Question

A4 pole DC generator with wave wound armature has 51 slots, each having 24 conductors The flux/pole is 0.01 W. The speed
with which it must be rotated to give an induced emf of 220V is
- гpm​

Answer 1

Answer:

Direct current machines (DC

machines)

• DC Machines: Rotating electrical machine.

• Example: Generator and motor.

• DC generator: Converts mechanical power to

electrical power of DC nature.

• DC motor: Converts electrical power into

mechanical power (converse of DC generator).

• Working principle: Faraday’s law of EM

induction (as already discussed during

transformer lectures).

Construction

• Armature (Rotatory part): Comprising of a number of

conductors suitably placed and connected so as to form a

closed winding.

• Field system (stationary part): To produce magnetic field.

• Airgap: Used to separate stator and rotor.

• Commutator: Comprise of large number of commutator

segments, properly insulated from each other.

• Commutator is used to convert an AC wave in the armature

winding into a DC wave at the output terminals (in case of DC

generator), and inverts the DC input wave into an AC wave in

the armature winding (in case of DC motor).

Construction

Armature core (rotor)

DC motor: Working principle

• When a current carrying

conductor is placed in a

magnetic field, it experiences

a torque and has a tendency

to move. This is known as

motoring action.

• If the direction of current in

the wire is reversed, the

direction of rotation also

reverses.

• When magnetic field and

electric field interact they

produce a mechanical force,

and based on that the

working principle of DC motor

is established.

DC motor

DC generator

General expression for resistance

of DC armature winding

• If Z- number of conductors, each of length L.

• S- Cross-section area arranged in ‘a’ pairs of

circuit.

• ‘p’- Pole pairs.

• ρ- Resistivity of the winding material.

• Resistance of any conductor is given as:

L R

S





• ρ- Resistivity of the conducting material in ohm-cm.

• L- Total length of the conductor including overhang.

• S- Cross-section area of the conductor.

• Armature winding consists of Z conductors arranged

in “a” pairs of the circuits.

• Thus, the number of conductors in each parallel

circuit= Z/a

• All the conductors in each parallel circuit are

connected in series. As a result, the resistance of Z/a

conductors connected in series is given by:

L

R

S





L Z R

S a

  

     

• There are “a” parallel circuits in the whole of

the armature winding. Thus the resistance of

armature winding is:

2

LZ LZ 1 1 R

Sa a a S

       

             

Numerical Problem: Calculate the resistance of a 6-pole,

lap connected armature winding using the following data:

Number of slots: 150

Conductor per slot: 8

Mean length of one turn: 250 cm

Cross-section of each conductor: 10 mm × 2.5 mm

Value of resistivity (ρ): 2.1 × 10-6 Ohm-cm

General expression for the

resistance of a DC armature

winding

Solution

• Total number of conductors of armature

winding (Z)= 150 × 8= 1200.

• Number of turns = 1200/2= 600

Explanation: