---
title: "Power by two generators, 1.1∠15° and 1.1∠0° pu"
subject: "Power System Analysis I"
chapter: 3
key: "semester_4/power_system_analysis_i:two-generators-1-1-pu"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/two-generators-1-1-pu
site: BCT Engineering
---

# Power by two generators, 1.1∠15° and 1.1∠0° pu

Power System Analysis I · Chapter 3: Computational Technique

- Asked in: 2070 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-generators-1-1-pu
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3

## Question

Determine the amount of power generated by each of the generators in the power system network below. Also specify the direction of power flow in the network. [Figure: E1 = 1.1∠15° pu behind reactance j0.1 pu, connected through a line of resistance 0.03 pu and reactance j0.1 pu to E2 = 1.1∠0° pu behind reactance j0.1 pu.]

## Short answer

Total impedance is 0.03 + j0.3 pu; current I = 0.952∠13.2° pu. G1 generates 1.047 pu real power, G2 receives 1.020 pu; line losses 0.027 pu. Both supply reactive power: G1 0.033 pu, G2 0.239 pu. Real power flows from G1 to G2.

## Answer

Total series impedance between the two emfs:

$$
Z = j0.1 + (0.03 + j0.1) + j0.1 = 0.03 + j0.3\ \text{pu}
$$

Let I flow from $E_1$ to $E_2$. Then $S_1 = E_1I^*$ is the power generated by G1 and $S_2 = E_2I^*$ is the power received by G2.

```text
  j0.1     0.03+j0.1     j0.1
 +-/\/\/-----/\/\/-------/\/\/-+
 |       I -->                 |
(E1) 1.1/15 deg         1.1/0 deg (E2)
 |                             |
 +-----------------------------+
```

#### Current
$$
\begin{aligned}
E_1 &= 1.1\angle15^\circ = 1.0625 + j0.2847\\
E_2 &= 1.1\angle0^\circ = 1.1\\
I &= \frac{E_1 - E_2}{Z} = \frac{-0.0375 + j0.2847}{0.03 + j0.3}\\
&= 0.9272 + j0.2177 = 0.9524\angle13.21^\circ\ \text{pu}
\end{aligned}
$$

#### Powers
$$
\begin{aligned}
S_1 &= E_1I^* = (1.1\angle15^\circ)(0.9524\angle-13.21^\circ) = 1.0472 + j0.0327\ \text{pu}\\
S_2 &= E_2I^* = (1.1)(0.9524\angle-13.21^\circ) = 1.0200 - j0.2394\ \text{pu}\\
S_{loss} &= |I|^2Z = 0.9071(0.03 + j0.3) = 0.0272 + j0.2721\ \text{pu}
\end{aligned}
$$

| Generator | Real power | Reactive power |
|---|---|---|
| G1 | **generates 1.0472 pu** | generates 0.0327 pu |
| G2 | receives 1.0200 pu (acts as motor) | $Q_2$ received = −0.2394, so it **generates 0.2394 pu** |

Check: $1.0472 - 1.0200 = 0.0272$ pu and $0.0327 + 0.2394 = 0.2721$ pu.

#### Direction of power flow
- **Real power** flows from G1 to G2, because $E_1$ leads $E_2$ by 15°. G1 is generating and G2 is motoring.
- Both machines supply reactive power to the network, and it is all consumed in the series reactances ($I^2X = 0.2721$ pu).

**Answer:** G1 generates 1.047 pu of real power. G2 receives 1.020 pu. The line and reactances absorb 0.027 pu. Reactive power supplied is 0.033 pu by G1 and 0.239 pu by G2. P flows G1 → G2.

## Check yourself

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

1. What is the total series impedance?
   A. 0.03 + j0.1 pu
   B. 0.03 + j0.3 pu
   C. j0.3 pu only
   D. 0.06 + j0.3 pu
   Answer: B. j0.1 + (0.03 + j0.1) + j0.1. (Key point: Total impedance)

2. Which way does the real power flow and why?
   A. G1 to G2, because E1 leads E2 by 15°
   B. G2 to G1 since E2 is the reference
   C. G2 to G1 since G2 has more reactance
   D. No flow since magnitudes are equal
   Answer: A. Real power flows from the leading emf. (Key point: Direction of P)

3. What is G2's role?
   A. It generates both P and Q to the network without receiving P
   B. It absorbs both P and Q
   C. It does neither
   D. It receives real power (motor) but supplies reactive power
   Answer: D. S2 = 1.02 - j0.2394, so P is received and Q is generated. (Key point: Role of G2)

4. What is the loss check?
   A. P1 + P2 = I^2R
   B. Q1 - Q2 = I^2X
   C. P1 - P2 = 0.0272 pu = I^2R and Q1 + Q2 = 0.2721 pu = I^2X
   D. P1 = P2
   Answer: C. The difference in real powers is the resistive loss; the reactive powers supplied sum to I^2X. (Key point: Check of results)

## Close variants set in other exams

- [Power by two generators, 1.05∠5° and 1.05∠20° pu](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Atwo-generators-1-05-pu) — Power System Analysis I, chapter 3, asked 2×

## Related in Power System Analysis I

- [Reactance diagram: two generators, two motors, j50 Ω line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-two-gens-two-motors) — 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
- [Reactance diagram: 90 MVA generator (15%), two motors](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-90mva-generator-15) — Power System Analysis I, chapter 3, asked 2×
- [Reactance diagram: three generators, two 220 kV lines](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-three-gens-two-lines) — Power System Analysis I, chapter 3
- [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

## Related in other subjects

- [N-R first iteration, line reactance 0.1](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Anr-first-iteration-x0-1) — Power System Analysis II, chapter 2
- [G-S two iterations, lines 0.02+j0.11](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Ags-two-iterations-2080) — Power System Analysis II, chapter 2
- [G-S two iterations, lines 0.026+j0.11](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Ags-two-iterations-2075) — Power System Analysis II, chapter 2
- [Node equations and Ybus, three-source network](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aybus-node-equations-three-sources) — Power System Analysis II, chapter 1, asked 2×