Please show all work Thumbs up The armature of a dc machine

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The armature of a dc machine contains a lap winding with 240 conductors arranged in 12 equal loops. The effective length of the conductor (under the magnetic field) is 35 cm. The average radius of the armature is 10 cm. The flux density in the machine airgap provided by the stator is 0.65 T. The stator has four poles. How many slots are on the rotor? How many sectors are on the commutator and how many brushes are required? How many parallel current paths are formed in the armature? Compute the average flux under the poles of the machine and the machine constant. What is the back-emf of the machine (voltage induced on the armature) at 2000 rpm (ignoring armature reaction)? The armature current is 5 A. What is the peak current of each conductor? What is the machine electric torque?

Solution

1. no. of lops = 240*12= 2880

2. 6 sectors, 2 brushes

3. parallel path = 240/12 = 20

4.flux density = flux/area = B

average flux density - 2B/Pi

= 1.3/pi

5. E = k*flux*speed

flux = B*a =.0.65*0.035

speed= 2000 rpm

6.average current = 2I/pi = 10/pi