DC Motor Problem Assignment Help

Electrostatics - DC Motor Problem

DC Motor

A D.C. motor converts direct current energy from a battery into mechanical energy of rotation.

Principle. It is based on the fact that when a coil carrying current is held in a magnetic field, it experiences a torque which rotates the coil.

Constructing it consists of the following five parts
Armature the armature coil ABCD consists of a large number of turns of insulated copper wire wound over a soft iron core.
Field magnet the magnetic field is supplied by a permanent magnet NS.
Split-rings or commentator. These are two halves of the same ring the ends of the armature coil are connected to these halves which also rotate with the armature. 
Brushes these are tow flexible metal plates or carbon rods B1 and B2 which are so fixed that they constantly touch the revolving rings.
Battery the battery consists or a few cells of voltage V connected across the brushes the brushes convey the current to the rings from where it is carried to the armature.

Working the battery sends current through the armature coil in the direction shown in applying Fleming left hand rule CD experiences a force directed inwards and perpendicular to the plane of the coil similarly AB experiences a force directed outwards and perpendicular to the plane of the coil these two forces being equal unalike and parallel form a couple. The couple rotates the armature coil in the anticlockwise direction. After the coil has rotated through 180 the direction of the current in AB and CD is reversed fig 1 now CD experiences an outwards force and AB experiences an inwards force. The armature coil thus continues rotating in the same anticlockwise direction.

Back as the armature rotates in the magnetic field the amount of magnetic flux linked with the coil changes therefore an is induced in the coil. The direction of the induced is such that it opposes the battery current in the circuit. This is called the back and its magnitude goes on increasing with the speed of the armature. 

Let V = applied across B1 and B2

R = resistance of the armature coil,

I = current flowing through the armature coil at any instant t

E = back at that instant.

As V and E are acting in the opposite direction, 

∴ Effective across B1 and B2 = V –E

According to ohm’s law,

I = V – E / R

Or V – IR = E

Efficiency of the motor. Since the current I is being supplied to the armature coil by the external source of V, therefore, 

Input electric power = VI

According to joule’s law of heating 

Power lost in the form of heat in the coil = fR

If we assume that there is no other loss of power then

Power converted into external work

Output mechanical power = VI – fR = (V – IR) I = EI [from (5a)]

 Efficiency of the D.C. motor 

Output mechanical power / input electric power

N = EI / VI = E/V = back / applied 

Maximum efficiency when E = V from (6)

N = 1 or 100%. Therefore for n to be maximum 100% back E should be equal to the applied v. but in that close the current I flowing through the armature coil becomes zero as is clear form. Obviously the motor in this case will just cease to work.

This is an anomaly practically the efficiency of the D.C. motor will be maximum when output mechanical power is maximum.

EI = maximum

Using E . (V – E) / R = maximum 

Now E (V-E) / R will be maximum constant), when its differential coefficient is zero.

D / dE [ E (V – E)/ R] = 0

Or d / dE[1/R (VE – E)] = 0 

Or V – 2E = 0

Or E = V/2 

From (6) n = E/V

When E = V/2 

N = V/2 / V = ½ = ½ X 100% = 50%

Hence a d.c motor delivering maximum output has an efficiency of only 50%

Further when E = V/2 then from current I in the coil may become too large as R is low. Thence in proactive we do not try to get maximum output mechanical power.


1. The D.C. motors are used in D.C. fans (exhaust ceiling or table) for cooling and ventilation 

2. They are used for pumping water.

3. Big D.C. motors are used of running tram-cars and even train.

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