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Terms and conditions

A field excitation of 20 A in certain alternator results in an armature current of 400 A in a short circuit and a terminal voltage of 2000 V on an open circuit. The magnitude of the internal voltage drop within the machine at a load current of 200 A is

Given parameters:

  • Field excitation If = 20 A
  • Short circuit armature current Isc = 400 A
  • Open circuit terminal voltage Voc = 2000 V
  • Load current IL = 200 A

The generated voltage per phase of an alternator is given by:

Ef = Vt + ILRa + ILXL

If we neglect the armature resistance, then the generated voltage becomes:

Ef = VPH + ILXL

Under short circuit conditions,

Ia = Isc and Vt = 0.

Thus, equation (1) becomes:

XL = Ef / Isc

For open circuit condition,

Ia = 0 and Ef = Voc.

Substituting the value of Ef in equation (1), we get:

XL = Voc / Isc = 2000 / 400 = 5 Ω

The internal voltage drop is given by the product of armature current and synchronous reactance:

Internal Voltage Drop = Ia × XL = 200 × 5 = 1000 V