# A motor takes a large current at starting because

A motor takes a large current at starting because

### Right Answer is: Back e.m.f. is low

#### SOLUTION

The Voltage equation of DC Motor is given by

V = Eb + IaRa

Where,

V is the supply voltage,

Ia is the armature current,

Ra is the armature resistance.

And the back emf is given by Eb.

Now the back emf, in case of a DC motor, is very similar to the generated emf of a DC generator as it’s produced by the rotational motion of the current carrying armature conductor in presence of the field. This back emf of DC motor is given by

${E_b} = \dfrac{{P\Phi ZN}}{{60A}}$

Where

P – Number of poles of the machine

ϕ – Flux per pole in Weber.

Z – Total number of armature conductors.

N – Speed of armature in revolution per minute (r.p.m).

A – Number of parallel paths in the armature winding.

From this equation, we can see that Eb is directly proportional to the speed N of the motor. Now since at starting N = 0, Eb is also zero, and under this condition, the voltage equation is modified to

V = 0 + IaRa

Ia = V/Ra

For all practical practices to obtain optimum operation of the motor the armature resistance is kept very small usually in the order of 0.5 Ω − 0.10Ω and the bare minimum supply voltage being 240 volts.

Even under this circumstance the starting current, Ia is as high as 240/0.5 amp = 480 amp.
Such high starting current of DC motor creates two major problems.

1. Firstly, the current of the order of 400 A has the potential of damaging the internal circuit of the armature winding of DC motor at the very onset.
2. Very high electromagnetic starting torque of DC motor is produced by virtue of the high starting current, which has the potential of producing the high centrifugal force capable of flying off the rotor winding from the slots.
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