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Two machines 100∠0° and 120∠30° V via j5 Ω

Short answer

Current I = 12.03∠176.3° A. Real power 1200 W per phase flows from B to A (B generates, A motors). Reactive power: B supplies 801.5 var, line absorbs 723.1 var, A receives 78.5 var per phase.

Two ideal 3-phase synchronous machines A and B are connected by a series impedance of 0 + j5 Ω per phase. The voltages at the two machines are 100∠0° and 120∠30° V respectively. Calculate and analyze the direction of power flow in the system.

Take the voltages as per-phase values. Let I flow from A to B. Then SA=VAI∗S_A = V_AI^* is the power supplied by A and SB=VBI∗S_B = V_BI^* is the power received by B.

Current

VA=100∠0∘=100 VVB=120∠30∘=103.92+j60 VI=VA−VBj5=−3.92−j60j5=−12.0+j0.785=12.026∠176.26∘ A\begin{aligned} V_A &= 100\angle0^\circ = 100\ \text{V}\\ V_B &= 120\angle30^\circ = 103.92 + j60\ \text{V}\\ I &= \frac{V_A - V_B}{j5} = \frac{-3.92 - j60}{j5} = -12.0 + j0.785\\ &= 12.026\angle176.26^\circ\ \text{A} \end{aligned}

Powers per phase

SA=VAI∗=100(−12.0−j0.785)=−1200−j78.46 VASB=VBI∗=−1200−j801.54 VASlink=∣I∣2(j5)=144.62×j5=j723.08 VA\begin{aligned} S_A &= V_AI^* = 100(-12.0 - j0.785) = -1200 - j78.46\ \text{VA}\\ S_B &= V_BI^* = -1200 - j801.54\ \text{VA}\\ S_{link} &= |I|^2(j5) = 144.62\times j5 = j723.08\ \text{VA} \end{aligned}

Check with the lossless-line formula: P=∣VA∣∣VB∣Xsin⁡(δA−δB)=100×1205sin⁡(−30∘)=−1200P = \frac{|V_A||V_B|}{X}\sin(\delta_A - \delta_B) = \frac{100\times120}{5}\sin(-30^\circ) = -1200 W.

Analysis

MachineReal powerReactive power
Asupplies −1200 W, so it receives 1200 W (motor)supplies −78.46 var, so it receives 78.46 var
Breceives −1200 W, so it generates 1200 Wreceives −801.54 var, so it supplies 801.54 var
Link (j5 Ω)0 W (no resistance)absorbs 723.08 var
  • Real power flows from B to A because VBV_B leads VAV_A by 30°. There is no resistance, so A receives exactly what B sends.
  • Reactive power flows from B (higher voltage, 120 V) to A (lower voltage, 100 V). B supplies 801.54 var: 723.08 var is used in the link and 78.46 var reaches A.
  • Three-phase totals (×3): 3600 W from B to A; B supplies 2404.6 var; the link absorbs 2169.2 var; A receives 235.4 var.

Answer: B is the generator and sends 1200 W per phase (3.6 kW total) to A, which acts as a motor. Q flows from B to A: B supplies 801.5 var, the link uses 723.1 var, and A receives 78.5 var (per phase).

✓ Answer checked by Claude Opus 5.5 (high effort)

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