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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 1611Given, dc series motor, TαI2 Eb1=400−20(1.4)=372V N1=2000rpm when load torque increases to 49% T2T1=I22I21 1.49T1T1=I22202 I2=24.41A At same voltage, Eb2=V−Ia2(Ra+Rac) =400−24.41(1.4) =365.82V For series motor, NαEbϕ N2N1=Eb2Eb1×Ia1Ia2 N22000=365.82372×2024.41 N2=1611rpm

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