---
title: "Two sources through Z = 0.5 + j5 Ω"
subject: "Power System Analysis I"
chapter: 3
key: "semester_4/power_system_analysis_i:two-sources-100v-30deg"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/two-sources-100v-30deg
site: BCT Engineering
---

# Two sources through Z = 0.5 + j5 Ω

Power System Analysis I · Chapter 3: Computational Technique

- Asked in: 2074 Bhadra, 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-100v-30deg
- 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 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.]

## 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.

## Answer

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

Current I flows from machine 1 to machine 2. With this direction:
- $S_1 = E_1 I^*$ is the power **supplied** by machine 1 (current leaves its + terminal).
- $S_2 = E_2 I^*$ is the power **absorbed** by machine 2 (current enters its + terminal).

#### Current
$$
\begin{aligned}
E_1 - E_2 &= 100 - (86.60 + j50) = 13.40 - j50\\
I &= \frac{13.40 - j50}{0.5 + j5} = 10.30\angle-159.29^\circ\ \text{A}\\
&= -9.635 - j3.643\ \text{A}
\end{aligned}
$$

#### (i) and (ii) Machine powers
$$
\begin{aligned}
S_1 &= E_1I^* = 100(-9.635 + j3.643) = -963.57 + j364.31\ \text{VA}\\
S_2 &= E_2I^* = -1016.63 - j166.29\ \text{VA}
\end{aligned}
$$

| Machine | P | Q |
|---|---|---|
| 1 ($S_1$ = supplied) | $P_1 = -963.57$ W, so it **consumes** 963.57 W (motor) | $Q_1 = +364.31$ var, so it **supplies** 364.31 var |
| 2 ($S_2$ = absorbed) | $P_2 = -1016.63$ W absorbed, so it **generates** 1016.63 W | $Q_2 = -166.29$ var absorbed, so it **supplies** 166.29 var |

#### (iii) Power absorbed by the impedance
$$
\begin{aligned}
|I|^2 &= 10.30^2 = 106.12\\
S_Z &= |I|^2 Z = 106.12(0.5 + j5) = 53.06 + j530.59\ \text{VA}
\end{aligned}
$$

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

Real power flows from machine 2 to machine 1 because $E_2$ leads $E_1$. 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.

## Check yourself

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

1. Which machine generates real power and why?
   A. Machine 1, lagging
   B. Machine 2, because E2 leads E1
   C. Machine 1, since I leaves it
   D. Neither
   Answer: B. Real power flows from the leading to the lagging source; S2 = -1016.6 W absorbed means it generates. (Key point: Real power direction)

2. What do the machines do with reactive power?
   A. Machine 1 absorbs, machine 2 supplies
   B. Both absorb
   C. Both supply, and the line reactance uses it all
   D. Both supply 0
   Answer: C. Q1 = +364.3 var supplied, Q2 = -166.3 absorbed means supplied; total 530.6 var to the impedance. (Key point: Reactive power)

3. How is power in Z found and what is it?
   A. |I|^2 Z = 53.06 + j530.59 VA
   B. |I| Z
   C. E1^2/Z
   D. 1016.6 + j0
   Answer: A. S_Z = |I|^2 (0.5 + j5) and equals the sum of the machine powers. (Key point: Impedance power)

## Related in Power System Analysis I

- [Two sources 120∠10° and 120∠30° V via 1+j5 Ω](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Atwo-sources-120v-1-j5) — 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