---
title: "P and Q of a 100 V single-phase source"
subject: "Power System Analysis I"
chapter: 3
key: "semester_4/power_system_analysis_i:single-phase-source-power-100v"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/single-phase-source-power-100v
site: BCT Engineering
---

# P and Q of a 100 V single-phase source

Power System Analysis I · Chapter 3: Computational Technique

- Asked in: 2075 Baisakh, 4 marks
- Video: [Active, Reactive & Apparent Power | You’ll not get an easier explanation than this| TheElectricalGuy](https://www.youtube.com/watch?v=0f7YkVorOmY) — Gaurav J - TheElectricalGuy
- 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/single-phase-source-power-100v
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3

## Question

A single phase voltage source with V = 100∠0° volts delivers a current I = 10∠10° A, which leaves the positive terminal of the source. Calculate the source real and reactive power and state whether the source delivers or absorbs each of these.

## Short answer

Complex power S = VI* = 1000∠-10° VA = 984.8 - j173.65 VA. The source delivers 984.8 W real power and absorbs 173.6 var reactive power.

## Answer

Complex power delivered by a source whose current **leaves** its positive terminal is $S = VI^*$ (generator convention). A positive P or Q then means the source delivers it; a negative value means it absorbs it.

$$
\begin{aligned}
S &= VI^* = (100\angle0^\circ)(10\angle-10^\circ)\\
&= 1000\angle-10^\circ\ \text{VA}\\
&= 1000\cos10^\circ - j\,1000\sin10^\circ\\
&= 984.81 - j\,173.65\ \text{VA}
\end{aligned}
$$

- $P = +984.81$ W: positive, so the source **delivers** real power.
- $Q = -173.65$ var: negative, so the source **absorbs** 173.65 var (equivalently it delivers $-173.65$ var).

The current leads the voltage by 10°, so the load connected to the source is capacitive; a capacitive load supplies reactive power, which flows back into the source.

**Answer:** P = 984.8 W delivered; Q = 173.6 var absorbed by the source.

## Check yourself

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

1. With the current leaving the + terminal, which expression gives power delivered by the source?
   A. S = V I
   B. S = V I*
   C. S = V* I*
   D. S = I^2 / V
   Answer: B. S = VI* with generator convention gives delivered P and Q. (Key point: Formula)

2. What is P and is it delivered or absorbed?
   A. 984.8 W absorbed
   B. 984.8 W delivered
   C. 173.6 W delivered
   D. 1000 W absorbed
   Answer: B. P = 1000 cos10° = 984.8 W > 0, so delivered. (Key point: Real power)

3. Current leads voltage by 10°. What is true of Q?
   A. Q = +173.6 var delivered
   B. Q = -173.6 var, absorbed by source
   C. Q = 0
   D. Q = 984.8 var absorbed
   Answer: B. φ = -10° gives Q = -173.6 var; the capacitive load returns reactive power to the source. (Key point: Reactive power)

## Related in Power System Analysis I

- [Line as source and sink of reactive power, effects](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aline-source-sink-reactive-effects) — Power System Analysis I, chapter 6
- [Line as source and sink of reactive power](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aline-source-sink-reactive) — Power System Analysis I, chapter 6
- [Reactive power proportional to voltage drop](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactive-power-voltage-drop) — Power System Analysis I, chapter 6
- [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
- [Source P and Q, Ean = -120∠210° V](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Asource-power-ean-120-210) — 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

## Related in other subjects

- [Generator reactive power at constant terminal voltage](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Agenerator-reactive-power-infinite-bus) — Power System Analysis II, chapter 1
- [Single-phase current source inverter](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Asingle-phase-csi) — Power Electronics, chapter 5
- [Single-phase current source inverter with AC motor](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Acsi-ac-motor) — Power Electronics, chapter 5
- [Design: 100 kVA 2200/480 V single phase transformer](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-100kva-single-phase-ducts) — Electric Machine Design, chapter 3, asked 3×