# 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 = E _{b} + I_{a}R_{a}**

Where,

**V** is the supply voltage,

**I _{a}** is the armature current,

**R _{a}** is the armature resistance.

And the back emf is given by **E _{b}**.

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, E_{b} is also zero, and under this condition, the voltage equation is modified to

**V = 0 + I _{a}R_{a}**

**I _{a} = V/R_{a}**

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.

- 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. - 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.