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A 200 V dc motor takes a field current of 2 A to generate a back emf of 180 V. If the field winding resistance is 80 Ω, the resistance of the shunt field regulator is (A) 10 Ω (B) 20 Ω (C) 100 Ω (D) 180 Ω?
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A 200 V dc motor takes a field current of 2 A to generate a back emf o...
Solution:
Given data:
Field current (If) = 2 A
Back emf (Eb) = 180 V
Field winding resistance (Rf) = 80 Ω
DC Motor voltage (V) = 200 V
We know that the voltage equation of a dc motor is given by:
V = Eb + (If × Rf)
Let's substitute the given values in the above equation:
200 = 180 + (2 × 80)
200 = 180 + 160
200 = 340
The above equation is not satisfied for the given data. Hence, there must be some mistake in the given data.
Let's assume that the back emf (Eb) is 80 V instead of 180 V. Then, the voltage equation becomes:
V = Eb + (If × Rf)
200 = 80 + (2 × 80)
On solving the above equation, we get:
R = (V - Eb) / If
R = (200 - 80) / 2
R = 60 Ω
Therefore, the resistance of the shunt field regulator is 60 Ω.
Answer: (Unknown as the given data is incorrect)
Given data:
Field current (If) = 2 A
Back emf (Eb) = 180 V
Field winding resistance (Rf) = 80 Ω
DC Motor voltage (V) = 200 V
We know that the voltage equation of a dc motor is given by:
V = Eb + (If × Rf)
Let's substitute the given values in the above equation:
200 = 180 + (2 × 80)
200 = 180 + 160
200 = 340
The above equation is not satisfied for the given data. Hence, there must be some mistake in the given data.
Let's assume that the back emf (Eb) is 80 V instead of 180 V. Then, the voltage equation becomes:
V = Eb + (If × Rf)
200 = 80 + (2 × 80)
On solving the above equation, we get:
R = (V - Eb) / If
R = (200 - 80) / 2
R = 60 Ω
Therefore, the resistance of the shunt field regulator is 60 Ω.
Answer: (Unknown as the given data is incorrect)
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A 200 V dc motor takes a field current of 2 A to generate a back emf of 180 V. If the field winding resistance is 80 Ω, the resistance of the shunt field regulator is (A) 10 Ω (B) 20 Ω (C) 100 Ω (D) 180 Ω?
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A 200 V dc motor takes a field current of 2 A to generate a back emf of 180 V. If the field winding resistance is 80 Ω, the resistance of the shunt field regulator is (A) 10 Ω (B) 20 Ω (C) 100 Ω (D) 180 Ω? 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 200 V dc motor takes a field current of 2 A to generate a back emf of 180 V. If the field winding resistance is 80 Ω, the resistance of the shunt field regulator is (A) 10 Ω (B) 20 Ω (C) 100 Ω (D) 180 Ω? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 200 V dc motor takes a field current of 2 A to generate a back emf of 180 V. If the field winding resistance is 80 Ω, the resistance of the shunt field regulator is (A) 10 Ω (B) 20 Ω (C) 100 Ω (D) 180 Ω?.
A 200 V dc motor takes a field current of 2 A to generate a back emf of 180 V. If the field winding resistance is 80 Ω, the resistance of the shunt field regulator is (A) 10 Ω (B) 20 Ω (C) 100 Ω (D) 180 Ω? 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 200 V dc motor takes a field current of 2 A to generate a back emf of 180 V. If the field winding resistance is 80 Ω, the resistance of the shunt field regulator is (A) 10 Ω (B) 20 Ω (C) 100 Ω (D) 180 Ω? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 200 V dc motor takes a field current of 2 A to generate a back emf of 180 V. If the field winding resistance is 80 Ω, the resistance of the shunt field regulator is (A) 10 Ω (B) 20 Ω (C) 100 Ω (D) 180 Ω?.
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