01․ When the supply terminal of a DC shunt motor are interchanged
The directionof eletromagnetic torque developed by armature depends upon the direction of flux or magnetic field and the direction of flow of current in armature conductors. If either of the two is reversed the direction of torque developed and direction of rotation will be reversed. When both (the direction of field as well as that of armature current) are reversed the direction of rotation will not change. So in case of DC shunt motor, if the supply terminals are interchnaged, there are no change in direction of rotation.
02․ The direction of rotation of a DC compound motor can be reversed by interchanging
The direction of rotation of a DC compound motor may be conveniently reversed by reversing the connection involve both series and shunt field winding may reversed the direction of rotation.
03․ The speed regulation of a DC motor is given as
The speed regulation of a DC motor is defined as the change in speed when the load on the motor is reduced from full load to zero and is expressed of rated full load speed.
∴ Speed regulation = (NnL - Nfl)/NfL
04․ If the load current and flux of a DC motor are held constant and voltage applied across it armature is increased by 5%, the speed of the motor will
If the load current is constant then the armature current is also constant. Therefore back emf of DC motor is proportioned to the terminal voltage. So increase 5% of terminal voltage means that increase in 5% of back emf. The speed of DC motor is directly proportional to back emf and inversely proportional to flux per pole. Since the flux is constant. Thereby the speed is only depends on the back emf and the speed isis proportional to the back emf. Hence speed is also increased by 5% with in increase 5% of back emf.
05․ If the flux of DC motor approaches zero, its speed will
In DC motor, the relation of speed and flux is given by N ∝ (1/φ). If the flux is zero then the speed will be approached.
06․ The current drawn by a 220 V DC series motor of armature resistance 0.5Ω and back back emf 200 V is
Use Eb = v - IaRa.
Where Eb = back emf.
Ia = armature current or line current.
Ra = Armature resistance.
07․ A 220 V DC machine has an armature resistance 1Ω. If the full load current is 20 A, the difference of induced voltage and generator is
In case of generator, induced emf
Eg = v + IRa.
In case of motor, the induced emf or back emf
Em = V - IRa.
The difference of induced voltage when the machine is running as a motor.
08․ Consider a separately excited DC motor. If the field current is increased while the armature terminal voltage is held constant, then speed of the motor
In case of separately excited DC motor,
N ∝ (1/If).
Where N = Speed and If = field current.
09․ If a DC motor is connected across the AC supply it will
If a DC motor is connected to AC supply, an alternating current passes through the brushes and commutator to the armature winding, while it passes thorugh the commutator it is converted into DC so that the group of conductors under successive field poles carry curent in same direction. As it result the flux per pole remain constant and not vary. So there will produced heat due to flow of eddy current in field winding and the motor will be burned.
10․ The speed of a DC series motor is
In case of DC series motor,
Ia = IL and N ∝ (Eb/φ ).
It is obvious that speed is proportional to back emf Eb and inversely proportional to the flux per pole φ. Now flux per pole is directly proportional to armature current i.e. φ ∝ Ia. So speed is inversely proportional to the armature current.
<<<6768697071>>>
