A 4-pole, lap-wound armature has 144 slots with two coil sides per slot, each coil having two turns. If the flux per pole is 20 mWb and the armature rotates at 720 rpm, what is the induced voltage?
Answers
Answer:
I don't know sorry
Given:
Number of poles of the armature = 4
Number of slots of the armature = 144
Number of coil per slot of the armature = 2
The flux per pole is = 20 m Wb
The number of rotations of the armature = 720 rpm
To Find:
The induced voltage.
Solution:
Here, the given number of poles is 4 which would be equal to the number of parallel paths in the armature since the armature is Lap-wound, as we know in lab-wound DC machines number of parallel paths denoted by A is equal to the number of poles denoted by P.
Therefore, A = P = 4
Now, given the flux per pole, Φ = 20mWb = 20 × 10⁻¹³ Wb and the number of rotations, N = 720 RPM.
Also, the number of slots is 144 with two coils sides per slot.
Hence, the total number of conductors in the armature,
Z = 144 × 2 × 2 = 576.
Therefore, induced emf across the armature is given by the formula,
Eg = ΦZNP/ 60A
= 20 × 10⁻¹³ Wb × 576 × 720 RPM × 4/ 60 × 4
= 138. 24V
Hence, the induced voltage is 138. 24V.