---
title: "3-phase power in symmetrical components"
subject: "Power System Analysis II"
chapter: 4
key: "semester_5/power_system_analysis_ii:three-phase-power-symmetrical-components"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/three-phase-power-symmetrical-components
site: BCT Engineering
---

# 3-phase power in symmetrical components

Power System Analysis II · Chapter 4: Unbalanced System Analysis

- Asked in: 2078 Kartik, 8 marks; 2076 Asoj, 4 marks; 2073 Chaitra, 6 marks (asked 3 times)
- Video: [PS87 3 Phase Power Using Symmetrical Components](https://www.youtube.com/watch?v=5lKWHhe4Xb4) — Lectures in Electrical Engineering
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/three-phase-power-symmetrical-components
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-4

## Question

Derive an expression to determine 3-phase (complex) power of an unbalanced system in terms of symmetrical components of voltages and currents.

## Short answer

Expressing phase quantities in symmetrical components, S = V_a I_a* + V_b I_b* + V_c I_c* = V_s^T A^T A* I_s*. Since A^T A* = 3U, the result is S = 3(V_a0 I_a0* + V_a1 I_a1* + V_a2 I_a2*). Cross-sequence products contribute no average power.

## Answer

The total complex power in a three-phase (balanced or unbalanced) system equals **three times the sum of the complex powers of the zero, positive and negative sequence components**: $S = 3(V_{a0}I_{a0}^* + V_{a1}I_{a1}^* + V_{a2}I_{a2}^*)$.

#### Phase-domain power
With phase-to-neutral voltages $V_a, V_b, V_c$ and line currents $I_a, I_b, I_c$:
$$
S = P + jQ = V_a I_a^* + V_b I_b^* + V_c I_c^*
$$
In matrix form, with $V_p = [V_a\ V_b\ V_c]^T$ and $I_p = [I_a\ I_b\ I_c]^T$:
$$
S = V_p^T I_p^*
$$

#### Symmetrical component relations
$$
V_p = A V_s, \qquad I_p = A I_s, \qquad
A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & a^2 & a \\ 1 & a & a^2 \end{bmatrix}, \quad a = 1\angle 120^\circ
$$
where $V_s = [V_{a0}\ V_{a1}\ V_{a2}]^T$ and $I_s = [I_{a0}\ I_{a1}\ I_{a2}]^T$.

#### Substitution
$$
\begin{aligned}
S &= (A V_s)^T (A I_s)^* \\
&= V_s^T A^T A^* I_s^*
\end{aligned}
$$
Since $A$ is symmetric, $A^T = A$. Also $a^* = a^2$ and $(a^2)^* = a$, so
$$
A^* = \begin{bmatrix} 1 & 1 & 1 \\ 1 & a & a^2 \\ 1 & a^2 & a \end{bmatrix}
$$
Multiplying, using $1 + a + a^2 = 0$ and $a^3 = 1$:
$$
A^T A^* = \begin{bmatrix} 1+1+1 & 1+a+a^2 & 1+a^2+a \\ 1+a^2+a & 1+a^3+a^3 & 1+a^4+a^2 \\ 1+a+a^2 & 1+a^2+a^4 & 1+a^3+a^3 \end{bmatrix}
= \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix} = 3U
$$
Therefore
$$
\begin{aligned}
S &= 3\,V_s^T I_s^* \\
&= 3\left(V_{a0} I_{a0}^* + V_{a1} I_{a1}^* + V_{a2} I_{a2}^*\right)
\end{aligned}
$$

#### Remarks
- The transformation is **power invariant except for the factor 3**; if the transformation is defined with $\frac{1}{\sqrt3}A$ (unitary form), the factor 3 disappears.
- There are **no cross terms** such as $V_{a1}I_{a2}^*$: a voltage of one sequence and a current of another sequence produce no net average power.
- For a balanced system $V_{a0}=V_{a2}=0$ and $I_{a0}=I_{a2}=0$, so $S = 3V_{a1}I_{a1}^*$, the familiar $3V_{ph}I_{ph}^*$.

## Check yourself

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

1. The derivation starts from the phase-domain power
   A. S = V_a I_a
   B. S = V_a I_a^* + V_b I_b^* + V_c I_c^* = V_p^T I_p^*
   C. S = |V||I|
   D. S = V_p I_p
   Answer: B. Total complex power is the sum of phase-to-neutral voltage times conjugate line current for each phase. (Key point: Starting point)

2. What is A^T A^* for the symmetrical component matrix A?
   A. The unit matrix U
   B. A itself
   C. 3U (a diagonal matrix with 3 on the diagonal)
   D. Zero
   Answer: C. Using 1 + a + a^2 = 0 and a^3 = 1, all off-diagonal terms vanish and the diagonal terms equal 3. (Key point: Key matrix product)

3. The resulting expression is
   A. S = V_a0 I_a0^* + V_a1 I_a1^* + V_a2 I_a2^*
   B. S = 3(V_a0 I_a0^* + V_a1 I_a1^* + V_a2 I_a2^*)
   C. S = sqrt(3)(...)
   D. S = 3(V_a1 I_a2^* + V_a2 I_a1^*)
   Answer: B. S = 3 V_s^T I_s^*; the factor 3 comes from A^T A^* = 3U. (Key point: Final result)

4. What does the result show for cross terms and a balanced system?
   A. Cross terms such as V_a1 I_a2^* appear and add power
   B. There are no cross terms; balanced gives S = 3 V_a1 I_a1^*
   C. Balanced systems give zero power
   D. Only the zero-sequence term remains
   Answer: B. Because A^T A^* is diagonal, only same-sequence products exist; for balanced systems V_a0, V_a2, I_a0, I_a2 are zero. (Key point: Cross terms and balanced case)

## Related in Power System Analysis II

- [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
- [Symmetrical components of unbalanced voltages](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aunbalanced-voltages-components-2070) — Power System Analysis II, chapter 4
- [Derive symmetrical components of currents](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Asymmetrical-components-currents-derivation) — Power System Analysis II, chapter 4
- [Representing unbalanced voltages by components](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aunbalanced-voltages-symmetrical-components) — Power System Analysis II, chapter 4
- [Significance of components; phasor voltages](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Asymmetrical-components-significance-voltages) — Power System Analysis II, chapter 4
- [L-G fault through Zf: currents and voltages](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Alg-fault-with-zf-derivation) — Power System Analysis II, chapter 5

## Related in other subjects

- [Unbalanced delta loads currents and power](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Adelta-unbalanced-loads-31j59-200v) — Basic Electrical Engineering, chapter 8
- [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
- [Delta load phase currents, power, line currents](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Adelta-load-100v-6-8-4-3-ohm) — Basic Electrical Engineering, chapter 8
- [Voltages and currents vary along a long line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Along-line-voltage-varies-along) — Power System Analysis I, chapter 5, asked 2×