**Significance of Back EMF in a DC Motor**

As soon as the armature starts rotating inside a magnetic field due to the current in the conductor, the following conditions are set in

a. The conductor is in motion

b. The flux of the main pole exists

c. While the conductor is in motion, it also cuts the main pole flux

This condition creates a situation for inducing e.m.f. in the conductor. The direction of induced e.m.f. as found by the Fleming's Right hand rule, is in direct opposition to the applied voltage across the armature conductor. It is therefore termed as back e.m.f. (Eb). The magnitude of back e.m.f. (back emf in dc motor formula) is given by :

E

_{b}= ZΦNP / 60A
Where,

Z is the number of conductors in the armature

Φ is the flux of the main pole in Webers

N is the speed of rotation of the shaft in r.p.m

P is the number of poles of the machine

A is the number of parallel path of the winding

A = P for Lap winding

A = 2 for Wave winding

This back e.m.f. acts in opposite direction to the applied voltage in the armature as shown in Figure. Hence, the net voltage across the armature would be the difference between the applied voltage and back e.m.f.

i.e., Net voltage across armature = (V — E

_{b}) Volts
Where,

V is the applied voltage across the armature

E

_{b}is the back e.m.f.
If the resistance of the armature is R

_{a}, then the armature current is given by
Armature current = Net voltage across armature /Armature resistance

I

_{a}=(V – E_{b})/ R_{a}amperes
It should be remembered that the magnitude of back e.m.f. is dependent upon the speed of rotation of the shaft (N). If the speed is zero, the back e.m.f. is also zero and as the speed increases the magnitude of back e.m.f, also increases. At stand still or starting position, the shaft is stationary and therefore the back e.m.f. is zero. Rewriting the equation for applied voltage

l

_{a}R_{a}= V—E_{b}or
V= E

_{b}+I_{a}R_{a}
The voltage V applied across the armature has to

a. overcome the back e.m.f. E

_{b}and
b. supply the armature ohmic drop l

_{a}R_{a}
This is known as the voltage equation of d.c. motor. Multiplying the equation on both sides by l

_{a}R_{a}, power equation is obtained.
Power = VI

_{a}= E_{b}I_{a}+ l_{a}^{2}R_{a}
Where,

VI

_{a}is the electrical input to the armature
E

_{b}l_{a}is the electrical equivalent of mechanical power developed in the armature
I

_{a}^{2}R_{a}is the copper loss in the armature
Hence, it would be seen that some power is wasted in the armature out of the input and the rest only is converted into mechanical power within the armature. The gross mechanical power developed by the motor is given by

P

_{m}= VI_{a}— I_{a}^{2}R_{a}
Differentiating both sides with respect to l

_{a}and equating the result to zero,
dP

_{m}/dI_{a}= V — 2I_{a}R_{a}=0
Therefore

I

_{a}R_{a}= V/2
Thus, the gross mechanical power developed by the d.c. motor will be maximum when the back e.m.f. is equal to the half of the applied voltage. If this condition is taken into account, then half the input would be wasted in the form of heat and taking into other losses like frictional and magnetic losses, the motor efficiency would fall to 50%. As such this condition is not realised in practice.

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