-
A field excitation of 20 A in a certain alternator results in an armature current of 400 A in short circuit and a terminal voltage of 2000 V on open circuit. The magnitude of the internal voltage drop within the machine at a load current of 200 A is
-
- 1 V
- 10 V
- 100 V
- 1000 V
Correct Option: D
Synchronus impedance,
| Zs = | = | = 5 Ω | ||
| ISC | 400 |
∴ Voltage drop = IZs
= 200 × 5 = 1000 V
