Question

I need help with all of these questions because they are related to one question. Please show all working so I fully understand.

Answers:

4c) T = 11.4 Nm

4. a) What is mcant by self-excitation. Explain the difference between a self excited shunt wound and series wound motor. Use a diagram if necessary. Explain in detail the working principle of a DC motor using clearly labelled diagrams. b) 10 c The armature resistance of a shunt wound DC generator is 1 2 and that of 10 the field winding 150 Ω. When the generator runs at 1200 rpm it supplies a curent of 20 A with an output volage of so v Neglecting mechanical losses what torque is required to turn the generator armature.

Answer 1

Ans)

4

a) The DC motor works on the principle which need interaction of field flux and armature flux for that we need two winding's field winding and armature winding ,the field can be magnet , instead when winding is used we need DC supply then it acts as electromagnet ,We can avoid the need of separate DC supply for field winding if it has residual magnetism and connected in parallel with armature winding and supply voltage from that we can build voltage this process is called self excitation

or

We can say in self excited motor some of the power from rotor is used to power the field winding's to produce flux

null

In shunt wound motor field winding is connected in parallel with the armature and its speed is almost constant with the mechanical load

Whereas in series wound motor field winding is connected in series with the armature winding in this motor speed varies with the mechanical load

==============

b)

The  working principle of a DC motor is that whenever a current carrying conductor is placed in a magnetic field, it experiences a mechanical force which is called Lorentz force,The Flemming left-hand rule gives the direction of the force which is If the thumb, middle finger and the index finger of the left hand are displaced from each other by an angle of 90°, the middle finger represents the direction of the magnetic field. The index finger represents the direction of the current, and the thumb shows the direction of forces acting on the conductor

=======================

c)

The generator and its equivalent is shown below

​​​​​​​

The field current is

$I_{sh}=\frac{V}{R_{sh}}=\frac{50}{150}=0.333\: \: A$

The current from armature i s

$I_{a}=I+I_{sh}=20+0.333=20.333\: \: A$

$I_{a}=20.333\: \: A$

The back emf is given as

$E=V+I_{a}R_{a}=50+20.333*1=70.333\: \: V$

$E=70.333\: \: V$

The speed is given as N=1200 RPM

The torque required is given as

$T=\frac{EI}{2\pi \frac{N}{60}}=\frac{70.333*20.333}{2\pi \frac{1200}{60}}=11.4\: \: N-m$

$\mathbf{T=11.4\: \: N-m}$