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Two sources through Z = 0.5 + j5 Ω

Short answer

With E2=100∠30° V, current I=10.30∠-159.3° A. Machine 1 consumes 963.6 W and supplies 364.3 var; machine 2 generates 1016.6 W and supplies 166.3 var; impedance absorbs 53.1 W and 530.6 var.

Two ideal voltage sources designated as machine 1 and 2 are connected as shown in figure below. If E1 = 100∠0° V and E2 = ∠30° V and Z = 0.5 + j5 Ω, determine (i) whether each machine is generating or consuming real power and the amount, (ii) whether each machine is receiving or supplying reactive power and the amount, and (iii) the P and Q absorbed by the impedance. [Figure: source E1 on the left and source E2 on the right, connected through series impedance Z = 0.5 + j5 Ω; current I flows from E1 towards E2.]

The magnitude of E2E_2 is not printed; it is taken as 100 V, i.e. E2=100∠30∘E_2 = 100\angle30^\circ V (the usual textbook data).

Current I flows from machine 1 to machine 2. With this direction:

  • S1=E1I∗S_1 = E_1 I^* is the power supplied by machine 1 (current leaves its + terminal).
  • S2=E2I∗S_2 = E_2 I^* is the power absorbed by machine 2 (current enters its + terminal).

Current

E1−E2=100−(86.60+j50)=13.40−j50I=13.40−j500.5+j5=10.30∠−159.29∘ A=−9.635−j3.643 A\begin{aligned} E_1 - E_2 &= 100 - (86.60 + j50) = 13.40 - j50\\ I &= \frac{13.40 - j50}{0.5 + j5} = 10.30\angle-159.29^\circ\ \text{A}\\ &= -9.635 - j3.643\ \text{A} \end{aligned}

(i) and (ii) Machine powers

S1=E1I∗=100(−9.635+j3.643)=−963.57+j364.31 VAS2=E2I∗=−1016.63−j166.29 VA\begin{aligned} S_1 &= E_1I^* = 100(-9.635 + j3.643) = -963.57 + j364.31\ \text{VA}\\ S_2 &= E_2I^* = -1016.63 - j166.29\ \text{VA} \end{aligned}
MachinePQ
1 (S1S_1 = supplied)P1=−963.57P_1 = -963.57 W, so it consumes 963.57 W (motor)Q1=+364.31Q_1 = +364.31 var, so it supplies 364.31 var
2 (S2S_2 = absorbed)P2=−1016.63P_2 = -1016.63 W absorbed, so it generates 1016.63 WQ2=−166.29Q_2 = -166.29 var absorbed, so it supplies 166.29 var

(iii) Power absorbed by the impedance

∣I∣2=10.302=106.12SZ=∣I∣2Z=106.12(0.5+j5)=53.06+j530.59 VA\begin{aligned} |I|^2 &= 10.30^2 = 106.12\\ S_Z &= |I|^2 Z = 106.12(0.5 + j5) = 53.06 + j530.59\ \text{VA} \end{aligned}

Check: P: 1016.63−963.57=53.061016.63 - 963.57 = 53.06 W. Q: 364.31+166.29=530.6364.31 + 166.29 = 530.6 var. Both balance.

Real power flows from machine 2 to machine 1 because E2E_2 leads E1E_1. Both machines supply reactive power, and all of it is used up in the line reactance.

Answer: Machine 1 consumes 963.6 W and supplies 364.3 var. Machine 2 generates 1016.6 W and supplies 166.3 var. The impedance absorbs 53.1 W and 530.6 var.

✓ Answer checked by Claude Opus 5.5 (high effort)

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