Shunt motor 480 V, 740 conductors

The armature and field resistances are not readable in the question. Assumed values: R_a = 0.2 Ω and R_f = 240 Ω (so the field current is 2 A). Winding type is also not stated, so a lap winding (A = P = 6) is assumed.

Equivalent circuit

 I_L = 120 A --+--> Ia --[Ra=0.2]--(Eb)--+
               |                          |
               +--> If --[Rf=240]---------+--- V = 480 V

(i) Armature current

• I_f = 480 / 240 = 2 A
• I_a = 120 − 2 = 118 A

Back emf

E_b = V − I_aR_a = 480 − 118 × 0.2 = 480 − 23.6 = 456.4 V

(ii) Speed

E_b = PΦZN / (60A), so N = 60 A E_b / (P Φ Z)

N = (60 × 6 × 456.4) / (6 × 0.08 × 740) = 164 304 / 355.2

N = 462.6 rpm

(iii) Torque developed by the armature

ω = 2πN/60 = 2π × 462.6 / 60 = 48.44 rad/s

T = E_b I_a / ω = (456.4 × 118) / 48.44 = 53 855 / 48.44

T = 1112 N·m

Check: T = PΦZ I_a / (2πA) = (6 × 0.08 × 740 × 118) / (2π × 6) = 1112 N·m ✓

If the armature is wave wound (A = 2), the speed would be 154.2 rpm and the torque 3335 N·m.

Answer: I_a = 118 A, N ≈ 463 rpm, T ≈ 1112 N·m (with the assumed R_a = 0.2 Ω, R_f = 240 Ω, lap winding).