---
title: "Source P and Q, Ean = -120∠210° V"
subject: "Power System Analysis I"
chapter: 3
key: "semester_4/power_system_analysis_i:source-power-ean-120-210"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/source-power-ean-120-210
site: BCT Engineering
---

# Source P and Q, Ean = -120∠210° V

Power System Analysis I · Chapter 3: Computational Technique

- Asked in: 2082 Kartik (new course), 4 marks
- Video: [44 - Power in AC Circuits 1 | Power Triangle - Apparent, Real and Reactive Power](https://www.youtube.com/watch?v=LzjwM0DF5wg) — SkanCity Academy
- 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/source-power-ean-120-210
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3

## Question

A voltage source Ean = -120∠210° V and the current through the source Ina = 10∠60° A. Find the values of real power (P) and reactive power (Q) and state whether the source is delivering or receiving each.

## Short answer

Simplify the voltage to 120∠30° V, then S = (120∠30°)(10∠-60°) = 1200∠-30° = 1039.2 - j600 VA. The source delivers 1039.2 W and absorbs 600 var (receives reactive power).

## Answer

Use $S = EI^*$. The current $I_{na}$ flows from n to a inside the source, so it leaves the positive terminal a. A positive $P$ or $Q$ therefore means the source is delivering that power.

#### Simplify the voltage
$$
E_{an} = -120\angle210^\circ = 120\angle(210^\circ - 180^\circ) = 120\angle30^\circ\ \text{V}
$$

#### Complex power
$$
\begin{aligned}
S &= E_{an}I_{na}^* = (120\angle30^\circ)(10\angle-60^\circ)\\
&= 1200\angle-30^\circ\\
&= 1200(\cos30^\circ - j\sin30^\circ)\\
&= 1039.2 - j600\ \text{VA}
\end{aligned}
$$

#### Interpretation
| Quantity | Value | Meaning |
|---|---|---|
| $P$ | 1039.2 W | Positive, so the source **delivers** real power |
| $Q$ | −600 var | Negative, so the source **receives (absorbs)** 600 var |

The current leads the voltage by 30°, so the source sees a leading (capacitive) load. That load returns reactive power to the source.

**Answer:** $P = 1039.2$ W, delivered by the source. $Q = 600$ var, absorbed by the source (delivered $Q = -600$ var).

## Check yourself

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

1. -120 angle 210 degrees equals
   A. 120 angle 30 degrees
   B. 120 angle 390 degrees
   C. 120 angle -30 degrees
   D. 120 angle 150 degrees
   Answer: A. Negation subtracts 180 degrees: 210 - 180 = 30. (Key point: Simplify E)

2. Which expression is used for S?
   A. E I*
   B. E* I
   C. E I
   D. E/I
   Answer: A. S = E I* with current leaving the positive terminal. (Key point: Formula)

3. S = (120 angle 30)(10 angle -60) =
   A. 1039.2 - j600 VA
   B. 1039.2 + j600 VA
   C. 600 - j1039.2 VA
   D. 1200 angle 90 VA
   Answer: A. S = 1200 angle -30 = 1039.2 - j600. (Key point: Result)

4. The negative Q means the source
   A. Delivers 600 var
   B. Receives 600 var
   C. Delivers P only as Q
   D. Has no reactive power
   Answer: B. Positive means delivering, so negative Q means the source absorbs 600 var. (Key point: Interpretation)

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