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EMF equation of a DC generator

Short answer

E = PΦZN/(60A) volts. In one revolution each conductor cuts flux ΦP; average conductor emf = ΦPN/60. With Z/A series conductors per parallel path, the terminal emf becomes E = (ΦPN/60)×(Z/A). For lap winding A=P, E = ΦZN/60; for wave winding A=2, E = PΦZN/120.

Derive an emf equation for a dc generator.

Statement: the average emf generated in a DC generator is E_g = PΦZN / (60A) volts.

Derivation

Let

  • P = number of poles
  • Φ = flux per pole (Wb)
  • Z = total number of armature conductors
  • N = speed of the armature (rpm)
  • A = number of parallel paths (lap: A = P, wave: A = 2)

  1. In one revolution, a conductor cuts the flux of all poles: total flux cut = Φ × P (Wb).

  2. Time taken for one revolution = 60/N seconds.

  3. By Faraday's law, average emf induced in one conductor = flux cut / time:

    e = ΦP / (60/N) = ΦPN/60 volts

  4. The Z conductors are divided into A parallel paths, so each path has Z/A conductors in series.

  5. The emf of the generator equals the emf across one path:

E_g = e x (Z/A)
    = (ΦPN/60) x (Z/A)

E_g = PΦZN / (60A) volts

Special cases

  • Lap winding (A = P): E_g = ΦZN/60
  • Wave winding (A = 2): E_g = PΦZN/120

For a given machine the quantity PZ/(60A) is constant, so E_g = KΦN. Hence the emf is directly proportional to the flux per pole and the speed. Terminal voltage V = E_g − I_a R_a − brush drop.

✓ Answer checked by Claude Opus 5.5 (high effort)

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