It can easily be seen from
the slip torque characteristics of an induction motor that, there is some
finite torque when the slip s=1 i.e. speed is zero. This simply means that
Induction Motor is a self-starting motor and begins to rotate on its own when
connected to a 3 phase supply.

At the instant of starting,
a three phase Induction Motor behaves like a Transformer with its secondary
winding shorted. Therefore, Induction Motor during starting takes a high
current from the supply mains. To limit this high starting current of Induction
Motor, different starting methods are used. In this post we will have a look at
the Auto-Transformer Starting Method of Induction Motor.

**A schematic diagram for Auto-Transformer Starting of an Induction Motor is given below.**

*The main philosophy of starting any Induction Motor is to start it at a reduced voltage and as soon as the motor reaches its rated speed, full supply voltage is applied to the terminals of Induction Motor.*
It shall be observed that,
using Auto-Transformer we are only applying a reduced voltage xV

_{1}to the Stator terminal of Induction Motor. Here x is less than 1. As soon as the Induction Motor reaches its rated speed, full supply voltage is applied to the terminals of stator.
Therefore, per phase starting
current of Motor = xV

_{1}/Z_{sc }= xI_{sc}_{}

Here Isc is the current
through the stator during direct switching of motor.

Thus we observe that
starting current of Motor has reduced and is x times that of current during DOL
(Direct Online) starting.

Again,

Input VA of Auto-Transformer
= Output VA of the Auto-Transformer

I

_{st}.V_{1}= xV_{1}(xI_{sc})
Therefore,

Per phase starting current
from the Supply Mains I

_{st}= x^{2}I_{sc}_{}

Thus per phase starting
current from Supply Mains has now became x

^{2}times that of DOL current. Mind that it has reduced as x is less than 1 so x_{2}will be much less than 1. Thus the main advantage of using Auto-Transformer is that it reduces the starting current from the Supply Mains by x^{2}times.
Note that starting current
is the motor winding is x times while the starting current from the supply
mains has became x

^{2}time of DOL starting current.
Now we will have a look at
Torque of Induction Motor.

As the torque of an
Induction Motor is directly proportional of square of applied voltage at the
stator terminals, therefore

Starting Torque with
Auto-Transformer Test = K(xV

_{1})^{2}where K is constant of proportionality.
Starting Torque with DOL
Test = KV

_{1}^{2}where K is constant of proportionality.
Therefore,

Test with Auto-Transformer/
Test with DOL = x

^{2}^{}

Thus starting torque with
Auto Transformer is less than the starting torque with DOL starting by a factor
of x

^{2}.**Have question? Write in comment box. Thank you!**

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