Sequence voltages, currents and powers (2071)

Sequence components (phase A reference, sequence A-B-C, $a = 1\angle120^\circ$): $X_0 = \frac13(X_A+X_B+X_C)$, $X_1 = \frac13(X_A+aX_B+a^2X_C)$, $X_2 = \frac13(X_A+a^2X_B+aX_C)$; sequence power $S_k = 3V_kI_k^*$.

Sequence voltages

$$\begin{aligned} V_0 &= \tfrac13(220\angle120^\circ + 120\angle30^\circ + 240\angle210^\circ) = -71.31 + j43.51 = 83.53\angle148.61^\circ\ \text{V} \\ V_1 &= \tfrac13(220\angle120^\circ + 120\angle150^\circ + 240\angle90^\circ) = -71.31 + j163.51 = 178.38\angle113.56^\circ\ \text{V} \\ V_2 &= \tfrac13(220\angle120^\circ + 120\angle270^\circ + 240\angle330^\circ) = 32.62 - j16.49 = 36.55\angle{-26.82^\circ}\ \text{V} \end{aligned}$$

Sequence currents

$$\begin{aligned} I_0 &= \tfrac13(10\angle150^\circ + 5\angle0^\circ + 15\angle180^\circ) = -6.220 + j1.667 = 6.440\angle165.0^\circ\ \text{A} \\ I_1 &= \tfrac13(10\angle150^\circ + 5\angle120^\circ + 15\angle60^\circ) = -1.220 + j7.440 = 7.540\angle99.31^\circ\ \text{A} \\ I_2 &= \tfrac13(10\angle150^\circ + 5\angle240^\circ + 15\angle300^\circ) = -1.220 - j4.107 = 4.284\angle{-106.55^\circ}\ \text{A} \end{aligned}$$

Sequence powers

$$\begin{aligned} S_0 &= 3V_0I_0^* = 3(83.53\angle148.61^\circ)(6.440\angle{-165.0^\circ}) = 1548.2 - j455.3\ \text{VA} \\ S_1 &= 3V_1I_1^* = 3(178.38\angle113.56^\circ)(7.540\angle{-99.31^\circ}) = 3910.6 + j993.1\ \text{VA} \\ S_2 &= 3V_2I_2^* = 3(36.55\angle{-26.82^\circ})(4.284\angle106.55^\circ) = 83.8 + j462.2\ \text{VA} \end{aligned}$$

Check

$S = V_AI_A^* + V_BI_B^* + V_CI_C^* = 2200\angle{-30^\circ} + 600\angle30^\circ + 3600\angle30^\circ = 5542.6 + j1000$ VA ✓

Answer: $S_0 = 1548 - j455$ VA, $S_1 = 3911 + j993$ VA, $S_2 = 84 + j462$ VA; total $S = 5542.6 + j1000$ VA.