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A 220v shunt motor has armature resistance of 0.5ohm and takes a full load current of 40A. By what percentage should the main field flux be reduced to raise the speed by 50%, if the developed torque remains constant?
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A 220v shunt motor has armature resistance of 0.5ohm and takes a full ...
Given Data:
- Voltage (V) = 220V
- Armature Resistance (Ra) = 0.5 ohm
- Full Load Current (I) = 40A
Step 1: Calculating Speed at Full Load
- Speed at full load (N) can be calculated using the formula: N = (V - I*Ra)/K
- Where K is a constant for the motor
- Let's assume K = 1 for simplicity
- Substituting the values, we get N = (220 - 40*0.5)/1 = 200 rpm
Step 2: Calculating New Speed
- As per the question, the speed needs to be increased by 50%
- New speed (N_new) = 1.5 * 200 = 300 rpm
Step 3: Calculating New Main Field Flux
- The main field flux is directly proportional to the speed of the motor
- Therefore, to increase the speed by 50%, we need to reduce the main field flux by the same percentage
- Let the main field flux be X initially, then the new main field flux (X_new) = X * (1 - 50/100) = 0.5X
Step 4: Calculating New Voltage
- The back EMF (Eb) is given by the formula: Eb = K*N
- At full load, Eb = V
- At the new speed, Eb_new = V_new
- Therefore, V_new = K * N_new = 1 * 300 = 300V
Step 5: Calculating New Current
- Since torque remains constant, current (I_new) is inversely proportional to voltage
- I_new = V_new / Ra = 300 / 0.5 = 600A
Step 6: Calculating Percentage Change in Main Field Flux
- Percentage change = (X - X_new) / X * 100% = (X - 0.5X) / X * 100% = 50%
Therefore, the main field flux should be reduced by 50% to raise the speed by 50%, keeping the torque constant.
- Voltage (V) = 220V
- Armature Resistance (Ra) = 0.5 ohm
- Full Load Current (I) = 40A
Step 1: Calculating Speed at Full Load
- Speed at full load (N) can be calculated using the formula: N = (V - I*Ra)/K
- Where K is a constant for the motor
- Let's assume K = 1 for simplicity
- Substituting the values, we get N = (220 - 40*0.5)/1 = 200 rpm
Step 2: Calculating New Speed
- As per the question, the speed needs to be increased by 50%
- New speed (N_new) = 1.5 * 200 = 300 rpm
Step 3: Calculating New Main Field Flux
- The main field flux is directly proportional to the speed of the motor
- Therefore, to increase the speed by 50%, we need to reduce the main field flux by the same percentage
- Let the main field flux be X initially, then the new main field flux (X_new) = X * (1 - 50/100) = 0.5X
Step 4: Calculating New Voltage
- The back EMF (Eb) is given by the formula: Eb = K*N
- At full load, Eb = V
- At the new speed, Eb_new = V_new
- Therefore, V_new = K * N_new = 1 * 300 = 300V
Step 5: Calculating New Current
- Since torque remains constant, current (I_new) is inversely proportional to voltage
- I_new = V_new / Ra = 300 / 0.5 = 600A
Step 6: Calculating Percentage Change in Main Field Flux
- Percentage change = (X - X_new) / X * 100% = (X - 0.5X) / X * 100% = 50%
Therefore, the main field flux should be reduced by 50% to raise the speed by 50%, keeping the torque constant.
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A 220v shunt motor has armature resistance of 0.5ohm and takes a full load current of 40A. By what percentage should the main field flux be reduced to raise the speed by 50%, if the developed torque remains constant?
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A 220v shunt motor has armature resistance of 0.5ohm and takes a full load current of 40A. By what percentage should the main field flux be reduced to raise the speed by 50%, if the developed torque remains constant? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A 220v shunt motor has armature resistance of 0.5ohm and takes a full load current of 40A. By what percentage should the main field flux be reduced to raise the speed by 50%, if the developed torque remains constant? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 220v shunt motor has armature resistance of 0.5ohm and takes a full load current of 40A. By what percentage should the main field flux be reduced to raise the speed by 50%, if the developed torque remains constant?.
A 220v shunt motor has armature resistance of 0.5ohm and takes a full load current of 40A. By what percentage should the main field flux be reduced to raise the speed by 50%, if the developed torque remains constant? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A 220v shunt motor has armature resistance of 0.5ohm and takes a full load current of 40A. By what percentage should the main field flux be reduced to raise the speed by 50%, if the developed torque remains constant? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 220v shunt motor has armature resistance of 0.5ohm and takes a full load current of 40A. By what percentage should the main field flux be reduced to raise the speed by 50%, if the developed torque remains constant?.
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