Skip to main content

Unbalanced source feeding star load

Short answer

Sequence voltages from A⁻¹Vp: V0=47.77∠57.63°, V1=112.78∠-0.03°, V2=61.62∠45.88° V. Load sequence impedances: Z1=Z2=8+j24 Ω, Z0=8+j28.5 Ω. Sequence currents: I0=1.614∠-16.69°, I1=4.458∠-71.60°, I2=2.436∠-25.68° A.

A 3-phase unbalanced source with voltage given by Vabc = [200∠25° 100∠−155° 80∠100°]ᵀ V is feeding a 3-phase star connected load with series impedance per phase of Zs = 8 + j24 Ohms and load neutral impedance of Zn = j1.5 Ohms. Determine the: (i) Symmetrical components of voltage (ii) Symmetrical components of currents

Find the sequence voltages with Vs=A−1VpV_s = A^{-1}V_p, then divide each by its own sequence impedance of the load, since a balanced star load has a diagonal sequence impedance matrix.

Assumption: the given voltages are the phase-to-ground voltages applied to the load terminals (source neutral grounded), a=1∠120∘a = 1\angle120^\circ.

Phase voltages in rectangular form

Va=200∠25∘=181.26+j84.52 VVb=100∠−155∘=−90.63−j42.26 VVc=80∠100∘=−13.89+j78.78 V\begin{aligned} V_a &= 200\angle25^\circ = 181.26 + j84.52\ \text{V} \\ V_b &= 100\angle{-155^\circ} = -90.63 - j42.26\ \text{V} \\ V_c &= 80\angle100^\circ = -13.89 + j78.78\ \text{V} \end{aligned}

(i) Symmetrical components of voltage

Va0=13(Va+Vb+Vc)=25.58+j40.35=47.77∠57.63∘ VVa1=13(Va+aVb+a2Vc)=112.78−j0.07=112.78∠−0.03∘ VVa2=13(Va+a2Vb+aVc)=42.90+j44.24=61.62∠45.88∘ V\begin{aligned} V_{a0} &= \tfrac13(V_a + V_b + V_c) = 25.58 + j40.35 = 47.77\angle57.63^\circ\ \text{V} \\ V_{a1} &= \tfrac13(V_a + aV_b + a^2V_c) = 112.78 - j0.07 = 112.78\angle{-0.03^\circ}\ \text{V} \\ V_{a2} &= \tfrac13(V_a + a^2V_b + aV_c) = 42.90 + j44.24 = 61.62\angle45.88^\circ\ \text{V} \end{aligned}

Check: Va0+Va1+Va2=181.26+j84.52=VaV_{a0} + V_{a1} + V_{a2} = 181.26 + j84.52 = V_a.

Sequence impedances of the load

Z1=Z2=Zs=8+j24=25.30∠71.57∘ ΩZ0=Zs+3Zn=8+j(24+4.5)=8+j28.5=29.60∠74.32∘ Ω\begin{aligned} Z_1 = Z_2 &= Z_s = 8 + j24 = 25.30\angle71.57^\circ\ \Omega \\ Z_0 &= Z_s + 3Z_n = 8 + j(24 + 4.5) = 8 + j28.5 = 29.60\angle74.32^\circ\ \Omega \end{aligned}

(ii) Symmetrical components of current

Ia0=Va0Z0=47.77∠57.63∘29.60∠74.32∘=1.614∠−16.69∘ AIa1=Va1Z1=112.78∠−0.03∘25.30∠71.57∘=4.458∠−71.60∘ AIa2=Va2Z2=61.62∠45.88∘25.30∠71.57∘=2.436∠−25.68∘ A\begin{aligned} I_{a0} &= \frac{V_{a0}}{Z_0} = \frac{47.77\angle57.63^\circ}{29.60\angle74.32^\circ} = 1.614\angle{-16.69^\circ}\ \text{A} \\ I_{a1} &= \frac{V_{a1}}{Z_1} = \frac{112.78\angle{-0.03^\circ}}{25.30\angle71.57^\circ} = 4.458\angle{-71.60^\circ}\ \text{A} \\ I_{a2} &= \frac{V_{a2}}{Z_2} = \frac{61.62\angle45.88^\circ}{25.30\angle71.57^\circ} = 2.436\angle{-25.68^\circ}\ \text{A} \end{aligned}

For reference, the line currents are Ia=7.72∠−48.16∘I_a = 7.72\angle{-48.16^\circ} A, Ib=4.15∠136.40∘I_b = 4.15\angle136.40^\circ A, Ic=2.91∠30.98∘I_c = 2.91\angle30.98^\circ A, and the neutral current In=3Ia0=4.84∠−16.69∘I_n = 3I_{a0} = 4.84\angle{-16.69^\circ} A.

Answer: V0=47.77∠57.63∘V_0 = 47.77\angle57.63^\circ, V1=112.78∠−0.03∘V_1 = 112.78\angle{-0.03^\circ}, V2=61.62∠45.88∘V_2 = 61.62\angle45.88^\circ V; I0=1.614∠−16.69∘I_0 = 1.614\angle{-16.69^\circ}, I1=4.458∠−71.60∘I_1 = 4.458\angle{-71.60^\circ}, I2=2.436∠−25.68∘I_2 = 2.436\angle{-25.68^\circ} A.

✓ Answer checked by Claude Sonnet 5.5

Related questions

All Power System Analysis II chapter 4 questions →