3-phase power in symmetrical components

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 $3V_{ph}I_{ph}^*$.