---
title: "Two sources 120∠10° and 120∠30° V via 1+j5 Ω"
subject: "Power System Analysis I"
chapter: 3
key: "semester_4/power_system_analysis_i:two-sources-120v-1-j5"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/two-sources-120v-1-j5
site: BCT Engineering
---

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

Power System Analysis I · Chapter 3: Computational Technique

- Asked in: 2072 Asoj, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/two-sources-120v-1-j5
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3

## Question

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.]

## 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.

## Answer

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

#### Current
$$
\begin{aligned}
E_1 &= 120\angle10^\circ = 118.18 + j20.84\ \text{V}\\
E_2 &= 120\angle30^\circ = 103.92 + j60.00\ \text{V}\\
I &= \frac{E_1-E_2}{Z} = \frac{14.26 - j39.16}{1 + j5} = 8.173\angle-148.69^\circ\ \text{A}
\end{aligned}
$$

#### (i) Machine powers
$$
\begin{aligned}
S_1 &= E_1I^* = (120\angle10^\circ)(8.173\angle148.69^\circ) = -913.73 + j356.43\ \text{VA}\\
S_2 &= E_2I^* = (120\angle30^\circ)(8.173\angle148.69^\circ) = -980.53 + j22.42\ \text{VA}
\end{aligned}
$$

| Machine | Real power | Reactive power |
|---|---|---|
| 1 | supplies −913.73 W, so it **consumes 913.73 W** (motor) | **generates 356.43 var** |
| 2 | absorbs −980.53 W, so it **generates 980.53 W** (generator) | **consumes 22.42 var** |

#### (ii) Power absorbed by Z
$$
\begin{aligned}
|I|^2 &= 8.173^2 = 66.80\\
S_Z &= |I|^2(1 + j5) = 66.80 + j334.01\ \text{VA}
\end{aligned}
$$

Check: $980.53 - 913.73 = 66.80$ W and $356.43 - 22.42 = 334.01$ var.

#### (iii) Direction of power flow
- **Real power** flows from machine 2 to machine 1, because $E_2$ leads $E_1$ 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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which machine is the generator?
   A. Machine 1
   B. Machine 2, since E2 leads by 20°
   C. Both
   D. Neither
   Answer: B. Real power flows from leading E2 to lagging E1. (Key point: Real power flow)

2. Machine 1 has S1 = -913.73 + j356.43. What does it do?
   A. Consumes 913.7 W and generates 356.4 var
   B. Generates 913.7 W and consumes 356.4 var
   C. Consumes both
   D. Generates both
   Answer: A. Negative P supplied means consumed; positive Q supplied means generated. (Key point: Machine 1 reactive)

3. What is S_Z?
   A. 66.8 + j334.0 VA
   B. 913.7 + j356.4
   C. 980.5 + j22.4
   D. 0
   Answer: A. |I|^2 (1 + j5) = 66.80(1 + j5). (Key point: Impedance power)

4. How does P balance?
   A. 980.53 - 913.73 = 66.80 W absorbed by Z
   B. 980.5 + 913.7
   C. They are unrelated
   D. Z gives zero W
   Answer: A. Generated minus consumed equals line loss. (Key point: Balance)

## Related in Power System Analysis I

- [Two sources through Z = 0.5 + j5 Ω](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Atwo-sources-100v-30deg) — Power System Analysis I, chapter 3
- [Two sources via 0 - j5 Ω: power flow](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Atwo-sources-0-j5) — Power System Analysis I, chapter 3
- [Two machines 100∠0° and 120∠30° V via j5 Ω](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Atwo-machines-100v-120v-j5) — Power System Analysis I, chapter 3