---
title: "Two machines 100∠0° and 120∠30° V via j5 Ω"
subject: "Power System Analysis I"
chapter: 3
key: "semester_4/power_system_analysis_i:two-machines-100v-120v-j5"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/two-machines-100v-120v-j5
site: BCT Engineering
---

# Two machines 100∠0° and 120∠30° V via j5 Ω

Power System Analysis I · Chapter 3: Computational Technique

- Asked in: 2071 Magh, 5 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-machines-100v-120v-j5
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3

## Question

Two ideal 3-phase synchronous machines A and B are connected by a series impedance of 0 + j5 Ω per phase. The voltages at the two machines are 100∠0° and 120∠30° V respectively. Calculate and analyze the direction of power flow in the system.

## Short answer

Current I = 12.03∠176.3° A. Real power 1200 W per phase flows from B to A (B generates, A motors). Reactive power: B supplies 801.5 var, line absorbs 723.1 var, A receives 78.5 var per phase.

## Answer

Take the voltages as per-phase values. Let I flow from A to B. Then $S_A = V_AI^*$ is the power supplied by A and $S_B = V_BI^*$ is the power received by B.

#### Current
$$
\begin{aligned}
V_A &= 100\angle0^\circ = 100\ \text{V}\\
V_B &= 120\angle30^\circ = 103.92 + j60\ \text{V}\\
I &= \frac{V_A - V_B}{j5} = \frac{-3.92 - j60}{j5} = -12.0 + j0.785\\
&= 12.026\angle176.26^\circ\ \text{A}
\end{aligned}
$$

#### Powers per phase
$$
\begin{aligned}
S_A &= V_AI^* = 100(-12.0 - j0.785) = -1200 - j78.46\ \text{VA}\\
S_B &= V_BI^* = -1200 - j801.54\ \text{VA}\\
S_{link} &= |I|^2(j5) = 144.62\times j5 = j723.08\ \text{VA}
\end{aligned}
$$

Check with the lossless-line formula: $P = \frac{|V_A||V_B|}{X}\sin(\delta_A - \delta_B) = \frac{100\times120}{5}\sin(-30^\circ) = -1200$ W.

#### Analysis
| Machine | Real power | Reactive power |
|---|---|---|
| A | supplies −1200 W, so it **receives 1200 W** (motor) | supplies −78.46 var, so it **receives 78.46 var** |
| B | receives −1200 W, so it **generates 1200 W** | receives −801.54 var, so it **supplies 801.54 var** |
| Link (j5 Ω) | 0 W (no resistance) | absorbs 723.08 var |

- **Real power** flows from B to A because $V_B$ leads $V_A$ by 30°. There is no resistance, so A receives exactly what B sends.
- **Reactive power** flows from B (higher voltage, 120 V) to A (lower voltage, 100 V). B supplies 801.54 var: 723.08 var is used in the link and 78.46 var reaches A.
- Three-phase totals (×3): 3600 W from B to A; B supplies 2404.6 var; the link absorbs 2169.2 var; A receives 235.4 var.

**Answer:** B is the generator and sends 1200 W per phase (3.6 kW total) to A, which acts as a motor. Q flows from B to A: B supplies 801.5 var, the link uses 723.1 var, and A receives 78.5 var (per phase).

## Check yourself

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

1. Which direction does real power flow?
   A. From A to B since A has the lower voltage
   B. From B to A, because VB leads VA by 30°
   C. From A to B since VA is the reference
   D. No real power flows since the link is purely reactive
   Answer: B. A purely reactive link still transfers P when there is an angle difference; it flows from leading to lagging. (Key point: Direction of real power)

2. Which formula checks the real power?
   A. P = |VA||VB| sin(delta)/X = 100x120/5 x sin(30°) = 1200 W per phase
   B. P = (|VA| - |VB|)/X = 4 W
   C. P = |VA||VB| cos(delta)/X
   D. P = |VA||VB| X
   Answer: A. For a lossless link P = V1V2 sin(delta)/X. (Key point: Real power magnitude)

3. In which direction does reactive power flow?
   A. From A to B
   B. From A to B because A is the reference
   C. There is none
   D. From B (120 V) to A (100 V), the higher voltage magnitude to the lower
   Answer: D. Q flows from the higher voltage magnitude to the lower; B supplies 801.54 var. (Key point: Reactive power direction)

4. What does the link absorb?
   A. Real power of 1200 W
   B. Nothing since it is ideal
   C. Q = |I|^2 X = 723.08 var and no real power
   D. Q = 801.54 var
   Answer: C. The link has no resistance so P loss is zero; I^2X is absorbed. (Key point: Link reactive loss)

## Related in Power System Analysis I

- [Two sources via 0 - j5 Ω: power flow](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Atwo-sources-0-j5) — Power System Analysis I, chapter 3
- [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
- [Reactive power flow between machines via j10 Ω](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Atwo-machines-reactive-j10) — Power System Analysis I, chapter 3
- [Two machines 3.81∠0° and 3.81∠-18° kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Atwo-machines-3-81kv) — Power System Analysis I, chapter 3
- [Active power between machines, 200∠0° and 200∠15° V](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Atwo-machines-200v-15deg) — Power System Analysis I, chapter 3
- [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

## Related in other subjects

- [Series capacitor to limit power angle to 30°](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aseries-capacitor-30-degree) — Power System Analysis II, chapter 6