Question
Asked Nov 12, 2019
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A 500 V dc shunt motor has an armature resistance of Ra = 0.25 2 and field circuit resistance
of R 240 . The motor drives a mechanical load, which requires a torque proportional
to speed. When the motor-load system is connected to a 500 V supply, it takes 100 A and rotates
at 1100 rpm.
The speed is to be reduced to 900 rpm by inserting a resistance in series with the armature
The field current is kept the same.
Determine the value of the added series resistance.
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A 500 V dc shunt motor has an armature resistance of Ra = 0.25 2 and field circuit resistance of R 240 . The motor drives a mechanical load, which requires a torque proportional to speed. When the motor-load system is connected to a 500 V supply, it takes 100 A and rotates at 1100 rpm. The speed is to be reduced to 900 rpm by inserting a resistance in series with the armature The field current is kept the same. Determine the value of the added series resistance.

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Step 1
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At 1100 rpm: E1 500 V K x1 100 At 900 rpm: E2 K x 900 'b2

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Step 2
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Therefore 900 1100 900 = 500 V x 1100 = 409.091 V

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