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A 250-V DC shunt motor has a shunt field resistance of 150 Ω and an armature resistance of 0.3 Ω. For a given load. The motor runs at 1400 rpm, and drawing 20 A current. If a resistance of 75 Ω is added in series with the field, find the new armature current (in A). Assume load torque and flux and constant.
Correct answer is between '18,20'. Can you explain this answer?
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Verified Answer
A 250-V DC shunt motor has a shunt field resistance of 150 Ω and...
The field current corresponds to:
Now:
IA = IL - If = 20 – 1.66 = 18.34 Amps
Observe that the current of the field circuit is much smaller than the armature current. Thus, adding the speed control resistance leads to minor power dissipation.
Now:
IA = IL - If = 20 – 1.66 = 18.34 Amps
Observe that the current of the field circuit is much smaller than the armature current. Thus, adding the speed control resistance leads to minor power dissipation.
Ea = VS – IaRa = 250 – 5.502 = 244.5 V
EA = K ϕ ω
The new armature current corresponds to:
Ia = IL – If = 20 – 1.11 = 18.89 Amps
The new armature current corresponds to:
Ia = IL – If = 20 – 1.11 = 18.89 Amps
Most Upvoted Answer
A 250-V DC shunt motor has a shunt field resistance of 150 Ω and...
Ohms and an armature resistance of 0.5 ohms. The motor draws a current of 10 A when running at full load.
To find the power output of the motor, we can use the formula:
Power = Voltage x Current
At full load, the voltage is 250 V and the current is 10 A.
Power = 250 x 10
Power = 2500 watts
To find the power loss in the armature due to its resistance, we can use the formula:
Power Loss = Current^2 x Resistance
At full load, the current in the armature is also 10 A, and the armature resistance is 0.5 ohms.
Power Loss = 10^2 x 0.5
Power Loss = 50 watts
Therefore, the power output of the motor is 2500 watts and the power loss in the armature is 50 watts. The remaining power is used to drive the load.
To find the power output of the motor, we can use the formula:
Power = Voltage x Current
At full load, the voltage is 250 V and the current is 10 A.
Power = 250 x 10
Power = 2500 watts
To find the power loss in the armature due to its resistance, we can use the formula:
Power Loss = Current^2 x Resistance
At full load, the current in the armature is also 10 A, and the armature resistance is 0.5 ohms.
Power Loss = 10^2 x 0.5
Power Loss = 50 watts
Therefore, the power output of the motor is 2500 watts and the power loss in the armature is 50 watts. The remaining power is used to drive the load.
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Community Answer
A 250-V DC shunt motor has a shunt field resistance of 150 Ω and...
Flux is not constant as field current is reduced by 33 percent on adding Rext. Answer and question is wrong, phi1xIa1=phi2×Ia2
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A 250-V DC shunt motor has a shunt field resistance of 150 Ω and an armature resistance of 0.3 Ω. For a given load. The motor runs at 1400 rpm, and drawing 20 A current. If a resistance of 75 Ω is added in series with the field, find the new armature current (in A). Assume load torque and flux and constant.Correct answer is between '18,20'. Can you explain this answer?
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A 250-V DC shunt motor has a shunt field resistance of 150 Ω and an armature resistance of 0.3 Ω. For a given load. The motor runs at 1400 rpm, and drawing 20 A current. If a resistance of 75 Ω is added in series with the field, find the new armature current (in A). Assume load torque and flux and constant.Correct answer is between '18,20'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A 250-V DC shunt motor has a shunt field resistance of 150 Ω and an armature resistance of 0.3 Ω. For a given load. The motor runs at 1400 rpm, and drawing 20 A current. If a resistance of 75 Ω is added in series with the field, find the new armature current (in A). Assume load torque and flux and constant.Correct answer is between '18,20'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 250-V DC shunt motor has a shunt field resistance of 150 Ω and an armature resistance of 0.3 Ω. For a given load. The motor runs at 1400 rpm, and drawing 20 A current. If a resistance of 75 Ω is added in series with the field, find the new armature current (in A). Assume load torque and flux and constant.Correct answer is between '18,20'. Can you explain this answer?.
A 250-V DC shunt motor has a shunt field resistance of 150 Ω and an armature resistance of 0.3 Ω. For a given load. The motor runs at 1400 rpm, and drawing 20 A current. If a resistance of 75 Ω is added in series with the field, find the new armature current (in A). Assume load torque and flux and constant.Correct answer is between '18,20'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A 250-V DC shunt motor has a shunt field resistance of 150 Ω and an armature resistance of 0.3 Ω. For a given load. The motor runs at 1400 rpm, and drawing 20 A current. If a resistance of 75 Ω is added in series with the field, find the new armature current (in A). Assume load torque and flux and constant.Correct answer is between '18,20'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 250-V DC shunt motor has a shunt field resistance of 150 Ω and an armature resistance of 0.3 Ω. For a given load. The motor runs at 1400 rpm, and drawing 20 A current. If a resistance of 75 Ω is added in series with the field, find the new armature current (in A). Assume load torque and flux and constant.Correct answer is between '18,20'. Can you explain this answer?.
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