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Question

A 400 V dc series motor drives a load at rated voltage by drawing 20 A current running at the speed of 2000 rpm. The total series resistance is 1.4 Ω. Assuming magnetic circuit to be linear, if load torque is increased by 49%. The speed of the motor in rpm will be _____

A
1596
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B
1632
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C
1590
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D
1611
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Solution

The correct option is D 1611

Given, dc series motor,

TαI2

Eb1=40020(1.4)=372V

N1=2000rpm

when load torque increases to 49%
T2T1=I22I21

1.49T1T1=I22202

I2=24.41A

At same voltage,
Eb2=VIa2(Ra+Rac)
=40024.41(1.4)
=365.82V

For series motor,
NαEbϕ

N2N1=Eb2Eb1×Ia1Ia2

N22000=365.82372×2024.41

N2=1611rpm


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