Skip to main content

Two sources 120∠10° and 120∠30° V via 1+j5 Ω

Short answer

Take I from machine 1 to 2 through Z=1+j5 ohm; I=8.173 angle -148.69 A. Machine 1 absorbs 913.7 W and generates 356.4 var; machine 2 generates 980.5 W and absorbs 22.4 var. The impedance absorbs 66.8 W and 334.0 var, and real power flows from machine 2 to machine 1.

Two ideal voltage sources designated as machine 1 and 2 are connected as shown in figure. If E1 = 120∠10° V, E2 = 120∠30° V and Z = 1 + j5 Ω, determine: (i) whether each machine is consuming P and Q or generating P and Q (ii) P and Q absorbed by the impedance (iii) direction of power flow. [Figure: source E1 on the left and source E2 on the right, both with + terminal at top, connected through series impedance Z; current I flows from E1 towards E2.]

Take I flowing from machine 1 to machine 2 through Z. Then S1=E1I∗S_1 = E_1I^* is the power supplied by machine 1 and S2=E2I∗S_2 = E_2I^* is the power absorbed by machine 2.

Current

E1=120∠10∘=118.18+j20.84 VE2=120∠30∘=103.92+j60.00 VI=E1−E2Z=14.26−j39.161+j5=8.173∠−148.69∘ A\begin{aligned} E_1 &= 120\angle10^\circ = 118.18 + j20.84\ \text{V}\\ E_2 &= 120\angle30^\circ = 103.92 + j60.00\ \text{V}\\ I &= \frac{E_1-E_2}{Z} = \frac{14.26 - j39.16}{1 + j5} = 8.173\angle-148.69^\circ\ \text{A} \end{aligned}

(i) Machine powers

S1=E1I∗=(120∠10∘)(8.173∠148.69∘)=−913.73+j356.43 VAS2=E2I∗=(120∠30∘)(8.173∠148.69∘)=−980.53+j22.42 VA\begin{aligned} S_1 &= E_1I^* = (120\angle10^\circ)(8.173\angle148.69^\circ) = -913.73 + j356.43\ \text{VA}\\ S_2 &= E_2I^* = (120\angle30^\circ)(8.173\angle148.69^\circ) = -980.53 + j22.42\ \text{VA} \end{aligned}
MachineReal powerReactive power
1supplies −913.73 W, so it consumes 913.73 W (motor)generates 356.43 var
2absorbs −980.53 W, so it generates 980.53 W (generator)consumes 22.42 var

(ii) Power absorbed by Z

∣I∣2=8.1732=66.80SZ=∣I∣2(1+j5)=66.80+j334.01 VA\begin{aligned} |I|^2 &= 8.173^2 = 66.80\\ S_Z &= |I|^2(1 + j5) = 66.80 + j334.01\ \text{VA} \end{aligned}

Check: 980.53−913.73=66.80980.53 - 913.73 = 66.80 W and 356.43−22.42=334.01356.43 - 22.42 = 334.01 var.

(iii) Direction of power flow

  • Real power flows from machine 2 to machine 1, because E2E_2 leads E1E_1 by 20°. Real power flows from the leading-angle end to the lagging end.
  • Reactive power flows from machine 1 towards machine 2. Machine 1 supplies 356.43 var: 334.01 var is used in the line reactance and 22.42 var reaches machine 2. The magnitudes are equal here, so Q is decided by the angle and the R/X of the line.

Answer: Machine 1 is a motor drawing 913.7 W and supplying 356.4 var. Machine 2 generates 980.5 W and draws 22.4 var. Z absorbs 66.8 W and 334.0 var. P flows 2 → 1.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Power System Analysis I chapter 3 questions →