This is the simplest mode of
starting of Induction Motor where the stator is directly connected to the mains
supply as shown in figure below.

The motor starts with its
own characteristics. When it is switched on, the motor behaves like a
transformer with its secondary, formed by the very low resistance rotor cage,
in short circuit. There is a high induced current in the rotor which results in
a current peak in the mains supply:

Current on starting = 5 to 8 times of rated current.

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Such a large current of
short duration do not harm rugged squirrel cage induction motor, but the high
current may cause objectionable voltage drop in the power supply line feeding
to the Induction Motor. This large voltage drop causes undesirable dip in the
supply line voltage which will affect the operation of other equipment
connected to the same source of supply. If the supply system is of sufficient
power capacity and high starting current of squirrel cage induction motor do
not cause objectionable voltage drop in the supply line voltage then DOL
starting should be preferred.

Now have better
understanding of DOL starting of Squirrel Cage Induction Motor, we will
calculate the ratio of Starting Torque T

_{est}and Full Load Torque T_{fl}.
As we know that torque in
Induction Motor is given as,

T

_{e}= I_{2}^{2}r_{2}/ w_{s}s
Where I

_{2}is rotor current, r_{2}is rotor resistance, ws is synchronous speed and s is slip.
Therefore ratio of starting
torque and full load torque,

T

_{est}/ T_{fl}= (I_{2st}^{2}/ I_{2fl}^{2}) s_{fl }……..[As slip during starting, s =1] ……….(1)
Here I

_{2st}and I_{2fl}are per phase rotor current during starting and at full load respectively.
From the basic concept of
Induction Motor, we know that Induction motor behaves as a Transformer and
therefore the concept of transformer action can be applied here. Neglecting no
load current we can write as,

Stator mmf = Rotor mmf

I

_{st}x Effective Stator Turns = I_{2st}x Effective Rotor Turns
⇒
I

_{st}= (Effective rotor to stator turn ratio)xI_{2st}_{}

Hence for starting and full
load current we can write as,

I

_{st}/ I_{fl}= [(Effective rotor to stator turn ratio)xI_{2st}] / [(Effective rotor to stator turn ratio)xI_{2fl}]
⇒ I

_{st}/ I_{fl}= I_{2st }/ I_{2fl}………………………..(2)
From equation (1) and (2),

T

_{est}/ T_{fl}= (I_{st}^{2}r_{2}/ I_{fl}^{2}r_{2}) s_{fl}_{}

⇒ T

_{est}/ T_{fl}= (I_{st}/ I_{fl})^{2}s_{fl}_{}

But during starting an
induction motor behaves as a Transformer having secondary shorted, hence
starting current will be short circuit current I

_{sc}. Therefore,
T

_{est}/ T_{fl}= (I_{sc}/ I_{fl})^{2}s_{fl}_{}

From the above equation we
observe that starting torque in DOL or Direct-on-Line Starting is high. The
average starting torque is in DOL starting,

Torque on starting = 0.5 to
1.5 rated Torque.

In spite of advantages of
DOL or Direct-on-Line Starting (being simple equipment, high starting torque,
fast start, low cost), direct on-line starting is only suitable when for the
following cases:

1) The
power of the motor is low compared to that of the mains, which limits
interference from inrush current.

2) The
machine to drive does not need to speed up gradually or has a damping device to
limit the shock of starting.

3) The
starting torque can be high without affecting machine operation or the load
that is driven.

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