---
title: "Two generators 200∠30° and 210∠50° V"
subject: "Power System Analysis I"
chapter: 3
key: "semester_4/power_system_analysis_i:two-generators-200v-210v"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/two-generators-200v-210v
site: BCT Engineering
---

# Two generators 200∠30° and 210∠50° V

Power System Analysis I · Chapter 3: Computational Technique

- Asked in: 2072 Magh, 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-200v-210v
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3

## Question

Two 3-phase synchronous generators connected in star with neutral grounded are connected to each other through a line with series impedance per phase of 2 + j8 Ω. The terminal voltages of generators 1 and 2 are 200∠30° V and 210∠50° V respectively. Determine the active and reactive power generated/consumed by each generator.

## Short answer

Using per-phase values, I from G1 to G2 is 8.716 angle -133.84 A. Three-phase totals: G1 consumes 5.02 kW and generates 1.46 kvar; G2 generates 5.48 kW and 0.37 kvar; the line absorbs 0.46 kW and 1.82 kvar. Real power flows from G2 to G1.

## Answer

The given terminal voltages are taken as **per-phase (line-to-neutral)** values, since the generators are star connected with neutral grounded and the impedance is per phase. Three-phase power is 3 times the per-phase power.

Let I flow from G1 to G2 through $Z = 2 + j8\ \Omega$. Then $S_1 = V_1I^*$ is the power **delivered** by G1 and $S_2 = V_2I^*$ is the power **received** by G2.

#### Current
$$
\begin{aligned}
V_1 &= 200\angle30^\circ = 173.21 + j100.00\ \text{V}\\
V_2 &= 210\angle50^\circ = 134.99 + j160.87\ \text{V}\\
I &= \frac{V_1 - V_2}{Z} = \frac{38.22 - j60.87}{2 + j8}\\
&= -6.037 - j6.287 = 8.716\angle-133.84^\circ\ \text{A}
\end{aligned}
$$

#### Powers per phase
$$
\begin{aligned}
S_1 &= V_1I^* = (200\angle30^\circ)(8.716\angle133.84^\circ) = -1674.31 + j485.19\ \text{VA}\\
S_2 &= V_2I^* = (210\angle50^\circ)(8.716\angle133.84^\circ) = -1826.24 - j122.55\ \text{VA}\\
S_{line} &= |I|^2Z = 75.97(2 + j8) = 151.94 + j607.74\ \text{VA}
\end{aligned}
$$

#### Interpretation
| Quantity | Per phase | Three-phase |
|---|---|---|
| G1 real power | consumes 1674.31 W (motoring) | consumes 5022.92 W |
| G1 reactive power | generates 485.19 var | generates 1455.57 var |
| G2 real power | generates 1826.24 W | generates 5478.73 W |
| G2 reactive power | generates 122.55 var | generates 367.66 var |
| Line loss | 151.94 W, 607.74 var | 455.81 W, 1823.23 var |

Check: $1826.24 - 1674.31 = 151.93$ W and $485.19 + 122.55 = 607.74$ var.

Real power flows from G2 to G1 because $V_2$ leads $V_1$ by 20°. G1 behaves as a synchronous motor. Both machines supply reactive power, which is consumed by the line reactance.

**Answer:** G2 generates 5.48 kW and 0.37 kvar. G1 absorbs 5.02 kW (motor) and generates 1.46 kvar. The line absorbs 0.46 kW and 1.82 kvar (three-phase totals).

## Check yourself

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

1. The given voltages are line-to-neutral values. How is the total three-phase power found?
   A. sqrt3 times the per-phase power
   B. Three times the per-phase power
   C. Same as per-phase power
   D. Per-phase power divided by 3
   Answer: B. For balanced systems S(3-phase) = 3 x per-phase S. (Key point: Per-phase vs three-phase)

2. How is the line current found (flow from G1 to G2)?
   A. I = (V1 - V2)/(2 + j8)
   B. I = (V1 + V2)/(2 + j8)
   C. I = (V1 - V2) x (2 + j8)
   D. I = V1/(2 + j8)
   Answer: A. The voltage difference across the series impedance drives the current. (Key point: Current in the line)

3. With I taken from G1 to G2, which gives the power delivered by G1?
   A. S1 = V1* I*
   B. S1 = V1 I
   C. S1 = |I|^2 Z
   D. S1 = V1 I*
   Answer: D. Complex power is voltage times conjugate of current; |I|^2Z is the line loss. (Key point: Power formula)

4. Which machine acts as a motor and why?
   A. G2, because V2 has the larger magnitude only
   B. G1, because V1 leads V2
   C. G1, because V2 leads V1 by 20° so real power flows from G2 to G1
   D. Neither, since the line has resistance
   Answer: C. Real power flows from the leading voltage to the lagging one; S1 has negative real part. (Key point: Direction of real power)

## Related in Power System Analysis I

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- [Reactance diagram: 50 MVA, 13.8 kV base, three generators](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-50mva-13-8kv) — Power System Analysis I, chapter 3

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