Test: Electrical Machines- 3 - Electrical Engineering (EE) MCQ

We know that,

P = No. of pole
ϕ = Flux per pale
N = Speed
A = No. of parallel path
A = 2 for wave winding
A = P for lap winding

Since the coil sides lies under opposite poles the induced emf’s in them are additive. Hence it will be 20 V.

For a double layer winding (universally used) number of commutator segments = number of armature coils. The number of slots in double layer winding is equal to the number of coils and the number of commutator segments is also the same.

Field current If = 23045 = 5.11A
Armature current I_a = I_f + I_L = 5.11 + 50 = 55.11 A

Stray loss = 400 W
Total loss = 400 + 30.37 + 1175.55 = 1605.92 W
Output power = 230 × 50 = 11500

Pole pitch = distance between two adjacent poles


Area of pole face = pole arc times the axial length 

When working as generator, armature current, 
= 26.67A
E_g = 300 + (26.67 × 0.5)
= 313.33 V
As motor, E_m = V – I_aR_a
= 300 – (26.67 × 0.5)
= 286.67 V
Now, E α Nϕ

Given that, Number of Poles (P) = 6
Number of slots = 32
Number of conductors = 12 per slot
The conductors (z) = 32 × 12 = 384
Speed (N) = 1000 rpm
Useful flux (∅) = 50 mwb
emf generated

for lap wound machine, A = p

= 320 V
For wave wound machine, A = 2

Load resistance = 150/200 = 0.75Ω
E₁ = 150 + (200 × 0.25) + 2
= 202 V
N₁ = 1500 rpm, N₂ = 1000 rpm

= 134.67 V
V = 134.67 – 0.25 I – 2
I = Load current
V = 132.67 – 0.25 I

Here, the number of pole = 2
As the number of poles are 2, the number of parallel paths for any kind of winding (either lap or wave) = 2
No of conductors connected in a path = 500
Equivalent resistance of a parallel path = 500 × 2 mΩ
Equivalent armature resistance of two parallel paths,
∴ Armature resistance 
At 1055 rpm
E_a1 = V_t + (I_L + I_f) × 0.5 = 100 + (10 + I_f) 0.5
At 1105 rpm
E_a2 = 100 + (20 + I_f) × 0.5
E_a1 ​∝ N₁ ​∝ 1055
E_a2 ∝​ N₂ ​∝ 1105

105500 + 10550 + 527.5 I_f = 110500 + 5525 + 552.5 I_f
I_f = 1 A
∴ Field circuit resistance =