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Sequence voltages, currents and powers (2072)

Short answer

Apply symmetrical component transformation to voltages and currents. Sequence powers are Sk = 3Vk Ik*. V0=52.07∠86.33° V, I0=1.78∠-163.35° A, S0=-96.6-j260.7 VA; V1=213.33∠0° V, I1=9.25∠20.27° A, S1=5554.6-j2051.3 VA; V2=52.07∠-86.33° V, I2=5.28∠-57.03° A, S2=719.6-j403.7 VA. Total S = 6177.6 - j2715.7 VA.

A 3 phase system has following voltage and currents: Va = 220V∠0°, Vb = 120V∠−120°, Vc = 300V∠−240°; Ia = 10A∠−10°, Ib = 5A∠−100°, Ic = 15A∠−200°. Calculate sequence voltages, currents and powers.

Sequence components are found with X0=13(Xa+Xb+Xc)X_0 = \frac13(X_a+X_b+X_c), X1=13(Xa+aXb+a2Xc)X_1 = \frac13(X_a+aX_b+a^2X_c), X2=13(Xa+a2Xb+aXc)X_2 = \frac13(X_a+a^2X_b+aX_c) (a=1∠120∘a = 1\angle120^\circ), and sequence powers with Sk=3VkIk∗S_k = 3V_kI_k^*.

Sequence voltages

Note Vc=300∠−240∘=300∠120∘V_c = 300\angle{-240^\circ} = 300\angle120^\circ; aVb=120∠0∘aV_b = 120\angle0^\circ, a2Vc=300∠0∘a^2V_c = 300\angle0^\circ; a2Vb=120∠120∘a^2V_b = 120\angle120^\circ, aVc=300∠240∘aV_c = 300\angle240^\circ.

V0=13(220∠0∘+120∠−120∘+300∠120∘)=3.33+j51.96=52.07∠86.33∘ VV1=13(220+120+300)=213.33∠0∘ VV2=13(220∠0∘+120∠120∘+300∠240∘)=3.33−j51.96=52.07∠−86.33∘ V\begin{aligned} V_0 &= \tfrac13(220\angle0^\circ + 120\angle{-120^\circ} + 300\angle120^\circ) = 3.33 + j51.96 = 52.07\angle86.33^\circ\ \text{V} \\ V_1 &= \tfrac13(220 + 120 + 300) = 213.33\angle0^\circ\ \text{V} \\ V_2 &= \tfrac13(220\angle0^\circ + 120\angle120^\circ + 300\angle240^\circ) = 3.33 - j51.96 = 52.07\angle{-86.33^\circ}\ \text{V} \end{aligned}

Sequence currents

I0=13(10∠−10∘+5∠−100∘+15∠−200∘)=−1.705−j0.510=1.780∠−163.35∘ AI1=13(10∠−10∘+5∠20∘+15∠40∘)=8.679+j3.205=9.252∠20.27∘ AI2=13(10∠−10∘+5∠140∘+15∠−80∘)=2.874−j4.432=5.282∠−57.03∘ A\begin{aligned} I_0 &= \tfrac13(10\angle{-10^\circ} + 5\angle{-100^\circ} + 15\angle{-200^\circ}) = -1.705 - j0.510 = 1.780\angle{-163.35^\circ}\ \text{A} \\ I_1 &= \tfrac13(10\angle{-10^\circ} + 5\angle20^\circ + 15\angle40^\circ) = 8.679 + j3.205 = 9.252\angle20.27^\circ\ \text{A} \\ I_2 &= \tfrac13(10\angle{-10^\circ} + 5\angle140^\circ + 15\angle{-80^\circ}) = 2.874 - j4.432 = 5.282\angle{-57.03^\circ}\ \text{A} \end{aligned}

Sequence powers (Sk=3VkIk∗S_k = 3V_kI_k^*)

S0=3(52.07∠86.33∘)(1.780∠163.35∘)=−96.6−j260.7 VAS1=3(213.33∠0∘)(9.252∠−20.27∘)=5554.6−j2051.3 VAS2=3(52.07∠−86.33∘)(5.282∠57.03∘)=719.6−j403.7 VA\begin{aligned} S_0 &= 3(52.07\angle86.33^\circ)(1.780\angle163.35^\circ) = -96.6 - j260.7\ \text{VA} \\ S_1 &= 3(213.33\angle0^\circ)(9.252\angle{-20.27^\circ}) = 5554.6 - j2051.3\ \text{VA} \\ S_2 &= 3(52.07\angle{-86.33^\circ})(5.282\angle57.03^\circ) = 719.6 - j403.7\ \text{VA} \end{aligned}
SequenceV (V)I (A)P (W)Q (var)
Zero52.07∠86.33°1.780∠−163.35°−96.6−260.7
Positive213.33∠0°9.252∠20.27°5554.6−2051.3
Negative52.07∠−86.33°5.282∠−57.03°719.6−403.7
Total6177.6−2715.7

Check with phase quantities

S=VaIa∗+VbIb∗+VcIc∗=2200∠10∘+600∠−20∘+4500∠−40∘=6177.6−j2715.7S = V_aI_a^* + V_bI_b^* + V_cI_c^* = 2200\angle10^\circ + 600\angle{-20^\circ} + 4500\angle{-40^\circ} = 6177.6 - j2715.7 VA ✓

Answer: total S=6177.6−j2715.7S = 6177.6 - j2715.7 VA (P ≈ 6.18 kW, Q ≈ −2.72 kvar), equal to the sum of the three sequence powers.

✓ Answer checked by Claude Sonnet 5.5

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