## Effect of Change in Rotor Resistance on Torque

It is shown that in slip

ring induction motor, externally resistance can be added in the rotor.

Let us see the effect of change in rotor resistance on the torque

produced.

_{2 }= Rotor resistance per phase

_{2}

^{2}R

_{2})/?(R

_{2}

^{2}+(s X

_{2})

^{2})

_{2}

**‘**= New rotor resistance per phase

**‘**?

**(s E**

_{2}

^{2}R

_{2}

**‘**)/?(R

_{2}

**‘**

^{2}+(s X

_{2})

^{2})

_{2 }and R

_{2}

**‘**can be written as

T

_{st }

**? (E**

_{2}

^{2}R

_{2})/?(R

_{2}

^{2}+(X

_{2})

^{2})

**‘**st ? (E

_{2}

^{2}R

**‘**

_{2})/?(R

**‘**

_{2}

^{2}+(X

_{2})

^{2})

_{m}

**? (E**

_{2}

^{2})/(2X

_{2})

**Key Point**: It can be observed that T

_{m}

**is independent of R**

_{2}

hence whatever may be the rotor resistance, maximum torque produced

never change but the slip and speed at which it occurs depends on R

_{2}.

_{2}, s

_{m}= R

_{2}/X

_{2}where T

_{m}

**occurs**

_{2}

**‘**, s

_{m}

**‘**= R

_{2}

**‘**/X

_{2}

**‘**where same T

_{m}

**occurs**

_{2}

**‘**> R

_{2}, the slip s

_{m}

**‘**> s

_{m}.

Due to this, we get a new torque-slip characteristics for rotor

resistance . This new characteristics is parallel to the characteristics

for with same but T

_{m}

**occurring at s**

_{m}

**‘**. The effect of change in rotor resistance on torque-slip characteristics shown in the Fig. 1.

**‘**st for R

_{2}

**‘**is more than Tst for R

_{2}. Thus by changing rotor resistance the starting torque can be controlled.

_{2}

**‘**and so on, it can be seen that T

_{m}

**remains same but slip at which it occurs increases to s**

_{m}

**‘**and so on. Similarly starting torque also increases to T

**‘**st

**and so on.**

Fig. 1 Effect of rotor resistance on torque-slip curves |

If maximum torque Tm is required at start then s_{m} = 1 as at start slip is always unity, so

s_{m }= R_{2}/X_{2} = 1

R_{2} = X_{2} Condition for getting Tst = T_{m}** **

**Key Point**: Thus by adding external resistance to rotor till it becomes equal to X

_{2}, the maximum torque can be achieved at start.

It is represented by point A in the Fig. 1.

^{2}

R) and hence efficiency of the motor will be very poor. Hence such

added resistance is cut-off gradually and finally removed from the rotor

circuit, in the normal running condition of the motor. So this method

is used in practice to achieve higher starting torque hence resistance

in rotor is added only at start.

Thus good performance at start and in the running condition is ensured.

**Key Point**: This is possible

only in case of slip type of induction motor as in squirrel cage due to

short circuited rotor, extra rotor resistance can not be added.

**Example**:

Rotor resistance and standstill reactance per phase of a 3 phase

induction motor are 0.04 ? and 0.2 ? respectively. What should be the

external resistance required at start in rotor circuit to obtain.

i) maximum torque at start ii) 50% of maximum torque at start.

**Solution **:

R_{2} = 0.04 ?, X_{2} = 0.2 ?

i) For T_{m}** **= T_{st}** ** , s_{m}** ** = R_{2}**‘**/X_{2} = 1

**. ^{.}.** R

_{2}

**‘**= X

_{2}= 0.2

Let R

_{ex}= external resistance required in rotor.

R

_{2}

**‘**= R

_{2}+ R

_{ex}

**.**R

^{.}._{ex}= R

_{2}

**‘**– R

_{2}= 0.2 – 0.04 = 0.16 ? per phase

ii) For T

_{st}

**= 0.5 T**

_{m},

Now T

_{m}= (k E

_{2}

^{2})/(2 X

_{2}) and

T

_{st}

**= (k E**

_{2}

^{2}R

_{2})/(R

_{2}

^{2}+ X

_{2}

^{2})

But at start, external resistance R

_{ex}is added. So new value of rotor resistance is say R

_{2}

**‘**.

R

_{2}

**‘**= R

_{2}+ R

_{ex}

**.**T

^{.}._{st}

**= (k E**

_{2}

^{2}R

_{2}

**‘**)/(R

_{2}

**‘**

^{2}+ X

_{2}

^{2}) with added resistance

but T

_{st}

**= 0.5T**

_{m}required.

Substituting expressions of T

_{st}

**and T**

_{m}, we get

(k E

_{2}

^{2}R

_{2}

**‘**)/(R

_{2}

**‘**

^{2}+ X

_{2}

^{2}) = 0.5 (k E

_{2}

^{2})/ (2X

_{2})

**.**4 R

^{.}._{2}

**‘**X

_{2}= (R

_{2}

**‘**

^{2}+ X

_{2}

^{2})

**.**(R

^{.}._{2}

**‘**

^{2}) – 4 x 0.2 x R

_{2}

**‘**+ 0.2

^{2}= 0

**.**(R

^{.}._{2}

**‘**

^{2}) – 0.8 R

_{2}

**‘**+ 0.04 = 0

**.**R

^{.}._{2}

**‘**=

**{**0.8 + ?(0.8

^{2}– 4 x 0.04)

**}**/2

**.**R

^{.}._{2}

**‘**= 0.0535 , 0.7464 ?

But R

_{2}

**‘**can not greater than X

_{2}hence,

R

_{2}

**‘**= 0.0535 = R

_{2}

**+ R**

_{ex}

**.**0.0535 = 0.04 + R

^{.}._{ex}

**.**R

^{.}._{ex }= 0.0135 ? per phase

This is much resistance is required in the rotor externally to obtain T

_{st}

**= 0.5 T**

_{m}.

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