive reactance
   C. A pure resistance
   D. A short circuit
   Answer: B. Negative reactance is capacitive. (Key point: Nature of the link)

2. Which way does real power flow?
   A. From machine 1 to machine 2, even though E2 leads
   B. From machine 2 to machine 1 because E2 leads
   C. None, because the link is lossless
   D. From 2 to 1 because it is capacitive
   Answer: A. With a capacitive link the usual leading-to-lagging rule is reversed; P = 100x100/(-5) x sin(-30°) = +1000 W. (Key point: Direction of P)

3. What happens to reactive power?
   A. Each machine supplies 267.95 var to an inductive line
   B. Machine 1 supplies all the vars
   C. No reactive power flows
   D. Each machine absorbs 267.95 var, supplied by the capacitive line (535.9 var)
   Answer: D. S1 = 1000 - j267.95 and S2 = 1000 + j267.95 (received); the line generates 535.9 var. (Key point: Reactive power)

4. Why is the real power sent equal to that received (1000 W)?
   A. Because machine voltages are equal
   B. Because the angle is 30°
   C. The link has no resistance
   D. Because I is 10 A
   Answer: C. With R = 0, no I^2R loss exists. (Key point: Real power loss)

## Related in Power System Analysis I

- [Two sources 120∠10° and 120∠30° V via 1+j5 Ω](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Atwo-sources-120v-1-j5) — Power System Analysis I, chapter 3
- [Two machines 100∠0° and 120∠30° V via j5 Ω](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Atwo-machines-100v-120v-j5) — Power System Analysis I, chapter 3
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- [Two machines 11∠0° and 11∠40° kV via j15 Ω](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Atwo-machines-11kv-j15) — Power System Analysis I, chapter 3

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