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Question

# A dc series motor of total resistance 0.5 Ω runs at a speed of 800 rpm when taking 60 A from 500 V dc mains. Now another identical series motor is mechanically coupled, but connected in series with the first motor. If the set runs at 400 rpm with load torque equal to 2.5 times the original torque. Find the current drawn from 500 V dc source

A
50 A
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B
80 A
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C
90 A
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D
85 A
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Solution

## The correct option is D 85 AEb=V−IaRa=500−60(0.5)=470V ωTe=EbIa 2π×80060×Te=470×60 Te=336.61Nm New load torque =2.5×336.61=841.52Nm Torque developed by the set. 841.52=2(250−Ia×0.5)×Ia2π×40060 I2a−500Ia+35250=0 Ia=500±√5002−4×352502=500±330.152 Ia=84.92(or)415 Ia=85A (Choosing smaller value)

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