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A 200V DC shunt motor has an armature resistance 0.4 ohm and field resistance 200 ohm. The motor runs at 750 rpm and takes a current Is to be 25 Amp. Assuming that the load tongue remain constant, find the reduction in the field resistance necessary to reduce the speed to 500 rpm?
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A 200V DC shunt motor has an armature resistance 0.4 ohm and field res...
Problem Statement

A 200V DC shunt motor has an armature resistance 0.4 ohm and field resistance 200 ohm. The motor runs at 750 rpm and takes a current Is to be 25 Amp. Assuming that the load tongue remains constant, find the reduction in the field resistance necessary to reduce the speed to 500 rpm?


Solution

Step 1: Find the back emf and armature current at 750 rpm

Back emf, Eb = V - IaRa (where V = 200V, Ia = 25A, Ra = 0.4 ohm)

Eb = 200 - (25 x 0.4) = 190V

At 750 rpm, Eb = KΦN (where K = constant, Φ = flux, N = speed)

190 = KΦ x 750

Φ = 0.253 Wb

Armature current, Ia = Is - If (where If = Vf/Rf, Vf = Eb)

If = Vf/Rf = 190/200 = 0.95A

Ia = 25 - 0.95 = 24.05A


Step 2: Find the new field resistance for 500 rpm

At 500 rpm, Eb = KΦN (where K = constant, Φ = flux, N = speed)

190 = KΦ x 500

Φ = 0.38 Wb

New armature current, Ia = Is - If (where If = Vf/Rf, Vf = Eb)

If = Vf/Rf = 190/Rf

Ia = 25 - (190/Rf)

From the speed torque characteristic of the motor, torque is proportional to flux

Thus, the ratio of the torque at 500 rpm to the torque at 750 rpm is equal to the ratio of the flux at 500 rpm to the flux at 750 rpm

Torque at 500 rpm/Torque at 750 rpm = Φ at 500 rpm/Φ at 750 rpm

Torque is proportional to current

Thus, (25 - (190/Rf)) x 500/(25 - 0.95 x 750) = 0.38/0.253

Rf = 154.7 ohm


Step 3: Find the reduction in field resistance

Reduction in field resistance = 200 - 154.7 = 45.3 ohm


Conclusion

The reduction in field resistance necessary to reduce the speed of the motor from 750 rpm to 500 rpm is 45.3 ohm.
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A 200V DC shunt motor has an armature resistance 0.4 ohm and field res...
A 200V DC shunt motor has an armature resistance 0.4 ohm and field resistance 200 ohm. The motor runs at 750 rpm and takes a current Is to be 25 Amp. Assuming that the load tongue remain constant, find the reduction in the field resistance necessary to reduce the speed to 500 rpm
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A 200V DC shunt motor has an armature resistance 0.4 ohm and field resistance 200 ohm. The motor runs at 750 rpm and takes a current Is to be 25 Amp. Assuming that the load tongue remain constant, find the reduction in the field resistance necessary to reduce the speed to 500 rpm?
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A 200V DC shunt motor has an armature resistance 0.4 ohm and field resistance 200 ohm. The motor runs at 750 rpm and takes a current Is to be 25 Amp. Assuming that the load tongue remain constant, find the reduction in the field resistance necessary to reduce the speed to 500 rpm? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A 200V DC shunt motor has an armature resistance 0.4 ohm and field resistance 200 ohm. The motor runs at 750 rpm and takes a current Is to be 25 Amp. Assuming that the load tongue remain constant, find the reduction in the field resistance necessary to reduce the speed to 500 rpm? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 200V DC shunt motor has an armature resistance 0.4 ohm and field resistance 200 ohm. The motor runs at 750 rpm and takes a current Is to be 25 Amp. Assuming that the load tongue remain constant, find the reduction in the field resistance necessary to reduce the speed to 500 rpm?.
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