---
title: "Derivation and phasor diagram of complex power"
subject: "Power System Analysis I"
chapter: 3
key: "semester_4/power_system_analysis_i:complex-power-derivation-phasor"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/complex-power-derivation-phasor
site: BCT Engineering
---

# Derivation and phasor diagram of complex power

Power System Analysis I · Chapter 3: Computational Technique

- Asked in: 2071 Magh, 6 marks
- Video: [Power System Analysis - Complex Power and the Power Triangle](https://www.youtube.com/watch?v=axS7MncCQ84) — Romero Engineering Company
- 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/complex-power-derivation-phasor
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3

## Question

Starting from voltage and current phasors, derive the expression for complex power. Construct the phasor diagram of complex power and explain the significance of the components.

## Short answer

Complex power S = VI* = P + jQ, with P = |V||I|cosφ (real power, W) and Q = |V||I|sinφ (reactive power, var). P does useful work; Q is energy exchanged with magnetic/electric fields and sets voltage levels.

## Answer

**Complex power** S is the phasor product of the voltage and the conjugate of the current. Its real part is the average (active) power and its imaginary part is the reactive power.

#### Derivation
Let the instantaneous voltage and current be
$$
v = V_m\cos(\omega t + \theta_v), \qquad i = I_m\cos(\omega t + \theta_i)
$$

The instantaneous power is
$$
\begin{aligned}
p = vi &= \frac{V_mI_m}{2}\left[\cos(\theta_v - \theta_i) + \cos(2\omega t + \theta_v + \theta_i)\right]\\
&= |V||I|\cos\phi\,[1 + \cos 2(\omega t + \theta_v)] + |V||I|\sin\phi\,\sin 2(\omega t + \theta_v)
\end{aligned}
$$

Here $|V| = V_m/\sqrt2$, $|I| = I_m/\sqrt2$ and $\phi = \theta_v - \theta_i$.
- The first term has average $P = |V||I|\cos\phi$. This is the **active power**.
- The second term has zero average and peak $Q = |V||I|\sin\phi$. This is the **reactive power**.

In phasor form, $V = |V|\angle\theta_v$ and $I = |I|\angle\theta_i$. Then
$$
\begin{aligned}
S = VI^* &= |V|\angle\theta_v\cdot|I|\angle-\theta_i = |V||I|\angle(\theta_v - \theta_i)\\
&= |V||I|\cos\phi + j|V||I|\sin\phi = P + jQ
\end{aligned}
$$

For a load $Z = R + jX$: $S = |I|^2Z = |I|^2R + j|I|^2X$.

#### Phasor diagram and power triangle
```text
 V reference (lagging load)       Power triangle
  ---------------> V                  /|
  \  phi                         S  /  | Q = |V||I|sin(phi)
    \                             /    |
      \  I (lags V)             / phi  |
        v                      +-------+
                                P = |V||I|cos(phi)
```

Multiplying V by the conjugate of I turns the V–I phasor diagram into a right-angled triangle. The hypotenuse is $|S|$, the base is P, the height is Q, and the angle is $\phi$.

#### Significance of the components
| Component | Unit | Meaning |
|---|---|---|
| P (real part) | W | Average power converted to useful work and losses. It is what energy meters record. |
| Q (imaginary part) | var | Power exchanged back and forth with magnetic and electric fields. It is needed for magnetising and sets the voltage levels, but does no net work. |
| $\lvert S\rvert$ | VA | Sets the current and the rating of generators, transformers and cables |
| $\phi$, $\cos\phi$ | — | Power factor: the share of S that is useful power |

- Q > 0 means a lagging (inductive) load and Q < 0 a leading (capacitive) load.
- A low power factor means more current for the same P. This causes more $I^2R$ loss, more voltage drop and larger equipment. So utilities add capacitors to reduce Q.

Example: $V = 230\angle0^\circ$ V and $I = 10\angle-36.87^\circ$ A give $S = 2300\angle36.87^\circ = 1840 + j1380$ VA, with pf 0.8 lagging.

## Check yourself

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

1. In the instantaneous power expression, what is the average value of the second term |V||I| sin(phi) sin 2(wt+theta_v)?
   A. P
   B. Zero; its peak value is Q
   C. S
   D. |V||I|
   Answer: B. This oscillating term averages to zero, with amplitude Q. (Key point: Instantaneous power terms)

2. Which phasor expression gives S = P + jQ?
   A. S = V I* = |V||I| angle(theta_v - theta_i)
   B. S = V I = |V||I| angle(theta_v + theta_i)
   C. S = V*I* only
   D. S = V/I
   Answer: A. Conjugating I gives angle phi = theta_v - theta_i. (Key point: Phasor derivation)

3. In the power triangle, what are the base, height and hypotenuse?
   A. Base Q, height P, hypotenuse |S|
   B. Base |S|, height P, hypotenuse Q
   C. Base P, height |S|, hypotenuse Q
   D. Base P, height Q, hypotenuse |S|
   Answer: D. P = |S| cos phi and Q = |S| sin phi. (Key point: Power triangle)

4. Why do utilities improve a low power factor?
   A. Q does useful work
   B. Low pf increases P
   C. More current is needed for the same P, causing more I^2R loss and voltage drop
   D. It raises the supply frequency
   Answer: C. For the same P, I = P/(V cos phi) increases as pf falls. (Key point: Significance of low pf)

## Related in Power System Analysis I

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- [Nominal-T derivation, phasor, losses, regulation](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Anominal-t-derivation-losses) — Power System Analysis I, chapter 5
- [Complex power and its significance](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Acomplex-power-significance) — Power System Analysis I, chapter 3, asked 2×
- [Receiving end complex power and power circle](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areceiving-power-circle) — Power System Analysis I, chapter 6
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## Related in other subjects

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- [Symmetrical components and complex power](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Asymmetrical-components-complex-power) — Power System Analysis II, chapter 4
- [Importance and derivation of current components](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Asymmetrical-components-importance-currents) — Power System Analysis II, chapter 4
- [Reactive power and power phasor diagram](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Areactive-power-phasor-diagram) — Basic Electrical Engineering, chapter 7