Question

A 250 d.c shunt motor has armature resistance of 0.25ohm,on load it takes an armature current of 50A and runs at 750r.p.m. If the flux of motor is reduced by 10% without changing the load torque, find the new speed of the motor

Answer 1

Answer:

70 +67 +67 ok I think it is right

Answer 2

The new speed of the motor is 828 rpm.

Step-by-step explanation:

Given:

A voltage of the D.C shunt motor $=250$

Armature resistance$R_{a} =0.25$ ohm.

Armature current $I_{a}=50$ A.

D.C motor runs at the speed $N_{1}= 750$

To find:

To find the new speed of the motor $N_{2}$.

Formula used:

$\frac{N_{2} }{N_{1} } =\frac{Eb_{2} }{Eb_{1} }=\frac{\theta_{1} }{\theta_{2} }$

Solution:

$Eb_{1} =V-Ia_{1}R_{a}$

$=250-(50\times0.25)=237.5V$

$Eb_{2} =V-Ia_{2}R_{a}$

$\theta_{1} Ia_{1} =\theta_{2} Ia_{2}$

So, $Ia_{2}=55.55A$

$Eb_{2} =250-55.55\times0.25=236.12V$

$\frac{N_{2} }{750_{} } =\frac{236.12_{} }{_{237.5} }=\frac{50}{55.55}$

$N_{2}= 828$

Hence, the new speed of the motor is $N_{2}= 828 rpm$