---
title: "Sync generator (6600 V) leading voltage"
subject: "Electrical Machine"
chapter: 6
key: "semester_4/electrical_machine:sync-generator-1200kva-6600v-leading"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/sync-generator-1200kva-6600v-leading
site: BCT Engineering
---

# Sync generator (6600 V) leading voltage

Electrical Machine · Chapter 6: Three Phase Synchronous Machines

- Asked in: 2070 Bhadra, 8 marks
- Video: [Alternator | Voltage regulation | Synchronous Machines | Lec-17](https://www.youtube.com/watch?v=trgOIXW1018) — Education 4u
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/sync-generator-1200kva-6600v-leading
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-6

## Question

A 1200 KVA, 6600 V, 3-phase star connected stator of a synchronous generator has an armature resistance of 0.4 Ω/phase and synchronous reactance of 6 Ω/phase. The generator delivers full load current at pf of 0.8 lagging at normal rated voltage. Calculate the terminal voltage for the same excitation and load current at 0.8 pf leading.

## Short answer

Full-load current I=1200kVA/(√3×6600V)=105 A/phase, Vph=3810 V. For 0.8 lagging, E=4249 V. For 0.8 leading, solving gives Vt=4560 V/phase, line voltage=7.90 kV.

## Answer

**Given:** 1200 kVA, 6600 V (line), 3-phase, star; R_a = 0.4 Ω/phase, X_s = 6 Ω/phase; full-load current at 0.8 pf lagging at rated voltage. Find the terminal voltage for the same excitation and current at 0.8 pf leading.

#### Step 1: Rated quantities
I = 1 200 000 / (√3 × 6600) = 104.97 A
V_ph = 6600/√3 = 3810.5 V
I R_a = 104.97 × 0.4 = 41.99 V; I X_s = 104.97 × 6 = 629.8 V

#### Step 2: Induced emf (0.8 lagging, cos φ = 0.8, sin φ = 0.6)
E = V + I (R_a + jX_s), where the current lags V by 36.87°:

```text
 E_real = V + IR cosφ + IX sinφ
        = 3810.5 + 41.99(0.8) + 629.8(0.6)
        = 3810.5 + 33.59 + 377.9 = 4222.0 V
 E_imag = IX cosφ - IR sinφ
        = 629.8(0.8) - 41.99(0.6) = 503.9 - 25.2 = 478.7 V
 E = √(4222.0² + 478.7²) = 4249.1 V per phase
```

#### Step 3: Terminal voltage at 0.8 leading (same E, same I)
For a leading current:
- E_real = V_t + IR cos φ − IX sin φ = V_t + 33.59 − 377.9 = V_t − 344.3
- E_imag = IX cos φ + IR sin φ = 503.9 + 25.2 = 529.1

```text
 4249.1² = (Vt - 344.3)² + 529.1²
 18 054 000 = (Vt - 344.3)² + 279 950
 (Vt - 344.3)² = 17 774 000   ->   Vt - 344.3 = 4216.0
 Vt = 4560.3 V per phase
 VL = √3 × 4560.3 = 7898.7 V
```
| Condition | Phase voltage | Line voltage |
|---|---|---|
| 0.8 lagging | 3810.5 V | 6600 V |
| 0.8 leading, same excitation | 4560 V | 7899 V |

**Answer: terminal voltage ≈ 7.90 kV (line), 4560 V per phase.**

## Check yourself

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

1. What is full-load current for 1200 kVA at 6600 V?
   A. 104.97 A
   B. 115.47 A
   C. 181.8 A
   D. 200 A
   Answer: A. 1,200,000/(sqrt(3) x 6600) = 104.97 A. (Key point: Rated current)

2. What is E per phase at 0.8 lagging?
   A. 3810 V
   B. 4249 V
   C. 6600 V
   D. 4560 V
   Answer: B. E = sqrt(4222^2 + 478.7^2) = 4249 V. (Key point: Induced emf)

3. Which equation is solved for Vt with the same E?
   A. E^2 = (Vt - 344.3)^2 + 529.1^2
   B. E = Vt
   C. E^2 = (Vt + 344.3)^2 + 529.1^2
   D. E = Vt - IZ
   Answer: A. For leading current, the real part is Vt + IR cos(phi) - IX sin(phi). (Key point: Leading load equation)

4. What is the line terminal voltage?
   A. 6600 V
   B. 7899 V
   C. 4560 V line
   D. 5000 V
   Answer: B. Vt = 4560 V per phase, 7899 V line. (Key point: Result)

## Close variants set in other exams

- [Sync generator terminal voltage, leading pf](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Async-generator-1200kva-leading-voltage) — Electrical Machine, chapter 6, asked 2×

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