Answer: Back EMF \(E_b=V-I_aR_a=220-20\times0.5=210\) V.
Sketch (describe in the exam with labels): outer cylindrical yoke; field poles with pole shoes bolted to the yoke and carrying field coils; a slotted laminated armature core on the shaft in the centre with armature winding in the slots; a commutator of copper segments at one end of the armature; carbon brushes pressing on the commutator; bearings and end covers.
| Part | Function |
|---|
| Yoke | Mechanical support and the return path for magnetic flux |
| Field poles and pole shoes | Carry the field winding; spread the flux uniformly in the air gap |
| Field winding | Produces the main magnetic field when excited |
| Armature core | Laminated to reduce eddy loss; houses the armature conductors |
| Armature winding | Current-carrying conductors on which torque is produced |
| Commutator | Reverses current in armature conductors so torque stays unidirectional |
| Brushes | Carbon contacts that connect the external supply to the rotating commutator |
Back EMF: when the armature rotates in the magnetic field, its conductors cut flux and an EMF is induced in them (generator action). By Lenz's law this EMF opposes the applied voltage, so it is called back (counter) EMF: \(E_b=\dfrac{\Phi ZNP}{60A}\). The armature current is \(I_a=\dfrac{V-E_b}{R_a}\). Significance: \(E_b\) makes the motor self-regulating (at light load the speed and \(E_b\) rise and \(I_a\) falls; at heavy load the reverse happens). At starting \(E_b=0\), so a starter is needed to limit the large current.
Numerical: \(E_b=V-I_aR_a=220-20\times0.5=210\) V.
