---
title: "Two machines 3.81∠0° and 3.81∠-18° kV"
subject: "Power System Analysis I"
chapter: 3
key: "semester_4/power_system_analysis_i:two-machines-3-81kv"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/two-machines-3-81kv
site: BCT Engineering
---

# Two machines 3.81∠0° and 3.81∠-18° kV

Power System Analysis I · Chapter 3: Computational Technique

- Asked in: 2069 Poush, 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-machines-3-81kv
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3

## Question

Two 50 Hz, 3-phase synchronous machines are connected to each other via a line with series impedance of 3 + j10 Ohm/phase. Determine the direction of active and reactive power flow in the line and power generated/consumed by each of these machines. The terminal phase voltages of machines 1 and 2 are 3.81∠0° kV and 3.81∠-18° kV respectively.

## Short answer

Use I=(V1-V2)/Z; active power flows from machine 1 to machine 2. Three-phase: machine 1 generates 1293.3 kW, machine 2 consumes 1175.9 kW, line loss 117.3 kW. Reactive power flows from machine 2 to machine 1: machine 2 supplies 565.9 kvar, machine 1 receives 174.8 kvar, line absorbs 391.1 kvar.

## Answer

Let I flow from machine 1 to machine 2 through $Z = 3 + j10\ \Omega$. Then $S_1 = V_1I^*$ is the power **delivered** by machine 1 and $S_2 = V_2I^*$ is the power **received** by machine 2 (per phase; three-phase = 3 × per phase).

#### Current
$$
\begin{aligned}
V_1 &= 3810\angle0^\circ\ \text{V}\\
V_2 &= 3810\angle-18^\circ = 3623.53 - j1177.35\ \text{V}\\
I &= \frac{V_1 - V_2}{Z} = \frac{186.47 + j1177.35}{3 + j10}\\
&= 113.15 + j15.30 = 114.18\angle7.70^\circ\ \text{A}
\end{aligned}
$$

#### Powers per phase
$$
\begin{aligned}
S_1 &= V_1I^* = 3810(113.15 - j15.30) = 431.09 - j58.28\ \text{kVA}\\
S_2 &= V_2I^* = (3810\angle-18^\circ)(114.18\angle-7.70^\circ) = 391.98 - j188.64\ \text{kVA}\\
S_{line} &= |I|^2Z = 13036.1(3 + j10) = 39.11 + j130.36\ \text{kVA}
\end{aligned}
$$

Check: $431.09 - 391.98 = 39.11$ kW and $188.64 - 58.28 = 130.36$ kvar.

#### Results
| Quantity | Per phase | Three-phase |
|---|---|---|
| Machine 1, P | **generates** 431.09 kW | 1293.26 kW |
| Machine 1, Q | $Q_1 = -58.28$, so it **absorbs** 58.28 kvar | 174.84 kvar |
| Machine 2, P | **consumes** 391.98 kW (motor) | 1175.94 kW |
| Machine 2, Q | receives −188.64, so it **generates** 188.64 kvar | 565.92 kvar |
| Line loss | 39.11 kW, 130.36 kvar | 117.33 kW, 391.08 kvar |

#### Direction of flow
- **Active power** flows from machine 1 to machine 2, because $V_1$ leads $V_2$ by 18°.
- **Reactive power** flows from machine 2 towards machine 1. Machine 2 supplies 188.64 kvar: 130.36 kvar is used in the line reactance and 58.28 kvar reaches machine 1. Both voltage magnitudes are equal, so the flow of Q depends on the angle and on the line resistance.

**Answer:** Machine 1 is the generator (1293.3 kW, three-phase) and machine 2 the motor (1175.9 kW). The line loss is 117.3 kW. Q flows from machine 2 (565.9 kvar) to machine 1 (174.8 kvar received). The line absorbs 391.1 kvar.

## Check yourself

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

1. Which way does active power flow?
   A. From 2 to 1
   B. From machine 1 to 2, since V1 leads V2 by 18°
   C. None, magnitudes are equal
   D. From 2 to 1 as the line has resistance
   Answer: B. Real power flows from the leading to the lagging machine. (Key point: Direction of P)

2. Which way does reactive power flow, even though magnitudes are equal?
   A. From machine 2 towards machine 1, because of the angle and the line resistance
   B. From 1 to 2
   C. None, because the magnitudes are equal
   D. Reactive power is zero
   Answer: A. S1 = 431.09 - j58.28 kVA shows machine 1 absorbs vars; machine 2 generates 188.64 kvar. (Key point: Reactive power direction)

3. How are three-phase values obtained?
   A. Multiply by sqrt3
   B. Divide by 3
   C. Use as is
   D. Multiply the per-phase result by 3
   Answer: D. Voltages are phase values. (Key point: Total vs per-phase)

## Related in Power System Analysis I

- [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
- [Direction of active and reactive power flow](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Adirection-active-reactive-flow) — Power System Analysis I, chapter 6
- [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
- [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 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 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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