DIRECT CURRENT MOTOR
An electric motor is a device for converting electrical energy into mechanical energy.
It is based on the fact that a current carrying coil experiences a torque in a magnetic field
A simple d.c motor consists of an armature ABCD (a coil of a large number of turns) wound on an iron core. It is placed in the magnetic field produced due to flow of current through the field winding. The ends of the armature are connected to two split rings R1 and R2 ( two halves of a single ring). The two split rings alternately come in contact with carbon brushes B1 and B2 during during the rotation of the armature. A battery is connected across the two brushes, so as to send d.c. through the armature.
DIRECT CURRENT MOTOR
Suppose that initially the armature is horizontal, such that the split ring R1 connected to arm AB is in contact with brushe B1,while the split ring R2 connected to arm CD is in contact to brush B2,the conventional current flows through the armature in the direction DCMA.According to Fleming’s left hand rule,an upward force equal to N BI l acts on arm AB,while an equal downward force N BI l acts on arm CD.The two equal and opposite forces acting on arms AB and CD of the armature constitute a torque and as a result of it,the armature starts rotating in clockwise direction.As such,during rotation of the coil,the perpendicular distance between the forces acting on the two arms goes on decreasing and becomes zero,when the armature becomes vertical.As a result,the armature tends to come to rest on reaching the vertical position.However,due to inertia,the armature is carried past the vertical position.As it happens,the split ring R1 comes in contact with brush B2,while the split ring R2 assumes contact with brush B1.Due to this,the direction of the flow of current through the armature gets reversed(earlier arm AB was moving up and CD down,but now arm AB moves down,while CD moves up),the direction of rotation of the armature remains unchanged leading to an uninterrupted rotatory motion of the armature in one direction.
Back e.m.f The armature of a motor continuously rotates inside the magnetic field.As the magnetic flux linked with the armature changes during its rotation,and induced e.m.f,say e is produced in the armature.According to the Lenz’s law,this induced e.m.f.If E is the applied battery e.m.f. and R,the resistance of the armature,then current through the armature is given by I= effective e.m.f/resistance of the armature=E-e/R (3.25)
It may be pointed out that when electric motor is just switched on,the magnitude of back e.m.f is quite small due to the low speed of the motor.As such,the battery e.m.f sends current through the armature,which may be large enough to burn the motor by breaking the insulation of the coils.To avoid damage to the motor on this account,a motor starter is used in series with the motor.
The motor starter also saves the motor from damage,if the electric supply fails by chance,while the motor is running.It is because the magnitude of back e.m.f is very large at break.
Efficiency of an electric motor.The effeciency of a motor decreases both due to the production of back e.m.f. and the production of heat in the armature coil.
The efficiency of an electric motor is defined as the ratio of the output power to the input power.
Thus, n=output power/input power (3.26)
Now,input power=battery e.m.f. *current through amature=eL,where I is current through the armature and is given by the equation
Heat produced per second in the armature=I²R
Therefore,output power=EI -I²R=(E-IR) *I
From the equation,substituting for (E-IR),we have
Now the equation form as by substituting out power and input power
Efficiency= eI/EI = e/ E
Thus, efficiency of a motor is equal to the ratio of the back e.m.f. produced in the applied battery e.m.f.
DIRECT CURRENT MOTOR
Condition for maximum efficiency:
From the above equation , it follows tha efficiency of a motor will be 1 (maximum), if back e.m.f setup in the armature is equal to applied battery e.m.f. In that event , the current through the armature will be zero and the motor will not run at all. Therefore, condition for maximum efficiency of the motor is obtained by finding the condition, so the output power is maximum.
Now , output power = eI = e(E-e)/R
The output power varies with back e.m.f. Therefore, output power wil be maximum if
Thus ,an electric motor will be maximum efficient I.e. the motor will deliver maximum output power, when the back e.m.f. set up in the armature is half the value of the applied battery e.m.f.
When the motor is delivering maximum output power, then
Efficiency =( E/2)/E = 1/2 (I.e 50%)
Thus , while delivering maximum output power , the motor has an efficiency of 50%.